Schaum’s Outline of College Physics

Eugene Hecht

Chapter 37

Refraction of Light - all with Video Answers

Educators

Chapter Questions

Problem 1

The speed of light in water is $(3 / 4) \mathrm{c}$. What is the effect, on the frequency and wavelength of light, of passing from vacuum (or air, to good approximation) into water? Compute the refractive index of water.

The same number of wave peaks leave the air each second as enter into the water. Hence, the frequency is the same in the two materials. But because Wavelength $=($ Speed $) /($ Frequency), the wavelength in water is three-fourths that in air.
The (absolute) refractive index of water is
$$
n=\frac{\text { Speed in vacuum }}{\text { Speed in water }}=\frac{\mathrm{c}}{(3 / 4) \mathrm{c}}=\frac{4}{3}=1.33
$$

Suzanne W.

Suzanne W.

Numerade Educator

Problem 2

A glass plate is $0.60 \mathrm{~cm}$ thick and has a refractive index of $1.55 .$ How long does it take for a pulse of light incident normally to pass through the plate?
$$
t=\frac{x}{v}=\frac{0.0060 \mathrm{~m}}{\left(2.998 \times 10^{8} / 1.55\right) \mathrm{m} / \mathrm{s}}=3.1 \times 10^{-11} \mathrm{~s}
$$

Suzanne W.

Numerade Educator

Problem 3

As is drawn in Fig. $37-4$, a ray of light in air strikes a glass plate $(n=1.50)$ at an incidence angle of $50^{\circ}$. Determine the angles of the reflected and transmitted rays.
The law of reflection applies to the reflected ray. Therefore, the angle of reflection is $50^{\circ}$, as shown.
For the refracted ray, $n_{i} \sin \theta_{i}=n_{t} \sin \theta_{t}$ becomes
$$
\sin \theta_{t}=\frac{n_{i}}{n_{t}} \sin \theta_{i}=\frac{1.0}{1.5} \sin 50^{\circ}=0.51
$$
from which it follows that $\theta_{t}=31^{\circ}$.

Suzanne W.

Numerade Educator

Problem 4

The refractive index of diamond is $2.42$. What is the critical angle for light passing from diamond to air?
We use $n_{i} \sin \theta_{i}=n_{t} \sin \theta_{t}$ to obtain
$$
(2.42) \sin \theta_{c}=(1) \sin 90.0^{\circ}
$$
from which it follows that $\sin \theta_{c}=0.413$ and $\theta_{c}=24.4^{\circ}$.

Suzanne W.

Numerade Educator

Problem 5

What is the critical angle for light passing from glass $(n=1.54)$ to water $(n=1.33)$ ?
$$
n_{i} \sin \theta_{i}=n_{t} \sin \theta_{t} \quad \text { becomes } \quad n_{i} \sin \theta_{c}=n_{t} \sin 90^{\circ}
$$
from which we get $\quad \sin \theta_{c}=\frac{n_{t}}{n_{i}}=\frac{1.33}{1.54}=0.864 \quad$ or $\quad \theta_{c}=59.7^{\circ}$

Suzanne W.

Numerade Educator

Problem 6

A layer of oil $(n=1.45)$ floats on water $(n=1.33)$. A ray of light shines onto the oil with an incidence angle of $40.0^{\circ}$. Find the angle the ray makes in the water. (See Fig. $37-5 .$ )
At the air-oil interface, Snell's Law gives
$$
n_{\text {air }} \sin 40^{\circ}=n_{\text {oil }} \sin \theta_{\text {oil }}
$$
At the oil-water interface, we have (using the equality of alternate angles)
$$
n_{\text {oil }} \sin \theta_{\text {oil }}=n_{\text {water }} \sin \theta_{\text {water }}
$$
Thus, $n_{\text {air }} \sin 40.0^{\circ}=n_{\text {water }} \sin \theta_{\text {water }}$; the overall refraction occurs just as though the oil layer were absent. Solving gives
$$
\sin \theta_{\text {water }}=\frac{n_{\text {air }} \sin 40.0^{\circ}}{n_{\text {water }}}=\frac{(1)(0.643)}{1.33} \quad \text { or } \quad \theta_{\text {water }}=28.9^{\circ}
$$

Khoobchandra Agrawal

Khoobchandra Agrawal

Numerade Educator

Problem 7

As shown in Fig. $37-6$, a small luminous body, at the bottom of a pool of water $(n=4 / 3) 2.00 \mathrm{~m}$ deep, emits rays upward in all directions. A circular area of light is formed at the surface of the water. Determine the radius $R$ of the circle of light.
The circular area is formed by rays refracted into the air. The angle $\theta_{c}$ must be the critical angle, because total internal reflection, and hence no refraction, occurs when the angle of incidence in the water is greater than the critical angle. We have, then,
$$
\sin \theta_{c}=\frac{n_{a}}{n_{w}}=\frac{1}{4 / 3} \quad \text { or } \quad \theta_{c}=48.6^{\circ}
$$
From the figure,
$$
R=(2.00 \mathrm{~m}) \tan \theta_{c}=(2.00 \mathrm{~m})(1.13)=2.26 \mathrm{~m}
$$

Khoobchandra Agrawal

Khoobchandra Agrawal

Numerade Educator

Problem 8

What is the minimum value of the refractive index for a $45.0^{\circ}$ prism which is used to turn a beam of light by total internal reflection through a right angle? (See Fig. 37-7.)
The ray enters the prism without deviation, since it strikes side $A B$ normally. It then makes an incidence angle of $45.0^{\circ}$ with normal to side $A C$. The critical angle of the prism must be smaller than $45.0^{\circ}$ if the ray is to be totally reflected at side $A C$ and thus turned through $90^{\circ} .$ From $n_{i} \sin \theta_{c}=n_{t} \sin 90^{\circ}$ with $n_{t}=1.00$,
$$
\text { Minimum } n_{i}=\frac{1}{\sin 45.0^{\circ}}=1.41
$$

Suzanne W.

Numerade Educator

Problem 9

The glass prism shown in Fig. $37-8$ has an index of refraction of $1.55 .$ Find the angle of deviation $D$ for the case shown.
No deflection occurs at the entering surface, because the incidence angle is zero. At the second surface, $\theta_{i}=30^{\circ}$ (because its sides are mutually perpendicular to the sides of the apex angle). Then, Snell's Law becomes
$$
n_{i} \sin \theta_{i}=n_{t} \sin \theta_{t} \quad \text { or } \quad \sin \theta_{t}=\frac{1.55}{1} \sin 30^{\circ}
$$
from which $\theta_{t}=50.8^{\circ} .$ But $D=\theta_{t}-\theta_{i}$ and so $D=21^{\circ}$.

Suzanne W.

Numerade Educator

Problem 10

As shown in Fig. $37-9$, an object is at a depth $d$ beneath the surface of a transparent material of refractive index $n$. As viewed from a point almost directly above, how deep does the object appear to be?
The two rays from $A$ that are shown emerging into the air both appear to come from point- $B$. Therefore, the apparent depth is $C B$. We have
$$
\frac{b}{\overline{C B}}=\tan \theta_{t} \quad \text { and } \quad \frac{b}{\overline{C A}}=\tan \theta_{i}
$$
If the object is viewed from nearly straight above, then angles $\theta_{t}$ and $\theta_{i}$ will be very small. For small angles, the sine and tangent are nearly equal. Therefore,
$$
\frac{\overline{C B}}{\overline{C A}}=\frac{\tan \theta_{i}}{\tan \theta_{t}} \approx \frac{\sin \theta_{i}}{\sin \theta_{t}}
$$
But $n \sin \theta_{t}=(1) \sin \theta_{t}$ from which
$$
\frac{\sin \theta_{i}}{\sin \theta_{t}}=\frac{1}{n}
$$
Hence,
$$
\text { Apparent depth } \overline{C B}=\frac{\text { actual depth } \overline{C A}}{n}
$$
The apparent depth is only a fraction $1 / n$ of the actual depth $d$.

Suzanne W.

Numerade Educator

Problem 11

A glass plate $4.00 \mathrm{~mm}$ thick is viewed from above through a microscope. The microscope must be lowered $2.58 \mathrm{~mm}$ as the operator shifts from viewing the top surface to viewing the bottom surface through the glass. What is the index of refraction of the glass? Use the results of Problem $37.10$.
We found in Problem $37.10$ that the apparent depth of the plate will be $1 / n$ as large as its actual depth. Hence,
or $\begin{aligned} \text { (Actual thickness) }(1 / n) &=\text { Apparent thickness } \\(4.00 \mathrm{~mm})(1 / n) &=2.58 \mathrm{~mm} \end{aligned}$
This yields $n=1.55$ for the glass.

