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: $({\text{given co}}{{\text{s}}^0}41 = 0.75)$
Answer
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Hint: In this question use the direct relationship between the angle of prism (A), angle of minimum deviation (${\theta _m}$) and refractive index of the prism, that is ${\mu _p} = \dfrac{{\sin \left( {\dfrac{{A + {\theta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$. 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 ${\theta _m}$.
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
$ \Rightarrow {\mu _p} = \dfrac{{\sin \left( {\dfrac{{A + {\theta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$, where ${\mu _p}$ 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.
$ \Rightarrow {\theta _m} = A$ And ${\mu _p} = 1.5$
So substitute these values in the given equation we have,
$ \Rightarrow 1.5 = \dfrac{{\sin \left( {\dfrac{{A + A}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
$ \Rightarrow 1.5 = \dfrac{{\sin \left( A \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
Now as we know that $\sin 2x = 2\sin x\cos x$ so use this property in above equation we have,
$ \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)$
$ \Rightarrow \cos \dfrac{A}{2} = \dfrac{{1.5}}{2} = 0.75$
$ \Rightarrow \dfrac{A}{2} = {\cos ^{ - 1}}\left( {0.75} \right)$
Now it is given that 0.75 = $\cos {41^o}$ so use this value we have,
$ \Rightarrow \dfrac{A}{2} = {\cos ^{ - 1}}\left( {\cos {{41}^o}} \right) = {41^o}$
$ \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 $\sin 2x = 2\sin x\cos x$. The other includes $\cos 2x = {\cos ^2}x - {\sin ^2}x$, ${\sin ^2}x = 1 - {\cos ^2}x$, $1 + {\tan ^2}x = {\sec ^2}x$ etc.
Complete step-by-step answer:
Let the angle of minimum deviation for a prism be ${\theta _m}$.
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
$ \Rightarrow {\mu _p} = \dfrac{{\sin \left( {\dfrac{{A + {\theta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$, where ${\mu _p}$ 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.
$ \Rightarrow {\theta _m} = A$ And ${\mu _p} = 1.5$
So substitute these values in the given equation we have,
$ \Rightarrow 1.5 = \dfrac{{\sin \left( {\dfrac{{A + A}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
$ \Rightarrow 1.5 = \dfrac{{\sin \left( A \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
Now as we know that $\sin 2x = 2\sin x\cos x$ so use this property in above equation we have,
$ \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)$
$ \Rightarrow \cos \dfrac{A}{2} = \dfrac{{1.5}}{2} = 0.75$
$ \Rightarrow \dfrac{A}{2} = {\cos ^{ - 1}}\left( {0.75} \right)$
Now it is given that 0.75 = $\cos {41^o}$ so use this value we have,
$ \Rightarrow \dfrac{A}{2} = {\cos ^{ - 1}}\left( {\cos {{41}^o}} \right) = {41^o}$
$ \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 $\sin 2x = 2\sin x\cos x$. The other includes $\cos 2x = {\cos ^2}x - {\sin ^2}x$, ${\sin ^2}x = 1 - {\cos ^2}x$, $1 + {\tan ^2}x = {\sec ^2}x$ etc.
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