Question

A transparent object with an isosceles right triangular cross-section has an index of refraction n2 = 1.2, as shown in the diagram below.

A light beam in air is incident on this object, making an angle ?_in = 75? with respect to the x-axis, as shown. At what angle (with respect to the x-axis), ?_out, does the observer see the light beam exit the object?

Answer 1

Concepts and reason

The concept required to solve this problem is Snell’s law.

Initially, use the geometry of the figure made by the light ray passing through the prism. Use the Snell’s law to calculate the angle of refraction from first side of the prism.

Later, calculate the angle of incidence on emergent side by using the property of angles of triangle. Then use the Snell’s law to calculate the angle of emergence to air from prism.

Finally, calculate the angle of emergence from the horizontal axis ding the angle of inclination of normal to the horizontal.

Fundamentals

A prism spreads a visible light or white light i.e. a combination of all visible wavelengths incident on it to a series of colored spectrum called visible spectrum.

Snell’s law of refraction gives the relation between the refractive index of the materials and extent of bending of light ray. When a light ray travels from medium 1 to medium 2 is the equation of Snell’s law is given as,

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

Here, ${n_1}$ and ${n_2}$ are the refraction indices in the two media, ${\theta _1}$ is the angle of incidence, and ${\theta _2}$ is the angle of refraction.

The property of triangle used is that sum of two opposite angles of the triangle is equal to the outer angle of the triangle.

Draw the ray diagram and use the geometry of the figure to determine the angles. The angle of refraction from the first side is ${\theta _2}$ . The angle of incidence is same angle of reflection for other side. The angle of incidence for emerging side is ${\theta _3}$ and angle of emergence is ${\theta _4}$ . The normal is inclined at $45^\circ$ from the horizontal. The angle between emerging ray and horizontal is ${\theta _{{\rm{out}}}}$ .

${\theta _{{\rm{out}}}}$ ${\theta _4}$ $45^\circ$ $90^\circ$ $45^\circ$ ${\theta _3}$ ${\theta _{{\rm{in}}}} = 75^\circ$ ${\theta _2}$ ${\theta _2}$ ${\theta _2}$

Diagram 1: The geometry of the light ray travelling through the prism

Use the Snell’s law for incident side of the prism.

Substitute 1 for ${n_1}$ , $75^\circ$ for ${\theta _1}$ , and 1.2 for ${n_2}$ in Snell’s law of refraction ${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$ and solve for ${\theta _2}$ .

$\begin{array}{c}\\1\left( {\sin 75^\circ } \right) = 1.2\left( {\sin {\theta _2}} \right)\\\\\sin {\theta _2} = \frac{{0.966}}{{1.2}}\\\\{\theta _2} = {\sin ^{ - 1}}\left( {0.805} \right)\\\\ = 53.6^\circ \\\end{array}$

The sum of interior angles is equal to outer angle of the triangle. The angle of incidence on emerging side of the prism can be calculated from this theorem.

$\begin{array}{c}\\{\theta _2} + 45^\circ = {\theta _3} + 90^\circ \\\\{\theta _3} = {\theta _2} - 45^\circ \\\end{array}$

Substitute $53.6^\circ$ for ${\theta _2}$ in the equation ${\theta _3} = {\theta _2} - 45^\circ$ and calculate the angle ${\theta _3}$ .

$\begin{array}{c}\\{\theta _3} = 53.6 - 45^\circ \\\\ = 8.6^\circ \\\end{array}$

Use the Snell’s law for emergent side of the prism.

Substitute ${\theta _4}$ for ${\theta _2}$ , 1.2 for ${n_1}$ , $8.6^\circ$ for ${\theta _1}$ , and 1 for ${n_2}$ in Snell’s law of refraction ${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$ and solve for ${\theta _4}$ .

$\begin{array}{c}\\1.2\left( {\sin 8.6^\circ } \right) = 1\left( {\sin {\theta _4}} \right)\\\\\sin {\theta _4} = \frac{{0.179}}{1}\\\\{\theta _4} = {\sin ^{ - 1}}\left( {0.179} \right)\\\\ = 10.33^\circ \\\end{array}$

Angle of emergence from the $x$ axis is,

${\theta _{{\rm{out}}}} = {\theta _4} + 45^\circ$

Substitute $10.33^\circ$ for ${\theta _4}$ in the equation ${\theta _{{\rm{out}}}} = {\theta _4} + 45^\circ$ and calculate ${\theta _{{\rm{out}}}}$ .

$\begin{array}{c}\\{\theta _{{\rm{out}}}} = 10.33^\circ + 45^\circ \\\\ = 55.33^\circ \\\end{array}$

Ans:

The observer sees the light beam exit the object at angle $55.33^\circ$ .