Question

A ray of light is incident on the surface of a sphere of refractive index $\mu =\sqrt{\frac{5}{2}}$ . After refraction, it is total internally reflected and refracted out of the sphere again such that the total deviation is minimum. The angle of incidence of the ray is

• A. ${53}^0$
• B. ${30}^0$
• C. ${60}^0$
• D. ${45}^0$

Answer 1

The correct option is D ${45}^0$


We know: $\mu =\frac{sini}{sinr}$
Given:
$\mu =\sqrt{\frac{5}{2}}$

Deviation of the ray AB after refraction
$=(i-r)$

After reflection at C, deviation of the ray $BC=({180}^0-2r)$, deviation of ray CD after refraction at $D=(i-r)$

Total deviation $\delta =2(i-r)+({180}^0-2r)$
=${180}^0+2i-4r$

By applying condition of minimum deviation and by applying Snell’s law

Therefore total deviation is minimum

For minimum value of $\delta ,\frac{d\delta }{di}=0$

$\delta ={180}^0+2i-4r$

Differentiate total deviation w.r.t ′i′,

we get ,

$\frac{d\delta }{di}=+2-\frac{4dr}{di}=0$

$\Rightarrow \frac{4dr}{di}=2$

$\Rightarrow \frac{dr}{di}=\frac{2}{4}=\frac{1}{2}$

$\mu =\frac{sini}{sinr}\Rightarrow \mu cosr\frac{dr}{di}=cosi$

$\Rightarrow co{s}^{2}i={\mu }^2(1-si{n}^{2}r){\left(\frac{dr}{di}\right)}^2=({\mu }^2-{\mu }^{2}si{n}^{2}r)\frac{1}{4}$

$\Rightarrow 4co{s}^{2}i={\mu }^2-si{n}^{2}i$

$3co{s}^{2}i={\mu }^2-1$

For $\mu =\sqrt{\frac{5}{2}},cosi=\sqrt{\frac{(\frac{5}{2}-1)}{3}}=\frac{1}{\sqrt{2}}$

FINAL ANSWER: (B)