A light ray composed of two monochromatic components passes through a trihedral prism with refracting angle `theta = 60^(@)`. Find the angle `Delta alpha` between the components of the ray after its passage through the prism if their respective indies of refraction are equal to `1.515` and `1.520`. the prism is oriented to provide the least deflection angle.

Explanation

From the Cauch's fromula and also experimetally the `R.I` of a medium depends upon the wavelength o fthe mochromatic ray i.e., `n = f (lambda)`. In the case of least deviation of a monochromtic ray the passage a prism, we have:
`n sin(theta)/(2) = sin (alpha + theta)/(2)` (1)
The above equation tales us that we have `n = n (alpha)`, so we may write
`Deltan = (dn)/(d alpha) Delta alpha` (2)
From Eqn. (1)
`d n sin.(theta)/(2) =(1)/(2) cos.(alpha + theta)/(2) d alpha`
or, `(dn)/(d alpha) = cos.((alpha+ theta)/(2))/(2sin.((theta)/(2)))` (3)
From Eqn (2) and (3)
`Deltan =cos.((alpha+ theta)/(2))/(2sin((theta)/(2))) Delta alpha`
or, `Delta n = sqrt(1-sin^(2)((a+ alpha)/(2)))/(2sin.((theta)/(2))) delta alpha = sqrt(1- n^(2)sin^(2)((theta)/(2)))/(2sin((theta)/(2))) Delta alpha` (Using Eqn. 1)
Thus `Delta alpha = (2 sin.((theta)/(2)))/(sqrt(1-n^(2)sin^(2)((theta)/(2)))) Delta n = 0.44`