Question 54: Ray Optics

Question

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is :

Accepted Answer

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,
$$ = {{2R} \over 5} - {{2R} \over 7} = 0.4R - 0.2857R$$
$$ \approx 0.4R - 0.286R = 0.114R$$.

Answer

0.124R

Answer

0.114R

Answer

0.411R

Answer

0.214R

Ray Optics

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is :

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