Problem 1 (5 Points)

In the first application of
interferometric methods in radio
astronomy, Australian astronomers
observed the interference between a
radio wave arriving at their antenna
directly from the Sun and on a path involving reflection from the surface of the sea.
Assume that radio waves have a frequency of 6.0 ¡Á 107 Hz, and that the radio receiver is
25 m above the surface of the sea. What is the smallest angle ¦È above the horizon that
will give destructive interference of the waves at the receiver?
Problem 2 (8 Points)
The figure to the right shows a Mach-
Zehnder interferometer. Invented over 100
years ago, it is still used for many optical
measurements.
Light form a point source in the lower-left
strikes a half-silvered mirror, which
reflects half the light incident on it and
transmits half the incident light. The two
split beams, U and D, strike good mirrors which directs the rays to another half silvered
mirror. This second half-silvered mirror splits the combined beams, directing half to
detector 1 and the other half to detector 2. The detectors are indicated as half-circles.
The device is constructed so that the light rays travel the same distance from where they
are split by the 1st ¡°beam splitter¡± until they are combined at the 2nd one.
What fraction of the light from the source enters detector 1?
Problem 3 (4 Points)
A converging lens has a focal length f, and a diverging
lens has a focal length ¨Cf, which has the same magnitude
as the converging lens. They are separated by a distance
D which is greater than f, as shown. Parallel light enters
from the left. Will the light be brought to a focus, and if
so where?
Problem 4 (3 Points)
Imagine that the parallel light rays in Problem 3 enter from the right. Explain why the
light will be brought to focus without using any equations.

I am not inclined to work these out for you.. Indicate your thinking in words, and I will critique. Reminder: on the first one, is there a phase shift upon reflection?

Jimmy goes to uoft!

Problem 1: To find the smallest angle above the horizon that will give destructive interference of the waves at the receiver, we can use the concept of path difference and destructive interference in waves.

The path difference between the direct wave and the reflected wave can be calculated by considering the extra distance traveled by the reflected wave. In this case, the extra distance is due to the height of the radio receiver above the surface of the sea (25 m).

Since the radio waves have a frequency of 6.0 × 10^7 Hz, we can use the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency. Rearranging the equation gives us λ = c/f.

Now, we can find the wavelength of the radio waves by substituting the given frequency into the equation. Let's assume the speed of light is 3 × 10^8 m/s:

λ = (3 × 10^8 m/s) / (6.0 × 10^7 Hz) = 5 meters

The extra distance traveled by the reflected wave is 2 × 25 meters because the wave has to go up to the receiver and then come back down to the surface. Therefore, the path difference is 2 × 25 = 50 meters.

For destructive interference to occur, the path difference should be equal to an integer multiple of the wavelength. Thus, we can set up the following equation:

50 meters = n × 5 meters

Solving for n, we get:

n = 50 meters / 5 meters = 10

Therefore, the smallest angle above the horizon that will give destructive interference of the waves at the receiver is given by the equation:

sin(θ) = opposite/hypotenuse = 25 meters / 25 meters = 1

Taking the inverse sine of 1, we find:

θ = arcsin(1) = 90°

So, the smallest angle above the horizon that will give destructive interference of the waves at the receiver is 90°.

Problem 2: To find the fraction of the light from the source that enters detector 1 in the Mach-Zehnder interferometer, we need to consider the paths taken by the two split beams.

Since the instrument is symmetric, the two split beams travel the same distance from where they are split by the first half-silvered mirror until they are combined at the second one.

The first half-silvered mirror reflects half the light and transmits half the incident light. Therefore, each split beam carries half the intensity (or power) of the original light.

When the two split beams reach the second half-silvered mirror, each beam is split again, with one-half of each beam directed towards detector 1 and the other half towards detector 2.

Since the second half-silvered mirror splits the combined beams in half, each beam has its intensity reduced by a factor of 2. Therefore, each split beam contributes 1/4th of the original intensity (or power) of the source.

Hence, 1/4th of the light from the source enters detector 1.

Problem 3: In the setup where a converging lens and a diverging lens are separated by a distance greater than the focal length of the converging lens, parallel light entering from the left will not be brought to a focus.

The converging lens has a positive focal length (f) and will converge the light. On the other hand, the diverging lens has a negative focal length (-f) and will diverge the light. When the two lenses are separated by a distance (D) greater than the focal length of the converging lens, the diverging lens will overpower the convergence of the converging lens.

As a result, the light rays will be diverging after passing through the diverging lens and will not come to a focused point.

Problem 4: When parallel light rays enter from the right in the setup where a converging lens and a diverging lens are separated, the light will be brought to a focus without using any equations because of the properties of the lenses.

When parallel light rays enter from the right, the diverging lens will cause the rays to diverge. However, the converging lens is designed to converge parallel rays. The diverging lens will spread the rays out, but the converging lens will bring them back together, ultimately focusing them to a point.

By using lenses with appropriate focal lengths and separation distances, the converging lens compensates for the divergence caused by the diverging lens. Hence, the light rays will converge and be brought to a focus point.