## Waves and Optics, solutions

### Problem 1: (C)

Speed of waves on a string

v = (T/m)^{½}, T = v^{2}μ = 90000/100 = 900.

### Problem 2: (A)

Sinusoidal traveling waves

### Problem 3: (E)

Sound waves, Doppler shift

Let v = speed of sound.

f = f_{0}(v - v_{obs})/(v -
v_{s}), where v_{obs} and v_{s}
are not the speeds, but the components of the observer's and the source's
velocity in the direction of the velocity of the sound reaching the observer.

f increases if the source and the observer approach each other and decreases if
they recede from each other.

f = f_{0} v/(v - v_{s}) = [1/(1 - 0.9)]f_{0} = 10 kHz.

### Problem 4: (E)

Images formed by mirrors

For the concave mirror: 1/x_{o }+ 1/x_{i }= 1/f, f = R/2 =
positive. If x_{o} < f then x_{i} is negative, we have a
virtual image.

### Problem 5: (D)

Sinusoidal traveling waves

v = ω/k = (5*10^{14}/5*10^{6}) m/s = 10^{8} m/s.
n = c/v = 3.

### Problem 6: (C)

Interference

We have constructive interference, if the optical path length is the same or
differs by an integer number of wavelength.

n_{1}L_{1} = n_{2}L_{2}. L_{1}/L_{2}
= (n_{2}/n_{1}).

### Problem 7: (C)

Resolving power

Δθ ~ λ/D = c/(fD) = 3*10^{8}/(30*10^{9}*10) = 10^{-3}.

[Uncertainty principle: ΔxΔp_{x} ~ h, Δp_{x}/p = Δθ
~ h/(p Δx). Δx = D, p = h/λ, Δ θ ~ λ/D.]

### Problem 8: (E)

Polarization

For a polarizer I_{transmitted }= I_{0}cos^{2}θ.

A polarizer always absorbs half the intensity of unpolarized light.

I = ½I_{0}cos^{2}θ cos^{2}(π/2
- θ) = (I_{0}/8)sin^{2}(2θ).

### Problem 9: (C)

Single slit diffraction

w sinθ = nλ for the diffraction minima.

### Problem 10: (B)

Thin film interference

When a light wave reflects from a medium with a larger
index of refraction, then the phase shift of the reflected wave with respect to
the incident wave is π (180^{o}). When a light wave reflects
from a medium with a smaller index of refraction, then the phase shift of the
reflected wave with respect to the incident wave is zero.

Here we have a 180^{o} phase shift upon reflection on both
interfaces of the coating. For destructive interference we therefore need

2nd = (k + ½)λ, k = 0, 1, 2, ... .

d = (k + ½)*4000/(2*1.25) = (2k + 1)*800

(2k + 1) =
odd integer