A prism of refractive index μ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of μ is

Question

A prism of refractive index μ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of μ is (A)  sin -1((μ/2)) (B)  sin -1(√(μ-1/2)) (C) 2 cos -1((μ/2)) (D)  cos -1((μ/2))

Tardigrade Admin · Accepted Answer

μ=( sin ((A+δm/2))/ sin ((A)2)) Here, δm=A ∴ μ =( sin A/ sin ((A)2))=(2 sin ((A/2)) cos ((A/2))/ sin ((A)2))=2 cos ((A/2)) A =2 cos -1((μ/2))

Q. A prism of refractive index $μ$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is $A$ , then the value of $A$ in terms of $μ$ is

$sin^{- 1} (\frac{μ}{2})$

$sin^{- 1} (\frac{μ - 1}{2})$

$2 cos^{- 1} (\frac{μ}{2})$

$cos^{- 1} (\frac{μ}{2})$

$μ = \frac{s i n ( \frac{A + δ _{m}}{2} )}{s i n ( \frac{A}{2} )}$
Here, $δ_{m} = A$
$∴ μ = \frac{s i n A}{s i n ( \frac{A}{2} )} = \frac{2 s i n ( \frac{A}{2} ) c o s ( \frac{A}{2} )}{s i n ( \frac{A}{2} )} = 2 cos (\frac{A}{2})$
$A = 2 cos^{- 1} (\frac{μ}{2})$

Q. A prism of refractive index μ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of μ is

sin−1(2μ​)

sin−1(2μ−1​​)

2cos−1(2μ​)

cos−1(2μ​)