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$\omega =\dfrac{{{\delta }_{y}}-{{\delta }_{r}}}{\dfrac{{{\mu }_{y}}+{{\mu }_{r}}}{2}}$

Let us write the formula for the angular dispersion for a light ray of a particular colour in terms of the refractive index of the prism for the light and the refracting angle of the prism.

The angle of deviation $\delta $ for a light ray for which the refractive index of the prism is $\mu $ and $A$ is the refracting angle of the prism is given by

$\delta =\left( \mu -1 \right)A$ --(1)

Now, let us analyze the question.

Let the angular deviation for yellow and red colour light rays be ${{\delta }_{y}}$ and ${{\delta }_{r}}$ respectively.

The refractive index of the prism for red colour is ${{\mu }_{r}}=1.620$.

The refractive index of the prism for yellow colour is ${{\mu }_{y}}=1.635$

The refracting angle of the prism is $A={{8}^{0}}$.

Now, using (1), we get

${{\delta }_{y}}=\left( 1.635-1 \right){{8}^{0}}=0.635\times {{8}^{0}}={{5.08}^{0}}$ --(2)

Also, using (1), we get

${{\delta }_{r}}=\left( 1.620-1 \right){{8}^{0}}=0.620\times {{8}^{0}}={{4.96}^{0}}$ --(3)

The required angular dispersion of the prism will be

${{\delta }_{y}}-{{\delta }_{r}}={{5.08}^{0}}-{{4.96}^{0}}={{0.12}^{0}}$ --(4)

Now, the dispersive power of the prism $\omega $ will be given by

$\omega =\dfrac{{{\delta }_{y}}-{{\delta }_{r}}}{\dfrac{{{\mu }_{y}}+{{\mu }_{r}}}{2}}$ --(5)

Using (4) in (5) and putting the values of the rest of the variables we get

$\omega =\dfrac{0.12}{\dfrac{1.635+1.620}{2}}=\dfrac{0.12}{\dfrac{3.255}{2}}=\dfrac{0.12}{1.6275}=0.0737$