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

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

Step-by-step 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)=(1.33\sin42)/(1.6)`

`\theta_(2)=\sin^(-1)((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°

Answer 2

62.8 degree

Step-by-step 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=(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* sin\theta_1=1.6sin(33.8)`

`sin\theta_1=1.6* 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