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# A circular beam of light of diameter $d = 2cm$ falls on a plane surface of the glass. The angle of incident is ${60^0}$ and the refractive index of glass is $\mu = \dfrac{3}{2}$ . The diameter of the refracted beam is:$\left( A \right)2.00cm$$\left( B \right)1.50cm$$\left( C \right)1.63cm$$\left( D \right)2.52cm$

Last updated date: 12th Aug 2024
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Hint: The ratio of the speed of light in a vacuum to the speed of light in the denser medium is termed as a refractive index. Refractive index of a material depends on the wavelength of light that penetrates. Refractive index increases with decreasing wavelength of light. Apply the snell's law to find the refracted angle. Then, find the relation between the diameter of the beam.

Formula used:
${\mu _1}\sin \left( i \right) = {\mu _2}\sin \left( r \right)$
${\mu _1},{\mu _2}$ are the refractive indexes.

Complete step by step solution:
The bending of a light wave passing through one medium to another medium due to the change in wave speed is the Refraction. The ray bends towards the normal when light travels from a rarer medium to a denser medium.
For lenses the sign convention:
The focal length of a convex lens is positive and for the concave lens is negative.
The optical center of a lens lies on the origin of the x-y axis.
The magnification is defined as the ratio of the height of the image to that of the height of the object. Lens formula gives the relationship between object distance, image distance, and the focal length.
The Refractive index of a material depends on the wavelength of light that penetrates. Refractive index increases with decreasing wavelength of light.
$\Rightarrow \sin \left( i \right) = \mu \sin \left( r \right)$
$\Rightarrow \sin 60^\circ = \dfrac{3}{2} \times \sin \left( r \right)$
$\Rightarrow r = \sin \left( {\dfrac{1}{{\sqrt 3 }}} \right)$
$\Rightarrow \cos r = \sqrt {\dfrac{2}{3}}$
The relation for the diameter of the beam,
$\Rightarrow \dfrac{{{D_i}}}{{\cos \left( i \right)}} = \dfrac{{{D_r}}}{{\cos \left( r \right)}}$
$\Rightarrow {D_r} = 2\sqrt {\dfrac{2}{3}}$
$\Rightarrow {D_r} = 1.63cm$

Hence, the option $(C)$ is the right option.

Note: If an object is virtual then that means the object is at infinity. The focal length of the plane mirror is infinity. Mirror will always produce a virtual image of an object that is real whether it is a convex or a concave mirror. In a plane mirror always, a real object will produce a virtual image and that is because the reflected rays diverge. Plane mirrors usually form virtual images.