Calculate the critical angle for a glass-air | vedclass

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(A) Given: Angle of incidence $i = 45^{\circ}$,Angle of deviation $\delta = 15^{\circ}$.Since the ray travels from air to glass,the angle of refractio

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$45$

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(A) Given: Angle of incidence $i = 45^{\circ}$,Angle of deviation $\delta = 15^{\circ}$.Since the ray travels from air to glass,the angle of refraction $r$ is given by $r = i - \delta = 45^{\circ} - 15^{\circ} = 30^{\circ}$.Using Snell's law,the refractive index of glass with respect to air is $\mu = \frac{\sin i}{\sin r} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.The critical angle $	heta_{c}$ is given by $\sin 	heta_{c} = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$.Therefore,$	heta_{c} = 45^{\circ}$.

Calculate the critical angle for a glass-air surface,if a ray of light incident in air on the glass surface is deviated by $15^{\circ}$ when the angle of incidence is $45^{\circ}$. (in $^{\circ}$)

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