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AP Physics 2 : Mirrors

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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?

Possible Answers:

Upright and real

Inverted and virtual

Upright and virtual

Inverted and real

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

Correct answer:

Inverted and real

Explanation:

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

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

Where:

$d_{o}$ is the distance the object is from the mirror

$d_{i}$ 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:

$\frac{1}{d_{i}}=\frac{1}{f}-\frac{1}{d_{o}}=\frac{1}{10cm}-\frac{1}{15cm}=\frac{1}{30cm}$

Therefore, the image distance is:

$d_{i}=30cm$

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:

$M=\frac{h_{i}}{h_{o}}=-\frac{d_{i}}{d_{o}}$

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

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

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Example Question #1 : Optics

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

Possible Answers:

There is not enough information to determine the image distance

17.0cm

15.8cm

19.3cm

13.1cm

Correct answer:

13.1cm

Explanation:

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, d_o is the object distance, and d_i is the image distance. Because we're trying to solve for d_i , 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 d_o :

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

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

$(7.631)^{-1}&= d_1$

0.131= d_1

Therefore, the image distance is 13.1cm.

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Example Question #1 : Mirrors

An object is set near a mirror like below.

Mirror2

At which point would the image be?

Possible Answers:

Correct answer:

Explanation:

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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Example Question #4 : Mirrors

You place a candle 4.12m away from a concave mirror with a radius of curvature of 0.53m . What is the image distance?

Possible Answers:

1.34m

0.98m

0.61m

1.61m

0.30m

Correct answer:

0.61m

Explanation:

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

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

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

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

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:

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

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

i= 0.61

Therefore, the image distance is 0.61m 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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Example Question #1 : Mirrors

An object is 42.1cm from a convex mirror with a radius of 11.3cm . What is the image distance?

Possible Answers:

-8.91cm

11.98cm

8.91cm

-14.23cm

14.23cm

Correct answer:

-8.91cm

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.

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

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

i= -0.0891m

Therefore, the image is at -8.91cm , which is on the concave side of the mirror.

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Example Question #431 : Ap Physics 2

An object of height 5 cm is placed 10 cm in front of a convex mirror that has a radius of curvature of 45.0 cm .

Determine the size of the image.

Possible Answers:

h_i=12cm

h_i=18cm

h_i=9cm

h_i=3cm

h_i=6cm

Correct answer:

h_i=9cm

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 :

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

i=18cm

Use the equation for magnification:

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

In this equation:

is magnification

h_i is image height

h_p is object height

Plug in values and solve for h_i :

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

h_i=9cm

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Example Question #1 : Mirrors

An object of height 7 cm is placed 4 cm in front of a concave mirror that has a radius of curvature of 25 cm . Determine the focal length of the mirror.

Possible Answers:

20cm

18cm

10cm

12.5cm

14.5cm

Correct answer:

12.5cm

Explanation:

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 :

f=12.5cm

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Example Question #8 : Mirrors

An object of height 5 cm is placed 10 cm in front of a convex mirror that has a radius of curvature of 45.0 cm .

Determine the distance of the image.

Possible Answers:

i=28cm

i=18cm

i=5cm

i=11cm

i=9cm

Correct answer:

i=18cm

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 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:

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

i=18cm

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Example Question #9 : Mirrors

An object of height 5 cm is placed 10 cm in front of a convex mirror that has a radius of curvature of 45.0 cm .

Determine the focal length of the mirror.

Possible Answers:

f=22.5cm

f=-22.5cm

f=-18.5cm

f=18.5cm

f=-26.5cm

Correct answer:

f=-22.5cm

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 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:

f=-22.5cm

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Example Question #10 : Mirrors

An object of height 5 cm is placed 10 cm in front of a concave mirror that has a radius of curvature of 37 cm .

Determine the magnification of the image.

Possible Answers:

M=5.25

M=2.5

M=6.5

M=3.75

M=1.75

Correct answer:

M=6.5

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:

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

i=6.5cm

Use the magnification equation:

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

Where

is magnification

h_i is image height

h_p is object height

Plug in values and solve for magnification:

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

M=6.5

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AP Physics 2 : Mirrors

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Mirrors

Example Question #1 : Optics

Example Question #1 : Mirrors

Example Question #4 : Mirrors

Example Question #1 : Mirrors

Example Question #431 : Ap Physics 2

Example Question #1 : Mirrors

Example Question #8 : Mirrors

Example Question #9 : Mirrors

Example Question #10 : Mirrors

All AP Physics 2 Resources

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