Vishal Gupta

Numerade Educator

Problem 12

As shown in Fig. $37-10$, a ray enters the flat end of a long rectangular block of glass that has a refractive index of $n_{2}$. Show that all entering rays can be totally internally reflected only if $n_{2}>1.414 .$
The larger $\theta_{1}$ is, the larger $\theta_{2}$ will be, and the smaller $\theta_{3}$ will be. Therefore, the ray is most likely to leak out through the side of the block if $\theta_{1}=90^{\circ} .$ In that case,
$$
n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2} \quad \text { becomes } \quad(1)(1)=n_{2} \sin \theta_{2}
$$
For the ray to just escape, $\theta_{4}=90^{\circ} .$ Then
$$
n_{2} \sin \theta_{3}=n_{1} \sin \theta_{4} \quad \text { becomes } \quad n_{2} \sin \theta_{3}=(1)(1)
$$
We thus have two conditions to satisfy: $n_{2} \sin \theta_{2}=1$ and $n_{2} \sin \theta_{3}=1$. Their ratio gives
$$
\frac{\sin \theta_{2}}{\sin \theta_{3}}=1
$$
But we see from the figure that $\sin \theta_{3}=\cos \theta_{2}$, and so this becomes
$$
\tan \theta_{2}=1 \quad \text { or } \quad \theta_{2}=45.00^{\circ}
$$
Then, because $n_{2} \sin \theta_{2}=1$, we have
$$
n_{2}=\frac{1}{\sin 45.00^{\circ}}=1.414
$$
This is the smallest possible value the index can have for total internal reflection of all rays that enter the end of the block. It is possible to obtain this answer by inspection. How?

Khoobchandra Agrawal

Khoobchandra Agrawal

Numerade Educator

Problem 13

The speed of light in a certain glass is $1.91 \times 10^{8} \mathrm{~m} / \mathrm{s}$. What is the refractive index of the glass?

Prabhu Ramji

Numerade Educator

Problem 14

What is the frequency of light which has a wavelength in air of $546 \mathrm{~nm}$ ? What is its frequency in water $(n=1.33) ?$ What is its speed in water? What is its wavelength in water?

Suzanne W.

Numerade Educator

Problem 15

A beam of light strikes the surface of water at an incidence angle of $60^{\circ}$. Determine the directions of the reflected and refracted rays. For water, $n=1.33$.

Suzanne W.

Numerade Educator

Problem 16

The critical angle for light passing from rock salt into air is $40.5^{\circ}$. Calculate the index of refraction of rock salt.

Suzanne W.

Numerade Educator

Problem 17

What is the critical angle when light passes from glass $(n=1.50)$ into air?

Suzanne W.

Numerade Educator

Problem 18

The absolute indices of refraction of diamond and crown glass are $5 / 2$ and $3 / 2$, respectively. Compute
(a) the refractive index of diamond relative to crown glass and $(b)$ the critical angle between diamond and crown glass.

Suzanne W.

Numerade Educator

Problem 19

A pool of water $(n=4 / 3)$ is $60 \mathrm{~cm}$ deep. Find its apparent depth when viewed vertically through air.

Suzanne W.

Numerade Educator

Problem 20

In a vessel, a layer of benzene $(n=1.50) 6 \mathrm{~cm}$ deep floats on water $(n=1.33) 4 \mathrm{~cm}$ deep. Determine the apparent distance of the bottom of the vessel below the upper surface of the benzene when viewed vertically through air.

Suzanne W.

Numerade Educator

Problem 21

A mirror is made of plate glass $(n=3 / 2) 1.0 \mathrm{~cm}$ thick and silvered on the back. A man is $50.0 \mathrm{~cm}$ from the front face of the mirror. If he looks perpendicularly into it, at what distance behind the front face of the mirror will his image appear to be?

Suzanne W.

Numerade Educator

Problem 22

A straight rod is partially immersed in water $(n=1.33)$. Its submerged portion appears to be inclined $45^{\circ}$ with the surface when viewed vertically through air. What is the actual inclination of the rod?

Suzanne W.

Numerade Educator

Problem 23

The index of refraction for a certain type of glass is $1.640$ for blue light and $1.605$ for red light. When a beam of white light (one that contains all colors) enters a plate of this glass at an incidence angle of $40^{\circ}$, what is the angle in the glass between the blue and red parts of the refracted beam?

Suzanne W.

Numerade Educator