Question

Dispersive power of a prism decreases with the increase in prism angle.
(A) True
(B) False

Answer 1

Hint: The difference in refraction of the light of the highest wavelength and the light of the lowest wavelength that enters the prism is known as the prism's dispersive power. The angle between the two extreme wavelengths can also be used to express the prism's dispersive power.

Complete step by step solution:
When light travels from one medium to another, it bends according to the refractive indices of the two media it passes through. Different wavelengths of light, on the other hand, bend differently due to the fact that each wavelength encounters a different index in the material. This phenomenon causes the rainbow effect on the output as light is shone through a prism. The visible wavelengths refract and split into their respective colors at different rates.
The difference in the refraction of the highest and lowest wavelengths that reach the prism is known as dispersive power. The angle formed by the two extreme wavelengths is used to express this. The greater the angle between them, the greater the dispersive force, and vice versa.
We will now try to derive an expression for the dispersion power of a prism.
We know that the refractive index of the material of a prism is given by, $ n = \dfrac{{\sin \left( {\dfrac{{A + D}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}} $ where $ A $ is the prism angle and $ D $ is the angle of minimum deviation. If $ A $ is the angle of a small angle prism and $ \delta $ is the angle of deviation of such a prism. Now for small angles of $ A $ and $ \delta $ , $ \sin \left( {\dfrac{{A + \delta }}{2}} \right) \approx \dfrac{{A + \delta }}{2} $ and $ \sin \left( {\dfrac{A}{2}} \right) \approx \dfrac{A}{2} $ .
We now get, $ n = \dfrac{{\dfrac{{A + \delta }}{2}}}{{\dfrac{A}{2}}} = \dfrac{{A + \delta }}{A} = 1 + \dfrac{\delta }{A} $
On further simplification we get, $ \delta = (n - 1)A $ . Remember that $ \delta $ is mean deviation.
We know that angular dispersion is $ {\delta _v} - {\delta _r} = ({n_v} - 1)A - ({n_r} - 1)A $ .
 $ \Rightarrow {\delta _v} - {\delta _r} = ({n_v} - {n_r})A $
We have $ {\text{Dispersive power = }}\dfrac{{{\text{Angular dispersion}}}}{{{\text{Mean deviation}}}} $
 $ \Rightarrow {\text{Dispersive Power = }}\dfrac{{({n_v} - {n_r})A}}{{(n - 1)A}} $
 $ \Rightarrow \omega {\text{ = }}\dfrac{{({n_v} - {n_r})}}{{(n - 1)}} $
Here, we can observe that the dispersive power depends only on the nature of the material and is independent of the angle of the prism.
So, the statement that the dispersive power of a prism decreases with the increase in prism angle is false.
Hence, the correct option is B. False.

Note:
Note that the angular dispersion $ {\delta _v} - {\delta _r} = ({n_v} - {n_r})A $ produced by a prism depends on both the nature of the material and the angle of the prism. Angular dispersion increases with increase in the prism angle.