AP Physics 2 : Magnification

Study concepts, example questions & explanations for AP Physics 2

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Example Questions

Example Question #421 : Ap Physics 2

An object \(\displaystyle 11cm\) tall is \(\displaystyle 9cm\) from a mirror. If the image distance is \(\displaystyle -3cm\) from the mirror, what is the image height?

Possible Answers:

\(\displaystyle -33.0cm\)

\(\displaystyle 33.0cm\)

\(\displaystyle 11.0cm\)

\(\displaystyle 3.67cm\)

\(\displaystyle -3.67cm\)

Correct answer:

\(\displaystyle 3.67cm\)

Explanation:

The Magnification Equation is as follows:

\(\displaystyle M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}\)

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 \(\displaystyle M\) for our problem. We want \(\displaystyle h_i\), so we just need to multiply both sides by \(\displaystyle h_o\).

\(\displaystyle h_i=-\frac{d_i}{d_o}\cdot h_o\)

Now, we can plug in our numbers.

\(\displaystyle h_i&=-\frac{d_i}{d_o} \cdot h_o\)

\(\displaystyle h_1=-\frac{-3\cdot 10^{-2}}{9\cdot 10^{-2}}\cdot (11\cdot 10^{-2})\)

\(\displaystyle h_1= 0.0367\)

Therefore, the image height is \(\displaystyle 3.67cm\)

Example Question #2 : Magnification

You have a concave mirror and a candle. The candle is \(\displaystyle 14.7cm\) tall, and the image is \(\displaystyle 21.9cm\) tall. What is the the magnification of the mirror?

Possible Answers:

\(\displaystyle 0.66\)

\(\displaystyle 1.79\)

\(\displaystyle 0.75\)

\(\displaystyle 1.49\)

\(\displaystyle 1.68\)

Correct answer:

\(\displaystyle 1.49\)

Explanation:

The magnification equation is 

\(\displaystyle M=\frac{h_i}{h_o}\)

where \(\displaystyle h_i\) is the height of the image, and \(\displaystyle h_o\) is the height of the object. We can plug these given values into the equation.

\(\displaystyle M&=\frac{h_i}{h_o}\)

\(\displaystyle M=\frac{21.9}{14.7}\)

\(\displaystyle M= 1.49\)

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 : Optics

The image produced by a concave mirror is at \(\displaystyle -16.2m\), and the magnification is \(\displaystyle 1.79\). What is the object distance?

Possible Answers:

\(\displaystyle 9.05m\)

\(\displaystyle -11.56m\)

\(\displaystyle -9.05m\)

\(\displaystyle 29.00m\)

\(\displaystyle -29.00m\)

Correct answer:

\(\displaystyle 9.05m\)

Explanation:

The equation for magnification is

\(\displaystyle M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}\)

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

\(\displaystyle M&=-\frac{d_i}{d_o}\)

\(\displaystyle M\cdot d_o&=-d_i\)

\(\displaystyle d_o&=-\frac{d_i}{M}\)

Now, we can plug in our numbers.

\(\displaystyle d_o&=-\frac{d_i}{M}\)

\(\displaystyle d_o= -\frac{-16.2}{1.79}\)

\(\displaystyle d_o= 9.05\)

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 \(\displaystyle 9.05m\). Note that the object distance is always positive.

Example Question #4 : Magnification

An object of height \(\displaystyle 2.9 cm\) object is placed \(\displaystyle 10 cm\) in front of a convex mirror that has a radius of curvature of \(\displaystyle 67.0 cm\). Determine the magnification of the image.

Possible Answers:

\(\displaystyle 1.88\)

\(\displaystyle .925\)

None of these

\(\displaystyle 6.11\)

\(\displaystyle 1.425\)

Correct answer:

\(\displaystyle 1.425\)

Explanation:

Using the relationship:

\(\displaystyle \frac{1}{p}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{r}\)

Where:

\(\displaystyle p\) is the object distance from the mirror

\(\displaystyle i\) is the image distance from the mirror

\(\displaystyle f\) is the focal length of the mirror

\(\displaystyle r\) is the radius of curvature of the mirror

Plugging in values:

\(\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{67}\)

Solving for \(\displaystyle i\):

\(\displaystyle i=\frac{1}{-\frac{2}{67}+\frac{1}{10}}\)

\(\displaystyle i=14.25cm\)

Using the equation for magnification:

\(\displaystyle M=\frac{h_i}{h_p}=-\frac{i}{p}\)

Where:

\(\displaystyle M\) is magnification

\(\displaystyle M=-\frac{14.25}{-10}\)

\(\displaystyle M=1.425\)

Example Question #5 : 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 \(\displaystyle 1.5cm\) and is actually a picture of a tree of size \(\displaystyle 10m\). What can we say about the magnification of this alcohol solution? 

Possible Answers:

\(\displaystyle -1.5*10^{-3}\)

\(\displaystyle 1.5*10^{-3}\)

\(\displaystyle -6.7*10^{2}\)

\(\displaystyle 6.7*10^{2}\)

Correct answer:

\(\displaystyle -1.5*10^{-3}\)

Explanation:

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

\(\displaystyle M=-\frac{1.5*10^{-2}m}{10m}=-1.5*10^{-3}\)

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

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