Question

A crown glass prism of angle $5^{\circ }$ is to be combined with a flint glass prism in such a way that the mean ray passes undeviated. The angle of flint glass prism is $A^{\prime }$ and angular deviation produced by the combination when white light goes through it is $\delta$. Refractive indices for red, yellow and violet light are $1.514,1.517$ and $1.523$ respectively for crown glass and $1.613,1.620$ and $1.632$ for flint glass. Then :

Answer 1

Given:-
$A=5^{\circ };{\mu }_r=1.514;{\mu }_y=1.517;{\mu }_v=1.523$
${\mu }_r^{{}^{\prime }}=1.613;{\mu }_y^{{}^{\prime }}=1.620;{\mu }_v^{{}^{\prime }}=1.632$

The deviation produced by crown prism is :
$\delta =({\mu }_y-1)A$

The deviation produced by flint prism is:
${\delta }^{\prime }=({\mu }_y^{\prime }-1)A^{\prime }$

Net deviation is given by :
$D=\delta -{\delta }^{\prime }=({\mu }_y-1)A-({\mu }_y^{\prime }-1)A^{\prime }$

If the net deviation is zero :

$\Rightarrow ({\mu }_y-1)A-({\mu }_y^{\prime }-1)A=0^{\prime }$

$A^{\prime }=\frac{({\mu }_y-1)}{({\mu }_y^{\prime }-1)}A$

$\Rightarrow A^{\prime }=\frac{(1.517-1)}{(1.620-1)}\times 5={4.2}^{\circ }$

Angular dispersion produced by the crown prism :

$\theta ={\delta }_v-{\delta }_r=({\mu }_v-{\mu }_r)A$

Angular dispersion produced by flint prism :

${\theta }^{\prime }={\delta }_v^{\prime }-{\delta }_r^{\prime }=({\mu }_v^{\prime }-{\mu }_r^{\prime })A^{\prime }$

The net angular dispersion is :
$\delta =({\mu }_v-{\mu }_r)A-({\mu }_v^{\prime }-{\mu }_r^{\prime })A^{\prime }$

$=(1.523-1.514)\times 5^{\circ }-(1.632-1.613)\times {4.2}^{\circ }=-{0.0348}^{\circ }$

$\therefore$ The magnitude of angular dispersion is ${0.0348}^{\circ }$.

Hence, options $\left(B\right)$ and $\left(C\right)$ are the correct answers.

Why this question : In this type of questions where two prisms are involved, put the values of the refractive index and angle of the prism carefully.