(a) The observer sees a virtual, erect, and magnified image of the star in the eyepiece.

(b) A large concave mirror (primary mirror) is used in the reflecting telescope to gather and focus incoming light.

(c) A convex lens is used at the eyepiece. The reason is that the concave mirror forms a real image of the distant star at its focal point. This real image acts as a virtual object for the convex lens (eyepiece), which then forms a magnified, virtual, and erect image for the observer to view comfortably. A convex lens thus acts as a magnifier.

Source: Light – Reflection and Refraction / The Human Eye and the Colourful World, Optical Instruments section

---

• (a): The eyepiece always acts as a simple microscope producing a virtual, erect, magnified image — examiners expect all three characteristics.
• (b): The key word "large concave mirror" must appear; "large" matters because it gathers more light (better resolution).
• (c): 2 marks means you need both the correct answer (convex lens) and the reason (it acts as a magnifying glass on the real image formed by the mirror). Stating only the type without reason will lose 1 mark.

(a) Less than 10 cm

A concave mirror forms a virtual, erect and enlarged image only when the object is placed between the pole (P) and the principal focus (F), i.e., less than the focal length (10 cm) from the mirror.

From Table 9.1 (Image formation by a concave mirror), when the object is placed between P and F, the image formed is virtual, erect, and enlarged — the only case where a concave mirror gives an erect image. Since f = 10 cm, the object must be placed at a distance less than 10 cm. Students often confuse this with convex mirrors; remember, a concave mirror gives an erect image only when the object is within the focal length.

Given: u = −10 cm, f = +15 cm (convex mirror)

Using mirror formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{+15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10} = \frac{2+3}{30} = \frac{5}{30}$$

$$v = +6 \text{ cm}$$

The image is formed 6 cm behind the mirror. It is virtual and erect.

Source: Chapter 9, Section 9.2.4 (Mirror Formula)

---

• For a convex mirror, both f and v are always positive (behind the mirror), and u is negative.
• The positive value of v confirms the image is virtual and behind the mirror — consistent with Table 9.2.
• Examiners award 1 mark for correct substitution/formula and 1 mark for the correct answer with nature of image. Always state the nature explicitly.

(i)

Type of image: When the object (candle) is between the pole and focus, the image formed by a concave mirror is virtual, erect, and enlarged.

Change as candle moves away from pole:

• Between P and F → Virtual, erect, enlarged image (behind mirror)
• At F → Image at infinity
• Between F and C → Real, inverted, enlarged
• At C → Real, inverted, same size
• Beyond C → Real, inverted, diminished

Ray diagram (object at C, i.e., u ≈ R):

```
C F P
←————|————|————|
\ /|
\/ |
Object ↑ ↑ Image (at C, same size, real, inverted)
```
(Two rays: one parallel to principal axis reflects through F; one through C reflects back on itself — image forms at C, real, inverted, same size.)

---

(ii)

Given: f = +3.0 m (convex mirror), u = –6.0 m

Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{3} - \frac{1}{-6} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$

$$v = +2 \text{ m}$$

The image is formed 2 m behind the mirror; it is virtual and erect.

---

• For (i), examiners expect the progressive change in image nature listed clearly as the object moves away, plus a neat labelled ray diagram at u = R showing image at C.
• For (ii), use sign convention: convex mirror → f positive; object distance u is negative. Substitute carefully and state the nature of the image. Show all steps for full marks.

Ray Diagrams for Convex Mirror:

(i) Object at infinity:

```
Parallel rays
───────────────────────►|
───────────────────────►| F P
|← →←→
```

Reflected rays appear to diverge from focus F behind the mirror.

• Position of image: At focus F, behind the mirror
• Size: Highly diminished, point-sized
• Nature: Virtual and erect

(ii) Object between infinity and pole P:

```
Object (AB)
A |
|────────────────────────| P
| F
```

Reflected rays diverge and appear to meet behind the mirror between P and F.

• Position of image: Between P and F, behind the mirror
• Size: Diminished
• Nature: Virtual and erect

(In both cases, use: (a) ray parallel to principal axis — reflects as if coming from F; (b) ray directed towards centre of curvature C — reflects back along same path.)

Source: Chapter 9, Section 9.2.2 and Table 9.2

---

• Examiners expect two labelled diagrams with at least two rays shown for each case. Since diagrams cannot be drawn here in text, describe them clearly and label P, F, C, principal axis, object, and image.
• The key result from Table 9.2 must be stated: image is always virtual, erect, and diminished for a convex mirror.
• For 3 marks: 1 mark per diagram + nature/position/size for each case. Don't skip the image characteristics — that is where marks are awarded.

(A) (i) and (ii)

Mirror X is a concave mirror (used in solar furnace); it forms a real, diminished image at its focus when the object is at infinity. Mirror Y is a convex mirror (rear-view mirror); it always forms a virtual, erect and diminished image.

• Mirror X = Concave mirror: Sunlight comes from infinity, so image forms at focus F — real, inverted, and highly diminished (point-sized). Statement (i) is correct.
• Mirror Y = Convex mirror: Always produces a virtual, erect, and diminished image regardless of object distance. Statement (ii) is correct.
• Statements (iii) and (iv) swap the properties incorrectly between the two mirrors.
• Key recall: Table 9.1 confirms concave mirror image at infinity → focus, real & diminished; convex mirror always gives virtual, erect, diminished image.

(d) +, +

For a convex mirror, the principal focus lies behind the mirror (to the right of pole), so focal length $f$ = positive. When object is at infinity, image forms at focus F, also behind the mirror, so image distance $v$ = positive.

Key points: Under New Cartesian Sign Convention, distances measured to the right of the pole are positive. A convex mirror's focus is behind the mirror (right side) → $f$ = +. Object at infinity gives image at F (behind mirror) → $v$ = +. This is directly supported by Example 9.1 where $f = +1.50$ m and $v = +1.15$ m for a convex mirror, and Table 9.2 which states the image forms "at focus F, behind the mirror" when object is at infinity.

(c) glass slab

When light passes obliquely through a glass slab, the emergent ray is parallel to the incident ray but shifted sideways — this sideward shift is called lateral displacement.

Source: Light – Reflection and Refraction, Section 9.3.1

---

The key phrase is "laterally displaced" — this is the defining property of refraction through a rectangular glass slab. The two parallel faces cause equal and opposite bending, so the ray emerges parallel but shifted. A prism disperses light; a lens converges/diverges it; a plane mirror reflects it. Only a glass slab produces lateral displacement without change in direction.

Ray Diagram — Refraction through a Triangular Prism:

```
A
/\
/ \
N / \
| / r₁ \
→→→→→•→ → → →•→→→→→
i / refracted \ e
/ ray OO' \
B________________C
```

A ray PQ strikes face AB at angle of incidence i, refracts towards the normal (angle of refraction r₁), travels through the prism as QR, exits face AC, and bends away from the normal as RS (emergent ray).

The angle of deviation (D) is the angle between the incident ray (extended) and the emergent ray RS, marked at their point of intersection.

(Diagram must show: incident ray, two normals, refracted ray inside prism, emergent ray, and angle D labelled between extended incident ray and emergent ray.)

Source: Chapter 9 (Light – Reflection and Refraction), Refraction of Light section.

---

• The question specifically asks to draw the diagram and mark angle of deviation — both are essential for full marks.
• Angle of deviation (D) is measured between the original direction of the incident ray (produced) and the final emergent ray.
• The ray bends towards the normal on entering the denser glass medium and away from the normal on exiting.
• Label: incident ray, normal at each face, refracted ray, emergent ray, and clearly mark ∠D — missing the label loses a mark.

(D) 1 and 6

The angle of incidence is the angle between the incident ray and the normal at the point of entry (angle 1), and the angle of deviation is the angle between the emergent ray and the original direction of the incident ray (angle 6).

In a glass slab diagram, the angle of incidence is always measured between the incident ray and the normal at the first surface. The angle of deviation is the angle by which the emergent ray deviates from the original incident ray's direction. Students often confuse the angle of deviation with the angle of refraction — remember, deviation measures the net shift of the ray, not the bending at a single surface.

(a)
(i) In the diagram, the ray bends towards the normal when entering medium B from medium A. This means medium B is optically denser than medium A. Therefore, the speed of light in medium A is greater than in medium B.

(ii) The ray bends away from the normal when entering medium C from medium B. This means medium C is optically rarer than medium B. Therefore, optical density of medium B is greater than optical density of medium C.

(b) When optical density of A = optical density of C, and optical density of B > both A and C:

```
Medium A → ray travels straight
─────────────────────────────
Medium B → ray bends towards normal (entering denser medium)
─────────────────────────────
Medium C → ray bends back, emerges parallel to original ray
```
The emergent ray in medium C is parallel to the incident ray in medium A (similar to refraction through a glass slab).

---

• Key rule: When light goes from rarer → denser medium, it bends towards the normal (speed decreases). From denser → rarer, it bends away (speed increases).
• For part (b), if density of A = density of C, the bending at the two interfaces is equal and opposite, so the emergent ray is parallel to the incident ray — this is the glass slab result from section 9.3.1.
• Examiners award marks for correct comparison AND correct ray diagram showing the parallel emergent ray.

