# College Physics : Electromagnetics, Waves, and Optics

## Example Questions

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### Example Question #1 : Fundamental Concepts

When a ray of light is reflected off of the surface of another medium, some of the energy from the light is transferred into this medium. As a result, which of the following is true about the reflected light ray?

It has a lower frequency and a higher wavelength

It has a lower frequency and a lower wavelength

It has a higher frequency and a lower wavelength

It has a higher frequency and a higher wavelength

It has a lower frequency and a higher wavelength

Explanation:

For this question, we're told that light is reflected off of the surface of another medium. As a result, the light loses some of its energy, which is transferred into the medium where it reflected.

Because the light has lost some of its energy, we need to determine how the wavelength and frequency of the wave will be affected. To do this, we can recall the equation for the energy of a wave.

Where  is Planck's constant and  represents the frequency. We can see that when energy is decreased, the frequency also decreases. Thus, we can eliminate two of the answer choices.

To see how the wavelength changes, it's important to recall the relationship that wavelength and frequency have with each other.

Where  is the speed of light and  is the wavelength. Since the light is still traveling in the same medium, its speed will not change. Thus, as the frequency of the wave decreases, the wavelength has to increase. We can alternatively show how all of these variables are related as follows.

So, in summary, when the energy of light decreases, the frequency will decrease and the wavelength will increase.

### Example Question #1 : Doppler Effect

Suppose that car A and car B are both traveling in the same direction. Car A is going  and is sounding a horn with a frequency of . If car B is traveling at a speed of  in front of car A, what frequency of sound does car B hear?

Note: The speed of sound in air is .

Explanation:

The important concept being tested in this question is the Doppler effect. When a source of sound waves (or any other wave) is emitted from a source and is moving relative to some observer, the actual frequency of the wave will be different for the observer.

For starters, we'll need to use the Doppler equation.

The trickiest part about this question is to decide whether to add or subtract in both the numerator and denominator. To find the right sign orientation, it's helpful to do a quick thought experiment. Imagine that only one of them is moving and the other is stationary. Decide how that will affect the observed frequency; will it increase it or decrease it? Then repeat for the other one.

First, let's consider the detector (aka observer). We're told that car B, the detector, is traveling ahead of car A and also in the same direction. This means that car B is driving away from car A. So in this situation, is the frequency that car B hears expected to increase or decrease? The answer is that it will decrease. Since car B is traveling away from car A, each successive wave will take longer to reach car B. Hence, we will use subtraction in the numerator because that will make the observed frequency smaller.

Now let's apply this same logic to the denominator, which deals with the source of the sound waves. We know that car A is traveling in the same direction as car B and is behind. This means that car A is traveling toward car B. So from this perspective, each successive wave is expected to get closer together, thus making the time between each wave smaller and the frequency bigger. In the denominator, will adding or subtracting make the observed frequency bigger? The answer is subtraction. By making a smaller number in the denominator, the entire fraction becomes larger.

Keeping this information in mind, we'll need to use subtraction in both the numerator and denominator. Once we plug in the values given in the question stem, we have everything we need to solve for the answer.

### Example Question #1 : Electromagnetics, Waves, And Optics

On planet Borg, a male insect is flying toward his mate at  while buzzing at . The female insect is stationary and is buzzing at . What is the speed of the sound on planet Borg?

Explanation:

Rearrange the equation by solving for  to get:

Finally we plug in our known values and solve for :

### Example Question #1 : Harmonics And Standing Waves

Suppose that the fifth harmonic of a standing wave contained within a pipe closed at both ends has a wavelength of . What is the length of the pipe?

Explanation:

For this question, we're told that a standing wave is contained within a pipe closed at both ends. We're also given the wavelength for the fifth harmonic, and are asked to find the length of the pipe.

The first step to solve this problem is to use the equation for a pipe closed at both ends.

Since we're told which harmonic the wave is on, as well as its wavelength, we have everything we need to solve for the length of the pipe.

### Example Question #1 : Waves

A  string under tension is oscillated at the 5th harmonic. What is the wavelength of this oscillation?

