A ray of light, incident on a equilateral glass prism $(μ_{medium} = \sqrt{3})$ moves parallel to the base of the prism inside it. What is the angle of incidence for this way?

Solution

Given, an equilateral glass prism.
Angle of prism = 60^o
Refractive inex of the medium, $μ$ = $\sqrt{3}$

When light ray moves parallel to the base of the prism inside it, then
r₁ = r₂so that ,the minimum deviation takes place.

For minimum deviation r = A/2 = $30 °$

Now, using Snell's law,
$μ = \frac{\sin i}{\sin r}$

$∴$ $\sqrt{3} = \frac{\sin i}{\sin 30 °} = \frac{\sin i}{1 / 2}$

$\Rightarrow$ $\sin i = \frac{\sqrt{3}}{2}$
$\Rightarrow$ $i = \sin^{- 1} (\frac{\sqrt{3}}{2}) = 60 °$
is the required angle of incidence so that, the ray travels parallel to the base ofthe prism.

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A ray of light, incident on a equilateral glass prism $(μ_{medium} = \sqrt{3})$ moves parallel to the base of the prism inside it. What is the angle of incidence for this way?

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A ray of light, incident on a equilateral glass prism (μmedium =3) moves parallel to the base of the prism inside it. What is the angle of incidence for this way?

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Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A ray of light, incident on a equilateral glass prism $(μ_{medium} = \sqrt{3})$ moves parallel to the base of the prism inside it. What is the angle of incidence for this way?