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Question: The dispersion of a medium for wavelength \(\lambda\) is D. Then the dispersion for the wavelength 2...

The dispersion of a medium for wavelength λ\lambda is D. Then the dispersion for the wavelength 2λ\lambda will be:
(A) (D/8)
(B) (D/4)
(C) (D/2)
(D) D

Explanation

Solution

Hint We know that dispersion is a term which is used to describe the situation in which the phase velocity of a wave depends on its frequency. Wave with a higher frequency is diffracted more than those which have a lower frequency.

Complete step by step answer
As we know that, Cauchy’s Dispersion formula is:
μ=A+Bλ2\mu = A + \dfrac{B}{{{\lambda ^2}}}
And the dispersion is given as:
D=dμdλD = - \dfrac{{d\mu }}{{d\lambda }}
Therefore, from the above 2 equations we can say that:
D=(2λ3)B=2Bλ3D = - ( - 2{\lambda ^{-3}})B = \dfrac{{2B}}{{{\lambda ^3}}}
D1λ3D\propto \dfrac{1}{{{\lambda ^3}}}
This is implied by the expression that:
D1λ3D\propto \dfrac{1}{{{\lambda ^3}}}
Hence, we can say that:
D/D=(λλ/)3\dfrac{{{D^/}}}{D} = {\left( {\dfrac{\lambda }{{{\lambda ^/}}}} \right)^3}
As the λ/=2λ{\lambda ^/} = 2\lambda

Therefore, the value of D/=D/8{D^/} = D/8.

Note : To avoid any confusion while solving such problems we should know that dispersion is defined as the property where the light is spread out of the according to its colour as it passes through an object. For instance, when we shine a while light into a, all of the different colours of light are bent different amounts, so they are spread out and make a rainbow.