In Fig. 33-71, two light rays pass from air through five layers of transparent plastic and then back into air. The layers have parallel interfaces and unknown thicknesses; their indexes of refraction are${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.7}\mathbf{,}\mathbf{}\mathbf{}{\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.6}\mathbf{,}\mathbf{}\mathbf{}{\mathit{n}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.5}\mathbf{,}\mathbf{}$. ${\mathit{n}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.4}\mathbf{,}\mathbf{}\mathbf{}{\mathit{n}}_{\mathbf{5}}\mathbf{=}\mathbf{1}\mathbf{.6}$Ray b is incident at angle${\mathit{\theta }}_{\mathbf{b}}\mathbf{=}\mathbf{20}\mathbf{°}$. Relative to a normal at the last interface, at what angle do (a) ray a and (b) ray b emerge? (Hint: Solving the problem algebraically can save time.) If the air at the left and right sides in the figure were, instead, glass with index of refraction 1.5, at what angle would (c) ray a and (d) ray b emerge?

.$n_1=1.7,n_2=1.6,n_3=1.5,n_4=1.4,n_5=1.6$

b) The refractive index of glass,$n_{glass}=1.5$

c) The ray b is incident at an angle,${\theta }_b=20°$

The special angle of incidence that produces a 90 degrees angle between the reflected and refracted ray is called the Brewster angle. If an unpolarized beam of light is incident on a boundary between two media at Brewster angle, the reflected wave will be fully polarized. For this situation, the tan of the Brewster angle is equal to the ratio of refractive indices of the two media.

Formula:

Snell’s law of refraction states that the ratio of refractive indices is given by the ratio of the sine angles of the incident and refracted ray, $n_{1}sin{\theta }_1=n_{2}sin{\theta }_2$ (i)

The ray a is incident along the normal to the boundary of the media. Hence it makes an angle.${\theta }_a=0°$

So, for ray a and media pair of air and layer 1, the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}{n}_{i}sin{\theta }_a=n_{1}sin{\theta }_1(where{n}_i=refractiveindexofair=1)\\ sin{\theta }_1=\frac{n_i}{n_1}sin{\theta }_a\\ =\frac{1}{n_1}\times \sin 0^0\\ =0^0\end{array}$

Thus, ray a emerges at an angle$0°$ in layer 1.

It will travel along the same line through all the layers. Hence its final angle of emergence in air will also be.$0°$

Hence, the ray a will pass un-deviated through all the layers.

So, for ray b and media pair of air and layer 1, the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}{n}_{i}sin{\theta }_b=n_{1}sin{\theta }_1(where{n}_i=refractiveindexofair=1)\\ sin{\theta }_1\\ =\frac{n_i}{n_1}sin{\theta }_b\\ =\frac{1}{n_1}\times \sin {20}^0\\ =\frac{1}{1.7}\text{sin}{20}^0\end{array}$

Thus, ray b emerges at an angle in layer 1.

ii) Now, the ray b is incident at an angleon the boundary between layer 1 and layer 2.

So the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_2=\frac{n_1}{n_2}sin{\theta }_1\\ =\frac{1.7}{1.6}\times \frac{1}{1.7}\times \sin {20}^0(where{n}_1=1.7and{n}_2=1.6)\\ =\frac{1}{1.6}\text{sin}{20}^0\end{array}$

iii) Now the ray b is incident at an angle${\theta }_2$on the boundary between layer 2 and layer 3.

So the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_3=\frac{n_2}{n_3}sin{\theta }_2\\ =\frac{1.6}{1.5}\times \frac{1}{1.6}\times \sin {20}^0(where{n}_2=1.6and{n}_3=1.5)\\ =\frac{1}{1.5}\text{sin}{20}^0\end{array}$

iv) Now, the ray b is incident at an angleon the boundary between layer 3 and layer 4. So the angle of emergence can be given using equation (i) as follows:

Thus, ray b emerges at an angle${\theta }_4$in layer 4.

v) Now, the ray b is incident at an angleon the boundary between layer 4 and layer 5.

So the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_5=\frac{n_4}{n_5}sin{\theta }_4\\ =\frac{1.4}{1.6}\times \frac{1}{1.4}\times \sin {20}^0(where{n}_4=1.4and{n}_5=1.6)\\ =\frac{1}{1.6}\text{sin}{20}^0\end{array}$

Thus, ray b emerges at an angle${\theta }_5$in layer 5.

vi) Now, the ray b is incident at an anglerole="math" localid="1663048809412" ${\theta }_5$on the boundary between layer 5 and air.

