- Draw the ray diagram showing refraction of light through a glass prism and hence obtain the relation between the refractive index $\mu$ of the prism, angle of prism and angle of minimum deviation.
- Determine the value of the angle of incidence for a ray of light travelling from a medium of refractive index $\mu_1=\sqrt{2}$ into the medium of refractive index $\mu_2=1$, so that it just grazes along the surface of separation.
CBSE FOREIGN - SET 1 2017
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From fig $\angle A+\angle QNR=180^\circ\text{ }\text{ }\text{ }\text{ }\text{ }\dots(1)$
From triangle $\triangle QNR \text{ }\text{ }r_1+r_2+\angle QNR=180^\circ\text{ }\text{ }\text{ }\text{ }\dots(2)$
Hence from equ (1) & (2)
$\therefore \angle A= r_1+r_2$
The angle of deviation
$\delta=(i-r_1)+(e-r_2)=\text{i + e - A}$
At minimum deviation i = e and $?_1 = ?_2$
$\therefore r=\frac{A}{2}$
And $\text{i}=\frac{A+\delta m}{2}$
Hence refractive index
$\mu=\frac{\sin i}{\sin r}=\frac{\sin\big(\frac{A+\delta}{2}\big)}{\sin\frac{A}{2}}$
- From Snell’s law $\mu_1 \sin i=\mu_2\sin r$
$\therefore\sqrt{2}\text{ }\sin \text{i}=1\sin90^\circ\Rightarrow \sin i\frac{1}{\sqrt{2}}$
$or \text{ }i=45^\circ$
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