# AP Physics 2 : Optics

## Example Questions

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### Example Question #1 : Thin Lens Equation

Suppose a student who was farsighted wears glasses that allows him to read at a distance of  from his eyes to the book. His near-point distance is . If his glasses are  from his eyes, what is the refractive power of his glasses lenses?

Explanation:

Write the thin-lens equation.

represents the object distance from the lens, subtracting the distance between the student's eyes and the lens.

The image distance is:

Substitute the givens to the equation.

Refractive power is the inverse of focal length in .

### Example Question #2 : Thin Lens Equation

A flower is placed  away from a converging lens. An upside down image of the flower is measured as  tall and at a distance of  from the lens. Find the focal length of the lens and the real height of the flower.

Explanation:

Let the object distance be  and the image distance be . The object distance is the distance the flower is from the lens. The image distance is the distance the image is located from the lens. The thin lens equation is

Alternatively you can use

To find the height, the equation for the magnification of an object is given by

it is also given as

The negative sign indicates an image that is upside down. Setting these equal gives

### Example Question #3 : Thin Lens Equation

You are passing a ray of light through a thin tank of alcohol to determine properties. Assume that the light ray is not effected by the tank's exterior. You find that if a piece of white paper is put underneath the alcohol tank, there is an image on the metal.

If the top of the image is  from the center of the tank, and the top of the object is located  from the center of the tank, what is the focal length of this lens?

Explanation:

Using object distance, image distance, and focal length formula:

Where  is object distance,  is image distance, and  is focal length.

Plugging everything in:

Solving for ,

### Example Question #1 : Magnification

An object  tall is  from a mirror. If the image distance is  from the mirror, what is the image height?

Explanation:

The Magnification Equation is as follows:

The negative on the last part is very important. If you don't include it (and it's easy to forget), you will get the wrong answer.

We'll use the parts without  for our problem. We want , so we just need to multiply both sides by .

Now, we can plug in our numbers.

Therefore, the image height is

### Example Question #421 : Ap Physics 2

You have a concave mirror and a candle. The candle is  tall, and the image is  tall. What is the the magnification of the mirror?

Explanation:

The magnification equation is

where  is the height of the image, and  is the height of the object. We can plug these given values into the equation.

The magnification is a scalar value, so it's unitless. We can verify this by examining what goes into the equation. Both heights have the same units, which cancel, so that leaves us with no units.

### Example Question #1 : Magnification

The image produced by a concave mirror is at , and the magnification is . What is the object distance?

Explanation:

The equation for magnification is

We're given the image distance and the magnification, so we'd use the second equality.

Now, we can plug in our numbers.

The negative in front of the equation is very important. If you forget it, the answer will be incorrect.

The distance from the mirror to the object is . Note that the object distance is always positive.

### Example Question #4 : Magnification

An object of height  object is placed  in front of a convex mirror that has a radius of curvature of . Determine the magnification of the image.

None of these

Explanation:

Using the relationship:

Where:

is the object distance from the mirror

is the image distance from the mirror

is the focal length of the mirror

is the radius of curvature of the mirror

Plugging in values:

Solving for :

Using the equation for magnification:

Where:

is magnification

### Example Question #1 : Magnification

You are passing a ray of light through a thin tank of alcohol to determine properties. Assume that the light ray is not effected by the tank's exterior. You find that if a piece of white paper is put underneath the alcohol tank, there is an image on the metal.

Assuming that the image is the size of  and is actually a picture of a tree of size . What can we say about the magnification of this alcohol solution?

Explanation:

To determine magnification, we simply divide the object length from the image length.

The negative sign is used to designate that the image is real.

### Example Question #1 : Mirrors

Suppose that an object is placed 15cm in front of a concave mirror that has a radius of curvature equal to 20cm. Will the resulting image be upright or inverted? Will it be a real or virtual image?

Inverted and virtual

There is no way to determine whether the image is real, virtual, inverted, or upright based on the information given

Inverted and real

Upright and real

Upright and virtual

Inverted and real

Explanation:

To solve this problem, it is essential to use the mirror equation:

Where:

is the distance the object is from the mirror

is the distance the image is from the mirror

is the focal length of the mirror

To calculate the focal length, we'll have to use the radius of curvature:

Plugging this value into the top equation and rearranging, we obtain:

Therefore, the image distance is:

Since the image distance calculated above is a positive value, this means that the image forms on the same side of the mirror that the reflected light traveled. Thus, the image is real.

To determine whether the image is inverted or upright, we'll need to use the magnification equation:

The magnification obtained above is a negative number. As a result, the image formed from this scenario will be inverted.

### Example Question #2 : Mirrors

You have an object 17.0cm away from a concave mirror with a focal length of 7.40cm. What is the image distance?

There is not enough information to determine the image distance

Explanation:

Because we're dealing with a mirror, appropriately we would use the Mirror Equation:

where  is the focal length,  is the object distance, and  is the image distance. Because we're trying to solve for , we need to rearrange the equation to solve for it.

Now, we can just plug in our values for  and :

Therefore, the image distance is 13.1cm.

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