(a) Ray diagram — light goes from diamond (denser) to water (less dense), bending away from the normal:

```
Diamond | Water
|
\ | /
\ | /
\ N /
\ | /
\ | /
-------\|/------- interface
*-------- refracted ray (bends away from normal)
|
N'
```

(Incident ray in diamond, normal at point of incidence, refracted ray in water bending away from normal — label all three)

---

(b) Refractive index of water w.r.t. diamond:

$$_{d}n_{w} = \frac{n_w}{n_d} = \frac{1.33}{2.42} \approx 0.55$$

• (a) Since diamond is optically denser than water, the ray bends away from normal on entering water. Label: incident ray, normal, refracted ray, diamond, water, interface.
• (b) Refractive index of medium 2 w.r.t. medium 1 = (absolute RI of medium 2) ÷ (absolute RI of medium 1). Here diamond is medium 1, water is medium 2. The value < 1 confirms water is rarer than diamond.

(a) Medium A is denser. Since the refracted ray in medium B bends away from the normal, the ray is moving from a denser medium to a rarer medium. A ray travelling from a denser medium to a rarer medium speeds up and bends away from the normal.

(b) Refractive index of B with respect to A:

$$n_{BA} = \frac{\text{Speed of light in medium A}}{\text{Speed of light in medium B}} = \frac{v_a}{v_b}$$

Source: Chapter 9, Section 9.3.2 — The Refractive Index

---

• Part (a): The key rule is: denser → rarer = bends away from normal; rarer → denser = bends toward normal. Since the ray bends away from normal in B, A must be the denser medium. Always justify using this rule.
• Part (b): Use formula $n_{21} = v_1/v_2$ from Eq. 9.5 — refractive index of medium 2 w.r.t. medium 1 = speed in medium 1 ÷ speed in medium 2. Here medium 1 = A, medium 2 = B, so $n_{BA} = v_a/v_b$.

Absolute Refractive Index: The absolute refractive index of a medium is the ratio of the speed of light in vacuum (or air) to the speed of light in that medium.
$$n_m = \frac{c}{v}$$

Calculation:
$$n_g = \frac{c}{v} \Rightarrow v = \frac{c}{n_g} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m/s}$$

The speed of light in glass is $2 \times 10^8$ m/s.

Source: Chapter 9, Section 9.3.2 — The Refractive Index

---

• Definition (1 mark): State clearly that it is the ratio of speed of light in vacuum to speed of light in the medium. The formula $n_m = c/v$ is expected.
• Calculation (1 mark): Rearrange the formula to find $v = c/n$, substitute values, and state the answer with correct units.
• Do not confuse absolute refractive index (medium w.r.t. vacuum) with relative refractive index (one medium w.r.t. another).

(a)

From the diagram, the ray bends towards the normal entering Medium B, then bends away from the normal entering Medium C, and the ray in C is parallel to the ray in A (like a glass slab).

(i) Speed of light: Speed in A > Speed in C > Speed in B

(ii) Optical density: Medium B is optically densest, Medium A is rarest, and C is intermediate. So: $n_B > n_C > n_A$

Justification (speed): The refractive index of medium 2 w.r.t. medium 1 is $n_{21} = \dfrac{v_1}{v_2}$. A higher refractive index means lower speed. Since $n_B$ is highest, light travels slowest in B. Since $n_C > n_A$, light travels faster in A than in C.

---

(b)

(i) When the ray falls obliquely on Medium B, which is optically denser than both A and C:

• At the A–B boundary: ray bends towards the normal (entering denser medium).
• At the B–C boundary: ray bends away from the normal (entering rarer medium).
• The emergent ray in C is not parallel to the incident ray in A (since A and C have different optical densities).

(Diagram: ray bends toward normal at top surface, bends away at bottom, with different angles in A and C.)

(ii) When the ray falls normally (perpendicularly) on Medium B:

• The ray passes straight through without bending at either surface.
• There is no change in direction at any boundary.

(Diagram: straight, undeviated ray passing through all three media.)

Source: Chapter 9, Sections 9.3, 9.3.1, 9.3.2

---

• For part (a), the key is reading the diagram: the ray bends toward the normal going into B (so B is densest), and the emergent ray in C is parallel to A (as in a glass slab), which means C's density is intermediate. Link speed and optical density using $n = v_1/v_2$.
• For part (b)(i), since A and C have different refractive indices, the emergent ray is not parallel to the incident ray — unlike a standard glass slab where both outer media are the same (air). Examiners expect this distinction.
• For part (b)(ii), a normally incident ray has angle of incidence = 0°, so angle of refraction = 0° — it goes straight through. This is a standard result from Snell's law.

(a) Optical density: Kerosene (n = 1.44) has a higher refractive index than water (n = 1.33), so kerosene is optically denser than water.

(b) Mass density: Kerosene has less mass density than water (kerosene floats on water).

(c) Speed of light: Since $n = c/v$, higher refractive index means lower speed. So light travels slower in kerosene than in water.

Inference: Optical density and mass density are not the same. A medium can be optically denser yet have lower mass density (e.g., kerosene).

Source: Chapter 9, Section 9.3.2 — The Refractive Index

---

• The key point examiners look for is the distinction between optical density (linked to refractive index) and mass density (physical property). This is explicitly stated in the textbook note.
• For speed: use $n = c/v$ — higher n → lower v. So kerosene slows light more than water does.
• The inference must state clearly that optical density ≠ mass density. This is the central learning from this comparison.
• Keep each part brief; this is a 2-mark question, so one line per part plus one inference line is enough.

(i) Refractive index, $n = \dfrac{c}{v}$

$$v = \frac{c}{n} = \frac{3 \times 10^8}{2.42} = 1.24 \times 10^8 \text{ m/s}$$

Speed of light in diamond = 1.24 × 10⁸ m/s

---

(ii) Higher refractive index → optically denser → lower speed → smaller angle of refraction.

Order of refractive indices: Carbon disulphide (1.62) > Glass (1.5) > Water (1.33)

∴ Increasing order of angle of refraction:
Carbon disulphide < Glass < Water

---

(iii) (The sub-question (A)/(B) options are not printed in the passage provided. Please include the full text of option A or B for a complete answer.)

Source: Light – Reflection and Refraction, Refraction of Light / Refractive Index

---

• (i) Use $n = c/v$, rearrange to $v = c/n$. Standard 1-mark numerical — show formula and substitution.
• (ii) From the passage: higher refractive index = optically denser = slower light. By Snell's law, denser medium refracts light closer to the normal (smaller refraction angle). So the medium with the highest n gives the smallest angle of refraction. Rank accordingly.
• (iii) The options A/B were not provided in the question; a student must copy those from their paper and answer accordingly. Do not leave it blank in the actual exam.

Answer: (A) 9/4 × 10⁸ m/s

Using $n = \frac{c}{v}$: Speed of light in water = $\frac{n_g \times v_g}{n_w} = \frac{\frac{3}{2} \times 2 \times 10^8}{\frac{4}{3}} = \frac{3 \times 10^8}{\frac{4}{3}} = \frac{9}{4} \times 10^8$ m/s.

Source: Chapter 9, Section 9.3.2

---

• Use $n = c/v$, so $v = c/n$. Since $n_g \cdot v_g = n_w \cdot v_w = c$ (constant), you can write $v_w = \frac{n_g \times v_g}{n_w}$.
• Substituting: $v_w = \frac{(3/2) \times 2 \times 10^8}{4/3} = \frac{3 \times 10^8}{4/3} = \frac{9}{4} \times 10^8$ m/s.
• Key concept: higher refractive index → lower speed of light in that medium. Water ($n = 4/3$) is rarer than glass ($n = 3/2$), so light travels faster in water — which checks out since $9/4 > 2$.

(A) 9/4 × 10⁸ m/s

Using $n = c/v$: speed in water $= c/n_w$ and speed in glass $= c/n_g$.

$$\frac{v_w}{v_g} = \frac{n_g}{n_w} = \frac{3/2}{4/3} = \frac{9}{8}$$

$$v_w = \frac{9}{8} \times 2 \times 10^8 = \frac{9}{4} \times 10^8 \text{ m/s}$$

Source: Chapter 9, Section 9.3.2 The Refractive Index

---

The key relation is $n = c/v$, so $v = c/n$. Rather than finding $c$ separately, take the ratio of speeds in the two media: $v_w/v_g = n_g/n_w$. This eliminates $c$ and gives the answer directly. Examiners expect you to show the ratio step clearly before substituting the given speed in glass.

(A) 10/9

Using the relation: $n_{YX} = \dfrac{n_Y}{n_X} = \dfrac{4/3}{6/5} = \dfrac{4}{3} \times \dfrac{5}{6} = \dfrac{20}{18} = \dfrac{10}{9}$

The refractive index of Y with respect to X equals the absolute refractive index of Y divided by the absolute refractive index of X (both measured with respect to air/vacuum). This follows from the relation $n_{21} = v_1/v_2$, where expressing both speeds relative to air gives $n_{YX} = n_Y / n_X$. Simply divide the two given values correctly.

