Two surfaces AB and CD of a thick lens have radii of curvature R and 2R respectively. Thickness at the centre is 3R . Refractive index of the material of the lens is 1.5 . The image of a point object placed on the axis of the lens at a distance 4R from the surface of smaller radius of curvature and to its left is formed at

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$\frac{8 R}{3}$ from CD to its right

$\frac{8 R}{3}$ from CD to its left

$\frac{4 R}{3}$ from CD to its right

$\frac{4 R}{3}$ from CD to its left

answer is A.

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Detailed Solution

Surface AB forms the image of O at v₁ from it. We have $\frac{1 .5}{v_{1}} - \frac{1}{- 4 R} = \frac{1 .5 - 1}{R}$ $\Rightarrow v_{1} = 6 R$
$∴$ For the surface $CD, u_{2} = 6 R - 3 R = + 3 R$ = object distance
Hence $u_{2}$ the image distance due to surface CD is got by
Giving $\frac{1}{v_{2}} = \frac{1}{4 R} + \frac{1}{2 R} = \frac{3}{4 R}$
or $v_{2} = \frac{4 R}{3}$
A real image is formed at a distance $\frac{4 R}{3}$ from the surface CD to its right