Question
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)
Solution
Hint: In this question use the direct relationship between the angle of prism (A), angle of minimum deviation (θm) and refractive index of the prism, that is μp=sin(2A)sin(2A+θm). 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.
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(2A)sin(2A+θm), where μp 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 And μp=1.5
So substitute these values in the given equation we have,
⇒1.5=sin(2A)sin(2A+A)
⇒1.5=sin(2A)sin(A)
Now as we know that sin2x=2sinxcosx so use this property in above equation we have,
⇒1.5=sin(2A)2sin(2A)cos(2A)=2cos(2A)
⇒cos2A=21.5=0.75
⇒2A=cos−1(0.75)
Now it is given that 0.75 = cos41o so use this value we have,
⇒2A=cos−1(cos41o)=41o
⇒A=2×41o=82o
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. The other includes cos2x=cos2x−sin2x, sin2x=1−cos2x, 1+tan2x=sec2x etc.