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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In Fig. 35-34, a light ray is an incident at angle ${{\mathbit{\theta }}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{50}}{\mathbf{°}}$on a series of five transparent layers with parallel boundaries. For layers 1 and 3 , ${{\mathbit{L}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{20}}{\mathbf{}}{\mathbit{\mu }}{\mathbit{m}}$ , ${{\mathbit{L}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{25}}{\mathbf{}}{\mathbit{\mu }}{\mathbit{m}}$ , ${{\mathbit{n}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{6}}$ and ${{\mathbit{n}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{45}}$ . (a) At what angle does the light emerge back into air at the right? (b) How much time does the light take to travel through layer 3?

(a) The angle of immerge of the light into air is ${50}^{o}$.

(b) The time taken by the light to travel through medium 3 is $0.142\text{ps}$.

See the step by step solution

## Step 1: Write the given data from the question.

The incident angle, ${\theta }_{1}=50°$

The length of the medium 1, ${L}_{1}=20\mu \text{m}$

The length of the medium 3, role="math" localid="1663069065664" ${L}_{3}=25\mu \text{m}$

The refractive index of medium 1, ${n}_{1}=1.6$

The refractive index of medium 3, ${n}_{3}=1.45$

## Step 2: Determine the required formulas:

Snell's law of refraction states that: The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a pair of given environments.

The expression for Snell’s law is given as follows.

$n\mathrm{sin}{\theta }_{1}={n}_{1}\mathrm{sin}{\theta }_{2}={n}_{2}\mathrm{sin}{\theta }_{3}.........$

The expression to calculate the time is given as follows.

$t=\frac{{n}_{3}L}{c}$ .......(1)

## Step 3: (a) Calculate the angle at which light emerges back into the air:

Let assume angle ${\theta }_{2},{\theta }_{3},{\theta }_{4},{\theta }_{5}$ and ${\theta }_{6}$ are the refracted angle in first, second, third, fourth, fifth, and air respectively.

Use Snell’s law for the air and first medium interface.

${n}_{air}\mathrm{sin}{\theta }_{1}={n}_{1}\mathrm{sin}{\theta }_{2}$

Use the Snell’s law for the first and second medium.

${n}_{1}\mathrm{sin}{\theta }_{2}={n}_{2}\mathrm{sin}{\theta }_{3}$

Substitute ${n}_{air}\mathrm{sin}{\theta }_{1}$ for ${n}_{2}\mathrm{sin}{\theta }_{3}$ into above equation.

${n}_{air}\mathrm{sin}{\theta }_{1}={n}_{3}\mathrm{sin}{\theta }_{4}$

Use the Snell’s law for the second and third medium.

${n}_{2}\mathrm{sin}{\theta }_{3}={n}_{3}\mathrm{sin}{\theta }_{4}$

Use the Snell’s law for the third and fourth medium

${n}_{3}\mathrm{sin}{\theta }_{4}={n}_{4}\mathrm{sin}{\theta }_{5}$

Substitute ${n}_{air}\mathrm{sin}{\theta }_{1}$ for ${n}_{3}\mathrm{sin}{\theta }_{4}$ into above equation.

${n}_{air}\mathrm{sin}{\theta }_{1}={n}_{4}\mathrm{sin}{\theta }_{5}$

Use the Snell’s law for the fourth and fifth medium.

${n}_{4}\mathrm{sin}{\theta }_{5}={n}_{5}\mathrm{sin}{\theta }_{6}$

Substitute ${n}_{air}\mathrm{sin}{\theta }_{1}$ for ${n}_{4}\mathrm{sin}{\theta }_{5}$ into above equation.

${n}_{air}\mathrm{sin}{\theta }_{1}={n}_{5}\mathrm{sin}{\theta }_{6}$

Use the Snell’s law for the fifth and air medium.

${n}_{5}\mathrm{sin}{\theta }_{6}={n}_{air}\mathrm{sin}{\theta }_{7}$

Substitute ${n}_{air}\mathrm{sin}{\theta }_{1}$ for ${n}_{5}\mathrm{sin}{\theta }_{6}$ into above equation

$\begin{array}{rcl}{n}_{air}\mathrm{sin}{\theta }_{1}& =& {n}_{air}\mathrm{sin}{\theta }_{7}\\ \mathrm{sin}{\theta }_{1}& =& \mathrm{sin}{\theta }_{7}\\ {\theta }_{1}& =& {\theta }_{7}\end{array}$

Substitute $50°$ for ${\theta }_{1}$ into above equation.

${\theta }_{7}=50°$

Therefore, the angle of immerge of the light into air is 50 degree.

## Step 4: (b) Calculate the time taken by the light to travel through layer  :

Consider the distance travelled by the light in third medium.

Recall the equation (2),

$\begin{array}{rcl}{n}_{air}\mathrm{sin}{\theta }_{1}& =& {n}_{3}\mathrm{sin}{\theta }_{4}\\ \mathrm{sin}{\theta }_{4}& =& \frac{{n}_{air}}{{n}_{3}}\mathrm{sin}{\theta }_{1}\end{array}$

Substitute 1 for ${n}_{air}$ , $1.45$ for ${\theta }_{3}$ and $50°$ for ${\theta }_{1}$ into above equation.

$\begin{array}{rcl}{\theta }_{4}& =& \frac{1}{1.45}\mathrm{sin}50°\\ \mathrm{sin}{\theta }_{4}& =& 0.528\\ {\theta }_{4}& =& {\mathrm{sin}}^{-1}\left(0.528\right)\\ {\theta }_{4}& =& 31.89°\end{array}$

From the figure,

$\begin{array}{rcl}{\theta }_{4}& =& \frac{{L}_{3}}{L}\\ L& =& \frac{{L}_{3}}{\mathrm{cos}{\theta }_{4}}\end{array}$

Substitute$25\mu \text{m}$ for ${L}_{3}$ and $31.89°$ for ${\theta }_{4}$ into above equation.

$L=\frac{25×{10}^{-6}}{\mathrm{cos}31.89°}\phantom{\rule{0ex}{0ex}}L=2.944×{10}^{-5}\text{m}$

Calculate the time taken by the light to travel through medium 3.

Substitute $2.944×{10}^{-5}\text{m}$ for $L$ , $1.45$ for ${n}_{3}$ and $3×{10}^{8}\text{m}}{\text{s}}$ for c into equation (1).

$\begin{array}{rcl}t& =& \frac{1.45×2.944×{10}^{-5}}{3×{10}^{8}}\\ & =& \frac{4.2688×{10}^{-13}}{3}\\ & =& 1.42×{10}^{-13}\\ & =& 0.142\text{ps}\end{array}$

Hence the time taken by the light to travel through medium 3 is $0.142\text{ps}$ .

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