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Question: The refractive index of a prism for a monochromatic wave is \(\sqrt 2\) and its refractive angle is ...

The refractive index of a prism for a monochromatic wave is 2\sqrt 2 and its refractive angle is 60{60^\circ }. For minimum deviation, the angle of incidence will be:
A) 30{30^\circ }
B) 45{45^\circ }
C) 60{60^\circ }
D) 75{75^\circ }

Explanation

Solution

In this question we have to find out the angle of incidence. The angle of incidence is defined as the angle between these two rays is angle of deviation denoted by ‘D’ and angle of incidence (i). Angle of deviation can be defined if ray passes different refractive indexes of different medium then the difference of incidence angle and refractive angle is equal to deviation of angle. Knowing all these things we can calculate the angle of incidence.

Formula used:
By using the formula we can find. The formula is μ=sinisin(A/2)\mu = \dfrac{\sin i}{\sin(A/2)}
Where, μ=\mu = Refractive index.

Complete step by step solution:
We know the formula sini=μ×sin(A/2)\sin i = \mu \times \sin(A/2)
Where, sini\sin i is angle of incidence, sinA\sin A is angle of reference, μ=\mu = refractive index of prism.
Refractive index- Refractive index is defined as that describes how the material affects the speed of light travelling through it. It is dimensionless. It also describes how fast light travels. Where cc is the speed of light in vacuum and vv is the velocity of light in medium.
Prism- Prism is defined as it is a transparent solid body. It is basically used for dispersing light into a spectrum, and also it can be used for reflecting rays of light. This is known as prism.
Minimum deviation can be defined as the angle of deviation decreases with increases in the angle of incidence up to a particular angle. This is known as minimum deviation.
Given that, μ=2\mu = \sqrt 2 , A=60A = {60^\circ }
Now, putting in the value.
sini=2×sin602\Rightarrow \sin i = \sqrt 2 \times \sin \dfrac{{{{60}^\circ }}}{2}
sini=2×sin30\Rightarrow \sin i = \sqrt 2 \times \sin {30^\circ }
Now, sin30=12\sin {30^\circ } = \dfrac{1}{2}
sini=22×22\Rightarrow \sin i = \dfrac{{\sqrt 2 }}{2} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}
Now, substituting the value
sini=222\Rightarrow \sin i = \dfrac{2}{{2\sqrt 2 }}
We know that 12\dfrac{1}{{\sqrt 2 }}
So, the correct answer is 45{45^ \circ }.

Note: For solving this question we need to know what is refractive index, angle of incidence, angle of deviation and also angle of incidence for prisms. The angle of incidence is defined as the angle between the normal and the ray of light is called the angle of incidence. The angle of deviation in a prism will be minimum if the incident rays and other existing rays will form the angle with the faces of the prism which will be equal; this is known as the angle of deviation in a prism. Also known as the unit. So, these are some important points that must be remembered while solving this question to get the accurate answer.