Question

The speed of light in air is 3.0 x 10^8 m/s. The speed of light in particular glass is 2.3 x 10^8 m/s. Use the information to determine the angle of refraction of light which travels from the glass into air, if the angle of incidence on the glass/air boundary is 25° . Draw a diagram.

Answer 1

33.61°

Step-by-step explanation:

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

v is the velocity in the medium (2.3 × 10⁸ m/s)

c is the speed of light in air = 3.0 × 10⁸ m/s

So,

n = 3.0 × 10⁸ / 2.3 × 10⁸

n = 1.31

Using Snell's law as:

`n_i* {sin\theta_i}={n_r}*{sin\theta_r}`

Where,

`{\theta_i}` is the angle of incidence ( 25.0° )

`{\theta_r}` is the angle of refraction ( ? )

`{n_r}` is the refractive index of the refraction medium (air, n=1)

`{n_i}` is the refractive index of the incidence medium (glass, n=1.31)

Hence,

`1.31* {sin25.0^0}={1}*{sin\theta_r}`

Angle of refraction =
`sin^(-1)0.5536` = 33.61°