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 : Index Of Refraction

You are passing a ray of light through clear alcohol to determine properties. You shine the light ray exactly 45^o to the surface of alcohol.

Determine the index of refraction required in the alcohol to have total internal reflection?

Possible Answers:

Correct answer:

Explanation:

To have total internal reflection, our equation will become:

1*sin(45^o)=n_2sin(90^o)

sin(45^o)=n_2=.71

Report an Error

Example Question #1 : Index Of Refraction

You are passing a ray of light through clear alcohol to determine properties. You shine the light ray exactly 45^o to the surface of alcohol.

Determine the index of refraction of alcohol if the light ray bends to 22^o to the normal. Assume index of refraction of air is .

Possible Answers:

2.1

1.9

Correct answer:

1.9

Explanation:

We can use our knowledge about the indices of refraction to come up with our equation:

n_1sin(45^o)=n_2sin(22^o) , where n_1 is the index of refraction of air and n_2 is the index of refraction for our alcohol.

Since n_1=1

$\frac{sin(45^o)}{sin(22^o)}=n_2=1.9$

Report an Error

Example Question #1 : Index Of Refraction

Suppose that a ray of light traveling through air strikes a new medium. Upon doing so, the light bends away from the normal. Which of the following could this new medium be?

Possible Answers:

Vacuum

Diamond

Glass

Carbon dioxide

Water

Correct answer:

Vacuum

Explanation:

For this question, we're told that light is passing from air into another medium. In doing so, the light is refracted such that it bends away from the normal. We're asked to identify a possible medium.

The most important thing to understand about refraction is that when light passes into a new medium at an angle with respect to the normal, that light will be refracted, either away from or toward the normal. This is because light will travel through different media at different speeds. The faster light travels, the more it will bend away from the normal.

Generally, the angle at which light is bent can be predicted by Snell's law. In doing so, this equation takes use of the refractive index, a value unique to each medium. The expression for refractive index is as follows.

$n=\frac{c}{v}$

Where refers to the refractive index, refers to the speed of light in a vacuum, and refers to the speed of light in a given medium.

Since the speed of light is fastest when in a vacuum, the refractive index can never be less than . Only when the light is in a vacuum is the refractive index equal to . In any other medium, the refractive index will be greater than , even if it is slight.

Light will always refract away from the normal when it passes into a medium with a lower refractive index (indicating the light is traveling faster). Starting from air, the only way this can happen is for the new medium to have an index of refraction that is less than air. Of the answer choices shown, the only one that fits that criteria is the vacuum.

Report an Error

Example Question #2 : Index Of Refraction

The speed of light in a vacuum, , is calculated to be $2.99 \cdot 10^8 \frac{m}{s}$ . The speed of light in a diamond is calculated to be $1.25 \cdot 10^8 \frac{m}{s}$ . What is the refractive index of diamond?

Possible Answers:

0.418

2.000

2.392

4.500

1.578

Correct answer:

2.392

Explanation:

The definition of refractive index of a medium is the speed of light in a vacuum divided by the speed of light in the medium:

$n = \frac{c}{v}$

We have values for and , so we can plug in our numbers into the equation.

$n&=\frac{2.99\cdot 10^8\ \frac{m}{s}}{1.25\cdot 10^8\ \frac{m}{s}}$

n= 2.392

Because we're dividing two values with the same units, our answer is unitless.

Report an Error

Example Question #1 : Other Optics Principles

Light rays encounter a mystery optical device, resulting in a new distribution of the light waves as shown. Assume the light travels from right to left.

Photo 2 1

What type of reflecting or refracting surface is depicted?

Possible Answers:

Diverging mirror

Plane (flat) mirror

Converging mirror

Diverging lens

Converging lens

Correct answer:

Converging lens

Explanation:

In a converging lens, the light waves pass through it and have their angles altered so that they point closer together than they did before they went through the lens. In the picture, the light waves are diverging from a point until they go into the lens, at which point they no longer diverge from each other. Therefore, this is a converging lens. Because the waves are travelling the same direction the whole time, it can't be the converging or diverging mirrors. If the lens were diverging, they'd be more separated. If it were a plane mirror, the waves would get polarized (they'd have the same phase angle).

Report an Error

Example Question #2 : Other Optics Principles

If a person with near point distance of 20cm observes a fine detailed coin with magnifying glass with an angular magnification of 5, what is the focal length?

Possible Answers:

0.04m

0.4m

0.0025m

0.25m

Correct answer:

0.04m

Explanation:

Use the formula for focal length.

$M=\frac{\frac{h_0}{f}}{\frac{h_0}{N}}$

Solve for focal length by substituting known values.

