(C) Virtual, erect and enlarged

When an object is placed between the pole (P) and the principal focus (F) of a concave mirror, the image formed is virtual, erect and enlarged.

Source: Light – Reflection and Refraction, Table 9.1

Table 9.1 in the textbook clearly states: when the object is placed between P and F, the image is formed behind the mirror, is enlarged, and is virtual and erect. This is also the principle behind concave mirrors used as shaving/makeup mirrors. Examiners expect you to recall this directly from the table.

(B) Bend away from the normal

Glass is a denser medium and air is a rarer medium. A light ray travelling from a denser medium to a rarer medium bends away from the normal.

The key rule to remember: denser → rarer = bends away from normal; rarer → denser = bends towards normal. Glass has a higher refractive index than air, making it the denser medium. The source passage explicitly states: "A light ray travelling obliquely from a denser medium to a rarer medium bends away from the normal." This is the direct fact the examiner tests here.

(B) u = –15 cm, measured to the left of the pole.

According to New Cartesian Sign Convention, distances measured to the left of the pole are negative; since the object is placed in front of (to the left of) the mirror, u = –15 cm.

The key rule: all distances are measured from the pole, and leftward distances are negative. The object is always placed to the left of the mirror, so object distance u is always negative. Options A, C, and D are incorrect — distances are not measured from the focus, and leftward is negative (not positive). The correct MCQ answer is (B); the correct option statement is (C) — "Distances measured to the left of the pole are negative."

Source: Chapter 9, Section 9.2.3 Sign Convention for Reflection by Spherical Mirrors

(B) At infinity

When an object is placed at the principal focus (f = 20 cm, u = 20 cm) of a convex lens, the image is formed at infinity.

From Table 9.4 (Ch. 9): when the object is placed at focus F₁, the image is formed at infinity. Here, object distance = focal length = 20 cm, so the object is at F₁. This is a standard result students must memorise from the image formation table for convex lenses.

(C) Concave, with radius of curvature 24 cm

A negative focal length indicates a concave mirror. Using $R = 2f$: $R = 2 \times 12 = 24$ cm.

In the New Cartesian Sign Convention, a negative focal length means the focus lies in front of the mirror — this is a concave mirror. The formula $R = 2f$ (magnitude only for R) gives $R = 24$ cm. Students often confuse sign of focal length with mirror type: remember, concave → negative $f$; convex → positive $f$.

(D) –25 cm, concave

$f = \frac{1}{P} = \frac{1}{-4.0} = -0.25\ \text{m} = -25\ \text{cm}$. Negative focal length indicates a concave (diverging) lens.

Use $f = \frac{1}{P}$, always convert to cm. A negative power → negative focal length → concave lens. Examiners expect the formula, substitution, and correct identification of lens type.

(D) Air

Light travels fastest in air among the given options, as air has the lowest refractive index (1.0003), meaning the speed of light in air is closest to its speed in vacuum ($3 \times 10^8$ m s$^{-1}$).

Source: Chapter 9, Section 9.3.2 (Table 9.3)

Higher refractive index = lower speed of light. From Table 9.3: Air (1.0003) < Water (1.33) < Crown glass (1.52) < Diamond (2.42). So light is slowest in diamond and fastest in air among these four options. Remember: speed and refractive index are inversely related.

(B) Virtual, erect and diminished

A concave lens always produces a virtual, erect and diminished image, irrespective of the position of the object.

Table 9.5 (Chapter 9) explicitly states that for all object positions, a concave lens gives a virtual and erect image that is diminished. The textbook concludes: "A concave lens will always give a virtual, erect and diminished image, irrespective of the position of the object." Remember: concave lens = diverging lens → rays never actually meet → no real image possible.

(C) A convex mirror, because it provides a wider field of view and always forms an erect image.

Convex mirrors always give a virtual, erect, and diminished image regardless of object distance, and have a wider field of view — hence preferred as rear-view mirrors.

Source: Chapter 9, Uses of convex mirrors

---

The key clues in the question are: image is smaller than actual (diminished) and always upright (erect) regardless of distance. Only a convex mirror satisfies both conditions for all object positions (see Table 9.2). The textbook explicitly states convex mirrors are used as rear-view mirrors because they always give an erect, diminished image and a wider field of view. Eliminate (A) — plane mirrors give same-size images; eliminate (B) and (D) — concave mirrors can form inverted/real images.

