JEE Main 2014 · 11 Apr · Shift 1 · Physics Q23

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An object is located in a fixed position in front of a screen. Sharp image is obtained on the screen for two positions of a thin lens separated by 10 cm. The size of the images in two situations are in the ratio 3 : 2. What is the distance between the screen and the object ?

124.5 cm

144.5 cm

65.0 cm

99.0 cm

Answer and worked solution

Official answer: D

In the displacement method the two lens positions swap $u$ and $v$ , so the magnifications are $\frac vu$ and $\frac uv$ . Their ratio is $\left(\frac vu\right)^2=\frac32$ , so $\frac vu=1.2247$ .

With $v-u=10$ cm: $u=44.5$ cm and $v=54.5$ cm. The object-screen distance is $u+v=99.0$ cm.

Topics:Thin lenses and lens maker's formula

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 23