A light ray falls on a glass surface of refractive index √3, at an angle 60°. The angle between the refracted and reflected rays would be:

Question

A light ray falls on a glass surface of refractive index √3, at an angle 60°. The angle between the refracted and reflected rays would be: (A) 120° (B) 30° (C) 60° (D) 90°

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f82cb4fb7a0633690674832d92c3e3c3-.png" /> Method (i) By Snell's law 1 sin 60°=√3 sin r (√3/2)=√3 sin r sin r=(1/2) r=30° Angle between refracted and reflected ray is 90° Method (ii) Because angle of incidence is Brewster's angle so that angle between reflected and refracted ray is 90° tan ip=μ=√3 ip=60°= i

Q. A light ray falls on a glass surface of refractive index $3$ , at an angle $6 0^{\circ}$ . The angle between the refracted and reflected rays would be:

$12 0^{\circ}$

$3 0^{\circ}$

$6 0^{\circ}$

$9 0^{\circ}$

Q. A light ray falls on a glass surface of refractive index 3​, at an angle 60∘. The angle between the refracted and reflected rays would be:

120∘

30∘

60∘

90∘

Method (i) By Snell's law 1sin60∘=3​sinr 23​​=3​sinr sinr=21​ r=30∘ Angle between refracted and reflected ray is 90∘ Method (ii) Because angle of incidence is Brewster's angle so that angle between reflected and refracted ray is 90∘ tanip​=μ=3​ ip​=60∘=i

Q. A light ray falls on a glass surface of refractive index $3$ , at an angle $6 0^{\circ}$ . The angle between the refracted and reflected rays would be:

$12 0^{\circ}$

$3 0^{\circ}$

$6 0^{\circ}$

$9 0^{\circ}$