A ray of light traveling in the air falls on the surface of a rectangular slab of a plastic material whose refractive index is $1.6$ . If the incident ray makes an angle of $53^{\circ}$ with the normal $(sin 53^{\circ} = 4 / 5)$ . Find the angle made by the refracted ray with the normal.

Open in App

Solution

We know that refractive index of air is 1
Here the incidence angle is $i = 53^{o}$
Using snell's law
$n_{a i r} s i n i = n_{g l a s s} s i n r$
$s i n 53 = 1.6 s i n r$
$\frac{4}{5} = 1.6 s i n r$
$r = 30^{o}$

Suggest Corrections

Similar questions

Q. A ray of light travelling in the air falls on the surface of a plastic slab at an angle 53° (

s i n 53^{0}

\frac{4}{5}

) with the normal. The refractive index of plastic slab is 1.6. Calculate the angle of refraction.

Q. A light ray passes from air through a thin rectangular plastic slab (refractive index = 1.3). If the ray in air makes an angle of

45^{\circ}

with the normal after leaving the slab, what is the angle of incidence for the light falling on the plastic slab?

A ray of light travelling in the air falls on the surface of a glass slab at an angle $45^{\circ}$ with the normal. The refractive index of glass is 1.5. Calculate the angle of refraction.

A ray of light traveling in air falls on the surface of water. The angle of incidence is $60^{0}$ with the normal to the surface. The refractive index of water = $\frac{4}{3}$ . Calculate the angle of refraction. [2 MARKS]

Q. A ray of light falls on the surface of spherical glass paperweight making an angle

α

with the normal and is refracted in the medium at an angle

β

. The angle of deviation of the emergent ray from the incident ray is

Join BYJU'S Learning Program

A ray of light traveling in the air falls on the surface of a rectangular slab of a plastic material whose refractive index is 1.6. If the incident ray makes an angle of 53∘ with the normal (sin53∘=4/5). Find the angle made by the refracted ray with the normal.

We know that refractive index of air is 1Here the incidence angle is i=53oUsing snell's lawnairsini=nglasssinrsin53=1.6sinr45=1.6sinrr=30o

A ray of light traveling in the air falls on the surface of a rectangular slab of a plastic material whose refractive index is $1.6$ . If the incident ray makes an angle of $53^{\circ}$ with the normal $(sin 53^{\circ} = 4 / 5)$ . Find the angle made by the refracted ray with the normal.

We know that refractive index of air is 1
Here the incidence angle is $i = 53^{o}$
Using snell's law
$n_{a i r} s i n i = n_{g l a s s} s i n r$
$s i n 53 = 1.6 s i n r$
$\frac{4}{5} = 1.6 s i n r$
$r = 30^{o}$