Dispersive power depends upon
A. The shape of prism
B. Material of prism
C. Angle of prism
D. Height of the prism
Answer
550.5k+ views
Hint: The ratio of the difference in refraction of the light of the highest wavelength and the light of the lowest wavelength and the refraction of the white light that enters the prism is known as the prism's dispersive power. Dispersive power of a prism is known as, $ \omega = \dfrac{\delta }{{{\delta _m}}} $ .
Complete step by step answer:
We know that, the refractive index of the material of a prism is given by, $ n = \dfrac{{\sin (\dfrac{{A + D}}{2})}}{{\sin (\dfrac{A}{2})}} $ where, $ A $ is the angle of prism and $ D $ is the angle of minimum deviation.
Now, for small angle prism we can write, $ \sin (\dfrac{{A + D}}{2}) \approx \dfrac{{A + D}}{2} $ and $ \sin (\dfrac{A}{2}) \approx \dfrac{A}{2} $ .So, we can write refractive index as, $ n = \dfrac{{\dfrac{{A + D}}{2}}}{{\dfrac{A}{2}}} $
up on simplifying we get,
$ n = (1 + \dfrac{D}{A}) $
So, angle of deviation becomes, $ D = (n - 1)A $
Now, we know, angular dispersion is the difference of deviation angle for highest and lowest wavelength of the visible light. So, $ {D_v} - {D_r} = ({n_v} - 1)A - ({n_r} - 1)A $ where, $ {n_v} $ is the refractive index for violet light in the prism and $ {n_r} $ is the is the refractive index for red light in the prism
Now we know dispersive power is the ratio of Angular dispersion and mean deviation of the prism
Hence, we can write dispersive power,
$ \omega = \dfrac{{({n_v} - 1)A - ({n_r} - 1)A}}{{(n - 1)A}} $
Or, $ \omega = \dfrac{{({n_v} - {n_r})}}{{(n - 1)}} $
So, we can see that the dispersive power solely depends on the refractive index of the material of the prism.
So, option (B) is the correct option.
Note:
Here the Refractive index of the material for white light is denoted by $ n $ and the refractive index for other colours of light is written in the suffix.
We can see that the angular dispersion of a prism is dependent on the angle of the prism but dispersive power does not depend on the geometry of the prism. Angular dispersion of a prism is given by $ \delta = ({n_v} - {n_r})A $ .
Complete step by step answer:
We know that, the refractive index of the material of a prism is given by, $ n = \dfrac{{\sin (\dfrac{{A + D}}{2})}}{{\sin (\dfrac{A}{2})}} $ where, $ A $ is the angle of prism and $ D $ is the angle of minimum deviation.
Now, for small angle prism we can write, $ \sin (\dfrac{{A + D}}{2}) \approx \dfrac{{A + D}}{2} $ and $ \sin (\dfrac{A}{2}) \approx \dfrac{A}{2} $ .So, we can write refractive index as, $ n = \dfrac{{\dfrac{{A + D}}{2}}}{{\dfrac{A}{2}}} $
up on simplifying we get,
$ n = (1 + \dfrac{D}{A}) $
So, angle of deviation becomes, $ D = (n - 1)A $
Now, we know, angular dispersion is the difference of deviation angle for highest and lowest wavelength of the visible light. So, $ {D_v} - {D_r} = ({n_v} - 1)A - ({n_r} - 1)A $ where, $ {n_v} $ is the refractive index for violet light in the prism and $ {n_r} $ is the is the refractive index for red light in the prism
Now we know dispersive power is the ratio of Angular dispersion and mean deviation of the prism
Hence, we can write dispersive power,
$ \omega = \dfrac{{({n_v} - 1)A - ({n_r} - 1)A}}{{(n - 1)A}} $
Or, $ \omega = \dfrac{{({n_v} - {n_r})}}{{(n - 1)}} $
So, we can see that the dispersive power solely depends on the refractive index of the material of the prism.
So, option (B) is the correct option.
Note:
Here the Refractive index of the material for white light is denoted by $ n $ and the refractive index for other colours of light is written in the suffix.
We can see that the angular dispersion of a prism is dependent on the angle of the prism but dispersive power does not depend on the geometry of the prism. Angular dispersion of a prism is given by $ \delta = ({n_v} - {n_r})A $ .
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