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

Angle of minimum deviation for a prism of refractive index 1.5, is equal to the angle of the prism. Then the angle of the prism is: (given cos041=0.75)({\text{given co}}{{\text{s}}^0}41 = 0.75)

Explanation

Solution

Hint: In this question use the direct relationship between the angle of prism (A), angle of minimum deviation (θm{\theta _m}) and refractive index of the prism, that is μp=sin(A+θm2)sin(A2){\mu _p} = \dfrac{{\sin \left( {\dfrac{{A + {\theta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}. Then use the constraints given in the question to find the value of the angle of the prism.

Complete step-by-step answer:
Let the angle of minimum deviation for a prism be θm{\theta _m}.
Let the angle of the prism = A.

Now as we know the relation of refractive index of the prism, angle of minimum deviation of the prism and angle of the prism is given as

μp=sin(A+θm2)sin(A2) \Rightarrow {\mu _p} = \dfrac{{\sin \left( {\dfrac{{A + {\theta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}, where μp{\mu _p} is the refractive index of the prism.
Now it is given that the angle of minimum deviation of the prism is equal to the angle of the prism.

θm=A \Rightarrow {\theta _m} = A And μp=1.5{\mu _p} = 1.5
So substitute these values in the given equation we have,
1.5=sin(A+A2)sin(A2)\Rightarrow 1.5 = \dfrac{{\sin \left( {\dfrac{{A + A}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}
1.5=sin(A)sin(A2)\Rightarrow 1.5 = \dfrac{{\sin \left( A \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}
Now as we know that sin2x=2sinxcosx\sin 2x = 2\sin x\cos x so use this property in above equation we have,
1.5=2sin(A2)cos(A2)sin(A2)=2cos(A2)\Rightarrow 1.5 = \dfrac{{2\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{A}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}} = 2\cos \left( {\dfrac{A}{2}} \right)
cosA2=1.52=0.75\Rightarrow \cos \dfrac{A}{2} = \dfrac{{1.5}}{2} = 0.75
A2=cos1(0.75)\Rightarrow \dfrac{A}{2} = {\cos ^{ - 1}}\left( {0.75} \right)
Now it is given that 0.75 = cos41o\cos {41^o} so use this value we have,
A2=cos1(cos41o)=41o\Rightarrow \dfrac{A}{2} = {\cos ^{ - 1}}\left( {\cos {{41}^o}} \right) = {41^o}
A=2×41o=82o\Rightarrow A = 2 \times {41^o} = {82^o}

So this is the required angle of the prism.
Hence option (C) is the correct answer.

Note: Sometimes the knowledge of trigonometric identities also helps in simplification of the problems of this kind, so it is advised to have a good gist of the trigonometric identities some of them are being mentioned above like sin2x=2sinxcosx\sin 2x = 2\sin x\cos x. The other includes cos2x=cos2xsin2x\cos 2x = {\cos ^2}x - {\sin ^2}x, sin2x=1cos2x{\sin ^2}x = 1 - {\cos ^2}x, 1+tan2x=sec2x1 + {\tan ^2}x = {\sec ^2}x etc.