A student conducts an experiment to study image formation by a convex lens. She places a burning candle at various distances from the lens (focal length = 10 cm) and records the image position and nature on a screen. When the candle is 30 cm from the lens, she gets a sharp image on the screen. When she moves the candle to 8 cm from the lens, she cannot get an image on the screen at all but sees a magnified image when she looks through the lens from the other side.
(i) Using the lens formula, verify the image distance when the object is at 30 cm from the lens. [1 mark]
(ii) Why can the student not obtain an image on the screen when the candle is at 8 cm? Describe the nature and position of the image formed. [1 mark]
(iii) Calculate the magnification when the object is at 30 cm. Is the image erect or inverted? [1 mark]
(iv) Name one common optical instrument that uses the principle of image formation observed when the candle is at 8 cm from the lens. [1 mark]
Generated by claude-sonnet-4-6 · 2026-06-26 01:13 · grounding rag
Model Answer
(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
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Explanation
- (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.