So the final angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_f=\frac{n_5}{n_f}sin{\theta }_5\\ =\frac{1.6}{1}\times \frac{1}{1.6}\times \sin {20}^0(where{n}_5=1.6and{n}_f=1)\\ =\frac{1}{1}\text{sin}{20}^0\\ sin{\theta }_f=\sin {20}^0\\ {\theta }_f=20°\end{array}$

The ray a is incident along the normal to the boundary of the media. Thus, it makes an angle.${\theta }_a=0°$

So, for ray a and media pair of glass and layer 1, the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}{n}_{i}sin{\theta }_a=n_{1}sin{\theta }_1\\ sin{\theta }_1=\frac{n_i}{n_1}sin{\theta }_a\\ =\frac{1}{n_1}\times \sin 0^0(where{n}_i=refractiveindexofglass=1.5)\\ =0^0\end{array}$

Thus, ray a emerges at an angle in layer 1.

It will travel along the same line through all the layers. Hence its final angle of emergence in air will also be. Thus, the ray a will pass un-deviated through all the layers.$0°$

i) The ray b is incident at an${\theta }_b=20°$angleon the boundary between glass and layer 1

So, for ray b and media pair of glass and layer 1, the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_1=\frac{n_i}{n_1}sin{\theta }_b\\ =\frac{1.5}{n_1}\times \sin {20}^0(where{n}_i=refractiveindexofglass=1.5)\\ =\frac{1.5}{1.7}\text{sin}{20}^0\end{array}$

Thus, ray b emerges at an angle${\theta }_1$in layer 1.

ii) The ray b is incident at an angle${\theta }_1$on the boundary between layer 1 and layer 2.

So the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_2=\frac{n_1}{n_2}sin{\theta }_1\\ =\frac{1.7}{1.6}\times \frac{1.5}{1.7}\times \sin {20}^0(where{n}_1=1.7and{n}_2=1.6)\\ =\frac{1.5}{1.6}\text{sin}{20}^0\end{array}$

Thus, ray b emerges at an angle${\theta }_2$in layer 2.

iii) The ray b is incident at an angle${\theta }_2$on the boundary between layer 2 and layer 3.

So the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_3=\frac{n_2}{n_3}sin{\theta }_2\\ =\frac{1.6}{1.5}\times \frac{1.5}{1.6}\times \sin {20}^0(where{n}_2=1.6and{n}_3=1.5)\\ =\text{sin}{20}^0\end{array}$

Thus, ray b emerges at an angle${\theta }_3$in layer 3.

iv) The ray b is incident at an angle${\theta }_3$on the boundary between layer 3 and layer 4.

So the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_4=\frac{n_3}{n_4}sin{\theta }_3\\ =\frac{1.5}{1.4}\times \sin {20}^0(where{n}_3=1.5and{n}_4=1.4)\\ =\frac{1.5}{1.4}\text{sin}{20}^0\end{array}$

Thus, ray b emerges at an angle${\theta }_4$in layer 4.

v) The ray b is incident at an angle${\theta }_4$on the boundary between layer 4 and layer 5.

So the angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_5=\frac{n_4}{n_5}sin{\theta }_4\\ =\frac{1.4}{1.6}\times \frac{1.5}{1.4}\times \sin {20}^0(where{n}_4=1.4and{n}_5=1.6)\\ =\frac{1.5}{1.6}\text{sin}{20}^0\end{array}\mathrm{F}$

Thus, ray b emerges at an anglein layer 5.

vi) The ray b is incident at an angleon the boundary between layer 5 and glass.

So the final angle of emergence can be given using equation (i) as follows:

$\begin{array}{c}sin{\theta }_f=\frac{n_5}{n_f}sin{\theta }_5\\ =\frac{1.6}{1.5}\times \frac{1.5}{1.6}\times \sin {20}^0(where{n}_5=1.6and{n}_f=1.5)\\ =\frac{1}{1}\text{sin}{20}^0\\ sin{\theta }_f=\sin {20}^0\\ {\theta }_f=20°\end{array}$

Hence, ray b emerges at an angle ${\theta }_f=20°$in glass.