Question

A circular beam of light of diameter d=2cm falls on a plane surface of glass. The angle of incident is ${60^ \circ }$ and the refractive index of glass is $\mu = \dfrac{3}{2}$ . The diameter of the refracted beam is
A. 2.00cm
B. 1.50cm
C. 1.63cm
D. 3.26cm

Answer 1

Find the angle of refraction using the angle of incidence by Snell’s law. Using the obtained refracted angle, find the diameter of the refracted beam.

Complete step by step answer:

We are given a circular beam of light of diameter d=2cm falls on a plane surface of glass and the angle of incidence is ${60^ \circ }$ and refractive index of glass is $\mu = \dfrac{3}{2}$
We have to calculate the diameter of the refracted beam.
Snell’s law states that the ratio of the sine functions of the angles of incidence and refraction of a wave are constant when it passes between two given media.
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\sin i}}{{\sin r}}$ , where ${n_1},{n_2}$ are the refractive indices of the given two media glass and air; i, r are the angles of incidence and refraction.
Refractive index of air is approximately equal to 1.
Refractive index of the glass is $\dfrac{3}{2}$
Angle of incidence is ${60^ \circ }$
Therefore,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\sin i}}{{\sin r}} \\\ \Rightarrow \dfrac{1}{{\left( {\dfrac{3}{2}} \right)}} = \dfrac{{\sin {{60}^ \circ }}}{{\sin r}} \\\ \Rightarrow \dfrac{2}{3} = \dfrac{{\sin {{60}^ \circ }}}{{\sin r}} \\\ \Rightarrow \sin r = \dfrac{2}{3} \times \sin {60^ \circ } \\\ \Rightarrow \sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2} \\\ \Rightarrow \sin r = \dfrac{2}{3} \times \dfrac{{\sqrt 3 }}{2} \\\ \Rightarrow \sin r = \dfrac{1}{{\sqrt 3 }} \\\ \Rightarrow r = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) \\\ \Rightarrow \cos r = \sqrt {1 - {{\sin }^2}r} \\\ \Rightarrow {\sin ^2}r = {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} = \dfrac{1}{3} \\\ \Rightarrow \cos r = \sqrt {1 - \dfrac{1}{3}} \\\ \Rightarrow \cos r = \sqrt {\dfrac{2}{3}} \\\$
Relation to calculate the diameter of the beam is
$\dfrac{{{D_i}}}{{\cos i}} = \dfrac{{{D_r}}}{{\cos r}} \\\ {D_i} = 2cm \\\ \cos i = \cos {60^ \circ } = \dfrac{1}{2} \\\ \cos r = \sqrt {\dfrac{2}{3}} \\\ {D_r} = ? \\\ \Rightarrow \dfrac{{{D_i}}}{{\cos i}} = \dfrac{{{D_r}}}{{\cos r}} \\\ \Rightarrow \dfrac{2}{{\left( {\dfrac{1}{2}} \right)}} = \dfrac{{{D_r}}}{{\sqrt {\dfrac{2}{3}} }} \\\ \Rightarrow 4 = \dfrac{{{D_r}}}{{\sqrt {\dfrac{2}{3}} }} \\\ \Rightarrow {D_r} = 4 \times \sqrt {\dfrac{2}{3}} \\\ \Rightarrow {D_r} = 3.26cm \\\$
Therefore, the length of the diameter of the refracted beam is 3.26 cm.
The correct answer is Option D.

Note: The angle formed at the point of incidence between the incident ray and the normal is called the angle of incidence. The angle formed between the refracted ray and the normal is called the angle of refraction. Refractive Index is calculated from the ratio of the speed of light in a vacuum to that in a second medium of greater density. The refractive index of a medium depends on the wavelength of the light. If the wavelength of light is less then, the refractive index will be high.