(B) The image is virtual, erect and diminished.

Since m = +0.5, the positive value indicates the image is virtual and erect, and |m| = 0.5 < 1 indicates the image is diminished.

For spherical mirrors, m = −v/u. A positive value of m means the image is virtual and erect (as stated in section 9.2.4: "A positive sign in the value of the magnification indicates that the image is virtual"). A magnitude less than 1 (|m| = 0.5) means the image is diminished. This matches a convex mirror, which always forms virtual, erect, diminished images.

(D) Emerges without any deviation

A ray of light passing through the optical centre of a lens emerges without any deviation.

Source: Chapter 9, Section 9.3.5

This is a standard ray diagram rule for lenses. The three rules for ray diagrams are: (i) parallel ray → passes through F₂, (ii) ray through F₁ → emerges parallel, (iii) ray through optical centre → no deviation. Examiners expect option D directly; briefly quoting the rule earns full marks.

(C) +2.0 D

When two lenses are placed in contact, the net power is the algebraic sum of their individual powers.
$$P = P_1 + P_2 = +3.5\text{ D} + (-1.5\text{ D}) = +2.0\text{ D}$$

Source: Chapter 9, Section 9.3.8 Power of a Lens

---

The key formula is $P = P_1 + P_2 + \ldots$ (powers simply add algebraically). Remember that convex lens power is positive and concave lens power is negative. This is a straightforward one-step calculation — just add with signs carefully.

(B) Concave mirror, because when the object is placed within the focal length, it forms a magnified, virtual, and erect image — useful for examining the eye closely.

Source: Light – Reflection and Refraction, Section 9.2 (Uses of concave mirrors)

The key property tested here: a concave mirror produces a virtual, erect, enlarged image only when the object is placed between the pole and the principal focus (within focal length). This is exactly how a shaving/examination mirror works. Option (D) is wrong because a concave mirror does NOT always form a real enlarged image — it depends on object position. The correct answer matches Option (C) "Shaving mirror" from the options list, which uses a concave mirror for this reason.

Option A — Both A and R are true, and R is the correct explanation of A. Glass ($n$ = 1.52) has a higher refractive index than water ($n$ = 1.33), making it optically denser; so light travelling from water into glass bends towards the normal.

When light moves from a rarer medium (water, lower $n$) to a denser medium (glass, higher $n$), it slows down and bends towards the normal — this directly supports Assertion A. The Reason correctly states that glass is optically denser than water (1.52 > 1.33) and this is precisely why the bending towards the normal occurs, making R the correct explanation of A. Remember: optically denser = higher refractive index = light bends towards normal.

Option A is correct. Both A and R are true, and R is the correct explanation of A. When the object is between F₁ and optical centre O, a convex lens forms a virtual, erect, enlarged image — this case directly explains why a convex lens can form virtual images, supporting Assertion A.

Source: Chapter 9, Section 9.3.4 (Table 9.4)

• A is true: Table 9.4 confirms a convex lens forms real images (object beyond F₁) AND virtual images (object between F₁ and O).
• R is true: Table 9.4 explicitly states that when the object is between F₁ and O, the image is virtual and erect.
• R correctly explains A: The reason directly gives the specific condition under which a convex lens forms a virtual image, thus explaining why both types of images are possible — making it the correct explanation, not just an additional fact.

Answer: A

Both A and R are true, and R is the correct explanation of A. The bending at the air-glass surface (AB) is exactly reversed at the glass-air surface (CD), giving zero net angular deviation but a lateral displacement.

Source: Chapter 9, Section 9.3.1

The textbook explicitly states: "The extent of bending of the ray of light at the opposite parallel faces AB and CD is equal and opposite. This is why the ray emerges parallel to the incident ray." This directly confirms both the assertion (parallel but laterally displaced emergent ray) and the reason (equal and opposite bending at the two surfaces = zero net angular deviation). Since R correctly and completely explains A, option A is correct.

