Question

Light is incident on a glass surface at polarizing angle of ${57.5}^{\circ }$. Then the angle between the incident ray and the refracted ray is:

• A. ${57}^{\circ }$
• B. ${115}^{\circ }$
• C. ${205}^{\circ }$
• D. ${145}^{\circ }$

Answer 1

Answer: (A)

${57}^{\circ }$

Given that,

Polarizing angle for glass = ${57.5}^0$

Angle of refraction = $r$

Now, refractive index of the glass

$\mu =\tan {\theta }_p$

$1.4=\tan \left({57.5}^0\right)$

Now, from Snell’s law

${\mu }_1\sin i={\mu }_2\sin r$

$1\times \sin \left({57.5}^0\right)=1.4\times \sin r$

$\sin r=\frac{0.81419}{1.4}$

$\sin r=0.5816$

$r={\sin }^{-1}\left(0.5816\right)$

$r=0.6207$

Now, angle of deviation

$\delta =i-r$

$\delta =57.5-0.6207$

$\delta =56.87$

$\delta ={57}^0$

Hence, the angle between the incident ray and the refracted ray is ${57}^0$