From the ray diagram, the angle of incidence (i) = 30° and the angle of refraction (r) = 60°.

Using Snell's law:

$$n_{21} = \frac{\sin i}{\sin r} = \frac{\sin 30°}{\sin 60°} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{1}{1.732} \approx 0.577$$

Therefore, the refractive index of Medium 2 with respect to Medium 1 is 0.577.

Source: Chapter 9, Section 9.3.1 (Snell's Law)

---

• The standard approach is to apply Snell's law: $n_{21} = \frac{\sin i}{\sin r}$.
• The angles are read from the diagram. A common diagram for this question uses i = 30° and r = 60° (or vice versa). If your diagram shows i = 60° and r = 30°, then $n_{21} = \frac{\sin 60°}{\sin 30°} = \sqrt{3} = 1.732$.
• Write the formula, substitute values, and show the calculation step-by-step — examiners award marks for each step.
• The given hint (√3 = 1.732) confirms one of these two angle combinations appears in the diagram.

(a) The speed of light in a medium is inversely related to its optical density. A medium with higher optical density has a lower speed of light, and vice versa.

(b) Higher speed → lower optical density. Arranging in ascending order of optical density (lowest to highest):

• A = 2.25 × 10⁸ m/s (least dense)
• C = 2.08 × 10⁸ m/s
• B = 2.00 × 10⁸ m/s (most dense)

Ascending order of optical density: A < C < B

---

• Part (a) tests the concept that optical density is inversely proportional to speed of light — slower speed means optically denser medium.
• Part (b) requires reversing the speed order: highest speed = lowest optical density. Always arrange speed in descending order to get optical density in ascending order.
• These are standard 1-mark each sub-questions; keep answers brief and direct.

(c) At twice the focal length of the lens.

When the object is placed at 2f, the image formed is real, inverted, and of the same size as the object, also at 2f on the other side.

This is a standard ray-optics result for a convex lens. At u = 2f, magnification m = v/u = 1, giving an equal-sized image. Exercise Q.3 in the source directly confirms option (b) "At twice the focal length" as the correct answer. Don't confuse this with the principal focus (which gives an image at infinity).

(c) 40 cm

For magnification $m = -1$, we need $v/u = -1$, so $v = -u$. Using lens formula with $f = 20$ cm and $u = -40$ cm gives $v = +40$ cm. ✓

Magnification $m = v/u = -1$ means the image is real, inverted, and same size — object must be at $2f$. Since $f = 20$ cm, object distance $= 2 \times 20 = 40$ cm. According to Table 9.4, when object is at $2F_1$, image is at $2F_2$, same size, real and inverted ($m = -1$). Option (c) is correct.

(a) The lens used is a concave (diverging) lens. The image formed is virtual, erect, and enlarged on the same side as the object, which matches the behaviour of a concave lens when the object is between F₁ and O.

(b) Given: image distance v = 30 cm, magnification m = 2 (real image, so m = –2).

$$m = \frac{v}{u} \Rightarrow -2 = \frac{30}{u} \Rightarrow u = -15 \text{ cm}$$

The object was placed 15 cm from the lens (between F₁ and 2F₁).

Source: Chapter 9, Section 9.3.4

---

(a) The key clue is the ray diagram — emergent rays diverge and their extensions meet on the same side as the object, forming a virtual, erect, enlarged image. This is characteristic of a concave lens (when object is between O and F₁). A convex lens gives a virtual erect image only when object is between O and F₁, but the image in that case also appears on the same side — however, the rectangular lens symbol with diverging rays in the diagram points to a concave lens.

(b) For a real image formed by a convex lens, magnification m = –v/u (negative for real inverted image). Since size of image is twice the object, |m| = 2, so m = –2. Using m = v/u: u = v/m = 30/(–2) = –15 cm. This means the object is 15 cm from the lens on the same side as the incoming light (between F₁ and 2F₁). Always use sign convention carefully in board exams.

Option C – –30 cm and +30 cm from lens

When image size equals object size for a convex lens (f = 15 cm), the object must be placed at 2F₁, i.e., –30 cm, and the image forms at 2F₂, i.e., +30 cm from the lens.

From Table 9.4 (Chapter 9): when object is placed at 2F₁, the image forms at 2F₂ and is of the same size, real and inverted. Since f = 15 cm, 2f = 30 cm. Using sign convention, object distance = –30 cm (left side) and image distance = +30 cm (right side). Option D is wrong because +30 and –30 would place both on the same side, which is incorrect for a real image.

(i) Ray diagrams for a convex mirror:

Case 1: A ray parallel to the principal axis, after reflection, appears to diverge from the principal focus F (behind the mirror).

Case 2: A ray directed towards the principal focus F, after reflection, emerges parallel to the principal axis.

(Diagrams should show the convex mirror with centre of curvature C and focus F behind the mirror, with reflected rays diverging as described.)

---

(ii) Given: h = 1.5 cm, f = −12 cm (concave), u = −18 cm

Using mirror formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-12} - \frac{1}{-18} = \frac{-3+2}{36} = \frac{-1}{36}$$
$$v = -36 \text{ cm}$$

The image is formed 36 cm in front of the mirror (real and inverted).

Magnification:
$$m = -\frac{v}{u} = -\frac{-36}{-18} = -2$$

Size of image: $h' = m \times h = -2 \times 1.5 = -3$ cm

The image is 3 cm tall, real, inverted, and magnified.

---

• For convex mirror ray diagrams, always remember: F and C lie behind the mirror, and reflected rays appear to diverge from F. Use standard incident ray rules.
• For the numerical, use sign convention carefully: distances in front of the mirror are negative. The negative image height confirms an inverted image. Examiners award marks for correct substitution, calculation of v, and calculation of m and h′ — show each step clearly.

(i) Ray diagrams for concave lens:

Case 1 – Ray parallel to principal axis:
A ray parallel to the principal axis, after refraction through a concave lens, diverges and appears to come from the focus F₁ on the same side as the incident ray.

Case 2 – Ray directed towards principal focus:
A ray directed towards the focus F₂ on the other side of the lens, after refraction, emerges parallel to the principal axis.

(Diagrams should show the concave lens with both cases marked clearly with arrows and F₁, F₂ labeled.)

---

(ii) Given:

• Object height, $h = 4$ cm, $f = +24$ cm, $u = -16$ cm

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} - \frac{1}{-16} = \frac{1}{24}$$
$$\frac{1}{v} = \frac{1}{24} - \frac{1}{16} = \frac{2-3}{48} = \frac{-1}{48}$$
$$v = -48 \text{ cm}$$

Magnification: $m = \frac{v}{u} = \frac{-48}{-16} = +3$

Image height $= 3 \times 4 = 12$ cm

The image is formed 48 cm on the same side as the object (virtual), erect, and 12 cm tall (enlarged).

Source: Light – Reflection and Refraction, Chapter 9

---

• For part (i), examiners expect two clearly labelled ray diagrams. Remember: for a concave lens, both standard rays diverge after refraction — no real focus exists.
• For part (ii), since object is between F and optical centre (u = 16 cm < f = 24 cm), the image is virtual and erect — consistent with Table 9.4. The negative value of v confirms the image is on the same side as the object. Always state the nature, position, and size clearly to get full marks.

Ray Diagram:

(Draw a convex lens with its upper half shaded. Show the object AB at 2F₁. Draw two rays from the lower half only — one parallel to the principal axis refracting through F₂, and one through the optical centre — meeting at 2F₂ to form image A'B'.)

Position and Nature of Image:
When the object is placed at 2F₁, the image is formed at 2F₂ — on the other side of the lens, at a distance of 2F. The image is real, inverted, and of the same size as the object.

Observable Difference when Lens is Uncovered:
The image formed is the same in position, nature, and size — but it becomes brighter (more intense) when the full lens is uncovered.

Reason:
Covering the upper half blocks some light rays but the lower half still refracts enough rays to form a complete image. Uncovering the lens allows more light rays to pass through, increasing the brightness (intensity) of the image. The image remains complete in both cases.

Source: Chapter 9, Section 9.3.4 & 9.3.5

---

• Examiners specifically award marks for: (1) correct ray diagram with at least two rays from lower half, (2) correct position — at 2F₂, (3) nature — real, inverted, same size, (4) stating brighter image (not a different image), (5) reason linking more light → more brightness.
• Common mistake: students write the image disappears or changes position — it does not. Only brightness changes.
• The ray through the optical centre is the easiest second ray to draw since it passes undeviated even through the lower half.

(a) Speed of light is more in Medium 1, because the refracted ray bends away from the normal, indicating Medium 2 is optically denser (slower).

(b) The ray bends away from the normal because Medium 2 is optically rarer than Medium 1 — light speeds up when going from a denser to a rarer medium, causing it to bend away from the normal.

(c) Refractive index of Medium 2 with respect to Medium 1:

$$n_{21} = \frac{v_1}{v_2}$$

where $v_1$ = speed of light in Medium 1, $v_2$ = speed of light in Medium 2.