Answer: C

Assertion (A) is false — optical density and mass density do not always increase together. Reason (R) is true — kerosene has a higher refractive index (1.44) than water (1.33), making it optically denser, yet its mass density is lower than water.

The textbook explicitly states: "an optically denser medium may not possess greater mass density — kerosene having higher refractive index is optically denser than water, although its mass density is less than water." So Assertion A is false (the two densities are independent), while Reason R correctly states a fact from Table 9.3. Option C is correct.

Option A is correct. Both A and R are true, and R is the correct explanation of A. Under the New Cartesian Sign Convention, distances measured to the left of the pole (in front of the mirror) are negative. Since the principal focus of a concave mirror lies in front of the mirror, its focal length is negative.

The key is linking the sign convention rule (left of pole = negative) directly to the position of the principal focus of a concave mirror (in front = left side). R correctly and completely explains why f is negative for a concave mirror, making Option A the right choice.

Physical meaning: The absolute refractive index of diamond is 2.42 means that the speed of light in air is 2.42 times the speed of light in diamond (i.e., light travels 2.42 times faster in air than in diamond).

Calculation:

$$n = \frac{c}{v} \Rightarrow v = \frac{c}{n} = \frac{3 \times 10^8}{2.42}$$

$$v = 1.24 \times 10^8 \text{ m s}^{-1}$$

The speed of light in diamond is $1.24 \times 10^8$ m s⁻¹.

Source: Chapter 9, Section 9.3.2 – The Refractive Index

---

• For the meaning part (1 mark): state clearly that the refractive index is the ratio of speed of light in air to speed in the medium. Saying "light slows down by a factor of 2.42 in diamond" earns the mark.
• For the calculation part (1 mark): use $n = c/v$, substitute correctly, and show the final answer with units. Division: $3 \times 10^8 \div 2.42 \approx 1.24 \times 10^8$ m s⁻¹.
• Do not just write the formula without substituting values — examiners expect full working for numericals.

Position: At C (i.e., at 30 cm in front of the mirror)
Size: Same as the object
Nature: Real and inverted

Verification using mirror formula:
Given: f = −15 cm, u = −30 cm (object at C, so R = 2f = 30 cm)

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

$$\frac{1}{v} = \frac{1}{-15} - \frac{1}{-30} = \frac{-2+1}{30} = \frac{-1}{30}$$

$$v = -30 \text{ cm}$$

Image is formed at C, 30 cm in front of the mirror. ✓

Source: Table 9.1 and Section 9.2.4, Chapter 9

• From Table 9.1: when object is at C, image is also at C — same size, real, inverted.
• For mirror formula, use sign convention: distances in front of mirror are negative. f = −15 cm, u = −30 cm (since R = 2f).
• The calculated v = −30 cm confirms the image is at C, verifying the table result.
• Examiners award marks for: correct position/size/nature (1 mark) and correct mirror formula calculation (1 mark).

(i) The negative sign indicates that the image is real and inverted.

(ii) Using $m = -\dfrac{v}{u}$, with $m = -3$ and $u = -10$ cm:

$$-3 = -\frac{v}{-10} \Rightarrow -3 = \frac{v}{10} \Rightarrow v = -30 \text{ cm}$$

Image distance = –30 cm (30 cm in front of the mirror).

(iii) The mirror is concave. It forms a real, enlarged image (|m| = 3 > 1) of an object placed in front of it, which is a property of a concave mirror.

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

---

• A negative magnification always means real and inverted (stated explicitly in the textbook).
• Use $m = -v/u$; remember $u$ is negative by sign convention (object in front).
• A convex mirror always gives virtual, diminished images (m < 1 in magnitude), so a real, enlarged image can only be produced by a concave mirror — examiners expect this justification.

When light rays travel from water (denser medium) to air (rarer medium), they bend away from the normal at the surface. The refracted rays, when extended backwards, appear to meet at a point higher than the actual position of the coin. This is refraction of light, which makes the coin appear raised/shallower than its real depth.

Source: Chapter 9, Section 9.3 Refraction of Light

---

• The key concept is refraction — light bends when passing from one medium to another.
• Specifically state the direction: denser → rarer medium → bends away from normal.
• Mention that the apparent position is higher than the actual position because the eye traces refracted rays backward in a straight line.
• Examiners look for: (1) naming refraction, (2) explaining the bending direction/apparent shift. Both points earn the 2 marks.

