Physics 112 Formulae

area of circular ringof radius r∧thicknessdr=2π r drvolumeof meat of cylindrical shell of length L , radius r ,∧thickness dr=2 π r L dr

volumeof meat of spherical shellof radius r∧thicknessdr=4 π r2 dr

UNIT 2: Potential Energy, Voltage, Capacitors∆ PE=qo ∆V or PE=qo V

PE=U =k q0 q1

r01= 1

4 π εo

q0 q1

r01, V=

k q0

r0= 1

4 π εo

q0

r0(V → 0 asr → ∞);

PE=U=q0

4 π εo( q1

r1+

q2

r2+

q3

r3+…), V= 1

4 π εo(q1

r1+

q2

r2+

q3

r3+…)(V →0 asr→∞ );

PE=q0

4 π εo∫a

b dqr

, V= 14 π εo

∫i

f dqr

(V → 0 asr → ∞ );

PEb−PE a=−∫a

b

F⃗ ∙ d l⃗=−∫a

b

q0 E⃗ ∙ d l⃗ ; V b−V a=−∫a

b

E⃗ ∙ d l⃗;

Infinite sheet: V=|⃗E|dQ=CV ; for parallel plates, C=

A εo

d;

PE=12

C V 2= Q2

2 C=1

2QV ;

C eq=C1+C2+…;1

Ceq= 1

C1+ 1

C2+…;

Cnew=K Cold ;

Fundamental Constantsρair (density of air) = 1.2 kg/m3 (20oC, 1 atm); speed of sound in air = 344 m/s (20oC, 1 atm) e = 1.602176487 X 10-19 C k = 1/4πεo = 8.987551787 X 109 N m2/C2

me = 9.109382 X 10-31 kg εo = 8.854187817 X 10-12 C2/Nm2

mp = 1.672621 X 10-27 kg μo = 4π X 10-7 T m/A

mn = 1.674927 X 10-27 kg c= 1√ε o μo

=299,792,458 m /s (exact)

UNIT 1: Charges, Coulomb’s Law, Gauss’s Law

Coulomb’s Law: |F⃗12|=k q1 q2

r122 = 1

4 π ε o

q1 q2

r122 ; F⃗=q E⃗

Point: |E⃗|= kqr2 = 1

4 π εo

qr2

Line: |E⃗|=2 kλr

= 12 π ε o

λr

Plane: |E⃗|=2πkσ= σ2 εo

∨ σεo

if conducting plane

ΦE=∫ E⃗ ∙d A⃗=∫E⏊dA=∫E cos φ dA; Gauss’s Law: ∮ E⃗ ∙ d A⃗=qencl

ε o=4 πk qins

UNIT 3: Current, Resistance, Introduction to CircuitsI=dQ

dt=n|q|vd A; |J⃗|= I

A=n|q|vd; J⃗=nq v⃗d;

Ohm’s Law: E⃗=ρ J⃗ and V=IR; R=ρLA ;

P=IV =I 2 R=V 2

R;

Req=R1+R2+R3+…; 1

Req= 1

R1+ 1

R2+ 1

R3+…;

∑ I ¿=∑ I out; ∑closed loop

∆ V=0;

Charging: Q (t )=Qf(1−e

−tRC )=C V o

(1−e−tRC ); I ( t )=

V o

Re

−tRC ;

Discharging: Q ( t )=Qo e−tRC ; I (t )=

Q o

RCe

−tRC;

UNIT 4: Magnetic FieldsF⃗=q v⃗× B⃗ ¿F⃗=I L⃗ × B⃗ (F=ILB sinθ );ΦB=BA cos θ (in certain situations)d F⃗=I d L⃗× B⃗ (dF=I dL B sin θ ) ;ΦB=∮ B⃗⋅ d A⃗=0 (closed surface)|μ⃗|=NIA ; τ⃗= μ⃗× B⃗(τ=μB sinθ)

B⃗=μo

4 πq v⃗ ×r̂

r2 ; d B⃗=μo

4 πI d l⃗ ×r̂

r 2 ;

∮ B⃗ ⋅d l⃗=μo ( Ithru+ I d ) where I d=εo

d ΦE

dt;

B=μo I2πr

;B=μo∋¿μoNL

I ;B=μo I2 r

;

V=−d ΦB

dt; E=cB ;c= 1

√ε o μo

;

I=12

ε oc Emax2=

Emax Bmax

2 μo=

Emax2

2 μo c=1

2 √ εo

μoEmax

2;

UNIT 5: Intro to Waves

∂2 y

∂ x2 =1v2

∂2 y∂ t 2 ⇒ y ( x , t )=A sin (kx∓ωt∓φ )∨A cos ( kx∓ωt∓φ ) ;

k=2πλ

;ω=2 πf ; f = 1T

;v=fλ=ωk

;

v=√ FT

μ; v=√ B

ρ=√ γRT

M;

I= PA

; I=pmax

2

2 ρv=

pmax2

2√ ρB;

