Question

refractive index of a flint glass varies from 1.60 to 1.66 for visible range. Radii of curvature of a thin convex lens are 10 cm and 15 cm Calculate the chromatic aberration between the extreme colors​

Answer 1

Given

$n_r$ = The refractive index for red colour = 1.6

$n_v$ = The refractive index for violet colour = 1.66

$R_1$ = Radius of curvature of first layer thin lens = 10 cm

$R_2$ = Radius of curvature of second layer thin lens = 15 cm

To find

Chromatic aberration between the extreme colors

Solution

Focal length for the red colour

$\dfrac{1}{f_r}=(n_r-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})\\\Rightarrow \dfrac{1}{f_r}=(1.6-1)(\dfrac{1}{10}-\dfrac{1}{-15})\\\Rightarrow \dfrac{1}{f_r}=0.6\times\dfrac{1}{6}\\\Rightarrow \dfrac{1}{f_r}=0.1\\\Rightarrow f_r=10\ \text{cm}$

Focal length for the violet colour

$\dfrac{1}{f_v}=(n_v-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})\\\Rightarrow \dfrac{1}{f_v}=(1.66-1)(\dfrac{1}{10}-\dfrac{1}{-15})\\\Rightarrow \dfrac{1}{f_v}=0.66\times\dfrac{1}{6}\\\Rightarrow \dfrac{1}{f_v}=0.11\\\Rightarrow f_v=9.091\ \text{cm}$

Longitudinal chromatic aberation is given by

$f_r-f_v=10-9.091=0.909\ \text{cm}$

The chromatic aberration between the extreme colors is $0.909\ \text{cm}$.