Source: Chapter 9, Section 9.3.2 – The Refractive Index

---

• The key observation is: ray bends away from normal → it speeds up → Medium 2 is rarer → Medium 1 is denser → speed is higher in Medium 1.
• For (b), always link "bending away from normal" to "denser → rarer" transition and increased speed.
• For (c), the formula $n_{21} = v_1/v_2$ is directly from Eq. 9.5 — note it is speed in medium 1 over speed in medium 2, not the other way.

(C)

A ray parallel to the principal axis, after reflection from a concave mirror, passes through the principal focus F. Hence, option (C) is correct.

Source: Chapter 9, Section 9.2.2 — Representation of Images Formed by Spherical Mirrors Using Ray Diagrams

---

The first standard rule for ray diagrams of concave mirrors states: a ray parallel to the principal axis, after reflection, must pass through the principal focus F. Options (A), (B), and (D) show the reflected ray missing F or going in the wrong direction, violating this rule. Examiners specifically test this rule, so always remember: parallel ray → reflects through F (concave mirror).

Answer: B — Between pole and focus of the mirror.

When the teeth are placed between the pole and focus of a concave mirror, the image formed is virtual, erect, and enlarged, allowing the dentist to see a magnified image of the teeth.

From Table 9.1 (Chapter 9): when an object is placed between P and F of a concave mirror, the image is virtual, erect, and enlarged. This is the only position that gives a magnified image — which is why dentists use concave mirrors held close to the teeth. At focus (A), no image forms; between F and C (C), image is real and inverted; at C (D), image is same size. Examiners expect you to recall the table directly.

(a) A convex lens converges the parallel rays of sunlight to a sharp bright point on the paper. The concentration of sunlight at that point generates a large amount of heat, which causes the paper to burn.

(b) The point is called the principal focus (F) of the convex lens. The bright spot formed on the paper represents a real image of the Sun.

Source: Chapter 9, Section 9.3.3 (Activity 9.11)

---

• (a) The key idea is convergence of parallel rays + heat generation — both must be mentioned for full credit.
• (b) Examiners expect the exact term "principal focus" and the fact that the bright spot is a real image of the Sun (not just "an image"). These two facts are explicitly stated in Activity 9.11 of the textbook.

(C) The brightness of the image will reduce.

When the upper half is covered, fewer light rays pass through the lens, so the image is complete but less bright (dimmer).

A lens forms an image using all light rays passing through it. Covering half the lens reduces the number of rays, but the remaining half still refracts light to form a complete image of the whole object — just with reduced intensity/brightness. Size is unaffected, and both halves of the image are still formed. This is a standard NCERT concept (Q9 in Exercises).

(i) Ray Diagram:

```
F₁ O F₂
——————————>————————|————————————>——————
Incident ray Lens Emergent ray
(through F₁) (parallel to
principal axis)
```

A ray passing through the principal focus F₁ on the same side as the object, after refraction through the convex lens, emerges parallel to the principal axis (as per Fig. 9.14(a)).

(Students must draw: convex lens, principal axis, incident ray directed through F₁, refracted ray parallel to axis, with labels: Convex lens, F₁, F₂, Principal axis.)

(ii) The incident ray passes through the principal focus (F₁) of the convex lens.

Source: Chapter 9, Section 9.3.5

---

• Part (i) tests whether you know ray rule (ii) for lenses: a ray through the focus on the incident side emerges parallel to the principal axis. Always label F₁, F₂, O (optical centre), and the lens.
• Part (ii) directly names that point as the principal focus (F₁). Many students write "focus F₂" (wrong side) — remember F₁ is on the same side as the object.
• Since this is a 2-mark question (1 mark each), keep the diagram neat and the written answer to one line.

Concave lens will diverge a parallel beam of light falling on it. It is also called a diverging lens.

Ray Diagram — Principal Focus of a Concave Lens:

```
|
\ | /
\ | /
----\---|---/------ Principal axis
\ | /
\ | /
\ |/
F₁ O
(Virtual focus, same side as incident rays)
```

Parallel rays, after refraction through the concave lens, appear to diverge from the principal focus F₁, located on the same side as the incident light.

Source: Chapter 9, Section 9.3.3 — Refraction by Spherical Lenses

---

• The question has two parts: (i) identifying which lens diverges — concave (1 mark), and (ii) drawing the ray diagram showing principal focus (1 mark).
• Key examiner expectation: the diagram must show parallel rays appearing to diverge from F₁ on the same side as the incident light — this is what distinguishes a concave (virtual) focus from a convex (real) focus.
• Always label the optical centre O, principal focus F, and the principal axis in the diagram.
• Don't confuse: convex = converging (rays meet at F on the other side); concave = diverging (rays appear to come from F on the same side).

(i) Ray diagram — Refraction through a rectangular glass slab:

```
N
|
E | Air
\ |
\ | (incident ray)
\ |i
------O----------- AB (air-glass surface)
\
r\ Glass
\
O'
---------|-------- CD (glass-air surface)
\r'
\ Air
\
H
|
M'
```

• Ray EO is the incident ray (oblique, in air).
• OO′ is the refracted ray (bends towards normal inside glass; angle of refraction < angle of incidence).
• O′H is the emergent ray (bends away from normal; parallel to EO but laterally displaced).
• NN′ ⊥ AB and MM′ ⊥ CD are the normals at the two surfaces.

(ii) Snell's Law:
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given colour of light and a given pair of media:

$$\frac{\sin i}{\sin r} = \text{constant (refractive index)}$$

(iii) Differences between virtual images formed by convex and concave lenses:

| Basis | Convex Lens | Concave Lens |
|---|---|---|
| (I) Object distance | Object must be placed between the optical centre O and focus F₁ (u < f). | Object can be placed anywhere between optical centre O and infinity; virtual image is always formed. |
| (II) Magnification | Magnification is greater than 1 (image is enlarged). | Magnification is less than 1 (image is diminished). |

Source: Chapter 9, Section 9.3.1 (Refraction through a Rectangular Glass Slab) and Section 9.3.4 (Image Formation by Lenses)

---

• (i) Draw the diagram neatly and label: incident ray, refracted ray, emergent ray, normals (NN′, MM′), angles i and r, surfaces AB and CD, and media (Air/Glass). The emergent ray must be shown parallel to the incident ray (lateral displacement). Examiners award marks for correct bending direction (towards normal entering glass, away from normal leaving glass) and correct labelling.

• (ii) State both parts of Snell's law clearly and write the formula. Many students write only the formula and lose the statement mark.

• (iii) This is a 2-mark differentiation, so a table with two clear rows (one per basis) is the cleanest format. Key facts to remember: convex lens forms a virtual image only when the object is inside the focal length (m > 1); a concave lens always forms a virtual, erect, diminished image (m < 1) for any object position.

Focal length of lens = 25 cm (rays from sun converge at focus). For a virtual image in a convex lens, object must be placed between F and optical centre, i.e., at a distance less than 25 cm → (A) 20 cm.

Sunlight (parallel rays) converges at 25 cm → f = 25 cm. A convex lens forms a virtual, erect, magnified image only when the object is placed between the optical centre and F (u < f). Among the options, only 20 cm < 25 cm satisfies this condition. All other options (40, 50, >50 cm) place the object beyond F, giving a real image.

(a) Ray Diagram:

• Ray 1: Parallel to principal axis → after reflection, passes through F (focus).
• Ray 2: Directed towards centre of curvature C → reflects back along the same path.
• The two reflected rays diverge; extend them behind the mirror (dashed lines). They appear to meet behind the mirror between P and C, giving a virtual image A′B′.

(b) Nature, Position and Size of Image:

• Position: Behind the mirror, between P and C.
• Nature: Virtual and erect.
• Size: Enlarged (magnified), larger than the object.

(This occurs when object is placed between P and F of a concave mirror.)

(c) Sign of Image Distance:

The image is formed behind the mirror (on the same side as the object in a concave mirror's convention). By Cartesian sign convention, this distance is positive (+).

---

• (a) Examiners check that Ray 1 reflects through F and Ray 2 reflects back on itself through C; dashed lines behind the mirror show virtual image formation correctly.
• (b) This is the standard case: object between P and F of a concave mirror → virtual, erect, magnified image behind the mirror (as in Exercise Q2 of the chapter).
• (c) In Cartesian convention, distances measured in the direction of incident light (in front of mirror) are negative; behind the mirror is opposite to incident light, so image distance v is positive.

Given:

• Object-distance, $u$ = –30 cm (object on left of lens)
• Focal length, $f$ = –15 cm (concave lens, negative sign)
• Image-distance, $v$ = ?

Using Lens Formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$$

$$\frac{1}{v} = \frac{1}{-15} + \frac{1}{-30}$$

$$\frac{1}{v} = -\frac{2}{30} - \frac{1}{30} = \frac{-3}{30} = \frac{-1}{10}$$

$$\boxed{v = -10 \text{ cm}}$$

Result: The image is formed at a distance of 10 cm from the optical centre, on the same side as the object (negative sign).

The image is virtual, erect, and diminished.

Source: Light – Reflection and Refraction, Section 9.3.7

---

• Always apply sign convention first: for a concave lens, both $f$ and $u$ are negative.
• Rearrange the lens formula as $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ — this avoids sign errors.
• A negative $v$ confirms the image is virtual and on the same side as the object, which is always the case for a concave lens with a real object.
• Examiners award marks for: correct signs (1 mark), correct substitution (1 mark), correct calculation (1 mark), result with unit (1 mark), and nature of image (1 mark).

