Maximum value of the angle derivation of deflection of beams
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We know the following relation of refractive index (μ) of the prism with the angle of prism (A) and minimum deviation (∂m)
μ=sin(∂m+A2)sin(A2)
As per given condition we have ∂m=A
So we haμ=sin(A+A2)sin(A2)
⇒μ=sinAsin(A2)
⇒μ=2sin(A2)cos(A2)sin(A2)
⇒cos(A2)=μ2
⇒A=2cos−1(μ2)
If we take for glass μ=1.5
Then A=2cos−1(1.52)=82.8∘
If we think otherwise
This relation reveals that the value of A increases with decrease in value of μ and A will be maximum when μ is minimum.The medium of the material of the prism is optically denser than the surrounding medium. So the minimum value of the refractive (μ) of the medium with respect to the surrounding one may be taken as ≈1.
Imposing this condition we can have the maximum value of A.
Hence Amax=2cos−1(12)=2×60∘=120∘
μ=sin(∂m+A2)sin(A2)
As per given condition we have ∂m=A
So we haμ=sin(A+A2)sin(A2)
⇒μ=sinAsin(A2)
⇒μ=2sin(A2)cos(A2)sin(A2)
⇒cos(A2)=μ2
⇒A=2cos−1(μ2)
If we take for glass μ=1.5
Then A=2cos−1(1.52)=82.8∘
If we think otherwise
This relation reveals that the value of A increases with decrease in value of μ and A will be maximum when μ is minimum.The medium of the material of the prism is optically denser than the surrounding medium. So the minimum value of the refractive (μ) of the medium with respect to the surrounding one may be taken as ≈1.
Imposing this condition we can have the maximum value of A.
Hence Amax=2cos−1(12)=2×60∘=120∘
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