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Name: ________________________________________________________________

Phys1502 – Exam 3, Practice for 1502-2018 Instructions

This exam will assess:

your knowledge of the physics concepts covered in class to date;

your graphical reasoning ability; and

your physics problem solving skills.

Please read each question thoroughly.

Only calculators and pens/pencils are allowed on your desk. No cell phones or scrap paper.

I pledge my honor that I have neither given nor received aid on this work.

Signed _______________________________________________

Directions: The test contains multiple choice and short answer questions. For the MC,

make sure to circle the correct answer and explain your reasoning when asked. Note,

questions may have more than one correct answer. Show all of your work!

For short answer questions, partial credit will be awarded based on having a correct

picture, identifying the appropriate strategy (which physics concept is needed), and

progress towards the solution. Good luck!

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Total Score = _______

1.) (10 pts) Consider the circuit shown below. The switch was in the ‘a’ position for a long time and then

switched to the ‘b’ position. All of the statements below refer to times after this switch to the ‘b’ position. The

maximum charge stored on the capacitor is Qmax. The maximum current observed after switching to the ‘b’

position is Imax.

For each of the following statements indicate whether it is

true or false. Make sure to explain your reasoning.

a. The charge on the capacitor will decrease with an

exponential decay. False, the charge on the capacitor will

oscillate. The capacitor would certainly start discharging,

but the inductor fights the change providing the restoring

“force” to create an oscillation.

b. The total energy stored in the circuit oscillates from zero to

a maximum of

1

2

Qmax2

C1

2LImax

2 . False, the total energy is constant. The total energy=(Qmax)^2/(2C)

=L(Imax)^2/2

c. The maximum current through the inductor L occurs when the capacitor is fully charged. False, the

current and charge are out of phase. The maximum current through the inductor occurs with the charge

on the capacitor is 0

d. The time to reach zero charge on the capacitor would increase if you replaced the inductor with one with

twice as many turns. True. The frequency of oscillation would decrease with an increase in inductance.

Since period is 1/frequency, the period would increase.

2.) (4 pts) Which of the following graphs correctly shows the current passing through the resistance R as a

function of time? (The time when the switch is closed is defined as t=0.)

1 is the correct answer. Before the switch is closed, no

current travels through the resistor. Once the swtich is closed, the inductor fights the change in current

through the circuit. It never wins, so eventually the current will be equivalent to the circuit without an inductor

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3. (5 points) When light passes from water (index of refraction n = 1.33) into air (n = 1),

a. the wavelength increases and the frequency is unchanged.

b. the wavelength decreases and the frequency is unchanged.

c. the wavelength is unchanged and the frequency increases.

d. the wavelength is unchanged and the frequency decreases.

e. both the wavelength and the frequency change.

Going from high to low index of refraction => going from low speed to high speed. A is the correct answer.

4. (5 points) You hold a metal spoon about 25 cm from your face and look at your own reflection off the back of

the spoon (the surface you view is convex with focal length 5cm). Is the image you see …

virtual or real? (circle one)

inverted or upright? (circle one)

1/d0+1/di=1/f

1/25+1/di=-1/.05 (focus is behind the mirror, so considered -). The image will be virtual (image distance is

-) and upright (m=-di/d0>0).

5.) (5 pts) A parallel-plate capacitor is made of two square plates, each with an area of 1m2, that are separated

by 1mm. An external source is adding charge to the capacitor such that the charge on each plate increases with

time as Q=1(C/s2)*t2. What is the displacement current passing through an imaginary loop located between the

two plates with an area of 0.1m2?

Displacement current refers to the modification needed for Ampere’s Law.𝑰𝒅𝒊𝒔𝒑 = 𝜺𝟎𝒅𝚽𝑬

𝒅𝒕=

𝜺𝟎𝒅(∫ 𝑬𝒅𝑨)

𝒅𝒕=𝜺𝟎

𝑨𝒍𝒐𝒐𝒑𝒅(𝑬)

𝒅𝒕

E we know from the field due to a parallel plate capacitor

𝑬 =𝝈

𝜺𝟎=

𝑸

𝑨𝒑𝒍𝒂𝒕𝒆𝒔𝜺𝟎=

𝟏

𝜺𝟎∗ 𝒕𝟐=> 𝑰𝒅𝒊𝒔𝒑 =. 𝟏 ∗ 𝟐𝒕 =. 𝟐𝒕

6. (5 points) In an old style dot matrix printer, an array of dots is used to form printed characters. If the dots

are close enough together, they cannot be resolved individually by the eye and therefore appear to form solid

lines. Suppose that the pupil of the eye has a diameter of 2.0mm in bright light and that the printer ink is

yellow-green (wavelength 563 nm in air), that the material in the eye has an average refractive index of 1.36

and that the printed page is intended to be read at a distance of 0.5m. What is the maximum separation of the

dots that would still be perceived as a straight line?

