Mirrors - AP Physics 2

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Question

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?

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Answer

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:

$f=\frac{r}{2}$=\frac{20cm}{2}$=10cm$

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:

$M=-$\frac{30cm}{15cm}$=-2$

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

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Question

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

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Answer

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

$\frac{1}{f}$=\frac{1}{d_o}$+\frac{1}{d_i}$

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.

$\frac{1}{f}$&=\frac{1}{d_o}$+\frac{1}{d_i}$

$\frac{1}{f}$-$\frac{1}{d_o}$&=\frac{1}{d_i}$

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

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

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

$($\frac{1}{7.40\cdot $10^{-2}$$}-$\frac{1}{17.0\cdot $10^{-2}$$$})^{-1}$&=d_1$

Therefore, the image distance is 13.1cm.

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Question

An object is set near a mirror like below.

At which point would the image be?

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Answer

If you were to draw a line from an object to its image on a flat mirror, the line would be perpendicular to the plane of the mirror. The distance from the image to the mirror would be the same distance from the mirror to the object. Only point A fits both criteria, as all other points aren't perpendicular with the mirror, and point E isn't the same distance either.

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Question

You place a candle away from a concave mirror with a radius of curvature of . What is the image distance?

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Answer

For this problem, we use the mirror equation, rearraged to find .

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

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

is the image location, is the object location, and is the focal point which is equal to the radius of curvature. Plugging in our known values:

$i= $($\frac{1}{0.53}$-$\frac{1}{4.12}$)^{-1}$$

Therefore, the image distance is from the mirror. Remember, for concave mirrors, the image is on the same side as the object, and has a flipped orientation.

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Question

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

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Answer

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}__MATH_EXPR_1__=\frac{1}{o}$+\frac{1}{i}$$

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

Now, we can plug in our numbers.

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

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

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Question

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.

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Answer

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 :

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

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Question

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.

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Answer

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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Question

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

Determine the distance of the image.

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Answer

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 is taken to be negative for convex mirrors

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 the image distance:

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

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Question

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

Determine the focal length of the mirror.

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Answer

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 is taken to be negative for convex mirrors

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 the focal length:

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Question

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.

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Answer

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:

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

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Question

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.

Tap to reveal answer

Answer

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}$$

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Question

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.

Tap to reveal answer

Answer

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}$}$

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Question

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.

Tap to reveal answer

Answer

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}$

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Question

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?

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Answer

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.

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Question

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.

Tap to reveal answer

Answer

Use the relationship for concave mirrors:

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

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

Plug in values:

$\frac{1}{f}$=\frac{2}{25}$

Solve for :

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Question

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

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Answer

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

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Question

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?

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Answer

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.

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Question

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?

Tap to reveal answer

Answer

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:

$f=\frac{r}{2}$=\frac{20cm}{2}$=10cm$

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:

$M=-$\frac{30cm}{15cm}$=-2$

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

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Question

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

Tap to reveal answer

Answer

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

$\frac{1}{f}$=\frac{1}{d_o}$+\frac{1}{d_i}$

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.

$\frac{1}{f}$&=\frac{1}{d_o}$+\frac{1}{d_i}$

$\frac{1}{f}$-$\frac{1}{d_o}$&=\frac{1}{d_i}$

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

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

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

$($\frac{1}{7.40\cdot $10^{-2}$$}-$\frac{1}{17.0\cdot $10^{-2}$$$})^{-1}$&=d_1$

Therefore, the image distance is 13.1cm.

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Question

An object is set near a mirror like below.

At which point would the image be?

Tap to reveal answer

Answer

If you were to draw a line from an object to its image on a flat mirror, the line would be perpendicular to the plane of the mirror. The distance from the image to the mirror would be the same distance from the mirror to the object. Only point A fits both criteria, as all other points aren't perpendicular with the mirror, and point E isn't the same distance either.

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