Light is incident in the air at an angle ua(Fig.) on the upper surface of a transparent plate, the surfaces of the plate being plane and parallel to each other.

(a) Prove that ua= u_a.

(b) Show that this is true for any number of different parallel

plates.

(c) Prove that the lateral displacement dof the emergent beam is given by the relationship

$d=\frac{t\sin \left({\theta }_a-\theta {\text{'}}_b\right)}{\cos \theta {\text{'}}_b}$where tis the thickness of the plate. (d) A ray of light is incident at an angle of 66.0_ on one surface of a glass plate 2.40 cm thick with an index of refraction of 1.80. The medium on either side of the plate is air. Find the lateral displacement between the incident and emergent rays.

The refractive index of an optical material is expressed by n and represents the speed of light in the vacuum divided by the speed of light in the material Snell's law for Figure is given by an equation in the form

$n\sin {\theta }_a=n\text{'}\sin \theta {\text{'}}_b$ (1)

This equation for the first refraction at the top plane between the above air and the water. From the Figure, we conclude that $\theta {\text{'}}_a={\theta }_b$. So, equation (1) could be in the form

$nsin{\theta }_a=n\text{'}\sin {\theta }_b$ (2)

Again we apply Snell's law for the bottom plane, where the refraction occurs between the water and the bottom air

$n\sin \theta {\text{'}}_a=n\text{'}\sin {\theta }_b$ (3)

From equations (2) and (3) we could conclude that

$\begin{array}{l}n\sin {\theta }_a=n\sin \text{'}{\theta }_a\\ \sin {\theta }_a=\sin \theta {\text{'}}_a\\ {\theta }_a=\theta {\text{'}}_a\end{array}$

Hence, prove that${\theta }_a=\theta {\text{'}}_a$

From step 1 we get the next equation

$nsin{\theta }_a=n\sin \theta {\text{'}}_b$

As shown, the angle of refraction depends on the refractive index of air, not the water. So, if we increase the number of planes and make the same steps as in step 1, we will get the same result

Therefore, for any number o different parallel plates${\theta }_a=\theta {\text{'}}_a$

If we consider the path that the light travels through the water between the two planes L, we can get d from the triangle inside the water by

$\begin{array}{ll}\sin \left({\theta }_a-\theta {\text{'}}_a\right)=\frac{d}{L}& \\ d=L\sin ({\theta }_a-\theta {\text{'}}_a)& \end{array}$ (4)

The distance between the two planes is t, and we find a relationship between t and L by

$\begin{array}{l}\cos \theta {\text{'}}_b=\frac{t}{L}\\ L=\frac{t}{\cos \theta {\text{'}}_b}\end{array}$

Now, use the expression of L into equation (4) to get d by

$\begin{array}{c}d=L\left({\theta }_a-{\theta }_b^{\mathrm{\prime }}\right)\\ =\left(\frac{t}{\cos {\theta }_b^{\mathrm{\prime }}}\right)\sin \left({\theta }_a-{\theta }_b^{\mathrm{\prime }}\right)\\ =\frac{t\sin \left({\theta }_a-{\theta }_b^{\mathrm{\prime }}\right)}{\cos {\theta }_b^{\mathrm{\prime }}}\end{array}$

We found an expression for the lateral displacement d by

$d=\frac{t\sin \left({\theta }_a-\theta {\text{'}}_b\right)}{\cos \theta {\text{'}}_b}$ (5)

We missed the value of $\theta {\text{'}}_b$We can use Snell's law to get $\theta {\text{'}}_b$, as next

$$\begin{gathered} nsin{\theta }_a=n^{\text{'}}{\theta }_b^{\text{'}} \\ {\theta }_b^{\text{'}}=si{n}^{-1}\frac{n\sin {\theta }_a}{n^{\text{'}}} \end{gathered}$$

Substitute the values for n, n' and ${\theta }_a$to get ${\theta }_b$,

$$\begin{gathered} {\theta }_b^{\text{'}}=si{n}^{-1}\left(\frac{n\sin {\theta }_a}{n\text{'}}\right) \\ =si{n}^{-1}\left(\frac{\left(1\right)\sin {66}^{\circ })}{1.80}\right)=30.5^{\circ } \end{gathered}$$

Now, we put the values for $\theta {\text{'}}_b,{\theta }_a$and t into equation (5) to get d

$\begin{array}{r}d=\frac{t\sin \left({\theta }_a-{\theta }_b^{\mathrm{\prime }}\right)}{\cos {\theta }_b^{\mathrm{\prime }}}\\ d=\frac{\left(2.40\mathrm{cm}\right)\sin \left({66}^{\circ }-{30.5}^{\circ }\right)}{\cos {30.5}^{\circ }}\\ d=1.62\mathrm{cm}\end{array}$