$f=\frac{N}{M} = \frac{20 { {cm}}}{5}=4 { cm}=0.04{ m}$

Report an Error

Example Question #93 : Optics

Malus' law:

$I=I_0cos^2(\theta)$

Where is the intensity of polarized light that has passed through the polarizer, I_0 is the intensity of polarized light before the polarizer, and $\theta$ is the angle between the polarized light and the polarizer. Malus law

Unpolarized light passes through a polarizer. It then passes through another at angle $25^{\circ}$ to the first. What percentage of the original intensity was the light coming out of the second polarizer?

Possible Answers:

$66.6\%$

$45.7\%$

$82.1\%$

$100\%$

$35.5\%$

Correct answer:

$82.1\%$

Explanation:

Malus law

Using Malus' law.

$I=I_0cos^2(\theta)$

Since the initial light is unpolarized, there will be no intensity lost.

$I=I_0cos^2(25^{\circ})$

I=I_0*.821

$82.1\%=.821$

Report an Error

Example Question #1 : Other Optics Principles

Malus' law:

$I=I_0cos^2(\theta)$

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $35^{\circ}$ to the first, and then another at angle $25^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Possible Answers:

$75.8\%$

None of these

$69.2\%$

$23.6\%$

$55.1\%$

Correct answer:

$55.1\%$

Explanation:

Malus law

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. It is thus necessary to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where I_0 is the initial intensity after the first polarizer.

I_1 is the intensity after the second polarizer.

I_2 is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combining equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(25^\circ)cos^2(35^\circ)$

I_2=.551I_0

Report an Error

Example Question #513 : Ap Physics 2

Malus' law:

$I=I_0cos^2(\theta)$

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $55^{\circ}$ to the first, and then another at angle $65^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Possible Answers:

$15.99\%$

$65.2\%$

$5.88\%$

$4.21\%$

None of these

Correct answer:

$5.88\%$

Explanation:

Malus law

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where I_0 is the initial intensity after the first polarizer.

I_1 is the intensity after the second polarizer.

I_2 is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combining equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(65^\circ)cos^2(55^\circ)$

I_2=.0588I_0

Report an Error

Example Question #6 : Other Optics Principles

Malus' law:

$I=I_0cos^2(\theta)$

Unpolarized light passes through a polarizer. It then passes through another polarizer at angle $5^{\circ}$ to the first, and then another at angle $45^{\circ}$ to the second. What percentage of the original intensity was the light coming out of the third polarizer?

Possible Answers:

$79.7\%$

$59.7\%$

$49.6\%$

None of these

$89.2\%$

Correct answer:

$49.6\%$

Explanation:

Malus law

Use Malus' law.

$I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$I_1=I_0cos^2(\theta_1)$

$I_2=I_1cos^2(\theta_2)$

Where I_0 is the initial intensity after the first polarizer.

I_1 is the intensity after the second polarizer.

I_2 is the intensity after the third polarizer.

$\theta_1$ is the angle between the first and second polarizers.

$\theta_2$ is the angle between the second and third polarizers.

Combine equations:

$I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$I_2=I_0cos^2(45^\circ)cos^2(5^\circ)$

I_2=.496I_0

Report an Error

View Tutors

Rebekah
Certified Tutor

Swarthmore College, Bachelor, Psychology & Linguistics. Pace University, Master's/Graduate, School Psychology.

View Tutors

Joy
Certified Tutor

Boston University, Bachelor in Arts, Conservation Biology.

View Tutors

Maryanne Federici
Certified Tutor

Brown University, Bachelor in Arts, Biology, General. The University of Texas at Austin, Master of Arts, Music.

All AP Physics 2 Resources

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

Popular Subjects

Biology Tutors in Philadelphia, Algebra Tutors in Dallas Fort Worth, ACT Tutors in New York City, Physics Tutors in Denver, GRE Tutors in San Diego, Math Tutors in Houston, Algebra Tutors in San Diego, SAT Tutors in Dallas Fort Worth, Statistics Tutors in Dallas Fort Worth, English Tutors in Houston

Popular Courses & Classes

GMAT Courses & Classes in Boston, ISEE Courses & Classes in Chicago, SAT Courses & Classes in Phoenix, GRE Courses & Classes in Chicago, LSAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Atlanta, SAT Courses & Classes in Atlanta, SAT Courses & Classes in Denver, MCAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Chicago

Popular Test Prep

SAT Test Prep in Houston, LSAT Test Prep in Dallas Fort Worth, GRE Test Prep in Miami, LSAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Houston, GMAT Test Prep in Washington DC, GMAT Test Prep in Boston, MCAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Chicago, GRE Test Prep in Chicago

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 : Index Of Refraction

Example Question #1 : Index Of Refraction

Example Question #1 : Index Of Refraction

Example Question #2 : Index Of Refraction

Example Question #1 : Other Optics Principles

Example Question #2 : Other Optics Principles

Example Question #93 : Optics

Example Question #1 : Other Optics Principles

Example Question #513 : Ap Physics 2

Example Question #6 : Other Optics Principles

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: