PHYS 222 SI Exam Review

What to do to prepare

• Review all clicker questions, but more importantly know WHY

• Review quizzes

• Make sure you know what all the equations do, and when to use them

• These equations are used exclusively in LRC circuits

• These equations are what let you find the major constants that do not change with time.

• Remember, capital letters are not time dependent.

𝑉 𝑟𝑚𝑠=𝑉√2, 𝐼 𝑟𝑚𝑠=

𝐼√2

• These equations are used to determine the root-mean square voltage and current in an LRC circuit

𝑖=𝐼 𝑐𝑜𝑠𝜔𝑡• These two equations assume that the current

in an LRC circuit is a maximum at t=0.• These equations tell the voltage and current

as a function of time.• To find remember to add the appropriate

phase constants to the cos term.

𝐼 𝑟𝑎𝑣=2𝜋 𝐼

• Not mentioned in class really

AC Current section on the Equation Sheet

• All the capitalized letters do not change with time.

• For example…none of these change with time:– V, I,

• To find the time dependent voltage and current, multiply by the appropriate time dependent equation.

How to determine v(t), i(t),

• The current phasor is parallel with the phasor.• leads , and lags • The voltage – I.E. if then

What happens to a circuit at resonant frequency?

• The voltage phasor is parallel with the current phasor (this does not usually happen)

• (recall that both do not depend on time)• The sum of the voltages across the inductor

and the capacitor equals 0.

Example #1

• At t=0, the current in the circuit is a maximum of 3 A.

• Then…

• Also note that without doing any math you know that and – Make sure you understand why.

Example #2

• Let’s say that in an LRC circuit,

• Also suppose that you’ve calculated the phase angle to be, and

• Then

Example #3

• They tell you

• To calculate I, find Z, then use • To calculate , first find I, then use

• Finally, if you need time dependence, add the appropriate phase shift to the cos or sin.

𝑝=𝑖𝑣• Equation relating power, current, and voltage

𝑃𝑎𝑣𝑒𝑟𝑎𝑔𝑒=12 𝐼𝑉𝑐𝑜𝑠𝜙=𝐼 𝑟𝑚𝑠𝑉 𝑟𝑚𝑠𝑐𝑜𝑠𝜙=𝐼 𝑟𝑚𝑠2 𝑅

• Average power in an LRC AC circuit

𝑉 2

𝑉 1=𝑁 2

𝑁 1

• Equations used to convert a voltage inside a transformer

𝑉 1 𝐼 1=𝑉 2 𝐼2• Current and voltage in a transformer

𝐸 (𝑥 , 𝑡 )=𝐸𝑚𝑎𝑥 cos (𝑘𝑥−𝜔𝑡 ) �̂�• The equations for electromagnetic radiation,

or in other words light• Note that the direction of propagation is +x.• Also note that

𝑐=1

√𝜖0𝜇0• Speed of light related to two constants

𝑢=12𝜖0𝐸2+

12𝜇0

𝐵2=𝜖0𝐸2=𝐵2𝜇0

• Energy density

𝑃=𝐹𝐴=

1𝐴𝑑𝑝𝑑𝑡 =

𝑆𝑐=

𝐸𝐵𝜇0𝑐

• Equations relating the radiation pressure of an electromagnetic wave to the poynting vector and E and B.

𝑺=1𝜇0

𝑬×𝑩

• Poynting vector.• Note that the poynting vector is perpendicular

to both E and B

𝐼=𝑆𝑎𝑣𝑒𝑟𝑎𝑔𝑒=𝐸𝑚𝑎𝑥𝐵𝑚𝑎𝑥

2𝜇0=𝐸𝑚𝑎𝑥2

2𝑐𝜇0=12 √ 𝜖0𝜇0 𝐸𝑚𝑎𝑥

2 =12𝜖0𝑐 𝐸𝑚𝑎𝑥

2

• The intensity of electromagnetic radiation, related to the E field.

• Equations relating the speed of light c, the wavelength of light , the frequency of light the angular frequency , and the wave number .

𝑛=𝑐𝑣

• The speed of light in a medium of index of refraction .

• For example, in glass the speed of light is not equal to m/s, but instead it’s equal to m/s (

𝜃𝑖=𝜃 𝑟𝑅𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛• The equations for reflection and refraction

𝐼=𝐼𝑚𝑎𝑥cos (𝜙 )2

• The equation for intensity of light through a diffraction grating.