Case (i): Real, inverted and magnified image
For a convex lens to form a real, inverted and magnified image, the object must be placed between F and 2F, i.e., object distance $u$ must satisfy:
$$f < |u| < 2f \implies 10 \text{ cm} < |u| < 20 \text{ cm}$$

Ray Diagram (i):
(Object between F and 2F; image forms beyond 2F on the other side — real, inverted, magnified)

```
|
O | I
↑ F 2F 2F' ↑↑
---●--●----●-----●----●---
|
```

---

Case (ii): Virtual, erect and magnified image
For a virtual, erect and magnified image, the object must be placed between the optical centre and F, i.e.:
$$0 < |u| < f \implies 0 < |u| < 10 \text{ cm}$$

Ray Diagram (ii):
(Object between O and F; image forms on the same side as object — virtual, erect, magnified)

```
|
I↑↑ | O↑
F
---●--●--●-----------
|
```

Source: Chapter 9, Section 9.3.5 – Image Formation in Lenses Using Ray Diagrams

---

• Examiners look for both ranges stated clearly with reference to f = 10 cm, and two ray diagrams (one per case). Without the diagrams you lose marks.
• Key rule: real magnified → object between F and 2F (|m| > 1, m negative); virtual magnified → object between O and F (|m| > 1, m positive).
• State the range in numbers (10 cm to 20 cm, and 0 to 10 cm) since f = 10 cm is given — this shows you applied the concept.
• In board exams, even a rough labelled ray diagram with correct rays earns full credit for the diagram component.

(i)

When a ray of light passes through a rectangular glass slab, it refracts at two parallel faces — AB (air to glass) and CD (glass to air). The bending at these two faces is equal and opposite because the faces are parallel to each other. Hence, the emergent ray comes out parallel to the incident ray, though it is laterally displaced.

Ray Diagram:

```
E N
\ |
\i |
----O-------+---- AB (air-glass)
|r |
OO'(refracted ray)
----O'------+---- CD (glass-air)
| |
\ M'
H (emergent, parallel to EO)
```

When light falls normally on a face: The angle of incidence = 0°, so the ray passes straight through without any bending (no refraction). It continues in the same direction.

---

(ii)

Given: $u = -30$ cm, $f = -20$ cm (concave lens)

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-20} + \frac{1}{-30}$$

$$\frac{1}{v} = \frac{-3 - 2}{60} = \frac{-5}{60} = \frac{-1}{12}$$

$$v = -12 \text{ cm}$$

The image is formed 12 cm in front of the lens (on the same side as the object). It is virtual and erect.

Source: Chapter 9, Section 9.3.1 (Refraction through a Rectangular Glass Slab) and Sign Convention for Spherical Lenses

---

• (i) The key reason is "equal and opposite bending at two parallel faces" — state this clearly. Examiners award marks for the diagram showing incident ray, refracted ray inside slab, and emergent ray parallel to incident. For normal incidence, state that angle of incidence = 0°, so no bending occurs — the ray goes straight through.
• (ii) Always write the lens formula first, substitute with correct signs (both u and f are negative for a concave lens with real object), and state the nature of the image. A negative v confirms the image is on the same side as the object — virtual and erect, consistent with Table 9.5.

Given: u = –60 cm, f = –30 cm (concave lens)

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-30} + \frac{1}{-60}$$

$$\frac{1}{v} = \frac{-2-1}{60} = \frac{-3}{60} = \frac{-1}{20}$$

$$v = -20 \text{ cm}$$

The image is formed 20 cm in front of the lens (on the same side as the object). It is virtual and erect.

Source: Light – Reflection and Refraction, Chapter 9

---

• Always assign signs first: object distance u is negative (object on left), focal length f is negative for concave lens.
• Use $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ carefully — rearrange to find v.
• A negative v confirms the image is on the same side as the object → virtual and erect, which is consistent with a concave lens always forming such images.
• Examiners award 1 mark for correct substitution/working and 1 mark for the correct answer with sign and position statement.

(a) Focal length = 25 cm.

At S.No. 3, u = –50 cm and v = +50 cm, meaning the object is at 2F and the image is also at 2F. This happens only when u = 2f, so 2f = 50 cm → f = 25 cm.

(b) Observation S.No. 6 (u = –15 cm) is incorrect. Since u = –15 cm is less than f (25 cm), the object lies between F and the lens; a convex lens then forms a virtual image on the same side as the object (v should be negative), but the table shows v = +37.5 cm (positive/real), which is wrong.

(c)
$$m = \frac{v}{u} = \frac{+150}{-30} = -5$$

The image is real, inverted, and highly magnified (|m| > 1, m is negative). Negative magnification indicates a real and inverted image formed on the other side of the lens.

Source: Light – Reflection and Refraction, Convex Lens / Lens Formula section

---

• Part (a): The quickest no-calculation clue is S.No. 3 where |u| = |v| = 50 cm. For a convex lens, u = v only when both equal 2f, giving f = 25 cm.
• Part (b): When the object is inside the focal length (u < f), a convex lens always produces a virtual, erect image — v must be negative in the sign convention used here. The table's positive v for S.No. 6 contradicts this.
• Part (c): Use m = v/u. Negative m → real + inverted; |m| = 5 → highly magnified. Always state all three characteristics of the image.

(i) Case 1 (Mirror A) will form a diminished image.
Justification: Object distance (45 cm) > 2f (40 cm), so the object is beyond C. A concave mirror forms a diminished image when the object is placed beyond the centre of curvature.

(ii) In Case 2, f = 15 cm, object distance = 30 cm = 2f (at centre of curvature). Properties of image:

• Real and inverted
• Same size as the object (at C)

(iii) For Mirror C: f = 30 cm, object distance = 20 cm < f. The object is placed between the pole and focus.

Nature and size: Virtual, erect, and magnified image is formed behind the mirror.

Ray Diagram:

(Diagram showing object between P and F of concave mirror, with two rays — one parallel to principal axis reflecting through F, and one directed towards F reflecting parallel — diverging after reflection, with virtual image formed behind mirror, larger than object.)

---

• For (i), compare object distance with 2f: if u > 2f → diminished image (beyond C rule).
• For (ii), object at C (u = 2f) always gives real, inverted, same-sized image — a standard result to memorise.
• For (iii), object between P and F is the only case where a concave mirror gives a virtual, erect, magnified image. CBSE expects the ray diagram with at least two rays clearly showing the virtual image behind the mirror. Label P, F, C on the diagram.

(a) Focal length = 15 cm
Reason: In rows 4 and 5, when object distance = 30 cm, image distance = 30 cm (equal distances, so object is at 2F). This means 2F = 30 cm, hence F = 15 cm.

(b) S.No. 8 (u = −12 cm) is incorrect. Since the object is placed at 12 cm, which is less than the focal length (15 cm), a convex lens cannot form a real image on the screen; the image would be virtual.

(c) Ray diagram for S.No. 4 (object at 2F, u = −30 cm):

```
F 2F 2F F
-----|----|----|lens|----|-----
[obj] [img]
```
Two rays drawn: one parallel to principal axis refracting through F, and one through optical centre passing undeviated — image forms at 2F on the other side, real, inverted, same size.

Source: Science Lab Manual / Light – Reflection and Refraction, Convex Lens experiment

---

• For (a): The key observation is symmetry — when u = v, the object is at 2F. Rows 4 & 5 confirm this without any calculation.
• For (b): Any object within the focal length of a convex lens gives a virtual image (cannot be caught on screen), so u = −12 cm < f = 15 cm is impossible in this experiment.
• For (c): CBSE expects a neat, labelled ray diagram with at least two rays; examiners award 1 mark for the diagram, so keep it clear and label F, 2F, object, image.

(a) Focal length = 20 cm

In Observation 4, u = –40 cm and v = +40 cm, i.e., the image is formed at 2F on the other side. This means 2F = 40 cm, so f = 20 cm.
(Alternatively, applying lens formula to any observation gives f = +20 cm.)

(b) For Observation 3: u = –30 cm, v = +60 cm

$$m = \frac{v}{u} = \frac{+60}{-30} = \mathbf{-2}$$

The magnification is –2 (image is real, inverted and enlarged, twice the size of the object).

(c) In Observation 1, u = –15 cm (object between F and O), v = –60 cm (negative).
In Observation 2, u = –25 cm (object between F and 2F), v = +100 cm (positive).
Numerical magnification = 4 in both cases.

| | Observation 1 | Observation 2 |
|---|---|---|
| Nature | Virtual and erect | Real and inverted |
| Position | Same side as object | Opposite side of lens |

Source: Light – Reflection and Refraction, Sections 9.3.4, 9.3.7

---

• (a) When |u| = |v|, the object is at 2F, so the image forms at 2F on the other side — this is the quickest way to read f without calculation. Observation 4 directly gives this.
• (b) Use m = v/u with correct signs. A negative m means real and inverted.
• (c) The key contrast is virtual/erect vs real/inverted, and image on same side vs opposite side. "Numerical magnification same" means |m| = 4 in both, but the sign (and hence nature) differs. Examiners expect two clear, distinct differences stated as a comparison.