Given: h = +3 cm, u = –12 cm, f = –8 cm (concave mirror)

(i) Image distance:

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

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-8} - \frac{1}{-12} = -\frac{1}{8} + \frac{1}{12} = \frac{-3+2}{24} = \frac{-1}{24}$$

$$v = -24 \text{ cm}$$

The image is formed 24 cm in front of the mirror.

(ii) Height of the image:

$$m = -\frac{v}{u} = -\frac{(-24)}{(-12)} = -2$$

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

Nature of the image: Real, inverted, and enlarged (magnified 2 times).

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

---

• Always assign signs first using New Cartesian Sign Convention: distances in front of mirror are negative, so u = –12, f = –8.
• A negative v confirms the image is real (in front of mirror).
• A negative h′ confirms the image is inverted.
• Examiners award marks for: correct formula application (1 mark), correct v (1 mark), correct h′ and nature (1 mark). Missing the nature statement loses a mark.

(i) Ray Diagram:

Object is placed between F₁ and O (since object distance 6 cm < focal length 10 cm).

```
|
\ | /
\ | /
─────\──┼──/─────── Principal axis
\ | /
\|/
O F₁ (lens) F₂
(6cm) (10cm) (10cm)
```

Draw two rays from the top of the candle: one parallel to the principal axis (refracts through F₂), and one through the optical centre (passes straight). The refracted rays diverge; when extended backwards, they meet on the same side as the object, forming a virtual, erect, enlarged image.

(ii) The image is formed in front of the lens (same side as the object). It is virtual, erect, and enlarged (larger than the object).

(iii) No, this image cannot be obtained on a screen, because it is a virtual image. Virtual images are formed by the apparent intersection of diverging rays and cannot be projected onto a screen.

Source: Chapter 9, Section 9.3.4 & 9.3.5

---

• The key fact: object distance (6 cm) < focal length (10 cm), so the object lies between F and O. From Table 9.4, this always gives a virtual, erect, enlarged image on the same side as the object.
• Examiners award marks for: correct identification of object position → correct image characteristics → correct reason for screen question.
• Virtual images cannot be caught on a screen — this is the standard one-line reason required.
• For the ray diagram, show at least two rays and the image on the same side as the object (dotted lines for virtual rays).

(i) Angle of Refraction:

Using Snell's law: $n = \dfrac{\sin i}{\sin r}$

$$1.5 = \frac{\sin 45°}{\sin r} = \frac{0.707}{\sin r}$$

$$\sin r = \frac{0.707}{1.5} = 0.471$$

$$r = \sin^{-1}(0.471) \approx 28.1°$$

(ii) At the second parallel surface, the ray goes from glass (denser) to air (rarer), so it bends away from the normal. A ray travelling from a denser medium to a rarer medium speeds up and bends away from the normal.

(iii) Speed of light inside glass is less than in air. Since $n = \dfrac{c}{v}$, and $n = 1.5 > 1$, the speed in glass $v = \dfrac{3 \times 10^8}{1.5} = 2 \times 10^8$ m s$^{-1}$.

Source: Chapter 9, Sections 9.3.1 and 9.3.2

---

• Part (i): Apply Snell's law directly. Show the substitution step — examiners award marks for working, not just the final answer.
• Part (ii): The key reason is denser→rarer = bends away from normal. This is a one-line conceptual answer; state the direction AND the reason.
• Part (iii): Use $n = c/v$ to both explain and calculate. The answer $2 \times 10^8$ m s$^{-1}$ shows the speed is reduced to ⅔ of its value in air.

Given:
$n_{AB} = \frac{2}{3}$, $n_{B,\text{air}} = 1.5$

Finding absolute refractive index of A:

$$n_{B,\text{air}} = 1.5 \Rightarrow n_B = 1.5$$

$$n_{AB} = \frac{n_A}{n_B} = \frac{2}{3} \Rightarrow n_A = \frac{2}{3} \times 1.5 = 1.0$$

Summary of refractive indices:

• Air ≈ 1.0
• Medium A = 1.0
• Medium B = 1.5

(i) Highest optical density: Medium B, as it has the highest refractive index (1.5). A higher refractive index means greater optical density.

