
The refractive indices of the material of a prism for red and yellow colours are $1.620$ and $1.635$ respectively. Calculate the angular dispersion and dispersive power, if the refracting angle is ${{8}^{0}}$.
Answer
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Hint: This problem can be solved by using the direct formula for the angular dispersion of ray of light passing through a prism in terms of the refractive index of the prism and the refracting angle. The dispersive power can be found out from the direct formula involving the angles of dispersion for red and yellow light through a prism.
Formula used: $\delta =\left( \mu -1 \right)A$
$\omega =\dfrac{{{\delta }_{y}}-{{\delta }_{r}}}{\dfrac{{{\mu }_{y}}+{{\mu }_{r}}}{2}}$
Complete step by step answer:
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
$\omega =\dfrac{0.12}{\dfrac{1.635+1.620}{2}}=\dfrac{0.12}{\dfrac{3.255}{2}}=\dfrac{0.12}{1.6275}=0.0737$
Hence, we have got the required dispersive power.
Note: By solving this problem, students must have realized that the angle of deviation for red light is lesser than the angle of deviation for yellow light in the prism. This is the reason why when white light passes through a prism, the light splits up into the component colours with red at the top and violet at the bottom of the spectrum. The reason is that each colour has its specific different angle of deviation where red has the least and thus is least deviated.
Formula used: $\delta =\left( \mu -1 \right)A$
$\omega =\dfrac{{{\delta }_{y}}-{{\delta }_{r}}}{\dfrac{{{\mu }_{y}}+{{\mu }_{r}}}{2}}$
Complete step by step answer:
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
$\omega =\dfrac{0.12}{\dfrac{1.635+1.620}{2}}=\dfrac{0.12}{\dfrac{3.255}{2}}=\dfrac{0.12}{1.6275}=0.0737$
Hence, we have got the required dispersive power.
Note: By solving this problem, students must have realized that the angle of deviation for red light is lesser than the angle of deviation for yellow light in the prism. This is the reason why when white light passes through a prism, the light splits up into the component colours with red at the top and violet at the bottom of the spectrum. The reason is that each colour has its specific different angle of deviation where red has the least and thus is least deviated.
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