Question

In the visible region the dispersive powers and the mean angular deviations for crown glass and flint glass prism are $ \omega ,{\omega _1} $ and $ d,{d_1} $ respectively. The condition for getting dispersion with zero deviation when the two prisms are combined is
A) $ \sqrt {\omega d} + \sqrt {{\omega ^1}{d^1}} = 0 $
B) $ {\omega ^1}d + \omega {d^1} = 0 $
C) $ \omega d + {\omega ^1}{d^1} = 0 $
D) $ \omega {d^2} + {\left( {{\omega ^1}{d^1}} \right)^2} = 0 $

Answer 1

Hint: The dispersive power for any material is the ratio of its angular deviation to the mean deviation of a ray of light as caused by the medium. To get zero deviation, the deviation from the crown glass prism and the flint glass prism must cancel each other out.

Complete step by step answer:
When light passes through a prism, light is deflected by the prism. The angular displacement is the difference between the angle formed by the final refracted ray of light and the initial ray of light. We’ve been given that the mean angular deviation of the crown glass prism and the flint glass prism are $ d,{d_1} $ respectively.
Now, the dispersive power of the prism is the ratio of the angular deviation and the mean deviation of the prism. So, we can write
 $ \omega = \dfrac{{{\text{angular}}\,{\text{deviation}}}}{{{\text{mean}}\,{\text{deviation}}}} $
Hence the angular deviation will be
 $ {\text{angular}}\,{\text{deviation = }}\omega d $ where $ d $ is the mean deviation.
Now, when the two prisms are placed side by side, they must have a net-zero angular deviation to have zero deviation. Hence the angular deviation of the two prisms must cancel each other out. Hence, the condition can be written as
 $ {\text{angular deviation of crown glass + angular deviation of flint glass = 0}} $
So, we can write
 $ \omega d + {\omega _1}{d_1} = 0 $
Which corresponds to option (C).

Note:
A prism splits white light into its seven constituent colours as well as all the colours experience a different amount of angular displacement. The colour yellow’s angular displacement is taken as the average angular deviation of all the colours. A net-zero deviation will mean that the light after getting refracted by both these prisms will emerge in the same line vector it was initially going in.