•

sin (𝜃𝑐𝑟𝑖𝑡 )=𝑛2𝑛1

• The equation used to find the critical angle between two interfaces.

• At angles equal to or greater than the critical angle, refracted rays stop going through the second medium. Instead they undergo total internal reflection.

• Sometimes light coming from one direction onto an interface doesn’t have a critical angle, but if the light goes the other direction, then the critical angle exists.

𝑡𝑎𝑛𝜃𝑃=𝑛2𝑛1

• Equation used to find Brewster’s Angle, also known as the Polarization Angle .

• This angle is where reflection stops happening.

1𝑠 +

1𝑠′

=1𝑓

• The master equation for both lenses AND mirrors

• ALWAYS…s>0. (the distance from the real object to the vertex of the mirror, or from the real object to the lens)

• For MIRRORS…– f>0 if the mirror is concave, if the mirror is convex, then

f<0. Also f=R/2.– s’>0 if the image is on the same side as the outgoing rays

• For LENSES…– f>0 if the lens is more converging, otherwise f<0 if the

lens is more diverging

𝑚=𝑦 ′𝑦 =− 𝑠

′

𝑠• Magnification caused by a mirror or lens

𝑓 h𝑠𝑝 𝑒𝑟𝑖𝑐𝑎𝑙𝑚𝑖𝑟𝑟𝑜𝑟=𝑅2

• Focal point of a spherical mirror

1𝑓 =(𝑛−1 )( 1𝑅1 −

1𝑅2 )

• “Lensmaker’s Equation”

𝑛𝑎𝑠 +

𝑛𝑏𝑠′

=𝑛𝑏−𝑛𝑎𝑅

• Look up

𝑚= 𝑦 ′𝑦 =−

𝑛𝑎𝑠 ′

𝑛𝑏 𝑠

• Spherical fish bowls is the main application of this equation

𝑀=𝜃′𝜃

• M is the angular magnification of a telescope.• Recall that and

𝑑𝑠𝑖𝑛𝜃=𝑚 𝜆• This is the equation for two-source

interference, used to find where the bright fringes are.

• To find the dark fringes, replace m with (m+1/2)

𝑑𝑠𝑖𝑛𝜃=(𝑚+ 12 )𝜆

• Destructive interference.

𝑦=𝑚𝑅𝜆𝑑

• Assuming small angles the equation on the previous slide gives this result, where R is the distance between the slits and the screen, d is the separation of the slits, m is an integer, and y is the height above the central interference maximum.

𝐼=𝐼𝑚𝑎𝑥cos ( 𝜃2 )2

• In single slit diffraction, you can use this equation to find the intensity as a function of the angle.

𝜙=2𝜋 (𝑟2−𝑟1 )𝜆 =2𝜋 𝑑𝑠𝑖𝑛𝜃𝜆

• For two sources of waves, this equation finds the phase angle between them, depending on the location of the point where you measure the interference of the two waves.

2 𝑡=𝑚𝜆𝑛• Thin film destructive reflection

2 𝑡=(𝑚+ 12 )𝜆𝑛

• Thin film constructive reflection• Recall that is the wavelength in that medium

of index of refraction

𝑎𝑠𝑖𝑛 𝜃=𝑚 𝜆• Single-slit diffraction• a is the width of the slit.• This equation gives diffraction minima• To get maxima, replace m with (m+1/2)

𝛽=2𝜋 𝑎𝑠𝑖𝑛 𝜃𝜆• This equation gives for diffraction, which can

then be used to get the intensity of light at various points.

• Intensity difference caused by single-slit diffraction.

• is calculated from a different equation

• This equation combines the effects of two-slit interference and the diffraction caused by each of the slits independently.

𝑅=𝜆Δ 𝜆=𝑁𝑚

• Used to find the chromatic resolving power for a diffraction grating

𝑠𝑖𝑛𝜃1=1.22𝜆𝐷

• This is used to find the resolving power of a small circular hole of diameter D.

• is the location of the first minimum.

2𝑑𝑠𝑖𝑛𝜃=𝑚𝜆• Used to find the location of maxima for

diffraction gratings

Answer: D

B, A

C, A

A,B

• B

A,C

B

• C,B

A