QJEEM25Ph6ab9: JEE Main 2025 Physics - Class 12 JEEM25Ph6ab9 Solution: Two concave refracting surfaces of equal radii of curvature and …

Using lens maker formula,

${{{\mu _2}} \over v} - {{{\mu _1}} \over 4} = {{{\mu _2} - {\mu _1}} \over R}$

For the image formed by the surface on the right side.

${\mu _1} = 1,\,{\mu _2} = 1.5,\,u = {{ - R} \over 2},{R_1} = - R$

so, ${{1.5} \over v} - {1 \over { - R/2}} = {{1.5 - 1} \over { - R}}$

$\Rightarrow {{1.5} \over v} + {2 \over R} = {{ - 1} \over {2R}}$

$\Rightarrow {{1.5} \over v} = {{ - 5} \over {2R}} \Rightarrow v = {{ - 3R} \over 5}$

Here ' $-$ ' sign tells that image is formed left to B.

So, the distance of image from $P, = {{R - 3R} \over \Delta } = {{2R} \over 5}$

Now, for the surface on the left side,

${\mu _1} = 1,{\mu _2} = 1.5,\mu = {{3R} \over 2},{R_2} = R$

So, ${{1.5} \over v} - {1 \over {{{3R} \over 2}}} = {{1.5 - 1} \over R}$

$\Rightarrow {{1.5} \over v} - {2 \over {3R}} = {1 \over {2R}}$

$\Rightarrow {{1.5} \over v} = {1 \over R}\left( {{1 \over 2} + {2 \over 3}} \right)$

$\Rightarrow {{1.5} \over v} = {7 \over {6R}}$

$\Rightarrow v = {{9R} \over 7}$

So distance from $P = {{9R} \over 7} - R = {{2R} \over 7}$

So the distance between two images,

$= {{2R} \over 5} - {{2R} \over 7} = 0.4R - 0.2857R$

$\approx 0.4R - 0.286R = 0.114R$ .