(ii) Lowest optical density: Air (and Medium A), both have refractive index ≈ 1.0, making them equally and least optically dense.

Source: Chapter 9, Section 9.3.2

---

• The key formula used is $n_{AB} = \dfrac{n_A}{n_B}$, which lets you find the absolute refractive index of A.
• Examiners expect you to show the calculation of $n_A$ clearly.
• Optical density is directly proportional to refractive index — higher refractive index = optically denser medium (as stated in the "More to Know" box).
• Note: optical density ≠ mass density. Don't confuse the two.
• If $n_A = 1.0$ exactly equals air, full credit is given for naming both as lowest; mentioning Medium A alone is also acceptable.

(i) Focal length of each lens:

Using $P = \dfrac{1}{f}$, so $f = \dfrac{1}{P}$

• For +2.0 D: $f = \dfrac{1}{+2.0} = +0.5 \text{ m}$
• For –0.5 D: $f = \dfrac{1}{-0.5} = -2.0 \text{ m}$

(ii) Type of each lens:

• Lens with power +2.0 D is a convex (converging) lens (positive power).
• Lens with power –0.5 D is a concave (diverging) lens (negative power).

(iii) Power of combination:

$$P = P_1 + P_2 = (+2.0) + (-0.5) = +1.5 \text{ D}$$

The combination acts as a convex lens of power +1.5 D.

Source: Chapter 9, Section 9.3.8 Power of a Lens

---

• The key formula is $f = 1/P$ (with $f$ in metres and $P$ in dioptres).
• Examiners expect you to state the sign and type of lens — positive power = convex, negative power = concave.
• For the combination, simply add powers algebraically: $P = P_1 + P_2$. Always state what type of lens the resultant power represents.
• Common mistake: forgetting to express focal length in metres, not centimetres.

Change in image as object moves from infinity to pole:

When the object is at infinity, the image is formed at the focus F (behind the mirror) — it is highly diminished (point-sized), virtual and erect. As the object moves closer towards the pole, the image shifts from F towards the pole P, remains virtual and erect, but increases in size (though always diminished compared to the object). At the pole, the image is at P, same size as the object.

Property making it suitable for shopping malls and road bends:

A convex mirror has a wider field of view (curved outward) and always forms a virtual, erect, and diminished image regardless of object position. This allows a larger area to be viewed in a small mirror, making it ideal for surveillance in shopping malls and for monitoring traffic around road bends.

Source: Chapter 9, Section 9.2 — Image formation by a Convex Mirror / Uses of convex mirrors

---

• Examiners expect you to clearly state both positions (infinity and between infinity and pole) and the change in image (position shifts from F toward P, size increases but stays diminished).
• For the second part, the two key properties are: wider field of view + always erect & diminished image. Mention both for full marks.
• Table 9.2 is the direct source for image positions — memorise it.
• "Virtual and erect" must always be stated as the nature of image in a convex mirror.

(a) Mirror Formula:

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

• u = object distance (from pole of mirror)
• v = image distance (from pole of mirror)
• f = focal length (distance of principal focus from pole)

(b) Finding image position:

Given: f = –25 cm (concave mirror), u = –40 cm

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-25} - \frac{1}{-40} = -\frac{1}{25} + \frac{1}{40} = \frac{-8+5}{200} = \frac{-3}{200}$$

$$v = -\frac{200}{3} \approx -66.7 \text{ cm}$$

The image is formed 66.7 cm in front of the mirror. Since v is negative, the image is real.

(c) Magnification and image height:

$$m = -\frac{v}{u} = -\frac{(-66.7)}{(-40)} = -1.67$$

Height of image: $h' = m \times h = -1.67 \times 2 = -3.34 \text{ cm}$

The image is 3.34 cm tall. The negative sign indicates the image is inverted.

Source: Chapter 9, Section 9.2.4

---

• Always apply sign convention: distances in front of the mirror are negative; concave mirror focal length is negative.
• A negative value of v confirms a real image (formed in front of the mirror).
• A negative magnification confirms the image is inverted; magnitude > 1 means it is enlarged.
• Examiners award marks separately for formula, substitution, answer, and nature of image — show each step clearly.