For a virtual and magnified image, the object must be placed between the pole (P) and the principal focus (F) of the concave mirror, i.e., between 0 and 18 cm from the mirror.

Verification using mirror formula:

Given: $f = -18$ cm, $m = +2$

Since $m = -\dfrac{v}{u}$,

$$+2 = -\frac{v}{u} \implies v = -2u$$

Applying mirror formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{-2u} + \frac{1}{u} = \frac{1}{-18}$$

$$\frac{-1+2}{2u} = \frac{1}{-18}$$

$$\frac{1}{2u} = \frac{-1}{18} \implies u = -9 \text{ cm}$$

The object distance is 9 cm, which lies between P and F (0 to 18 cm). This confirms that placing the object between the pole and focus gives a virtual, magnified image.

Source: Chapter 9, Section 9.2.4 — Mirror Formula and Magnification

---

• The key concept: a concave mirror gives a virtual, erect, magnified image only when the object is between P and F (refer Table 9.1).
• Magnification is positive for virtual images in mirrors, so $m = +2$ is the right clue.
• Use $m = -v/u$ (mirror formula, not lens formula) to link $v$ and $u$, then substitute into the mirror formula.
• Sign convention: object distance $u$ is negative (object in front of mirror); focal length $f$ of concave mirror is negative.
• The calculated $u = -9$ cm (between 0 and 18 cm) justifies the initial statement — examiners look for this logical closure.

(i)
Given: h = 5 cm, u = −20 cm, f = −18 cm (concave lens)

Using lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-18} + \frac{1}{-20} = \frac{-10 - 9}{180} = \frac{-19}{180}$$

(1) Image distance: $v = \frac{-180}{19} \approx -9.47$ cm (virtual, same side as object)

(2) Magnification: $m = \frac{v}{u} = \frac{-180/19}{-20} = \frac{180}{380} \approx +0.47$

(ii)
A concave lens always forms a virtual image with magnification m < 1 (diminished).
A convex lens forms a virtual image (object between F and O) with magnification m > 1 (enlarged).

Source: Light – Reflection and Refraction, Section 9.3.4

---

• Always use sign convention: distances measured from optical centre; concave lens has negative focal length.
• For part (ii), examiners expect a direct comparison: concave → m < 1; convex → m > 1. This follows directly from Table 9.5 and Table 9.4 in the textbook.
• Show formula and substitution clearly for full marks.

Given: h = +10 cm, u = –25 cm, f = +15 cm (converging lens)

Using Lens Formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{15} + \frac{1}{(-25)} = \frac{1}{15} - \frac{1}{25} = \frac{5-3}{75} = \frac{2}{75}$$

$$v = +37.5 \text{ cm}$$

The image is formed 37.5 cm on the other side of the lens. It is real and inverted.

Height of image:

$$h' = h \times \frac{v}{u} = 10 \times \frac{+37.5}{-25} = -15 \text{ cm}$$

The image height is 15 cm (negative sign indicates it is inverted).

Source: Chapter 9, Section 9.3.7 – Lens Formula and Magnification

---

• Always apply sign convention: object on left → u is negative; converging lens → f is positive.
• The formula for lenses is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (not the mirror formula).
• A positive v confirms real image on the opposite side; negative h′ confirms inverted image.
• Examiners award marks for: correct substitution with signs (1 mark), correct v (1 mark), correct h′ (1 mark).

(a)

Ray Diagram: Object placed at 2F₁ (distance = 2f from lens):

```
|
O | I
↑ ----+---- ↓
2F₁ F₁ O F₂ 2F₂
```

Two rays are drawn: (1) parallel to principal axis, refracts through F₂; (2) through optical centre, passes undeviated. Image forms at 2F₂ — real, inverted, same size as object.

$$\text{Magnification} = -1$$

---

(b)

Given: h = 2 cm, f = 12 cm, u = −16 cm

(i) Distance of image from lens:

Using lens formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$$\frac{1}{v} - \frac{1}{-16} = \frac{1}{12}$$

$$\frac{1}{v} = \frac{1}{12} - \frac{1}{16} = \frac{4-3}{48} = \frac{1}{48}$$

$$v = +48 \text{ cm}$$

The image is formed 48 cm from the lens (on the opposite side).

(ii) Length of image:

$$m = \frac{v}{u} = \frac{48}{-16} = -3$$

$$h' = m \times h = -3 \times 2 = -6 \text{ cm}$$

Length of image = 6 cm (real and inverted).

Source: Light – Reflection and Refraction, Section 9.3.4 & 9.3.6

---

• Part (a): At u = 2f, image forms at 2f on the other side — same size. Magnification = −1 (negative sign shows inversion). The ray diagram must show two standard rays meeting at 2F₂.
• Part (b): Always use sign convention: u is negative (object on left). Apply lens formula carefully. Magnification = v/u; multiply by object height to get image height. Negative h′ confirms real, inverted image. Examiners award marks for formula, substitution, and correct final answer with units.

(i) Distance of the image from the mirror:

Given: h = +3 cm, f = −12 cm (concave), u = −18 cm

Using mirror formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} = \frac{1}{-12} - \frac{1}{-18} = -\frac{1}{12} + \frac{1}{18} = \frac{-3+2}{36} = \frac{-1}{36}$$
$$v = -36 \text{ cm}$$

The image is formed 36 cm in front of the mirror (real and inverted).

(ii) Height of the image:

$$m = -\frac{v}{u} = -\frac{-36}{-18} = -2$$
$$h' = m \times h = -2 \times 3 = -6 \text{ cm}$$

The height of the image is 6 cm (negative sign indicates it is real and inverted).

Source: Chapter 9, Section 9.2.4 — Mirror Formula and Magnification

---

• Sign convention: For a concave mirror, f and u are both negative (distances measured in front of the mirror). Always assign signs before substituting.
• Examiners look for: correct substitution with signs, the mirror formula working shown step-by-step, and both answers stated clearly with units and nature.
• The negative value of h′ confirms the image is real and inverted — mentioning this adds quality to the answer.
• Object is between C and F (u = 18 cm, f = 12 cm), so the image is expected beyond C — v = 36 cm confirms this, consistent with Table 9.1.

(i) Principal Axis: The straight line passing through the two centres of curvature of the two surfaces of a lens (i.e., through its optical centre) is called the principal axis.

Ray Diagram – Concave Lens:

```
|
\ | /
\ | /
----\---O---------/-----→ Principal Axis
F₁ | /
| /
```

A ray parallel to the principal axis, after refraction through a concave lens, appears to diverge from the principal focus (F₁) located on the same side as the incident ray.

---

(ii) Given:

• Focal length of concave lens, f = −20 cm
• Object height, h = 5 cm
• Image distance, v = −15 cm (image is virtual, on same side as object)
• Object distance, u = ?

Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{-15} - \frac{1}{u} = \frac{1}{-20}$$

$$\frac{1}{u} = \frac{1}{-15} + \frac{1}{20} = \frac{-4+3}{60} = \frac{-1}{60}$$

$$\therefore u = -60 \text{ cm}$$

The object should be placed 60 cm from the lens.

Size of image:
$$m = \frac{v}{u} = \frac{-15}{-60} = +\frac{1}{4}$$

$$h' = m \times h = \frac{1}{4} \times 5 = 1.25 \text{ cm}$$

The image is 1.25 cm tall, virtual, erect, and diminished.

Source: Chapter 9, Sections 9.3.5 and 9.3.6

---

• Part (i): Examiners expect a one-line definition and a ray diagram clearly showing the incident ray parallel to the principal axis diverging after refraction, with the diverged ray appearing to come from F₁ on the same side. Label the lens, principal axis, and F₁.
• Part (ii): For a concave lens, both f and v are negative by sign convention (real-is-positive/new Cartesian). Apply the lens formula correctly. The magnification formula $m = v/u$ gives the image height directly. A common error is using wrong signs — always state the signs used.
• The image formed by a concave lens is always virtual, erect, and diminished — confirm this matches your answer.

(i) Focal Length:
From observation 3, u = −50 cm and v = +50 cm. Using lens formula:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{50} - \frac{1}{-50} = \frac{2}{50}$$
f = +25 cm

Justification (without calculation): When object is placed at 2f, image is also formed at 2f on the other side. Here |u| = |v| = 50 cm, so 2f = 50 cm, giving f = 25 cm.

---

(ii) Incorrect Observation:
Observation 6 (u = −15 cm, v = +37.5 cm) is incorrect.

Since f = 25 cm and u = −15 cm, the object is placed within the focal length (between O and F). For such a position, a convex lens forms a virtual, erect, magnified image on the same side as the object, so v should be negative, not positive.

Ray Diagram: (Draw object between O and F; two rays — one parallel to principal axis refracted through F₂, one through optical centre straight — diverging rays are extended backward to meet on the same side, giving a virtual, erect, magnified image.)

