Problems

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Two light pulses are emitted simultaneously and hit a screen directly in front of them. If one light pulse passes through 6.2 m of ice on its way to the screen, what is the time difference between the arrival of the two pulses at the screen? (The index of refraction of ice is 1.309.)

Solution:
Suppose the distance to the screen is D . The time it takes for the first pulse to arrive is:

t₁ = $${D\over c}$$

The time of flight for the second pulse is:

t₂ = $${D-6.2m\over c}$$ + $${6.2m\over v_{{\:\rm ice}}}$$

where D - 6.2m is the distance travelled in air and v _ice = c/n = c/1.309 is the speed of light in the ice. The time difference is therefore:

$$\Delta$$ = t₂ - t₁

$${D\over c} {D-6.2\over c} {6.2 \times n \over c}$$ =

$${6.2\;{\:\rm m}\over 3\times 10^8\;{\:\rm m/s}} {6.2\;{\:\rm m}\times 1.309 \over 3\times10^8 \;{\:\rm m/s}}$$ =

= 2.705 x 10^{- 8}  s - 12.067 x 10^{- 8}  s

= 6.38 x 10^{- 9}  sec (4)

Note: The time difference is just the difference in time of flight for light in the ice compared to an equivalent distance in air.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A light ray of wavelength $\lambda$ = 589 nm is incident on glass with an angle of incidence of 30 ^o . The index of refraction of glass is 1.52. a)What is the angle of refraction? b) What are the speed and wavelength of the light inside the glass?

Solution:

  

Figure 22.11: Law of reflection

$$\theta_{2}^{} {n_1\over n_2} \theta_{1}^{}$$ =

$${1.00\over 1.52}$$ =

$$\theta_{2}^{}$$ = 19.2 ^o (5)

$$\lambda_{2}^{} {n_1\over n_2} \lambda_{1}^{} {1.00\over 1.52}$$ =

= 387 x 10^{- 9} m = 387 nm

$${c\over n_2} {3\times10^8{\:\rm m/s}\over 1.52}$$ v₂ = (6)

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Light passes through a flat slab of glass. The angle of incidence of the light onto the glass is 30 ^o . What is the angle with which the light emerges on the other side of the slab?

Solution:

  

Figure 22.12: Law of reflection

First calculate the angle of refraction inside the glass:

sin $$\theta_{2}^{}$$ = $${n_1\over n_2}$$sin $$\theta_{1}^{}$$

Then use $\theta_{2}^{}$ as the angle of incidence onto the boundary between the glass and the air on the other side, so the angle of refraction in the air is:

$$\theta_{3}^{} {n_2\over n_3} \theta_{2}^{}$$ =

$${n_2\over n_3} {n_2\over n_2} \theta_{1}^{}$$ =

$${n_1\over n_3} \theta_{1}^{}$$ = (7)

Since n₁ = n₃ because there is air on both sides of the slab, $\theta_{3}^{}$ = $\theta_{1}^{}$ , and the ray emerges parallel to the incoming ray.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A fish lives at the bottom of a lake 10m deep filled with water whose index of refraction 1.52. a) At what angle relative to the normal must the fish look up towards the surface of the water in order to see a fisherman who is sitting on a distant shore? b)What is the closest that another fish living at the bottom of the lake can approach in order that the first fish can see it by looking towards the surface?

