Light enters from air into a given medium at | vedclass

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(C) The angle of incidence $i$ is the angle between the incident ray and the normal. The given angle is with the interface,so $i = 90^{\circ} - 45^{\c

Vedclass · Accepted Answer

$1.414$

Vedclass · Answer

(C) The angle of incidence $i$ is the angle between the incident ray and the normal. The given angle is with the interface,so $i = 90^{\circ} - 45^{\circ} = 45^{\circ}$.The angle of deviation $D$ is given by $D = i - r$,where $r$ is the angle of refraction.Given $D = 15^{\circ}$,we have $15^{\circ} = 45^{\circ} - r$,which gives $r = 30^{\circ}$.Using Snell's law,$n_1 \sin i = n_2 \sin r$,where $n_1 = 1$ (for air) and $n_2 = \mu$ (refractive index of the medium).$1 \cdot \sin 45^{\circ} = \mu \cdot \sin 30^{\circ}$$\frac{1}{\sqrt{2}} = \mu \cdot \frac{1}{2}$$\mu = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$.

Light enters from air into a given medium at an angle of $45^{\circ}$ with the interface of the air-medium surface. After refraction,the light ray is deviated through an angle of $15^{\circ}$ from its original direction. The refractive index of the medium is:

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