Question

A light ray passes from air through a glass plate with refractive index 1.60 into water. The angle of the refracted ray in the water is 42.0°. Determine the angle of the incident ray at the air-glass interface?

Answer 1

Answer:

62.8 degree

Explanation:

Let the incident ray incident at an angle $\theta_1$ at air glass surface.

$\theta_3=42^{\circ}$=Angle of refraction when ray travel from glass to water

$\theta_2=$Angle of refraction when the ray travel from air to glass

Refractive index of glass,$n_2=1.6$

We know that

Refractive index of water=$n_3=1.33$

Snell's law

$n_1sin\theta_1=n_2sin\theta_2$

Where $\theta_1$=Angle of incidence

$\theta_2=$Angle of refraction

$n_1=$Refractive index of medium 1

$n_2=$Refractive index of medium 2

When the ray travel from glass to water

$n_2sin\theta_2=n_3sin\theta_3$

Where $n_2=$Refractive index of medium 1(Glass)

$n_3$=Refractive index of medium 2 (Water)

$\theta_2=$Angle of incidence

$\theta_3$=Angle of refraction

Substitute the values

$1.6sin\theta_2=1.33sin42$

$sin\theta_2=\frac{1.33sin42}{1.6}$

$sin\theta_2=0.556$

$\theta_2=sin^{-1}(0.556)=33.8^{\circ}$

Angle of refraction when ray travel from air to glass= Angle of incidence of when ray travel from glass to water

Angle of refraction when the ray travel from air to glass=33.8 degree

Refractive index of air=$n_1=1$

Again apply Snell's law

$n_1sin\thet_1=n_2sin\theta_2$

$1\times sin\theta_1=1.6sin(33.8)$

$sin\theta_1=1.6\times 0.556=0.8896$

$\theta_1=sin^{-1}(0.8896)=62.8^{\circ}$

Hence, the angle of the incident ray at the air-glass interface=62.8 degree

Answer 2

Answer:

The angle of the incident ray at the air-glass interface is 62.86°

Explanation:

Given that,

Refractive index of glass =1.60

Angle = 42°

We need to calculate the angle of the incident ray at glass-water interface

Using Snell's law

$n_{2}\sin\theta_{2}=n_{3}\sin\theta_{3}$

Where, $n_{2}$ = refractive index of glass

$n_{3}$ = refractive index of water

$\theta_{3}$ = angle of refraction

Put the value into the formula

$1.6\sin\theta_{2}=1.33\sin42$

$\sin\theta_{2}=\dfrac{1.33\sin42}{1.6}$

$\theta_{2}=\sin^{-1}(\dfrac{1.33\sin42}{1.6})$

$\theta_{2}=33.8^{\circ}$

We need to calculate the angle of the incident ray at the air-glass interface

Using Snell's law

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

Where, $n_{1}$ = refractive index of air

$n_{2}$ = refractive index of glass

$\theta_{1}$ = angle of incident

Put the value into the formula

$1\sin\theta_{1}=1.6\sin33.8$

$\theta=\sin^{-1}(1.6\sin33.8)$

$\theta=62.86^{\circ}$

Hence, The angle of the incident ray at the air-glass interface is 62.86°