---

(iii) Magnification for u = −30 cm:
From table, v = +150 cm.

$$m = \frac{v}{u} = \frac{+150}{-30} = \mathbf{-5}$$

The image is real, inverted and 5 times magnified.

Source: Chapter 9, Sections 9.3.6 and 9.3.7

---

• (i) The key insight is symmetry: when u = v numerically, the object is at 2f. No calculation needed — just state this logic clearly.
• (ii) Examiners expect you to identify that u = −15 cm < f = 25 cm (object inside focal length), so image must be virtual (v negative). Must also draw a ray diagram showing virtual image on same side.
• (iii) Use m = v/u directly from the table values with correct signs. The negative value confirms real and inverted image — mention this for full marks.

Given: Object size (h) = 20 cm, Image size (h') = 80 cm, Object distance (u) = –60 cm

Magnification: $m = \frac{h'}{h} = \frac{80}{20} = +4$

(i) Nature of the image: Since the image and object are on the same side of the lens, the image is virtual and erect (magnification is positive).

(ii) Position of the image:
$$m = \frac{v}{u} \Rightarrow +4 = \frac{v}{-60} \Rightarrow v = -240 \text{ cm}$$
The image is formed 240 cm from the optical centre, on the same side as the object.

(iii) Focal length: Using lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$$\frac{1}{f} = \frac{1}{-240} - \frac{1}{-60} = \frac{-1+4}{240} = \frac{3}{240}$$
$$f = +80 \text{ cm}$$

Source: Chapter 9, Section 9.3.6

---

• When object and image are on the same side of a convex lens, the image is virtual and erect — this matches Table 9.4 (object between F₁ and O).
• Apply sign convention: u is negative (object on left); virtual image also on the same (left) side, so v is negative too.
• Magnification $m = v/u = h'/h$; since the image is virtual and erect, m is positive, confirming both v and u have the same sign.
• Examiners award one mark each for (i) nature, (ii) position, and (iii) focal length. Show formula and substitution clearly.

(a) The magnification produced by a lens is given by:

$$m = \frac{v}{u}$$

where $v$ is the image distance and $u$ is the object distance from the optical centre.

(b) Given: Object height $h$ = 4 cm, $f$ = +20 cm, $u$ = −10 cm

Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$$\frac{1}{v} = \frac{1}{20} + \frac{1}{(-10)} = \frac{1}{20} - \frac{1}{10} = \frac{-1}{20}$$

So $v$ = −20 cm

Magnification: $m = \frac{v}{u} = \frac{-20}{-10} = +2$

Size of image: $h' = m \times h = 2 \times 4 = \textbf{8 cm}$

The image is virtual, erect and enlarged.

Source: Chapter 9, Section 9.3

---

• (a) is a standard 1-mark definition — write only the formula with symbols defined.
• (b) Since $u$ = 10 cm < $f$ = 20 cm, the object is between F and O; expect a virtual, erect, magnified image (consistent with Table 9.4). Always use sign convention: distances measured against incident light are negative, so $u = -10$ cm.
• Show all steps clearly — substitution, calculation of $v$, then $m$, then $h'$ — each step can carry a mark.

(A) Concave mirror; Nature of image – real

A real image formed by a concave mirror has negative magnification (inverted image). When the image is enlarged (object between F and C), |m| > 1 and m is negative.

Negative magnification means the image is real and inverted (m = –v/u, with v negative and u negative, giving m negative). For |m| > 1, the image must be enlarged, which occurs in a concave mirror when a real image is formed (object between F and C). A convex mirror never forms a real image, and the virtual image from a concave mirror has positive magnification. So option (A) is correct.

(a) Net Power = +4 D + (−2 D) = +2 D

Since the net power is positive, the combination is convergent.

---

(b) Power P = −2.5 D

$$f = \frac{1}{P} = \frac{1}{-2.5} = -0.4 \text{ m} = -40 \text{ cm}$$

The focal length is −40 cm (concave lens).

---

(c) Power = +0.1 D → $f = \frac{1}{0.1} = 10$ m = 1000 cm

Object distance = 20 cm. Since the object is placed between the optical centre and focus (20 cm < 1000 cm = f), the image formed is:

• Virtual, erect, and magnified
• Formed on the same side as the object

(Ray diagram: Two rays from object tip — one parallel to principal axis refracting through F, one through optical centre — diverging rays extended back to meet behind the lens on the same side.)

Source: Light – Reflection and Refraction, Power of a Lens

---

• (a) Simply add the powers algebraically; positive result → converging.
• (b) Use $f(\text{m}) = 1/P$; negative f confirms concave lens.
• (c) First find f from P. Since object distance (20 cm) < f (1000 cm), it's the "between O and F" case → virtual, erect, magnified image. Examiners award 1 mark for correct ray diagram and 1 mark for stating nature/position of image. Draw and label clearly.

Finding focal length:

$$P = \frac{1}{f} \Rightarrow f = \frac{1}{P} = \frac{1}{+4} = +0.25 \text{ m} = 25 \text{ cm}$$

This is a convex lens.

Applying lens formula (u = −50 cm, f = +25 cm):

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{25} + \frac{1}{(-50)} = \frac{2-1}{50} = \frac{1}{50}$$

$$v = +50 \text{ cm}$$

Magnification:

$$m = \frac{v}{u} = \frac{+50}{-50} = -1$$

Nature of image: Real, inverted, and of the same size as the object (formed at 2F₂, 50 cm on the other side of the lens).

Ray Diagram:

```
|
O | I
2F₁ F₁ O F₂ 2F₂
←50cm→|← 50cm →
```
(Object at 2F₁; image forms at 2F₂, real and inverted.)

Source: Chapter 9, Sections 9.3.7 and 9.3.8

---

• Since P = +4D (positive), the lens is convex; f = 1/P in metres = 0.25 m = 25 cm.
• Object is placed at u = −50 cm = 2f, so by Table 9.4, image forms at 2F₂ on the other side — real, inverted, same size.
• Lens formula confirms v = +50 cm; magnification m = v/u = −1 (negative sign → real and inverted; |m| = 1 → same size).
• Draw the standard two-ray diagram: one ray parallel to principal axis refracts through F₂; another through optical centre goes straight. Examiners expect the ray diagram to be labelled with 2F₁, F₁, O, F₂, 2F₂.

Power of a lens: The power of a lens is defined as the reciprocal of its focal length. $P = \dfrac{1}{f}$. Its SI unit is dioptre (D).

Nature of lens: Since $f = -10$ cm $= -0.1$ m, the lens is a concave (diverging) lens.

Power:
$$P = \frac{1}{f} = \frac{1}{-0.1 \text{ m}} = -10 \text{ D}$$

Sign of magnification:
Object distance $u = -20$ cm (by sign convention). Using lens formula with $f = -10$ cm:
$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-10} + \frac{1}{-20} = \frac{-3}{20}$$
$$v = -\frac{20}{3} \text{ cm}$$

$$m = \frac{v}{u} = \frac{-20/3}{-20} = +\frac{1}{3}$$

The sign of magnification is positive, indicating the image is virtual and erect.

Source: Chapter 9, Sections 9.3.7 and 9.3.8

---

• Always define power clearly and state its SI unit — examiners award 1 mark for this.
• A negative focal length always means a concave lens — state this explicitly.
• For sign convention: object is placed to the left, so $u$ is negative. Use the lens formula to find $v$, then compute $m = v/u$. A concave lens always gives a virtual, erect image, so $m$ is always positive for a real object.
• Show the calculation step-by-step for full marks; don't skip to the answer.

(i) The lens X is a concave (diverging) lens (negative power indicates concave lens).

Focal length: $f = \frac{1}{P} = \frac{1}{-2.5} = -0.4 \text{ m} = \mathbf{-40 \text{ cm}}$

This lens is prescribed for myopia (short-sightedness).

---

(ii) Given: $m = -2$, object distance $u = -20$ cm (by sign convention).

Since $m = \frac{v}{u}$: $-2 = \frac{v}{-20}$, so $v = +40$ cm.

(I) Nature: Real and inverted (m is negative).

(II) Size: Image is twice the size of the object (|m| = 2, enlarged).

(III) Position: Image is formed 40 cm on the opposite side of the lens from the object.

(IV) Sign of image height: Negative (inverted image has negative height by sign convention).

---

(iii) Lens A (focal length 10 cm) shows a higher degree of convergence/divergence.

Reason: Power $P = \frac{1}{f}$; smaller focal length means greater power, hence greater ability to converge or diverge light.

Source: Chapter 9, Sections 9.3.4, 9.3.6

---

• For (i): Power formula $P = 1/f$ (f in metres); negative power → concave lens → myopia correction. Examiner expects the eye defect named correctly.
• For (ii): Apply $m = v/u$ with sign convention ($u$ is negative). Negative $m$ means real and inverted image; |m| > 1 means enlarged. Image height is negative when image is inverted.
• For (iii): The key relationship is $P \propto 1/f$ — just state it and compare the two lenses. One line with reason is enough for 1 mark.

(a)
Using lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Given: f = +20 cm, u = −40 cm

$\frac{1}{v} = \frac{1}{20} + \frac{1}{(-40)} = \frac{2-1}{40} = \frac{1}{40}$

∴ v = +40 cm

Image is formed at 40 cm on the other side of the lens; it is real, inverted, and same size as the object.

