The refractive index for a prism is given as μ = cot (A/2). Then, angle of minimum deviation in terms of angle of prism is

Question

The refractive index for a prism is given as μ = cot (A/2). Then, angle of minimum deviation in terms of angle of prism is (A) 90° - A  (B) 2 A (C) 180° - A  (D) 180° - 2A

Tardigrade Admin · Accepted Answer

Using prism formula, μ = ( sin ((A + δm/2))/ sin ((A)2)) ......(i) where, A = angle of prism δm = angle of minimum deviation Given, μ = cot((A/2)) = ( cos((A/2))/ sin ((A)2)) So, from Eq. (i) ( cos((A/2)) / sin ((A)2)) = ( sin ((A+δ m/2))/ sin((A)2) ) ⇒ sin((π/2) - (A/2)) = sin((A/2) + (δm/2)) ⇒ δm = π - 2A = 180° - 2A

Q. The refractive index for a prism is given as $μ = cot \frac{A}{2}$ . Then, angle of minimum deviation in terms of angle of prism is

$9 0^{\circ} - A$

$2 A$

$18 0^{\circ} - A$

$18 0^{\circ} - 2 A$

Q. The refractive index for a prism is given as μ=cot2A​. Then, angle of minimum deviation in terms of angle of prism is

90∘−A

2A

180∘−A

180∘−2A

Q. The refractive index for a prism is given as $μ = cot \frac{A}{2}$ . Then, angle of minimum deviation in terms of angle of prism is

$9 0^{\circ} - A$

$2 A$

$18 0^{\circ} - A$

$18 0^{\circ} - 2 A$