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Question: Angle of minimum deviation of a prism of a refractive index 1.5 is equal to the angle of the prism o...

Angle of minimum deviation of a prism of a refractive index 1.5 is equal to the angle of the prism of prism. Then the angle of the prism is:
A. 4124{41^{^ \circ }}24'
B. 80{80^ \circ }
C. 60{60^ \circ }
D. 8248{82^ \circ }48'

Explanation

Solution

Hint In the question, the angle of minimum deviation of a prism of a refractive index is given. By using the trigonometric equations in the refractive index as per the given conditions and simplifying the equation, then we get the value of the angle of the prism.

Complete step by step solution
A prism is a wedge-shaped body made from a refracting medium bounded by two plane faces inclined to each other at some angle. The two plane faces are called the refracting faces and the angle included between these faces is called the angle of the prism or the angle of the refraction.
Let δm{\delta _m} be the angle of minimum deviation of the prism.
A=δmA = {\delta _m}
μ=sin(A+δm)2Sin(A2)\mu = \dfrac{{\sin \dfrac{{\left( {A + {\delta _m}} \right)}}{2}}}{{\operatorname{Sin} \left( {\dfrac{A}{2}} \right)}}
Substitute the parameter of δm{\delta _m}in the above equation, we get
μ=sin(A+A)2Sin(A2)\mu = \dfrac{{\sin \dfrac{{\left( {A + A} \right)}}{2}}}{{\operatorname{Sin} \left( {\dfrac{A}{2}} \right)}}
Substitute the known values in the above equation, we get
1.5=sin(A+A)2Sin(A2)1.5 = \dfrac{{\sin \dfrac{{\left( {A + A} \right)}}{2}}}{{\operatorname{Sin} \left( {\dfrac{A}{2}} \right)}}
Simplify the above equation, we get
1.5=sin2(A2)sin(A2)1.5 = \dfrac{{\sin \,2\left( {\dfrac{A}{2}} \right)}}{{\sin \,\left( {\dfrac{A}{2}} \right)}}
Performing the algebraic operation in the above equation, we get
1.5=2SinA2cosA2Sin(A2)1.5 = \dfrac{{2\operatorname{Sin} \dfrac{A}{2}\cos \dfrac{A}{2}}}{{\operatorname{Sin} \left( {\dfrac{A}{2}} \right)}}
Simplify the above equation, we get
1.5=2cosA21.5 = 2\cos \,\dfrac{A}{2}
Performing the arithmetic operation in the above equation, we get
0.75=cosA20.75 = \cos \dfrac{A}{2}
Convert the equation in terms of A, we get
A2=cos1(0.75)\dfrac{A}{2} = {\cos ^{ - 1}}\left( {0.75} \right)
Substitute the algebraic parameters in terms of the equation, we get
A=41×2A = 41 \times 2
A=82.A = {82^ \circ }.
Therefore, the angle of the prism is 82.{82^ \circ }.

Hence from the above options, option D is correct.

Note In the question, a refractive index is given. If here the angle of the prism is given. By substitute those values in the expression of the angle of the deviation of the prism. We get the value of the angle of the prism.