Question

Let v be the wave's speed, λ its wavelength, and f its frequency. These quantities are related via the equation v=λf. Note that, if the wave speed decreases, the wavelength must also decrease for the frequency to remain constant.

a) What is the wavelength λ of light in glass if its wavelength in air is λ0, its speed in air is c, and its speed in the glass is v?

Express your answer in terms of λ0, c, and v.

d) If light strikes the air/glass interface at an angle 32.0° to the normal, what is the angle of reflection,θt?

e) If light strikes the air/glass interface at an incidence angle of 32.0° , what is the angle of refraction, θb? Use 1.50 for the index of refraction of glass.

Answer 1

Concepts and reason

The concepts used to solve this problem are Planck-Einstein relation, law of reflection, and Snell’s law.

First, find the relationship between the wavelengths of light in glass and its wavelength in air, and speed in glass and air using Planck-Einstein relation.

First, find the angle of reflection of the light that is reflected from the surface of glass using the law of reflection.

Finally, find the angle of refraction of the light using Snell’s law that connects the wavelength of a light in a medium with the refractive index of the medium.

Fundamentals

The expression for Planck-Einstein relation is as follows:

$c = f\left( \lambda \right)$

Here, the speed of light is $c$, the frequency of light is $f$, and the wavelength of light is $\lambda$.

The expression for the frequency of light in air is as follows:

${f_{air}} = \frac{c}{{{\lambda _0}}}$

Here, the frequency of light in air is ${f_{air}}$, the wavelength of light in air is ${\lambda _0}$, and the speed of light in air is $c$.

The expression for the frequency of light in glass is as follows:

${f_{glass}} = \frac{v}{\lambda }$

Here, the frequency of light in glass is ${f_{glass}}$, the wavelength of light in glass is $\lambda$, and the speed of light in air is $v$.

The law of reflection states that the angle of incidence and angle of reflection will be equal for a light which gets reflected by a surface.

The expression for this condition is as follows:

${\theta _i} = {\theta _r}$

Here, the angle of incidence is ${\theta _i}$, and the angle of reflection is ${\theta _r}$.

Snell’s law states that the ratio between the sine of angles of incidence and the refraction of a light that is crossing a boundary between two media is equivalent to the inverse ratio between the refractive indices of those media.

The expression for Snell’s law for the light travelling from air to glass is as follows:

$\frac{{\sin {\theta _i}}}{{\sin {\theta _r}}} = \frac{{{n_g}}}{{{n_a}}}$

Here, the angle of incidence is ${\theta _i}$, the angle of refraction is ${\theta _r}$, the refractive index of air is ${n_a}$, and the refractive index of the glass is ${n_g}$.

(a)

The expression for the frequency of light in air is as follows:

${f_{air}} = \frac{c}{{{\lambda _0}}}$ …… (1)

The expression for the frequency of light in glass is as follows:

${f_{glass}} = \frac{v}{\lambda }$ …… (2)

The relationship between the frequency of light in air and frequency of light in glass is as follows:

${f_{air}} = {f_{glass}}$

Use the above condition in Equations (1) and (2).

$\frac{c}{{{\lambda _0}}} = \frac{v}{\lambda }$

Rearrange the above expression for the wavelength of light in glass.

$\lambda = \frac{{{\lambda _0}v}}{c}$

(d)

The law of reflection is expressed as follows:

${\theta _i} = {\theta _r}$

Substitute $32^\circ$ for ${\theta _i}$.

${\theta _r} = 32^\circ$

(e)

Snell’s law is expressed as follows:

$\frac{{\sin {\theta _i}}}{{\sin {\theta _r}}} = \frac{{{n_g}}}{{{n_a}}}$

The above expression can be rearranged to get the angle of refraction.

${\theta _r} = {\sin ^{ - 1}}\left[ {\left( {\frac{{{n_a}}}{{{n_g}}}} \right)\left( {\sin {\theta _i}} \right)} \right]$

Substitute $1.00$ for ${n_a}$, $1.50$ for ${n_g}$, and $32^\circ$ for ${\theta _i}$.

$\begin{array}{c}\\{\theta _r} = {\sin ^{ - 1}}\left[ {\left( {\frac{{1.00}}{{1.50}}} \right)\left( {\sin \left( {32^\circ } \right)} \right)} \right]\\\\ = 20.7^\circ \\\end{array}$

Ans: Part a

The expression for the wavelength of light in glass is $\left( {{\lambda _0}v} \right)/c$.

Part d

The angle of reflection of light from the surface of glass is ${\bf{32^\circ }}$.

Part e

The required angle of refraction of light is $20.7^\circ$.