A small object o is placed on the principal axis of an equi-convex lens at a distance of 30cm. The refractive index of glass of lens is 1.5 and its surfaces has radius of curvature R=20cm. Two glass slabs S₁ and S₂ have been placed behind the lense . Thickness of two slabs is 6cm & 4cm and their refractive indices are 3/2 and 2 respectively . Which of the following options are true?

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The distance of final image measured from lense is 62 cm

The magnification of final image is -1.5

The position of final image, changes if the slab S₂ is move to left so as put it in contact with S₁

The position of image does not change if the two slabs in contact are together move to right by a distance of 100cm

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{1}{f} = (μ - 1) \frac{2}{R} = (1 .5 - 1) \frac{2}{R} \Rightarrow f = R = 20 cm \\ \frac{1}{v} - \frac{1}{4} = \frac{1}{f} \\ \frac{1}{v} - \frac{1}{- 30} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30} \\ v = 60 cm \\ m = \frac{v}{u} = \frac{60}{- 30} = - 2 \end{array}$
Shift due to S₁ is $: - t_{1} (1 - \frac{1}{μ_{1}}) = 6 (1 - \frac{2}{3}) = 2 cm$
$t_{2} (1 - \frac{1}{μ_{2}}) = 4 (1 - \frac{1}{2}) = 2 cm$
$∴$ Final image will be at a distance of 64cm from the lens glass slab does not produce
magnification position of image will remain unchanged in both options(C)and (D)