Explanation:

Each harmonic has a wavelength that behaves according to the equation

, where  is the wavelength and  is the harmonic.

Since this is the 5th harmonic,  and . Solving for wavelength gives .

### Example Question #1 : Electromagnetics, Waves, And Optics

A pipe has a length of . If the pipe is open at both ends, what is the frequency of the third harmonic assuming that the velocity in which sound moves through the air is .

Explanation:

Since the pipe is open at both ends, we can use the equation where f is the frequency we are solving for,  is the number of the harmonic,  is the velocity at which sound moves through air, and L is the length of the pipe.

We know from the question that:

When we plug in these values into the frequency equation, we get  as the answer.

### Example Question #1 : Lenses

For most people, the nearest distance that objects can be located away from the eye and still seen clearly is . This is referred to as the near point; if the object comes any closer, the object cannot be seen clearly. Suppose that a person who needs glasses cannot see objects clearly if they are closer than  from the eye; that is to say, their near point is . A lens with what refractive power is needed in order to correct this person's vision to bring their near point to ?

Explanation:

For this question, we're given the definition of near point. We're told what the near point is in the average person, and also the near point for a certain person who can't see well and needs glasses. We're asked to find the refractive power of a lens that will bring this individual's near point to the average, healthy value.

As was stated in the question stem, the near point is the closest distance of an object from the eye where that object can still be seen clearly. In the question, we're told that the normal value for this is . Moreover, a person with a near point of  means that the object will need to be twice as far away, and no closer, to be seen clearly. Thus, in order to correct for this, a lens will be needed.

The idea is to be able to make the individual see things clearly when objects are located  away. To accomplish this, a lens will need to diffract the light coming from an object  away. This diffracted light will then need to form an image  away, which is where this particular individual's near point is.

With this information in hand, we can use the lens-makers equation to solve for the refractive power of the lens.

We know that the object will be located a distance of  away. Moreover, the image will need to form at a distance of  away. However, since the image is forming on the same side of the lens that the object is located, the image will be virtual. Thus, the value used in the equation will be .

Also, remember that to find refractive power, we'll need to have our units be in meters.

Furthermore, recall that refractive power is equal to the inverse of focal length.

Hence, the power of the refractive lens will need to be  diopters to correct this person's near point.

### Example Question #1 : Lenses

A convex lens connected to a projector projects an image onto a board of an object that is  away from the lens. If the object is  tall and the inverted image is  tall, what is the focal length of the projector lens?

Explanation:

To solve this one will require two equations. First, to find focal length, you will need the lens equation.

But we are only given the object distance, not the image distance. So to find that we need to know the relationship between magnification and object and image distances.

Plug in the quantities we know.

(remember it was inverted)

. Then solve for  which is . So the projector would sit that far from the board.

Now finally we can find the focal length of this lens.

Plugging in:

Solving for  gives a focal length of  or

### Example Question #1 : Lenses

An object located  away from a convex lens whose focal point is  produces an image located at what distance away from the lens?

Explanation:

For this question, we'll need to use the thin lens equation in order to determine the image distance from the lens.

Because this is a convex lens, the refracted light will form an image on the side of the lens where light is expected to go (real image). Thus, we know that the focal length should be a positive value. Moreover, the object distance is also positive.

Rearranging the above equation, we can isolate the term for image distance.

Then we can plug in the values given to us in the question stem.

And finally we take the inverse of this value to arrive at the correct answer.

### Example Question #1 : Snell's Law And Index Of Refraction

A light ray is traveling through air hits a transparent material at an angle of  from the normal. It is then refracted at . What is the speed of light in the material?

Explanation:

This problem requires Snell's Law and the corresponding equation:

We know that the index of refraction of air is:

We also know that:

and

Now we can plug in these values into the Snell's Law equation to find the index of refraction for the transparent material.

Finally, we need to calculate the speed of light moving through this transparent material now that we know the index of refraction for it. To do that, we need to use this equation:

Where  is the speed of light and  is the index of refraction. We plug in our known values and get:

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