Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : Optics

Study concepts, example questions & explanations for AP Physics 2

Create An Account

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Optics

Suppose a student who was farsighted wears glasses that allows him to read at a distance of 20cm from his eyes to the book. His near-point distance is 63cm . If his glasses are 1.5cm from his eyes, what is the refractive power of his glasses lenses?

Possible Answers:

3.846D

1.613D

2.342D

4.536D

6.346D

Correct answer:

3.846D

Explanation:

Write the thin-lens equation.

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

d_o represents the object distance from the lens, subtracting the distance between the student's eyes and the lens.

d_o=20cm-1.5cm=18.5cm

The image distance is:

d_i=1.5cm-63cm=-61.5cm

Substitute the givens to the equation.

$\frac{1}{f}= \frac{1}{18.5}+\frac{1}{-61.5}= 0.037794 cm^{-1}$

$f=\frac{1}{0.037794cm}=26.4592cm = 0.26 m$

Refractive power is the inverse of focal length in .

$\frac{1}{0.26m}= 3.846D$

Report an Error

Example Question #1 : Thin Lens Equation

A flower is placed 20cm away from a converging lens. An upside down image of the flower is measured as 6cm tall and at a distance of 30cm from the lens. Find the focal length of the lens and the real height of the flower.

Possible Answers:

f=0.08cm

$h_{o}=6.7cm$

f=20cm

$h_{o}=12cm$

f=12cm

$h_{o}=4cm$

f=30.5cm

$h_{o}=4.5cm$

Correct answer:

f=12cm

$h_{o}=4cm$

Explanation:

Let the object distance be $d_{0}$ and the image distance be $d_{i}$ . The object distance is the distance the flower is from the lens. The image distance is the distance the image is located from the lens. The thin lens equation is

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

$f=\left ( \frac{1}{d_{o}}+\frac{1}{d_{i}}\right )^{-1}=12cm$

Alternatively you can use

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

To find the height, the equation for the magnification of an object is given by

$m=-\frac{d_{i}}{d_{o}}$

it is also given as

$m=-\frac{h_{o}}{h_{i}}$

The negative sign indicates an image that is upside down. Setting these equal gives

$\frac{d_{i}}{d_{o}}=\frac{h_{i}}{h_{o}} \rightarrow h_{o}=\frac{d_{o}}{d_{i}}h_{i}=4cm$

Report an Error

Example Question #1 : Thin Lens Equation

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.

If the top of the image is 10cm from the center of the tank, and the top of the object is located 3mm from the center of the tank, what is the focal length of this lens?

Possible Answers:

f=5.5

f=340

f= 2.9

f= .0029

Correct answer:

f= .0029

Explanation:

Using object distance, image distance, and focal length formula:

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

Where is object distance, is image distance, and is focal length.

Plugging everything in:

$\frac{1}{.003}+\frac{1}{.1}=\frac{1}{f}$

Solving for ,

f= .0029

Report an Error

Example Question #1 : Optics

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

Possible Answers:

3.67cm

-3.67cm

11.0cm

33.0cm

-33.0cm

Correct answer:

3.67cm

Explanation:

The Magnification Equation is as follows:

$M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

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 for our problem. We want h_i , so we just need to multiply both sides by h_o .

$h_i=-\frac{d_i}{d_o}\cdot h_o$

Now, we can plug in our numbers.

$h_i&=-\frac{d_i}{d_o} \cdot h_o$

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

h_1= 0.0367

Therefore, the image height is 3.67cm .

Report an Error

Example Question #1 : Magnification

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

Possible Answers:

0.75

1.49

1.79

0.66

1.68

Correct answer:

1.49

Explanation:

The magnification equation is

$M=\frac{h_i}{h_o}$

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

$M&=\frac{h_i}{h_o}$

$M=\frac{21.9}{14.7}$

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.

Report an Error

Example Question #1 : Optics

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

Possible Answers:

-11.56m

9.05m

-29.00m

29.00m

-9.05m

Correct answer:

9.05m

Explanation:

The equation for magnification is

$M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

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

$M&=-\frac{d_i}{d_o}$

$M\cdot d_o&=-d_i$

$d_o&=-\frac{d_i}{M}$

Now, we can plug in our numbers.

$d_o&=-\frac{d_i}{M}$

$d_o= -\frac{-16.2}{1.79}$

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 9.05m . Note that the object distance is always positive.

Report an Error

Example Question #2 : Optics

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

Possible Answers:

1.425

None of these

1.88

6.11

Correct answer:

1.425

Explanation:

Using the relationship:

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

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

Plugging in values:

$\frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{67}$

Solving for :

$i=\frac{1}{-\frac{2}{67}+\frac{1}{10}}$

i=14.25cm

Using the equation for magnification:

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

Where:

is magnification

$M=-\frac{14.25}{-10}$

M=1.425

Report an Error

Example Question #5 : Magnification

Assuming that the image is the size of 1.5cm and is actually a picture of a tree of size 10m . What can we say about the magnification of this alcohol solution?

Possible Answers:

$6.7*10^{2}$

$1.5*10^{-3}$

$-1.5*10^{-3}$

$-6.7*10^{2}$

Correct answer:

$-1.5*10^{-3}$

Explanation:

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

$M=-\frac{1.5*10^{-2}m}{10m}=-1.5*10^{-3}$

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

Report an Error

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:

Inverted and real

Inverted and virtual

Upright and real

Upright and virtual

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.

Report an Error

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

15.8cm

17.0cm

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.

Report an Error

View Tutors

Sharif
Certified Tutor

NCCU, Bachelors, Physics and Mathematics.

View Tutors

Alejandro
Certified Tutor

Kennesaw State University, Bachelor of Science, Computer Software Engineering.

View Tutors

Joy
Certified Tutor

Boston University, Bachelor in Arts, Conservation Biology.

All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

LSAT Tutors in Philadelphia, English Tutors in New York City, GMAT Tutors in Dallas Fort Worth, Computer Science Tutors in Boston, English Tutors in Denver, French Tutors in Denver, Computer Science Tutors in Los Angeles, Biology Tutors in Phoenix, Statistics Tutors in Houston, LSAT Tutors in Los Angeles

Popular Courses & Classes

MCAT Courses & Classes in Seattle, Spanish Courses & Classes in Atlanta, MCAT Courses & Classes in Phoenix, SSAT Courses & Classes in Chicago, GRE Courses & Classes in Houston, SAT Courses & Classes in Boston, Spanish Courses & Classes in Washington DC, MCAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Houston, SAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

GMAT Test Prep in Los Angeles, MCAT Test Prep in Philadelphia, GRE Test Prep in Houston, GMAT Test Prep in Washington DC, GRE Test Prep in Phoenix, ISEE Test Prep in San Diego, SAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Chicago, SSAT Test Prep in Dallas Fort Worth, ACT Test Prep in Denver

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Physics 2 : Optics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Optics

Example Question #1 : Thin Lens Equation

Example Question #1 : Thin Lens Equation

Example Question #1 : Optics

Example Question #1 : Magnification

Example Question #1 : Optics

Example Question #2 : Optics

Example Question #5 : Magnification

Example Question #1 : Mirrors

Example Question #1 : Optics

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: