Question

(b) Calculate the angle of emergence (e) of the ray of
light incident normally on the face AC of a glass
prism ABC of refractive index 13. How will the
angle of emergence change qualitatively, if the
ray of light emerges from the prism into a liquid
of refractive index 1.3 instead of air ?
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Answer 1

Therefore, this is your solution.

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Answer 2

Answer:

The angle of emergence when the ray of light travels from air to glass is 60° and the angle of emergence when the ray of light emerges from the prism into liquid is 22.02°

Explanation:

Given:

Refractive index of prism ($n_{g}$) = √3

Refractive index of liquid ($n_{l}$) = 1.3

To find:

The angle of emergence (e)

Formula:

$\frac{sin(i)}{sin(e)}=\frac{n_{1} }{n_{2} }$

Step 1:

The angle of emergence for a ray of light traveling from air to prism.

$\frac{sin(i)}{sin(e)}=\frac{n_{a} }{n_{g} }$

The refractive index of the air medium is 1. Therefore,

$\frac{sin(i)}{sin(e)}=\frac{1 }{n_{g} }$

As the ray of light is incident normally, i = 30°.

Substituting the given values, we get

$\frac{sin(30)}{sin(e)}=\frac{{1} }{\sqrt{3} }$

$sin(e)=\sqrt{3} * sin (30)$

$sin(e) = \sqrt{3} * \frac{1}{2}$

$e = sin^{-1}(\frac{\sqrt{3} }{2})$

e = 60°

Step 2:

The angle of emergence for a ray of light traveling from prism to liquid.

$\frac{sin(i)}{sin(e)}=\frac{n_{g} }{n_{l} }$

As the ray of light is incident normally, i = 30°.

Substituting the given values, we get

$\frac{sin(30)}{sin(e)}=\frac{{\sqrt{3} } }{1.3 }$

$sin(e)=\frac{1.3 sin (30)}{\sqrt{3}}$

$sin(e) = 0.75* \frac{1}{2}$

$e = sin^{-1}(\frac{0.75 }{2})$

e = 22.02°

Therefore, the angles of emergence are 60° and 22.02° respectively, and the angle of emergence decreases if the ray of light emerges from prism into a liquid.

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