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:

Given that,

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

Angle of refraction = $r$

Now, refractive index of the glass

$\mu =\tan {{\theta }_{p}}$

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

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 $57^{0}$