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Question
The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00 cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32 cm below the top surface of the slab. Calculate the dispersive power of the material.
Solution
Given:-
Difference in the refractive indices of violet and red lights = 0.014
Let μv and μr be the refractive indices of violet and red colours.
Thus, we have:-
μv − μr = 0.014
Now,
Real depth of the newspaper = 2.00 cm
Apparent depth of the newspaper = 1.32 cm
\[\text{Refractive index }= \frac{\text{Real depth}}{\text{Apparent depth}}\]
Refractive index for yellow light \[\left( \mu_y \right)\] is given by
\[ \mu_y = \frac{2 . 00}{1 . 32} = 1 . 515\]
Also,
Dispersive power, \[\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}\]
\[= \frac{0 . 014}{1 . 515 - 1}\]
Or, \[\omega = \frac{0 . 014}{0 . 515} = 0 . 027\]
Thus, the dispersive power of the material is 0.027.
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