Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 2 : AP Physics 2

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 #12 : Optics

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=3cm

h_i=6cm

h_i=9cm

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

Report an Error

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:

12.5cm

20cm

14.5cm

10cm

18cm

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

Report an Error

Example Question #1 : 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=18cm

i=11cm

i=28cm

i=5cm

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

Report an Error

Example Question #1 : 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=-26.5cm

f=-18.5cm

f=18.5cm

f=22.5cm

f=-22.5cm

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

Report an Error

Example Question #2 : 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=6.5

M=1.75

M=3.75

M=5.25

M=2.5

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

Report an Error

Example Question #431 : Ap Physics 2

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 focal length of the mirror.

Possible Answers:

f=18.5cm

f=-10cm

f=-4cm

f=10cm

f=4cm

Correct answer:

f=18.5cm

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 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:

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

f=18.5cm

Report an Error

Example Question #11 : Optics

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 distance of the image.

Possible Answers:

i=8cm

i=1.5cm

i=6.5cm

i=4cm

i=10cm

Correct answer:

i=6.5cm

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 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:

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

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

i=6.5cm

Report an Error

Example Question #21 : Optics

An object of height 5 cm is placed 10 cm in front of a concave mirror that has a radius of curvature of 37.5 c m . Will the image be real or virtual?

Possible Answers:

There will be no image

Real

Virtual

Cannot be determined without knowing the index of refraction

Correct answer:

Virtual

Explanation:

Use the relationship between focal length and radius:

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

Plug in values.

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

f=18.75cm

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

Report an Error

Example Question #432 : 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 magnification of the image.

Possible Answers:

M=4.8

M=3.0

M=1.8

M=6.8

M=2.8

Correct answer:

M=1.8

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 and solve for magnification, :

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

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

M=1.8

Report an Error

Example Question #14 : 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 size of the image.

Possible Answers:

h_i=3.25cm

h_i=8.50cm

h_i=5.25cm

h_i=2.25cm

h_i=3.75cm

Correct answer:

h_i=3.25cm

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 formula:

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

Where

is magnification

h_i is image height

h_p is object height

Plug in values and solve for the image height:

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

h_i=3.25cm

Report an Error

View Tutors

Kisa
Certified Tutor

University of Toronto, Scarborough, Bachelor, Public Policy.

View Tutors

Cindy
Certified Tutor

Midwestern State University, Masters in Education, Educational Administration.

View Tutors

Melissa
Certified Tutor

The University of Texas at Austin, Bachelor of Science, Mechanical Engineering.

All AP Physics 2 Resources

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

Popular Subjects

SAT Tutors in Miami, Physics Tutors in Philadelphia, Spanish Tutors in Los Angeles, English Tutors in Dallas Fort Worth, LSAT Tutors in Washington DC, English Tutors in San Francisco-Bay Area, Computer Science Tutors in Houston, SAT Tutors in Houston, GMAT Tutors in New York City, Calculus Tutors in Philadelphia

Popular Courses & Classes

LSAT Courses & Classes in Denver, SSAT Courses & Classes in Washington DC, ACT Courses & Classes in Seattle, GMAT Courses & Classes in Seattle, ISEE Courses & Classes in Denver, LSAT Courses & Classes in Seattle, SSAT Courses & Classes in Houston, SAT Courses & Classes in New York City, SSAT Courses & Classes in Boston, SAT Courses & Classes in Denver

Popular Test Prep

SSAT Test Prep in Philadelphia, GRE Test Prep in San Diego, MCAT Test Prep in New York City, GMAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Dallas Fort Worth, ACT Test Prep in Houston, LSAT Test Prep in Denver, ACT Test Prep in Dallas Fort Worth, MCAT Test Prep in Philadelphia

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 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #12 : Optics

Example Question #1 : Mirrors

Example Question #1 : Mirrors

Example Question #1 : Mirrors

Example Question #2 : Mirrors

Example Question #431 : Ap Physics 2

Example Question #11 : Optics

Example Question #21 : Optics

Example Question #432 : Ap Physics 2

Example Question #14 : Mirrors

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: