Optics and Electricity
1. All of the electrical outlets in a room are connected in a single parallel circuit. The circuit is equipped with a 20-A fuse and the voltage is 120 V.A) What is the maximum power that can be supplied by the outlets without blowing the fuse? B)How many 1200 W appliances can be plugged into the sockets without blowing the fuse?
2.The light bulb used in an overhead projector has a resistance of 48 Ohm. What is the current through the bulb when it is operating on 120 V? Which costs more, running a 1,200 W hair dryer for 3 min or leaving a 60‐W lamp on overnight (1h)?
3. A current of 15 A flows through an electric heater operating on 120 V. What is the heater’s resistance? Compute the power consumption of the electric heater in kW? How much energy does it use in 600 min (=10 hours)? (Express the answers in both units: kWh and Joule)
4. A 5-Hz continuous wave travels on a Slinky. If the wavelength is 0.3m, what is the speed of waves on the Slinky?
5. A 1000 Hz sound travels through pure oxygen. The wavelength of the sound is measured to be 0.65m. What is the speed of sound in oxygen?
6. A steel cable with total length 20m and mass 100kg is connected to two poles. The tension in the cable is 2,000 N, and the wind makes the cable vibrate with a frequency of 2 Hz. Calculate the wavelength of the resulting wave on the cable.
7. The focal length of diverging lens is negative. If f= -20cm fro a particular diverging lens, where will the image be formed of an object located 50cm to the left of the lens on the optical axis? What is the magnification of the image?
8. The equation connecting s, p, and f for a simple lens can be employed for spherical mirrors, too. A concave mirror with a focal length of 8cm forms an image of a small object placed 10cm in front of the mirror. Where will this image be located?
9. What are the frequencies of the first four harmonics of middle C (261.6Hz)? Approximately how many times louder is a 110 dB sound than a 80dB sound?
1.
A) the electrical power is by definition the current multiplied by the voltage. Therefore
$P = U*I = 20*120 =2400 W$
B) The power as is defined is a scalar value. It means it is adding algebraically. The number of appliances that can be connected in parallel is
$N = P_{tot}/P =2400/1200 =2$
2.
A) The Ohm law says that the Resistance of a device is equal to the voltage (on that device) over the current (through the device).
$R = U/I, I = U/R = 120/48 = 2.5 A$
B) One is paying the energy needed to that particular device. (Power = Energy/time)
$E1 = P1*t1 =1200*3*60 =216000 J$
$E2 =P2*t2 =60*1*60*60 =216000 J$
The cost is the same for both devices.
3.
The ohm law says that the Resistance of a device is equal to the voltage (on that device) over the current (through that device).
$R = U/I = 120/15 = 8$ Ohm
The electrical power is by definition the current multiplied to the voltage, or the squared current multiplied by the resistance or the squared voltage over the resistance.
$P = U*I = R*I^2 =U^2/R$
$P = 120*15 = 1800 W$
The power is by definition the Energy over the time. ($P = E/t$) Hence
$E =P*t = 1800*600*60 = 64800000 J$
4.
The relation between the wavelength of a wave ($\lambda$) , its speed V, and its time period T (or its frequency F ) is given by
$\lambda = V*T = V/F$
$V =F*\lambda =5*0.3=1.5 m/s$
5.
The relation between the wavelength of a sound , its speed Vs, and its time period T (or its frequency F ) is given by
$\lambda = Vs*T = Vs/F$
$Vs =F*\lambda =1000*0.65 =650 m/s$
6.
The speed of waves in a string is directly proportional to the square root of the tension in the string and inversely proportional to the square root of the linear mass density of the string.
$V = \sqrt{T/(m/l}) =\sqrt{T/\rho} =\sqrt{200/(100/20)} =\sqrt{200/5}=6.32 m/s$
The relation between the wavelength ($\lambda$), and the speed of waves is
$\lambda = V*T = V/F$ where t is the time period and F the frequency
$\lambda = 6.32/2 = 3.16 m/s$
7.
a) The equation of the lens is
$1/x1 +1/x2 = 1/f$
where $x1$ is the position of the object, positive to the left of the lens
and $x2$ is the position if the image positive to the right positive to the lens
and f is positive for a converging lens and negative for a diverging lens
$1/x2 =1/f-1/x1$
$1/x2 =-1/20 -1/50 =-70/100$
$x2 =-100/70 =-1.429$ cm, therefore the image is to the left of the lens
The magnification is simply
$y2/y1 = -x2/x1 = 1.429/50 =0.02958$, the image is smaller and non-inverted
8.
b) the equation of a concave mirror is
$1/p1 +1/p2 =1/f$,
where p1 and p2 are the positions of the object, respectively image (positive if both to the left of the mirror) and f is positive for a concave mirror
$1/p2 =1/f -1/p1 =1/8-1/10 =2/80$
p2 =40 cm , to the left of the mirror, real image
the magnification is simply $Y2/Y1 =-p2/P1 =-40/10 = -4$, the image is bigger and inverted
9.
a) $F1 = 261.6$ Hz, fundamental or so called first harmonic
Second harmonic harmonic frequency $F2 = 261.6*2 =522.4$ Hz
Third harmonic frequency $F3 =261.2*3 = 783.6$ Hz
Fourth harmonic freq. $F4 = 261.2*4 = 1044.8$ Hz
Fifth harmonic freq. $F5 =261.2*4 = 1306$ Hz
b) By definition the intensity I of a sound in decibels is given by the relation
$dB = 10*log(I/I0)$, where $I0$ is the reference intensity and $I0=10^{-12} W/m^2$ and the log is taken in base 10.
Therefore$110 =10*log (I1/I0)$
$80 =10*log(I2/I0)$
$10^{11} = I1/I0$
$10^8 =I2/I0$
and by dividing one has
$10^{(11-8)} = I1/I2$
$I1/I2 =100$
$I1 =100*I2$
the 110 dB sound is 100 times louder than the 80 dB sound
