Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 35-34, a light ray is an incident at angle $θ_{1} = 50 °$ on a series of five transparent layers with parallel boundaries. For layers 1 and 3 , $L_{1} = 20 μ m$ , $L_{2} = 25 μ m$ , $n_{1} = 1.6$ and $n_{3} = 1.45$ . (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 $0.142 ps$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the given data from the question.

The incident angle, $θ_{1} = 50 °$

The length of the medium 1, $L_{1} = 20 μ m$

The length of the medium 3, role="math" localid="1663069065664" $L_{3} = 25 μ m$

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.

$n \sin θ_{1} = n_{1} \sin θ_{2} = n_{2} \sin θ_{3} . . . . . . . . .$

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 $θ_{2}, θ_{3}, θ_{4}, θ_{5}$ and $θ_{6}$ are the refracted angle in first, second, third, fourth, fifth, and air respectively.

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

$n_{a i r} \sin θ_{1} = n_{1} \sin θ_{2}$

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

$n_{1} \sin θ_{2} = n_{2} \sin θ_{3}$

Substitute $n_{a i r} \sin θ_{1}$ for $n_{2} \sin θ_{3}$ into above equation.

$n_{a i r} \sin θ_{1} = n_{3} \sin θ_{4}$

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

$n_{2} \sin θ_{3} = n_{3} \sin θ_{4}$

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

$n_{3} \sin θ_{4} = n_{4} \sin θ_{5}$

Substitute $n_{a i r} \sin θ_{1}$ for $n_{3} \sin θ_{4}$ into above equation.

$n_{a i r} \sin θ_{1} = n_{4} \sin θ_{5}$

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

$n_{4} \sin θ_{5} = n_{5} \sin θ_{6}$

Substitute $n_{a i r} \sin θ_{1}$ for $n_{4} \sin θ_{5}$ into above equation.

$n_{a i r} \sin θ_{1} = n_{5} \sin θ_{6}$

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

$n_{5} \sin θ_{6} = n_{a i r} \sin θ_{7}$

Substitute $n_{a i r} \sin θ_{1}$ for $n_{5} \sin θ_{6}$ into above equation

$\begin{array}{rcl} n_{a i r} \sin θ_{1} & = & n_{a i r} \sin θ_{7} \\ \sin θ_{1} & = & \sin θ_{7} \\ θ_{1} & = & θ_{7} \end{array}$

Substitute $50 °$ for $θ_{1}$ into above equation.

$θ_{7} = 50 °$

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),

$\begin{array}{rcl} n_{a i r} \sin θ_{1} & = & n_{3} \sin θ_{4} \\ \sin θ_{4} & = & \frac{n_{a i r}}{n_{3}} \sin θ_{1} \end{array}$

Substitute 1 for $n_{a i r}$ , $1.45$ for $θ_{3}$ and $50 °$ for $θ_{1}$ into above equation.

$\begin{array}{rcl} θ_{4} & = & \frac{1}{1.45} \sin 50 ° \\ \sin θ_{4} & = & 0.528 \\ θ_{4} & = & \sin^{- 1} (0.528) \\ θ_{4} & = & 31.89 ° \end{array}$

From the figure,

$\begin{array}{rcl} θ_{4} & = & \frac{L_{3}}{L} \\ L & = & \frac{L_{3}}{\cos θ_{4}} \end{array}$

Substitute $25 μ m$ for $L_{3}$ and $31.89 °$ for $θ_{4}$ into above equation.

$L = \frac{25 \times 10^{- 6}}{\cos 31.89 °} L = 2.944 \times 10^{- 5} m$

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

Substitute $2.944 \times 10^{- 5} m$ for $L$ , $1.45$ for $n_{3}$ and $3 \times 10^{8} \frac{m}{s}$ for c into equation (1).

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

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

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Short Answer

Step by Step Solution

Step 1: Write the given data from the question.

Step 2: Determine the required formulas:

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

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

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Fundamentals Of Physics

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Step by Step Solution

Step 1: Write the given data from the question.

Step 2: Determine the required formulas:

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

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

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