# MCAT Physical : Mirrors and Lenses

## Example Questions

### Example Question #151 : Ap Physics 2

The focal point for a mirror is 56cm behind the mirror. Is the mirror concave or convex, and what is its radius of curvature?

Concave with radius of curvature of 112cm

Convex with radius of curvature of 28cm

Convex with radius of curvature of 112cm

Concave with radius of curvature of 28cm

Convex with radius of curvature of 112cm

Explanation:

Since the focal point falls behind the mirror it must be convex. The radius of curvature can be found using the focal length equation.

### Example Question #2 : Mirrors And Lenses

A concave mirror has a radius of curvature of 0.85m. Where is the mirror's focal point?

The focal point cannot be found from the given information

Explanation:

Since the mirror is concave, the focal point will be in front of the mirror. The focal length is equal to one half of the radius of curvature.

Rc is the radius of curvature. Plugging in 0.85m for Rc allows us to solve for the focal length.

0.43m is equal to 43cm.

### Example Question #1 : Focal Length

A lens has a focal length of . What is the strength and type of lens?

, diverging lens

, converging lens

, diverging lens

, converging lens

, diverging lens

Explanation:

Since the focal length is negative, the lens is diverging.

The diopter of a lens is found through the following formula:

Since the focal length of the lens is :

### Example Question #4 : Mirrors And Lenses

A virtual image is formed  from a convex mirror with a focal length of . How far from the mirror is the object that created this image?

Explanation:

Use the equation:

Focal length is negative for convex mirrors, and image distance is negative for virtual images. We are given these values in the question, allowing us to calculate the object distance.

### Example Question #5 : Mirrors And Lenses

How far from a converging lens must an object be placed to produce an image that is NOT real and inverted? Given the answer as  in terms of the focal length, .

Explanation:

When an object is placed a distance from a converging lens or mirror that is equal to the focal length, no image is produced. To test this out, stand in front of a single concave mirror and continue to back up until you no longer see an image. Once you've reached this point, you will be standing one focal length away from the mirror.

When the object is less than one focal length away from the converging lens/mirror, the image will be virtual and upright.

If you don't have these trends committed to memory, you can derive them from the equation .

When  is a negative integer the image is virtual, and when it is a positive integer the image is real.

### Example Question #1 : Thin Lens Equation

A certain farsighted person cannot focus on objects closer to his eyes than . What focal length eyeglass lenses are needed in order to focus on a newspaper held at  from the person's eyes, if the glasses are worn  from his eyes?

Explanation:

First we need the object and image distances away from the eyeglass lenses. Here, the newspaper is the object and the  focal point is where the image needs to be located. Find the distance from the object to the lens, and the distance of the image to the lens, by subtracting out the distance from the lens to the eye.

Now apply the thin lens equation to determine focal length.

Recall that if the image is on the same side of the lens as the object, then image distance is negative.

### Example Question #2 : Thin Lens Equation

An object is placed 50cm in front of a concave mirror of radius 60cm. How far from the mirror is the image?

Explanation:

Relevant equations:

Step 1: Find the focal length of the mirror.

Step 2: Plug this focal length and the object distance into the lens equation.

### Example Question #8 : Mirrors And Lenses

An object that is 3cm tall is placed 30cm from a convex spherical mirror of radius 40cm, along its central axis. What is the height of the image that is formed?

Explanation:

Relevant equations:

Step 1: Find the focal length of the mirror (remembering that convex mirrors have negative focal lengths, by convention).

Step 2: Find the image distance using the thin lens equation.

Step 3: Use the magnification equation to relate the object distances and heights.

### Example Question #1 : Optics

A person approaches a plane mirror at 5m/s. How fast do they approach the mirror image?

10m/s

5m/s

7.5m/s

2.5m/s

10m/s

Explanation:

The image distance for a plane mirror is always equal to the object distance because the magnification is 1.

If the object and image are the same distance from the mirror and magnification is 1, then as the object approaches the mirror at a certain speed, the image is approaching the plane mirror at the same speed, therefore you approach the image more quickly than you approach the mirror, since you travel 5m/s toward the mirror and the image travels 5m/s toward the mirror.

### Example Question #10 : Mirrors And Lenses

In a microscope has a  tube length and objective lens with a focal length of . The viewer's eye is from the objective lens, and they desire a magnification of . What must the focal length of the eyepiece lens be to achieve this magnification?

Explanation:

The total magnification of a compound microscope is the product of the objective lens magnification and the eyepiece magnification.

Objective magnification and eyepiece magnification are given by the following equations:

We are given the tube length and focal length of the objective lens, allowing us to solve for its magnification.

We also know the distance of the viewer's eye. Use this value in the eyepiece magnification equation.

Finally, combine the eyepiece magnification and objective lens magnification into the original equation for total magnification.

Use the given value for total magnification to solve for the focal length of the eyepiece lens.