When a light ray moves from air to a transparent medium the angle of incidence and the angle of refraction are $45^{\circ}$ and $30^{\circ}$ respectively. Find the refractive index of this medium is

Question

When a light ray moves from air to a transparent medium the angle of incidence and the angle of refraction are

45^{\circ}

and

30^{\circ}

respectively. Find the refractive index of this medium is

Toppr Admin · Accepted Answer

Given-Angle of incidence -x2220-i-450Angle of refraction -x2220-r-300By Snell-apos-s lawn-sinisinr- where n is the refractive index of the medium-n-sin450sin300-x221A-2-1-42

When a light ray moves from air to a transparent medium the angle of incidence and the angle of refraction are $45^{\circ}$ and $30^{\circ}$ respectively. Find the refractive index of this medium is
$1.21$
$1.42$
$1.88$
$2.42$

Given,
Angle of incidence $∠ i = 45^{0}$
Angle of refraction $∠ r = 30^{0}$

By Snell's law
$n = \frac{s i n i}{s i n r}$ , where $n$ is the refractive index of the medium.

$n = \frac{s i n 45^{0}}{s i n 30^{0}} = \sqrt{2} = 1.42$

When a light ray moves from air to a transparent medium the angle of incidence and the angle of refraction are 45∘ and 30∘ respectively. Find the refractive index of this medium is1.211.421.882.42

Given,Angle of incidence ∠i=450Angle of refraction ∠r=300By Snell's lawn=sinisinr, where n is the refractive index of the medium.n=sin450sin300=√2=1.42

When a light ray moves from air to a transparent medium the angle of incidence and the angle of refraction are $45^{\circ}$ and $30^{\circ}$ respectively. Find the refractive index of this medium is
$1.21$
$1.42$
$1.88$
$2.42$

Given,
Angle of incidence $∠ i = 45^{0}$
Angle of refraction $∠ r = 30^{0}$

By Snell's law
$n = \frac{s i n i}{s i n r}$ , where $n$ is the refractive index of the medium.

$n = \frac{s i n 45^{0}}{s i n 30^{0}} = \sqrt{2} = 1.42$