CBSE Class 10 Science 2017 · Paper 531 · Question 23

(a) Lens A (convex, $f=+15$ cm): the candle, 20 cm away, is between F₁ and 2F₁, so the student obtains a sharp, real, inverted and enlarged image of the flame on the screen when the screen is 60 cm from the lens on the other side ($v=+60$ cm, $m=-3$: three times the size of the flame). The refracted rays actually meet there, so the image can be caught on the screen. Lens B (concave, $f=-15$ cm): he obtains no image of the flame on the screen at any position. A concave lens makes the rays spread out; they only appear to come from a virtual, erect, diminished image on the same side as the candle ($v=-\frac{60}{7}\approx -8.6$ cm, $m=+\frac{3}{7}\approx +0.43$), and a virtual image cannot be caught on a screen. Ray diagrams: for lens A, a ray from the tip of the flame parallel to the axis passes through F₂ and a ray through O goes straight on; they meet beyond 2F₂. For lens B, the parallel ray spreads out as if from F₁ and the ray through O goes straight on; the two diverge, and their backward extensions meet between F₁ and O. (b) $f=\frac{R}{2}=+1.0$ m and $u=-3.0$ m, so $\frac{1}{v}=\frac{1}{1.0}+\frac{1}{3.0}=\frac{4}{3}$ and $v=+0.75$ m. The image is 0.75 m behind the mirror (between P and F); it is virtual, erect and diminished, with $m=+0.25$, one-quarter of the size of the bus.