SIL∈dB= (10dB ) log10( I10−12W /m2 );

f n=n f 1=nv2 L

; f n=(2 n−1 ) v

4 L;PLD=nλ ;PLD=(n+ 1

2 ) λ ;

f L=v ± v L

v∓vsf s; f b=|f 1− f 2|

UNIT 6: Light and Optics

Emax=c Bmax ;c= 1√εo μo

; I=12

εo c Emax2=

Emax Bmax

2 μo=

Emax2

2μo c=1

2 √ εo

μoEmax

2

c=fλ ; v= cn

; λnew=λold

n;visible : λ≈ 400−700 nm;

specular reflection :θ i=θr ; Snel l ' s Law :n1 sin θ1=n2sin θ2 ;Law of Malus :I =I o cos2 ϕ;

1f= (n−1 )( 1

R1− 1

R2 );no

o+

ni

i=

ni−no

R;m=

−no ini o

;

1f=1

i+ 1

o; f = R

2;m=−i

o;

converging: f > 0; diverging: f < 0; “normal” near point = 20cm; “normal” far point = ∞;

w sin θ=mλ ;d sinθ=nλ ;d sin θ=(n+ 12 ) λ ;

2 t=nλ∨(n+ 12 ) λ ;

D sin θ=1.22 λ ;

A

Io

B20o

?

Physics 112 – Test #616 December 2016

Name _____________________________

Directions: This is a CLOSED BOOK test, and you may only use the calculator and cheat sheet provided. SHOW ALL OF YOUR REASONING! You can only get partial credit for an incorrect answer if you show your reasoning. You may have as much time as you like to finish this exam.

1. You stand 2 m away from a 100 W light bulb.a) (5 pts) How far away from the bulb must you stand so that the intensity at

your location is ½ what it was when you were 2 m away?b) (5 pts) How far away from the light bulb must you stand so that the maximum

electric field from the light waves is ½ what it was when you were 2 m away?

2. (10 pts) Light polarized vertically with intensity Io is incident on polarizer A, which has its polarization axis oriented at an angle of 20o with respect to the vertical. What angle with the vertical must the polarization axis of polarizer B be so that the light exiting B has an intensity of (1/8) Io?

coin looks like it is here

coin

cannot see coincan see coin

4.0 cm 1.5 cm

7.0 cm8.0 cm

0.5 cm

3. (16 pts) Looking into the empty cup (left), Gertrude cannot see the coin at the bottom of the cup. Fill the cup with an unknown fluid as shown (right), and Gertrude sees the coin in the place indicated due to light refraction. Determine the index of refraction of the fluid. The coin is in the same location in both pictures below.

GaP (n = 3.5)

AIR (n = 1.0)

AIR (n = 1.0)

4. (16 pts) White light reflects at normal incidence from the top and bottom surface of a gallium phosphide (GaP) plate (n = 3.5). There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 560 nm. What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 784 nm?

5. (16 pts) You shine laser light with wavelength 500.0 nm through two rectangular slits. The pattern that forms on the wall (4.00 meters from the slits) looks like this:

The distance between the center of one spot and the center of the spot next to it is 1.00 cm (the spots are evenly spaced, but note that there are some missing ones). Determine the center to center distance between the slits AND the width of the two slits (both have the same width).

central spot

6. (20 pts) An object is placed in front of two lenses as shown. Determine the location AND overall magnification of the final image. The answers might surprise you.

20.00 cm

lensf1 = +10 cmobject

lensf1 = –10 cm

4 13 cm

7. (12 pts) Robert Austin (Florida Polytechnic University) won a recent Physics Today contest with his story about what kind of physics projects we’d be doing 100 years from now. He guessed that we’d be using space probes equipped with lasers to smooth out shallow, converging mirror-like surfaces on small asteroids. If this is done to billions of asteroids, we could turn the entire asteroid belt into a gigantic telescope with an effective aperture diameter of 5 AU (6.0 × 108 km). Determine the smallest sized features that could possibly be discerned with this telescope on the (possibly habitable) exoplanet Gliese 832c, which is 16 light years (1.51×1014 km) away. You may use 500 nm for λ. (Another way to put this: How close can two lights on the surface of Gliese 832c be and still be resolved as two lights by this huge telescope?) The answer will probably surprise you.

Read about it and other wild ideas in the December 2016 issue of Physics Today.

dφ θ

BONUS BRAIN BUSTER!!! (+5 points) Laser light of wavelength λ is incident at an angle φ on a pair of slits separated by a distance d as indicated in the figure. Determine a formula that gives the angles θ which indicate where constructive interference takes place on a distant screen.

Credit will only be given for correct reasoning. If you are able to guess the correct answer but give no correct reasoning, you will receive no extra credit. However, if you give some correct reasoning that would (or does) lead to the correct answer, at least some credit will be given.

________________________________________________

By signing my name above, I affirm that this test represents my work only, without aid from outside sources. In all aspects of this course I perform with honor and integrity.