Question

A laser beam is incident at an angle of 33.0° to the vertical onto a solution of cornsyrup in water.

(a) If the beam is refracted to 24.84° to the vertical, what is the index ofrefraction of the syrup solution?
1

(b) Suppose the light is red, with wavelength 632.8 nm in a vacuum.Find its wavelength in the solution.
2 nm

(c) What is its frequencyin the solution?
3 Hz

(d) What is its speed in the solution?
4 m/s

Answer 1

Concepts and reason

The concepts used to solve this problem are Snell’s law, index of refraction, Planck–Einstein relation, and frequency of light.

First, use the expression for Snell’s law to determine the relation between the refractive index of a medium and the angle of incidence and refraction.

Use the relation between the wavelength and refractive index at the interface to determine the wavelength of the laser in corn syrup.

Use the Planck–Einstein relation to determine the relation between the frequency of laser and its wavelength.

Finally, use the relation between the refractive index and the speed of light to determine its speed in corn syrup solution.

Fundamentals

According to Snell’s law, the ratio of the sine of the angle of incidence and the sine of the angle of refraction is constant.

The expression for the Snell’s law is given below:

$\frac{{{n_1}}}{{{n_2}}} = \frac{{\sin {\theta _2}}}{{\sin {\theta _1}}}$

Here, the refractive index of vacuum is ${n_1}$ , the refractive index of the second medium is ${n_2}$ , the incident angle is ${\theta _1}$ , and the refracted angle is ${\theta _2}$ .

The expression for the relation between the refractive indices and the wavelength of light in a medium is given below:

$\frac{{{n_1}}}{{{n_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}}$

Here, the wavelength of light in the vacuum is ${\lambda _1}$ , and the wavelength of light in the second medium is ${\lambda _2}$ .

The expression for the Planck-Einstein relation is given below:

$f = \frac{c}{{{\lambda _1}}}$

Here, the frequency of light is $f$ , and the speed of light in vacuum is $c$ .

The expression relating to the speed of the light and the refractive index of a medium is given below:

$\frac{v}{c} = \frac{{{n_1}}}{{{n_2}}}$

Here, the speed of laser in the second medium is $v$ , the speed of light in vacuum is $c$ , the refractive index of the vacuum is ${n_1}$ , the refractive index of the second medium is ${n_2}$ .

(a)

The expression for Snell’s law in terms of refractive index, incident angle and refracted angle is given below:

$\frac{{{n_1}}}{{{n_2}}} = \frac{{\sin {\theta _2}}}{{\sin {\theta _1}}}$

Here, ${n_2}$ is the refractive index of the corn syrup.

Rearrange the above expression for the refractive index of the corn syrup,

${n_2} = {n_1}\left( {\frac{{\sin {\theta _1}}}{{\sin {\theta _2}}}} \right)$

Substitute $1$ for ${n_1}$ , $33^\circ$ for ${\theta _1}$ , $24.84^\circ$ for ${\theta _2}$

$\begin{array}{c}\\{n_2} = \left( 1 \right)\left( {\frac{{\sin \left( {33} \right)}}{{\sin \left( {24.84} \right)}}} \right)\\\\ = 1.296\\\end{array}$

[Part a

Part a

[Part a]

The relation between the wavelength in a medium and its refractive index is given below:

$\lambda \alpha \frac{1}{n}$

The ratio between the refractive indices between the vacuum and corn syrup is given below:

$\frac{{{n_1}}}{{{n_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}}$

Here, ${\lambda _2}$ is the wavelength of the light in the corn syrup.

Rearrange the above expression for the wavelength in the corn syrup,

${\lambda _2} = {\lambda _1}\left( {\frac{{{n_1}}}{{{n_2}}}} \right)$

Substitute $1$ for ${n_1}$ , $1.296$ for ${n_2}$ , and $632.8 \times {10^{ - 9}}\,{\rm{m}}$ for ${\lambda _1}$

$\begin{array}{c}\\{\lambda _2} = \left( {632.8 \times {{10}^{ - 9}}\,{\rm{m}}} \right)\left( {\frac{1}{{1.296}}} \right)\\\\ = 488.3 \times {10^{ - 9}}\,{\rm{m}}\\\\ = {\rm{488}}{\rm{.3}}\,{\rm{nm}}\\\end{array}$

(c)

The expression for Planck-Einstein relation is given below:

$f = \frac{c}{{{\lambda _1}}}$

Substitute $3 \times {10^8}\,{\rm{m/s}}$ for $c$ , and $632.8 \times {10^{ - 9}}\,{\rm{m}}$ for ${\lambda _1}$

$\begin{array}{c}\\f = \frac{{3 \times {{10}^8}\,{\rm{m/s}}}}{{632.8 \times {{10}^{ - 9}}\,{\rm{m}}}}\\\\ = 4.74 \times {10^{14}}\,{\rm{Hz}}\\\end{array}$

(d)

The expression relating to the refractive index and the speed of light is given below:

$\frac{{{n_1}}}{{{n_2}}} = \frac{{{v_2}}}{{{v_1}}}$

Rearrange the above expression for the speed of light in the solution as shown below:

${v_2} = {v_1}\left( {\frac{{{n_1}}}{{{n_2}}}} \right)$

Substitute $3 \times {10^8}\,{\rm{m}}{{\rm{s}}^{ - 1}}$ for ${v_1}$ , $1$ for ${n_1}$ , and $1.296$ for ${n_2}$

$\begin{array}{c}\\{v_2} = \left( {3 \times {{10}^8}\,{\rm{m}}{{\rm{s}}^{ - 1}}} \right)\left( {\frac{1}{{1.296}}} \right)\\\\ = 2.314 \times {10^8}\,{\rm{m}}{{\rm{s}}^{ - 1}}\\\end{array}$

Ans: Part a

The refractive index of the corn syrup is ${\bf{1}}{\bf{.296}}$ .

Part b

The wavelength of the red light in the corn syrup is ${\bf{488}}{\bf{.3}}\,{\bf{nm}}$ .

Part c

The frequency of the light in the corn solution is ${\bf{4}}{\bf{.74 \times 1}}{{\bf{0}}^{{\bf{14}}}}\,{\bf{Hz}}$ .

Part d

The speed of light in the corn syrup is ${\bf{2}}{\bf{.314 \times 1}}{{\bf{0}}^{\bf{8}}}\,{\bf{m}}{{\bf{s}}^{{\bf{ - 1}}}}$ .