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A ray falls on a prism, $ABC (AB = BC)$ and travels as shown in the figure. If the refractive index of glass with respect to air is $1.5$ , find $sin r$ .

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3 / \sqrt{2}

\frac{3}{2 \sqrt{2}}

\frac{\sqrt{2}}{3}

\frac{\sqrt{2}}{3}

answer is D.

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Detailed Solution

The value of

sin r = \frac{\sqrt{2}}{3}

.
From the fig. the incident angle,
It is given that,

AB = BC

So,

< A = < C = 45 °

\Rightarrow

i = 45 °

(The ray of light is parallel to the base of the prism).
The refractive index of the glass,

n = 1.5

.
By Snell’s law;

n = \frac{sin i}{sin r}

;
Where

r

is the angle of refraction.
Using the formula,

n = \frac{sin i}{sin r}

\Rightarrow 1.5 = \frac{\sin 45 °}{sin r}

\Rightarrow sin r = \frac{1 / \sqrt{2}}{3 / 2}

\Rightarrow sin r = \frac{2}{3 \sqrt{2}}

\Rightarrow sin r = \frac{\sqrt{2}}{3}

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