Q. The refractive indices of the crown glass for blue and red light are 1.51 and 1.49 respectively and those of the flint glass area 1.77 and 1.73 respectively. An isosceles prism of angle $6{^\circ}$ is made of crown glass. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident light (i)Determine the angle of the flint glass prism (ii)Calculate the net dispersion of the combined system
Explanation
(a) (i)When angle of prism is small and angle of incidence is also small, the deviation is given by $\delta = (\mu - 1)A.\ $Net deviation by the prism is zero. So,
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$\delta_{1} + \delta_{2} = 0$
Or ${\ (\mu}_{1} - 1)A_{1} + \left( \mu_{2} - 1 \right)A_{2} = 0$
Here, $\mu_{1}\ and\ \mu_{2}$ are the refractive indices for crown and flint glass respectively
Hence,${\ \mu}_{1} = \frac{1.51 + 1.49}{2} = 1.5\ $and
$\mu_{2} = \frac{1.77 + 1.73}{2} = 1.75$
This gives $A_{2} = - 4{^\circ}$
Hence, angle of flint glass prism is $4{^\circ}.\ $Negative sign shows that flint glass prism is inverted with respect to the crown glass prism
(ii)Net dispersion due to the two prism is
$= \left( {\mu_{b}}_{1} - {\mu_{r}}_{1} \right)A_{1} + \left( {\mu_{b}}_{2} - {\mu_{r}}_{2} \right)A_{2}$
$= (1.51 - 1.49)(6{^\circ}) + (1.77 - 1.73)( - 4{^\circ})$
$= - 0.04{^\circ}$
$\therefore\ $Net dispersion is $- 0.04{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.