Solution:

  

Figure 22.13:

a)

As indicated in the Fig(22.13), the incoming light from the fisherman must be coming almost parallel to the surface of the lake. Thus, the ray that hits the fish's eye must be coming at close to the critical angle for water: i.e. the angle of incidence is approximately 90 ^o :

$$\theta_{2}^{} {n_1\over n_2} {1\over 1.52}$$ =

$$\theta_{2}^{}$$ = 41 ^o (8)

b)

The closest approach will be when the first fish is looking up at the critical angle. Any closer and there will not be internal reflection. Thus:

$${10\;{\:\rm m}\over d/2}$$ = sin $$\theta_{c}^{}$$ = $${n_1\over n_2}$$ = $${1\over 1.52}$$

Thus

d = $${20 \;{\:\rm m}\over \sin\theta_c}$$ = 20  m x 1.52 x = 30  m

Note: that both the fisherman and the other fish have approximately the same apparent position to the first fish. Fish therefore have a pretty strange overhead view: if they look almost directly up, they see birds and clouds. If they look near the critical angle they see fisherman and trees on the distant shore. If they look at even steeper angles relative to the normal, they see other fish swimming ``overhead". I would like to sit in on an aquatic astronomer's conference to see how they explain it.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

White light approaches one side of a glass prism at an angle of incidence of 40 ^o as shown in the diagram below. The apex angle of the prism is 60 ^o .

  

Figure 22.14:

At what angles does red light emerge on the other side? At what angle does blue light emerge?

Solution:
First calculate the angle of refraction $\theta_{2}^{}$ inside the prism.

sin $$\theta_{2}^{}$$ = $${n_1\over n_2}$$sin $$\theta_{1}^{}$$

From Table 22.1, we find that red light has an index of refraction: n₂^R = 1.51 , while for blue light, n₂ = 1.53 . Thus:

$$\theta^{R}_{2} {1\over1.51}$$ =

$$\Rightarrow \theta^{R}_{2}$$ = 25.2 ^o

$$\theta^{B}_{2} {1\over1.53}$$

$$\Rightarrow \theta^{B}_{2}$$

Next we calculate the angle of incidence $\theta_{3}^{}$ on the far side using geometry.

  

Figure 22.15:

From the diagram above, we get the relationship:

$$\theta_{2}^{} \theta_{3}^{}$$ (9)

which implies that

$$\theta_{3}^{} \theta_{2}^{}$$ (10)

Thus

$$\theta^{R}_{3}$$ = 60 ^o - 25.2 ^o = 34.8 ^o

and

$$\theta^{B}_{3}$$ = 60 ^o - 24.8 ^o = 35.2 ^o

Finally we calculate the angle of refraction sin $\theta_{4}^{}$ of the emerging rays:

$$\theta_{4}^{} {n_2\over n_1} \theta_{3}^{}$$ (11)

Which gives:

$$\theta^{R}_{4} \sin^{-1}_{} \cdot$$ =

$$\theta^{B}_{4} \sin^{-1}_{} \cdot$$ =

The blue light therefore is bent (61.8 - 59.5) ^o = 2.3 ^o more than the red light.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Monochromatic light goes through two thin slits 0.03mm apart. It is found that the second bright fringe on a screen 1.2 m away is 4.5 cm from the center. What is the wavelength of the light?

Solution:
Using:

y₂ = n$$\lambda$$$${L\over d}$$

where n = 2 , we find:

$$\lambda$$ = $${d y_2\over 2 \lambda}$$ = $${0.03\times 10^{-3}m \times 4.5\times 10^{-2}m\over 2\times 1.2 m}$$ = 5.6 x 10^{- 7}m = 560 nm

Note: This problem shows that is possible to measure very short wavelengths using the double slit experiment quite accurately as long as L/d is very large.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A radio transmitter transmits signals to a receiver 500m away along two paths. One is directly from the transmitter to the receiver, the other is bounced off a mountain a distance x directly behind the receiver. (See Fig. (22.16))

  

Figure 22.16:

If the wavelength is 300m , what is the shortest value for x such that there is constructive interference at the receiver between rays traveling along the two different paths?

Solution:
Since the rays bounce off a more dense mountain, there is a phase change on reflection, so the condition for destructive interference is:

r₁ - r₂ = 2x = (m + $${1\over 2}$$)$$\lambda$$

The smallest value for x occurs when m = 0 , in which case:

x = $${\lambda \over 4}$$ = 75 m