Question

(a) Prove that a ray of light incident on the surface of a sheet of plate glass of thickness$\mathit{t}$emerges from the opposite face parallel to its initial direction but displaced sideways, as in Fig. 33-69. (b) Show that, for small angles of incidence $\mathit{\theta }$, this displacement is given by$\mathit{x}\mathbf{=}\mathit{t}\mathit{\theta }\frac{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{n}}$

where $\mathit{n}$is the index of refraction of the glass and$\mathit{\theta }$is measured in radians.

Short answer

a. A ray of light incident on the surface of a sheet of plate glass of thickness temerges from the opposite face parallel to its initial direction but displaced sideways is proved.

b. For small angles of incidence $\theta$, the displacement is given by $x=t\theta \frac{n-1}{n}$ is proved.

Step-by-step solution

Given Information

The thickness of the glass plate is $t$.

Understanding the concept of refraction

Apply the law of refraction to both faces of a glass plate.

The law of reflection states that the incident ray, the refracted ray, and the angle of incidence lie in the same plane.

Also, Snell's law of refraction states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.

Formula:

Snell’s law of refraction of a light ray through two different mediums,

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$...............(1)

(a) Calculation to prove that an emergent ray from the opposite plate is parallel to the incident ray

Suppose the angle of refraction in the glass plate is ${\theta }_2$, and the angle of emerging ray, from the opposite face is, ${\theta }_3$ as shown in the figure.

Let apply the law of refraction to the left face of the glass plate:

As known from the figure that

The reflective index of air, $n_1=1$

The reflective index of medium, $n_2=n$

The angle of incidence according to the figure is${\theta }_1=\theta$.

Substitute the above values into equation (1), and we get,

$\sin \theta =n\sin {\theta }_2$ ..............(2)

Similarly, applying the law of refraction from equation (1) at the right face, we get,

$n\sin {\theta }_2=\sin {\theta }_3$

From eq. (2) and (3), we get,

$\begin{array}{rcl}\sin \theta & =& \sin {\theta }_3\\ \theta & =& {\theta }_3\end{array}$

This proves that the emerging ray from the opposite face of the glass plate is parallel to the incident ray.

(b) Calculation of the displacement of emergent ray for small incident angles

The displacement by which the emerging ray is displaced is given from the figure as:

$x=d\sin \alpha$

$x=d\sin \left(\theta -{\theta }_2\right)$.........................(4)

Where, $d$ is the length of the ray in the glass plate.

Now, from the given data and figure, the thickness of the material in relationship with the lengthcan be given as:

role="math" localid="1662998769513" $\begin{array}{rcl}d\cos {\theta }_2& =& t\\ d& =& \frac{t}{\cos {\theta }_2}\end{array}$

Substituting this in the above equation (4) of displacement, you get the displacement value as:

$x=\frac{t\sin \left(\theta -{\theta }_2\right)}{\cos {\theta }_2}$

Now, using the small angle approximation, we have,

$\sin \theta ~\theta \text{and}\cos \theta ~1$

So, the displacement will become,

$x=t\left(\theta -{\theta }_2\right)$............................(5)

Applying the small angle approximation to equation (2), we get

$\theta =n{\theta }_2$

Using this, in equation (3), we get the displacement value for the emergent ray as follows;

$\begin{array}{rcl}x& =& t\left(n{\theta }_2-{\theta }_2\right)\\ & =& t{\theta }_2\left(n-1\right)\\ & =& t\theta \left(\frac{n-1}{n}\right)\\ & & \end{array}$

Hence, the displacement for small angles of incidence is $t\theta \left(\frac{n-1}{n}\right)$.