(a) Ray Diagrams:

Case (i) — Object beyond C:

```
Object C F P
↓ | | |
----[ray 1→ parallel to axis → reflects through F]
----[ray 2→ through C → reflects back along same path]
Image (between F and C, inverted, diminished)
```

Draw two rays: one parallel to the principal axis (reflects through F) and one through C (reflects back along the same path). They meet between F and C to form a real, inverted, diminished image.

Case (ii) — Object between P and F:

Draw one ray parallel to the axis (reflects through F) and one through C. The reflected rays diverge; when extended behind the mirror they meet to form a virtual, erect, enlarged image behind the mirror.

(b) Nature, position and size of image:

| Position of Object | Position of Image | Size | Nature |
|---|---|---|---|
| Beyond C | Between F and C | Diminished | Real, Inverted |
| Between P and F | Behind the mirror | Enlarged | Virtual, Erect |

(c) Calculation:

Given: u = −2 cm, f = −8 cm

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

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-8} - \frac{1}{-2} = -\frac{1}{8} + \frac{1}{2} = \frac{3}{8}$$

$$v = +\frac{8}{3} \approx +2.67 \text{ cm}$$

Magnification: $m = -\dfrac{v}{u} = -\dfrac{+8/3}{-2} = +\dfrac{4}{3} \approx +1.33$

The image is virtual, erect, and magnified (m > 1). Since the tooth is between P and F, the concave mirror forms an enlarged, erect, virtual image behind the mirror, allowing the dentist to see a clear, magnified view of the tooth. This makes it very suitable for dental examination.

Source: Chapter 9, Section 9.2.1 and 9.2.2

---

• Ray diagrams: Always use two standard rays. For concave mirror, the most convenient pair is (i) parallel to axis → reflects through F, and (ii) through C → reflects back along the same path.
• Table 9.1 from the textbook directly gives the nature, position, and size for both cases — memorise it.
• Part (c): Object at 2 cm, focal length 8 cm → object is between P and F (since 2 < 8). Mirror formula gives a positive v, confirming a virtual image behind the mirror. Positive magnification > 1 confirms virtual, erect, enlarged image — exactly what a dentist needs to examine teeth closely.
• Sign convention: distances measured from pole; distances in direction of incident light are positive, opposite are negative. Here u and f are both negative (real object, concave mirror).

(a)
Speed of light in X, $v_X = 2 \times 10^8$ m/s; in Y, $v_Y = 1.5 \times 10^8$ m/s

$$n_{YX} = \frac{v_X}{v_Y} = \frac{2 \times 10^8}{1.5 \times 10^8} = \frac{4}{3} \approx 1.33$$

Since $n_{YX} > 1$, medium Y is optically denser (higher refractive index means lower speed of light).

(b)
Since Y is optically denser than X, light slows down when passing from X into Y. According to Snell's law, the ray bends towards the normal at the interface.

(c)

• Convex lens: $f_1 = +15$ cm $= +0.15$ m → $P_1 = \dfrac{1}{0.15} = +6.67$ D
• Concave lens: $f_2 = -30$ cm $= -0.30$ m → $P_2 = \dfrac{1}{-0.30} = -3.33$ D

(ii) Net power: $P = P_1 + P_2 = 6.67 + (-3.33) = +3.33$ D

Focal length of combination: $f = \dfrac{1}{P} = \dfrac{1}{3.33} \approx +0.30$ m $= +30$ cm

(iii) Since net power is positive, the combination behaves as a converging (convex) lens.

Source: Chapter 9, Sections 9.3.2 and 9.3.8

---

• For refractive index of Y w.r.t. X, use $n_{YX} = v_X / v_Y$ (speed in the first medium divided by speed in the second).
• The medium with higher refractive index = lower speed = optically denser; light bends towards the normal entering a denser medium — a direct application of Snell's law logic.
• For lens powers: convex focal length is positive, concave is negative. Convert cm to metres before calculating power.
• Powers add algebraically; a positive net power → converging combination. Examiners expect all three sub-parts of (c) answered clearly with units.

