Mirrors

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AP Physics 2 › Mirrors

Questions 1 - 10

An object is from a convex mirror with a radius of . What is the image distance?

Explanation

For convex mirrors, the mirror equation is slightly different than for concave mirrors.

$\frac{1}{o}+\frac{1}{i}=-\frac{1}{f}$

Note the negative sign in front of the focal point. This is because a convex mirror is the same as a concave mirror pointing in the opposite direction.

Now, we rearrange the equation to solve for .

$-\frac{1}{f}&=\frac{1}{o}+\frac{1}{i}$

$\frac{1}{i}&=-\frac{1}{f}-\frac{1}{o}$

$i&=(-\frac{1}{f}-\frac{1}{o})^{-1}$

Now, we can plug in our numbers.

$i=(-\frac{1}{f}-\frac{1}{o})^{-1}$

$i= (-\frac{1}{0.113}-\frac{1}{0.421})^{-1}$

Therefore, the image is at , which is on the concave side of the mirror.

A ray of light is parallel to the principal axis and reflects from a concave mirror. The reflected ray will __________.

pass through the focal point

pass through the center of curvature

pass through both the focal point and the center of curvature

also be parallel to the principal axis

None of the other answers.

Explanation

Parallel rays that reflect off a concave mirror always pass through the focal point of the mirror.

An object of height is placed in front of a concave mirror that has a radius of curvature of .

Determine the magnification of the image.

Explanation

Use the mirror/lens equation:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}=\frac{2}{r}$

Where:

is the object distance from the mirror, which is taken to be negative

is the image distance from the mirror

is the focal length of the mirror, which for concave mirrors is taken to be positive

is the radius of curvature of the mirror

Plug in values:

$\frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=\frac{2}{37}$

Solve for the image distance:

$i=\frac{1}{-\frac{2}{37}+\frac{1}{10}}$

Use the magnification equation:

$M=\frac{h_i}{h_p}=-\frac{i}{p}$

Where

is magnification

is image height

is object height

Plug in values and solve for magnification:

$M=-\frac{6.5}{-10}$

If an object is situated at a distance farther than the radius of curvature of a concave mirror, what will be true about the image formed?

The image will be real and inverted

The image will be virtual and inverted

The image will be real and up-right

The image will be virtual and up-right

It is impossible to determine

Explanation

In this question, we're told that an object is positioned outside of the curvature radius of a concave mirror. We're asked to identify how the resulting image will appear.

One way to approach this problem is to draw a ray diagram. In such a diagram, a straight line is first drawn from the top of the object horizontally to the mirror. From there, the line is drawn to pass through the mirror's focal point.

A second line is then drawn from the top of the object and straight through the focal point to the mirror. Then, the line is drawn horizontally away from the mirror.

The point at which these two lines intersect will represent the top of the image. In the diagram shown below, we can see that the image will appear in front of the mirror. Hence, it is a real image. Furthermore, the image appears upside down, meaning that it has an inverted orientation. So all together, the image will be real and inverted, which makes this the correct answer.

Mirror light rays

An object of height is placed in front of a concave mirror that has a radius of curvature of .

Determine the distance of the image.

Explanation

Use the mirror/lens equation:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}=\frac{2}{r}$

Where:

is the object distance from the mirror, which is taken to be negative

is the image distance from the mirror

is the focal length of the mirror, which for concave mirrors is taken to be positive

is the radius of curvature of the mirror

Plug in values and solve for the image distance:

$\frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=\frac{2}{37}$

$i=\frac{1}{-\frac{2}{37}+\frac{1}{10}}$

An object of height is placed in front of a concave mirror that has a radius of curvature of .

Determine the size of the image.

Explanation

Use the mirror/lens equation:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}=\frac{2}{r}$

Where:

is the object distance from the mirror, which is taken to be negative

is the image distance from the mirror

is the focal length of the mirror, which for concave mirrors is taken to be positive

is the radius of curvature of the mirror

Plug in values:

$\frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=\frac{2}{37}$

Solve for the image distance:

$i=\frac{1}{-\frac{2}{37}+\frac{1}{10}}$

Use the magnification formula:

$M=\frac{h_i}{h_p}=-\frac{i}{p}$

Where

is magnification

is image height

is object height

Plug in values and solve for the image height:

$\frac{h_i}{5}=-\frac{6.5}{-10}$

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

Determine the size of the image.

Explanation

Use the mirror/lens equation:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{r}$

Where:

is the object distance from the mirror, which is taken to be negative number

is the image distance from the mirror

is the focal length of the mirror, which for convex mirrors is taken to be negative number

is the radius of curvature of the mirror

Plug in values:

$\frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{45}$

Solve for :

$i=\frac{1}{-\frac{2}{45}+\frac{1}{10}}$

Use the equation for magnification:

$M=\frac{h_i}{h_p}=-\frac{i}{p}$

In this equation:

is magnification

is image height

is object height

Plug in values and solve for :

$\frac{h_i}{5}=-\frac{18}{-10}$

An object of height is placed in front of a concave mirror that has a radius of curvature of .

Determine the focal length of the mirror.

Explanation

Use the mirror/lens equation:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}=\frac{2}{r}$

Where:

is the object distance from the mirror, which is taken to be negative

is the image distance from the mirror

is the focal length of the mirror, which for concave mirrors is taken to be positive

is the radius of curvature of the mirror

Plug in values and solve for the focal point:

$\frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=\frac{2}{37}$

An object of height is placed in front of a concave mirror that has a radius of curvature of . Will the image be real or virtual?

Virtual

Real

There will be no image

Cannot be determined without knowing the index of refraction

Explanation

Use the relationship between focal length and radius:

$f=\frac{r}{2}$

Plug in values.

$f=\frac{37.5cm}{2}$

Based on the properties of a concave mirror, objects inside the focal length of the mirror will generate virtual images.

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

Determine the magnification of the image.

Explanation

Use the mirror/lens equation:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{r}$

Where:

is the object distance from the mirror, which is taken to be negative number

is the image distance from the mirror

is the focal length of the mirror, which for convex mirrors is taken to be negative number

is the radius of curvature of the mirror

Plug in values:

$\frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{45}$

Solve for :

$i=\frac{1}{-\frac{2}{45}+\frac{1}{10}}$

Use the equation for magnification and solve for magnification, :

$M=\frac{h_i}{h_p}=-\frac{i}{p}$

$M=-\frac{18}{-10}$

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