Question

As shown in the figure, two concave refracting surfaces of equal radii of curvature (R) and refractive indices (n = 1.5) face each other in air (n = 1.0). A point object (0) is placed midway in between the centre and one of the vertices of the refracting surfaces. Find the distance between image O' formed by one surface and image O' ' formed by the other surface in terms of R.

Answer 1

Given two concave refracting surface,
radius of each concave refracting surface is R.
and refractive index , n = 1.5

for refraction at the first surface , we have
u = AO = -(R + R/2) = -3R/2
$\mu_2$ = 1.5
$\mu_1$ = 1
radius of curvature = -R

the distance $v_1$ of image $I_1$ from A is given by relation,
$\frac{\mu_2}{v_1}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$

$\frac{1.5}{v_1}-\frac{1}{-3R/2}=\frac{1.5-1}{-R}$

$\frac{1.5}{v_1}+\frac{2}{3R}=-\frac{1}{2R}$

$\frac{1.5}{v_1}=-\frac{2}{3R}-\frac{1}{2R}$

$\frac{1.5}{v_1}=-\frac{7}{6R}$

$v_1=-\frac{9R}{7}$

for the second surface , we have
u = BO = - R/2
the distance $v_2$ of image $i_2$ from the second surface is given by
$\frac{\mu_2}{v_2}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$

$\frac{1.5}{v_2}-\frac{1}{-R/2}=\frac{1.5-1}{-R}$

$\frac{1.5}{v_2}+\frac{2}{R}=-\frac{1}{2R}$

$\frac{1.5}{v_2}=-\frac{2}{R}-\frac{1}{2R}$

$\frac{1.5}{v_2}=-\frac{5}{2R}$

$v_2=\frac{-3R}{5}$

Answer 2

Answer:

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