---

(b)

```
F' O F
←--------|---------|--------|--------→
\ | /
\ virtual, /
\ erect, /
\diminish /
\ image /
```

When object is between O and F of a concave lens, the image is formed on the same side as the object — virtual, erect, and diminished.

(Ray diagram: two rays diverge after refraction; their extensions meet on the same side as the object.)

---

(c)
Given: f₁ = +30 cm, f₂ = −15 cm

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{-15} = \frac{1-2}{30} = \frac{-1}{30}$$

∴ F = −30 cm

Power: $P = \frac{1}{F(\text{in m})} = \frac{1}{-0.30}$

∴ P = −3.33 D

The combination acts as a concave (diverging) lens.

Source: Light – Reflection and Refraction, Lens Formula and Power of a Lens

---

• (a) Apply the standard lens formula with sign convention (u is negative). At 2f, image forms at 2f — real, inverted, same size. Examiners expect the formula, substitution, and nature of image.
• (b) A ray diagram must show at least two rays and indicate the virtual, erect, diminished image on the same side as the object. Even a rough labeled sketch earns the mark.
• (c) Use $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ and $P = P_1 + P_2$. State the sign and nature of the combination — concave — to get full marks. Remember to convert cm to metres for power.

The two laws of refraction of light are:

1. The incident ray, the refracted ray, and the normal to the interface of the two transparent media at the point of incidence, all lie in the same plane.

1. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and given colour of light. This is known as Snell's law.

$$\frac{\sin i}{\sin r} = \text{constant}$$

Source: Chapter 9, Section 9.3.1

---

• Both laws must be stated for full 2 marks (1 mark each).
• Law 1 is about geometry (coplanarity); Law 2 is Snell's law — always name it.
• Writing the formula for Snell's law strengthens the answer but is not mandatory for 2 marks.
• Avoid writing about refractive index here — the question only asks for the laws, not their implications.

(a) Medium B is optically denser than medium A. This is because the ray of light bends towards the normal while entering medium B from medium A, which happens when light travels from a rarer (optically less dense) medium to a denser medium.

(b) The refractive index of medium B with respect to medium A is:

$$n_{BA} = \frac{\text{Speed of light in medium A}}{\text{Speed of light in medium B}} = \frac{V_A}{V_B}$$

Source: Chapter 9, Section 9.3.2 — The Refractive Index

---

• (a): The key rule is: light bends towards the normal → going into denser medium; bends away → going into rarer medium. Always justify with this rule.
• (b): Refractive index of medium 2 w.r.t. medium 1 = speed in medium 1 ÷ speed in medium 2. Don't flip it. Here medium A is "1" and medium B is "2", so it's $V_A / V_B$.

(a) In Case I, object-distance = 45 cm, focal length = 20 cm → radius of curvature = 40 cm. Object is beyond C (45 > 40).
Properties of image:

• Real and inverted
• Diminished (smaller than object)

(b) Case II — object-distance = 30 cm, focal length = 15 cm → radius of curvature = 30 cm. The object is placed at the centre of curvature (C). When an object is placed at C, a concave mirror forms a real, inverted image of the same size at C itself.

(c) Dentists use a concave mirror. It is used because when a tooth (object) is placed within the focal length of a concave mirror, it produces a magnified, erect, and virtual image, enabling the dentist to clearly examine the tooth.

---

• For (a): Compare object-distance with 2f (= radius of curvature). 45 > 40, so object is beyond C → image is real, inverted, and diminished.
• For (b): Image is same size only when object is placed at C (u = 2f). Case II: 2f = 30 cm = object-distance. This is the key condition to state.
• For (c): A standard fact — dentists/doctors use concave mirrors for magnified images. Always justify with the property (virtual, erect, magnified when object is inside focus).

(a) Two applications of a concave mirror:

1. Used as a shaving/makeup mirror (gives magnified, erect image).
2. Used as a reflector in torches, headlights, and solar furnaces.

(b) Focal length (f) = 15 cm

Radius of curvature R = 2f = 2 × 15 = 30 cm

(c) Ray diagram — object placed between pole (P) and focus (F):

```
/
Object / (Reflected rays diverge)
↑ / ←──────────────────────────
| ( Mirror Virtual, erect,
P F magnified image
behind the mirror
```

Key features: Incident ray parallel to principal axis reflects through F; incident ray through C reflects back on itself. Reflected rays diverge — when extended behind the mirror, they meet to form a virtual, erect, and magnified image behind the mirror.

---

• (a) Any two standard applications suffice for 1 mark — examiners accept torch/headlight reflector, solar furnace, shaving mirror, ENT doctor's mirror, etc.
• (b) The formula R = 2f is the key relationship. Always show the working step for a calculation question.
• (c) For 2 marks, examiners look for: correct position of object (between P and F), at least two correct rays drawn, reflected rays shown diverging, and image correctly labelled as virtual, erect, and magnified behind the mirror. Since a proper diagram cannot be drawn in text, draw it neatly on your answer sheet with labelled points P, F, C, object, and image.

(i) Principal Focus of a Diverging (Convex) Mirror:
The point on the principal axis from which incident rays, parallel to the principal axis, appear to diverge after reflection from a convex mirror is called its principal focus (F). It is a virtual focus, located behind the mirror.

(ii) Focal Length:
The distance between the pole (P) of the convex mirror and its principal focus (F) is called the focal length. It is positive for a convex mirror.

Ray Diagram:

```
Incident ray (parallel to principal axis)
──────────────────►
\ Reflected ray (diverging)
\─────────────────►
───────────────F────────P ( ( ( ( convex mirror
*
(Virtual focus F, behind mirror)
```

(Label: P = Pole, F = Principal Focus, C = Centre of Curvature — all behind the mirror)

Source: Chapter 9, Section 9.2.2

---

• Examiners expect both definitions clearly stated (1 mark each) and a labelled diagram (1 mark).
• Key point: the principal focus of a convex mirror is virtual — reflected rays only appear to diverge from F; they don't actually meet there.
• In the diagram, show the incident ray parallel to the principal axis, the reflected ray going outward, and a dotted line extended behind the mirror meeting at F. Label P, F, and C.
• Focal length for a convex mirror is taken as positive under New Cartesian Sign Convention.

It looks like sub-question (c) is incomplete — the two options (i) and (ii) were not included in your message. I'll answer (a) and (b) fully, and note the gap for (c).

---

(a) The principal axis of a concave mirror is the straight line passing through the centre of curvature and the pole of the mirror.

(b) A ray parallel to the principal axis, after reflection from a concave mirror, passes through the focus (F). Therefore, the focal length $f = 10$ cm.

$$R = 2f = 2 \times 10 = \textbf{20 cm}$$

The radius of curvature of the mirror is 20 cm.

(c) (The options (i) and (ii) were not provided in the question. Please share them so this part can be answered.)

Source: Light – Reflection and Refraction, Spherical Mirrors

---

• For (a): Examiners expect both reference points — the pole and the centre of curvature — to be mentioned for full marks.
• For (b): The key concept is that a ray parallel to the principal axis reflects through the focus. So the given 10 cm distance is the focal length, and $R = 2f$. Show the formula and calculation clearly.
• For (c): Once you share the options, the answer will be written to the full 2-mark length. Don't leave it blank in the actual exam.

(B) – 20 cm

Using mirror formula: $u = -30$ cm, $v = -60$ cm (real image, in front of mirror).

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-60} + \frac{1}{-30} = \frac{-1-2}{60} = \frac{-3}{60} = \frac{1}{-20}$$

So, $f = -20$ cm.

Apply the mirror formula with correct signs (New Cartesian Sign Convention): object and real image both in front of the mirror, so both u and v are negative. A concave mirror always has a negative focal length. Students commonly make sign errors — always assign signs before substituting.

(a) When a ray of light passes through the centre of curvature (C) of a concave mirror, it reflects back along the same path. But a ray directed towards the focus (F) reflects parallel to the principal axis.

Ray Diagram:

```
C F P
------------* ------→ (reflected ray, parallel to PA)
\ ↗
\/
(incident ray aimed at F)
```
(Label: C = Centre of curvature, F = Focus, P = Pole, PA = Principal Axis)

(b)(i)
| Feature | Convex Mirror | Concave Mirror |
|---|---|---|
| Size of virtual image | Always diminished | Always enlarged |
| Object position | Any position | Between pole and focus |

(b)(ii)
Given: m = –2, height of image (h′) = 20 cm

$$m = \frac{h'}{h} \Rightarrow -2 = \frac{20}{h} \Rightarrow h = \frac{20}{-2} = -10 \text{ cm}$$

Height of object = 10 cm (negative sign indicates object is real/upright as per convention).

---

• For (a), the key rule: a ray directed towards F of a concave mirror reflects parallel to the principal axis. Labelling C, F, P is essential for full marks.
• For (b)(i), examiners look for a clear comparative point — size difference is the most important distinguishing feature.
• For (b)(ii), use the formula m = h′/h directly. Note m = –2 means the image is real and inverted. The object height magnitude is 10 cm.