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(B) For a thin prism,the angle of deviation $\delta$ is given by $\delta = (\mu - 1)A$. Given $\delta = 1.15^{\circ}$,we have $(\mu - 1)A = 1.15^{\cir

Vedclass · Accepted Answer

$1.575$

Vedclass · Answer

(B) For a thin prism,the angle of deviation $\delta$ is given by $\delta = (\mu - 1)A$. Given $\delta = 1.15^{\circ}$,we have $(\mu - 1)A = 1.15^{\circ}$ (Equation $1$).When the ray is incident normally on the first face,it strikes the second face at an angle of incidence equal to the prism angle $A$. The ray is reflected internally at the second face,so the angle of reflection is $A$. This reflected ray then strikes the first face at an angle $2A$ with the normal. Upon emerging from the first face,the angle of refraction $r'$ is given by $\sin(r') = \mu \sin(2A)$. For small angles,$r' \approx \mu(2A)$.The angle between the emerging ray and the incident ray is given as $6.3^{\circ}$. The emerging ray makes an angle $r'$ with the normal,and the incident ray is normal to the surface,so the angle between them is $r' = 6.3^{\circ}$.Thus,$2\mu A = 6.3^{\circ}$ (Equation $2$).Dividing Equation $2$ by Equation $1$: $\frac{2\mu A}{(\mu - 1)A} = \frac{6.3}{1.15} \implies \frac{2\mu}{\mu - 1} = 5.478$.$2\mu = 5.478\mu - 5.478 \implies 3.478\mu = 5.478 \implies \mu \approx 1.575$.

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