When light rays are incident on a prism at an angle of 45°, the minimum deviation is obtained. If refractive index of the material of prism is √2, then the angle of prism will be

Question

When light rays are incident on a prism at an angle of 45°, the minimum deviation is obtained. If refractive index of the material of prism is √2, then the angle of prism will be (A) 30° (B) 40° (C) 50° (D) 60°

Tardigrade Admin · Accepted Answer

( sin (A+δm/2)/ sin (A)2)=μ, But (A+δm/2)=i=45° So, ( sin 45°/ sin (A / 2))=√2 ⇒ (1/2)= sin (A/2) ⇒ A=60°

Q. When light rays are incident on a prism at an angle of $4 5^{\circ}$ , the minimum deviation is obtained. If refractive index of the material of prism is $2$ , then the angle of prism will be

$3 0^{\circ}$

$4 0^{\circ}$

$5 0^{\circ}$

$6 0^{\circ}$

$\frac{s i n \frac{A + δ _{m}}{2}}{s i n \frac{A}{2}} = μ$ , But $\frac{A + δ _{m}}{2} = i = 4 5^{\circ}$
So, $\frac{s i n 4 5 ^{\circ}}{s i n ( A /2 )} = 2$
$\Rightarrow \frac{1}{2} = sin \frac{A}{2} \Rightarrow A = 6 0^{\circ}$

Q. When light rays are incident on a prism at an angle of 45∘, the minimum deviation is obtained. If refractive index of the material of prism is 2​, then the angle of prism will be

30∘

40∘

50∘

60∘