Question

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° with respect to the normal to the surface. It passes into the second medium and refracts at an angle of θ2 = 69° with respect to the normal.Randomized Variablesn1 = 1.47θ1 = 59°θ2 = 69°Part (a) Write an expression for the index of refraction of the second material.Part (b) Numerically, what is this index?Part (c) Numerically, what is the light's velocity in medium 1, in meters per second?Part (d) Numerically, what is the light's velocity in medium 2, in meters per second?

Answer 1

Answer:

a) $n_{12}=0.918$

b) $n_2=1.349$

c) $v_1=2.041\times 10^{8}\ m.s^{-1}$

d) $v_2=2.223\times 10^8\ m.s^{-1}$

Explanation:

Given:

• refractive index of liquid medium 1 (with respect to air), $n_1=1.47$

• angle of incidence in medium 1, $\theta_1=59^{\circ}$

• angle of refraction in medium 2, $\theta_2=69^{\circ}$

(a)

According to Snell's Law:

refractive index of medium 2 with respect to medium 1:

$n_{12}=\frac{sin\ \theta_1}{sin\ \theta_2}$

$n_{12}=\frac{sin\ 59^{\circ}}{sin\ 69^{\circ}}$

$n_{12}=0.918$

(c)

Now the other form of Snell's law:

$n =\frac{c}{v}$ ..............................(2)

where:

c = speed of light in air

n = refractive index of the medium with respect to air

v = speed of light in medium

Using eq. (2) for medium 1:

$1.47=\frac{3\times 10^{8}}{v_1}$

$v_1=2.041\times 10^{8}\ m.s^{-1}$

(d)

Using eq. (2) for medium 2:

$n{12}=\frac{v_1}{v_2}$

$0.918 =\frac{2.041\times 10^{8}}{v_2}$

$v_2=2.223\times 10^8\ m.s^{-1}$

(b)

Now, refractive index of the medium 2 with respect to air

$n_2=\frac{3\times 10^8}{2.223\times 10^8}$

$n_2=1.349$

Answer 2

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(c) $v_{1} = 2.04\times 10^{8}\ m/s$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$