This is a question about the Rayleigh resolution criteria. Too close, and the two dots are indistinguishable. As

they separate, we can distinguish them. The resolution criteria finds that the central maximum of one spot falls

directly on the first destructive interference ring of the second spot. Sin(theta)=1.22*lambda in eye/pupil.

From geometry, the tan(theta)=separation distance/distance to eye. Since this is a small angle, we can set

tan(theta)~sin(theta). The maximum separation distance is then 0.17 mm.

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7. (5pts) The figure to the right shows two coils wound around an iron ring. Current appears in the second coil

the moment

a.) the battery is continually connected with the switch closed

b.) the battery is continually disconnected with the switch open

c.) the battery is disconnected by opening the switch

d.) the battery is connected by closing the switch

to drive current you need a changing B flux, both c and d would

cause a change in magnetic flux

8. (5 points) You are working with Vanderbilt Student

Volunteers for Science to set up new physics labs to take

to middle schools. You build a light tight box with a

narrow, 100 micron slit as the opening. You place a

screen inside the box 70 cm from the opening, such that

600 nm light passes through the slit is incident on the

screen. Which of the following diagrams best matches

what you would expect to see on the screen? On the

diagram of choice, indicate the vertical distances for the

feature(s) that you would expect to see.

You would see a diffraction pattern (A). The minimum occur at 0.42, 0.84, and 1.26 cm.

A thin oil film (thickness of 405 nm) floats on the surface of a water puddle. The indices of refraction of oil and

water are 1.5 and 1.333 respectively. The surface of the oil is illuminated from above with white light.

9.) (10 pts) Considering only light in the visible band (400nm to 800 nm as measured in air), what wavelengths

are strongly reflected. They are:

Interference problem:

Light travels from air to oil to water. At each surface there is a reflected wave and a transmitted wave. When we look at the surface, we see the combination of the two main reflected waves, since these are the waves that arrive at our eye. To make the diagram more clear, I draw the waves as separated horizontally and the reflected wave at an angle, but actually they are at the same position and the reflected waves follow the law of reflection so in this case, travel straight back

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Reflected wave 1 bounced off an interface going from

low to high index of refraction, Reflected wave 1

must be shifted by pi

Reflected wave2 bounced off an interface going from

high to low index of refraction, Reflected wave 2 is

not phase shifted from reflection, but it does have an

increase in path of 2D (it travels through the film

twice). This results in a phase change of

2*pi*2D/lambda, where lambda is the wavelength in

the film = n1/n2*lambda in air

For constructive interference (see a most strongly reflected wave, the phase difference is an integer multiple of

2*pi

m can equal any positive integer, so we have to check what m’s will give constructive interference in the visible.

M =0 corresponds to 2430 nm light. Not in the visible. It would be a strongly reflected wavelength, but we

wouldn’t see it. M=1 corresponds to 810 nm light. That is just outside the visible range. M=2 corresponds to

486 nm light – we see that!, M=3 corresponds to 347 nm light. So the best answer is 486 nm light

10.) (5pts) What wavelengths in the visible band are most weakly reflected?

For destructive interference (see a most strongly reflected wave, the phase difference is an odd multiple of pi or

(2m+1)*pi where m is any + integer

Next step, determined what phase changes occur for each reflected wave

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m can equal any positive integer, so we have to check what m’s will give constructive interference in the visible.

M =0 corresponds to 1215 nm light. M=1 corresponds to 607.5 nm light. That is inside visible range. M=2

corresponds to 405nm light – we see that!, M=3 corresponds to 303 nm light. So the best answer is 608 and

405 nm light will be missing from the reflection

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11. (10 points). The questions below refer to a very narrow beam of light (for example, a laser beam) that can

be represented by a single ray. The light is initially traveling from left to right in a transparent medium of index

of refraction n1, and incident on a second transparent medium of index of refraction n2. The reflected and

refracted rays are as shown in the diagrams below. (If either is missing, it means there is no reflected or no

refracted ray.) Answer each of the questions below with one of the following choices, A through F. Make sure

to explain why.

A. Only if n2 < n1,

B. Only if n2 > n1,

C. Only if n2 = n1,

D. Can happen with A

or C.

E. Never possible.

F. Always possible

regardless of the

relative sizes of the

indexes of refraction

a.) For which conditions, A through F could the rays be

as shown in the figure to the left?

A , transmitted wave is bent towards the normal

b.) For which conditions, A through F could the rays be

as shown in the figure to the left?

B, transmitted wave is bent away from the normal

c.) For which conditions, A through F could the rays be as shown in the

figure to the left?