(i) Given: u = −30 cm, f = +10 cm

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{(-30)} = \frac{3-1}{30} = \frac{2}{30}$$

$$v = +15 \text{ cm}$$

The image is formed 15 cm on the other side of the lens. ✓

(ii) The candle is at 8 cm, which is between the focus and the optical centre (f = 10 cm). A convex lens forms a virtual, erect and magnified image on the same side as the object in this case. Since the image is virtual, it cannot be collected on a screen.

(iii)
$$m = \frac{v}{u} = \frac{+15}{-30} = -0.5$$

The image is inverted (negative magnification) and diminished.

(iv) A magnifying glass (simple microscope) uses this principle (object placed between F and O of a convex lens).

Source: Chapter 9, Sections 9.3.4 and 9.3.7

---

• (i) Always apply sign convention: u is negative (object on left), f is positive for convex lens. Show the substitution step clearly for full marks.
• (ii) The key idea from Table 9.4 is: object between F₁ and O → virtual, erect, magnified image on the same side. "Virtual" means no screen image — examiners look for this reason explicitly.
• (iii) m = v/u directly. Negative m means inverted image. At 30 cm (between F and 2F), the image is real, inverted, and diminished — check this matches Table 9.4 ("between F₁ and 2F₁ → beyond 2F₂, enlarged"; here object is at 2F so image is at 2F, same size — but with u=30 cm > 2f=20 cm, object is beyond 2F, so image is diminished. m = −0.5 confirms this).
• (iv) Magnifying glass is the standard one-word answer expected here. A compound microscope's eyepiece also works but magnifying glass is safer for 1 mark.

(i) Using mirror formula: f = −15 cm, u = −40 cm

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-40} = \frac{-8+3}{120} = \frac{-5}{120}$$

∴ v = −24 cm. The image is formed 24 cm in front of the mirror. It is real and inverted.

(ii) The image is the same size as the object when the object is placed at C (centre of curvature), i.e., at 30 cm (= 2f) from the mirror.

(iii) When the candle is placed at F (15 cm), no image is formed on the screen. This is because the reflected rays become parallel to the principal axis and meet only at infinity.

(iv) Object at 10 cm is between P and F. The image is formed behind the mirror, is virtual, erect and enlarged.

Source: Chapter 9, Section 9.2.1 (Table 9.1) and Section 9.2.4

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• (i) Apply the mirror formula with sign convention (concave mirror: f negative, object distance negative). v = −24 cm confirms real image (negative v in concave mirror context).
• (ii) Table 9.1 directly states: object at C → image at C, same size. C = 2f = 30 cm.
• (iii) Object at F → reflected rays are parallel → image at infinity → no image on screen. Examiners expect the reason about parallel rays.
• (iv) Object between P and F → virtual, erect, enlarged image behind mirror. This is the standard result from Table 9.1. No calculation needed for 1 mark.

(i) Light travels at the lowest speed in diamond (n = 2.42). Higher refractive index means lower speed of light in the medium ($n = c/v$), so the highest n gives the lowest v.

(ii) No. Optical density (refractive index) is not the same as mass density. Kerosene is optically denser than water (higher n), but its mass density is actually less than that of water.

(iii) The light bends towards the normal. Crown glass (n = 1.52) is optically denser than water (n = 1.33). When light travels from a rarer to a denser medium, it slows down and bends towards the normal.

(iv)
$$n = \frac{c}{v} \implies v = \frac{c}{n} = \frac{3 \times 10^8}{1.52} \approx 1.97 \times 10^8 \text{ m s}^{-1}$$

Source: Light – Reflection and Refraction, Section 9.3.2

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• (i) The key formula is $n = c/v$; highest n → lowest v. Diamond has n = 2.42, the highest in the table.
• (ii) The textbook explicitly states: "an optically denser medium may not possess greater mass density" and uses kerosene vs water as the exact example. Quote or paraphrase this directly.
• (iii) Compare refractive indices: water (1.33) < crown glass (1.52), so water is rarer. Rarer → denser means bending towards the normal.
• (iv) Straightforward formula application. Show the formula, substitution, and answer with units. $3 \times 10^8 / 1.52 \approx 1.97 \times 10^8$ m s⁻¹ is the expected result.