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,

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

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