Question

A ray of light in air is incident at an angle of 25.0° on a glass slide with index of refraction 1.51. (a) At what angle is the ray refracted? ________° from the normal (b) If the wavelength of the light in vacuum is 580 nm, find its wavelength in the glass. ________ nm

Answer 1

Answer:

16.25° , 348.11 nm

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

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  (glass, n=1.51)

${n_i}$ is the refractive index of the incidence medium (air, n=1)

Hence,  

$1\times {sin25.0^0}={1.51}\times{sin\theta_r}$

Angle of refraction = $sin^{-1}0.2798$ = 16.25°

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.

The frequency of the light does not change but the wavelength of the light changes with change in the speed.

c = frequency × Wavelength

Frequency is constant,  

The formula can be written as:

n = λ / λn.

Where,  

λn is the wavelength in the medium  

λ is the wavelength in vacuum (580 nm)

So,

1.51 = 580 /  λn

λn = 348.11 nm