E, not possible n1=n2 if no refraction, but then there should be no

reflected wave

d.) For which conditions, A through F could the rays be as shown in

the figure to the left?

Here there is no refracted wave, so the incident angle >= critical angle.

This happens when n1>n2, A

e.) For which conditions, A through F could the rays be as shown in

the figure to the left?

Bent towards the normal, so n2>n1, B

n1

n2

incident

n1

n2

n1

n2

n1

n2

reflected

n1

n2

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12. (15 points) A green laser, 525 nm, traveling in air moving right to left down an optics bench. The beam has

a diameter of 4 mm. The laser beam is incident on a polarizer with a transmission axis exactly vertical. You

use a power meter with an opening diameter of 3 cm to measure the RMS power of the wave on the incident

side of the polarizer as 150 mW and after the polarizer as 115 mW. Write the complete E&B equations for the

light after the polarizer. How would you change these equations to describe the incident light (before the

polarizer)? Take the x axis to be the direction parallel to the optics bench and y axis to indicate the vertical

axis.

After the polarizer, the E field will be aligned with the vertical (j hat direction).

The wave is moving in –x (right to left) down the optics bench. Right hand rule gives that B must point into the

page.

The wavelength of light is 525 nm. It is traveling in air so speed is 3e8 m/s, hence the period of oscillation is

1.75 femtoseconds (10^-15s).

Power is 150mW, Intensity is Power /Area, so the Intensity of this beam is .150/(pi*.002^2)=11936 W/m^2

Intensity is related to field by

Irms=4*pi*10^-7/(2*3e8)*E^2 => E=2.4e9 N/C.

B=E/C=8T

E=(2.4e9 N/C jhat)*cos(2*pi*x/(525nm)+2*pi*t/1.75fs)

B=(-8 T khat)*cos(2*pi*x/(525nm)+2*pi*t/1.75fs)

Before the polarizer, the light has a greater intensity, so the electric and magnetic fields would be stronger. It

is also rotated by an angle. I could find the angle via Iafter=Ibefore*cos(theta)^2, then adjust the polarization

of E and B.

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13. (15 points). A patient comes to see you and says that she is having difficulty performing her job as a

robotics engineer because she can no longer focus on close-up objects for detail work. You find that she cannot

focus on objects closer than 40 cm from her eyes. Please show your work. Partial credit is available; full

work is needed to earn full credit.

A) Write a prescription for the glasses she would need (in Diopters) to be able to focus on objects as close as the

ideal/normal near point of 25 cm. Assume that the lenses would be worn 1.8 cm from the front surface of her

cornea and that the distance from her eye lens to retina is 2.0 cm.

The patient is far sighted, she has difficulty focusing close up.

near point (the closest she can see) is 40 cm instead of the typical near point of 25 cm. To see an object that

is 25cm from her eye, she needs the glasses to create a virtual image located where she can see (40 cm from

her eye). Then her eye will take over and focus that image on her retina.

1/d0+1/di=1/f

1/(.25-.018)-1/(.40-.018)=1/f

1/f=1.7 Diopters

B) Determine the power range for her eye without the glasses.

Can see far away clearly her far point is infinity.

1/infinity+1/.02=50 Diopters

Her nearpoint is 40 cm

1/.4+1/02=52.5 Diopters.

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14. (15 points) A 0.5 cm arrow is placed midway between a concave mirror with a radius of curvature of 10.0

cm and a convex (converging) lens with a focal length of 16.0 cm. The total distance between the mirror and

lens is 25.0 cm. You are looking through the lens. You see two predominant images of the arrow.

A.) Find the location of the two images seen by an observer looking through the lens. Measure the image

distances with respect to the center of the lens.

B.) What is the height and orientation of each of the images formed?

To get full credit, you must show your tracing and show that it matches up with the numbers you calculate for

image position and magnification. The next two pages have grids to help you lay out your ray tracing carefully.

Focus of the concave mirror f= R/2=5cm.

First image is formed by the light that reflects off the object and passes thru the lens

Image distance to lens 12.5 cm. focal length of lens is 16 cm.

1/(12.5)+1/di=1/16 => di=-57cm. Forms a virtual image that is 57 cm away from the lens. This would be

upright and magnified by 57/12.5=4.56

Second image is formed by the light that reflects off the object then reflects off the mirror and then encounters

the lens. This light is equivalent to emerging from the image of the object in the mirror then seen by the lens.

1/12.5+1/di=1/5 => di=8.33 forming a real image in front of the mirror which is 25-8.33=16.67 cm from the

lens, just outside of the focal length.

1/16.67+1/di=1/16

This light then forms a real image 398 cm in front of the lens

To Check with ray tracing…

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