A ray of light is incident at a polarising an | vedclass

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(B) When a ray of light is incident at the polarising angle $(i_p)$,the reflected and refracted rays are perpendicular to each other. Thus,$i_p + r =

Vedclass · Accepted Answer

$57$

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(B) When a ray of light is incident at the polarising angle $(i_p)$,the reflected and refracted rays are perpendicular to each other. Thus,$i_p + r = 90^{\circ}$,where $r$ is the angle of refraction.From the geometry of refraction,the angle of deviation $(\delta)$ is given by $\delta = |i_p - r|$.Given $\delta = 24^{\circ}$,we have $i_p - r = 24^{\circ}$ (since $i_p > r$ for glass-air interface).We have two equations:$1$) $i_p + r = 90^{\circ}$$2$) $i_p - r = 24^{\circ}$Adding these two equations:$2i_p = 114^{\circ}$$i_p = 57^{\circ}$Therefore,the angle of incidence is $57^{\circ}$.

$A$ ray of light is incident at a polarising angle such that its deviation is $24^{\circ}$. Then the angle of incidence is: (in $^{\circ}$)

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