Question

At what angle of incidence should light beam strike a glass slab of refractive index √3 such that the refracted and reflected rays are perpendicular to each other ?

Answer 1

Let incident angle is i.

according to law of reflection,

incident angle = reflection angle

so, reflected angle = i ......(1)


again, law of refraction,

$\mu_{air}sin\theta_1=\mu_2sin\theta_2$

here, $\mu_1=1,\theta_1=i,\mu_2=\sqrt{3}$ and $\theta_2=r$

so, 1 × sini = √3 × sinr ......(2)

a/c to question, reflected ray and refracted ray are perpendicular to each other .

so, i + r = 90° [ reflected angle = i from (1), ]


now, equation (2), sini = √3sin(90 - i)

sini = √3 cosi

tani = √3 = tan60° hence, i = 60°

hence, incidence angle should be 60°

Answer 2

Answer:

As we know that when light is incident with blisters angel, reflected and refracted ray are perpendicular to each other ...so

Tan Q=√3

==>Q = 60°

Explanation: