PHY222 Midterm #2 – Step-by-Step Physics Guidance

Q1A. Monochromatic light falls on two narrow slits that are 0.05 mm apart. On a screen 4.00 m away, the first order fringe (bright spot) is 6.00 cm from the middle of the interference pattern. What is the wavelength of the light?

Background

Topic: Double-Slit Interference

This question tests your understanding of the double-slit interference pattern and how to relate the geometry of the setup to the wavelength of light.

Key Terms and Formulas

• Double-slit equation: $d \sin\theta = m\lambda$

• Small angle approximation: $\sin\theta \approx \tan\theta \approx y/L$ (for small $\theta$)

• $d$ = slit separation, $\lambda$ = wavelength, $m$ = order of fringe, $y$ = distance from center to fringe, $L$ = distance to screen

Step-by-Step Guidance

1. Identify the known values: $d = 0.05\ \text{mm} = 5.0 \times 10^{-5}\ \text{m}$, $L = 4.00\ \text{m}$, $y = 6.00\ \text{cm} = 0.060\ \text{m}$, $m = 1$ (first order).

2. Write the double-slit equation: $d \sin\theta = m\lambda$.

3. Use the small angle approximation: $\sin\theta \approx y/L$.

4. Substitute the values into the equation to solve for $\lambda$ (but do not calculate the final value yet): $\lambda = \frac{d\,y}{mL}$.

Try solving on your own before revealing the answer!

Final Answer: 7.5 × 10-7 m

$\lambda = \frac{(5.0 \times 10^{-5}\ \text{m})(0.060\ \text{m})}{1 \times 4.00\ \text{m}} = 7.5 \times 10^{-7}\ \text{m}$

This matches choice [d].

Q1B. Monochromatic light of wavelength 500 nm falls on two narrow slits that are 0.04 mm apart and this generates an interference pattern on a distant screen. At what angle (in degrees) does the first minima occur in the interference pattern?

Background

Topic: Double-Slit Interference Minima

This question tests your understanding of the condition for minima (dark fringes) in a double-slit interference pattern.

Key Terms and Formulas

• Minima condition: $d \sin\theta = (m + 0.5)\lambda$ for minima, where $m = 0$ for the first minima.

• Convert all units to SI: $\lambda = 500\ \text{nm} = 5.0 \times 10^{-7}\ \text{m}$, $d = 0.04\ \text{mm} = 4.0 \times 10^{-5}\ \text{m}$.

Step-by-Step Guidance

1. Identify the known values: $\lambda = 5.0 \times 10^{-7}\ \text{m}$, $d = 4.0 \times 10^{-5}\ \text{m}$, $m = 0$ for the first minima.

2. Write the minima condition: $d \sin\theta = (m + 0.5)\lambda$.

3. For the first minima, $m = 0$, so $d \sin\theta = 0.5\lambda$.

4. Rearrange to solve for $\sin\theta$: $\sin\theta = \frac{0.5\lambda}{d}$.

Try solving on your own before revealing the answer!

Final Answer: 0.72°

$\sin\theta = \frac{0.5 \times 5.0 \times 10^{-7}}{4.0 \times 10^{-5}} = 0.00625$

$\theta = \arcsin(0.00625) \approx 0.72^\circ$

This matches choice [a].

Q1C. A light beam traveling through air (n=1.00) reflects perpendicularly off the top and bottom surfaces of a thin layer of oil (n=1.50) floating on water (n=1.33). What is the thinnest layer of oil that will give rise to constructive interference at a wavelength of 700 nm?

Background

Topic: Thin Film Interference

This question tests your understanding of constructive interference in thin films, considering phase changes upon reflection.

Key Terms and Formulas

• Constructive interference for thin film (with both reflections causing a $\pi$ phase shift): $2 n t = m \lambda$ (for $m=1$ for thinnest film)

• $n$ = refractive index of oil, $t$ = thickness, $\lambda$ = wavelength in vacuum

Step-by-Step Guidance

1. Identify the indices: $n_{\text{air}} = 1.00$, $n_{\text{oil}} = 1.50$, $n_{\text{water}} = 1.33$.

2. Both reflections (air-oil and oil-water) cause a $\pi$ phase shift, so use the constructive condition: $2 n t = m \lambda$.

3. For the thinnest film, use $m = 1$.

4. Rearrange to solve for $t$: $t = \frac{\lambda}{2n}$.

Try solving on your own before revealing the answer!

Final Answer: 233 nm

$t = \frac{700\ \text{nm}}{2 \times 1.50} = 233\ \text{nm}$

This matches choice [b].

Q1D. A soap film (n=1.35) of thickness 150 nm is surrounded by air (n=1.00) on both sides. If light of a certain wavelength is beamed at the soap film such that it hits the film perpendicular to its surface, what is the longest wavelength of light for which you’ll see no reflection?

Background

Topic: Thin Film Interference (Destructive)

This question tests your understanding of destructive interference in thin films, considering phase changes upon reflection.

Key Terms and Formulas

• Destructive interference for thin film (one $\pi$ phase shift): $2 n t = m \lambda$ (for $m=1$ for longest wavelength)

• $n$ = refractive index of film, $t$ = thickness, $\lambda$ = wavelength in vacuum

Step-by-Step Guidance

1. Identify the indices: $n_{\text{air}} = 1.00$, $n_{\text{film}} = 1.35$.

2. Since the film is surrounded by air, only the top reflection causes a $\pi$ phase shift. Use the destructive condition: $2 n t = m \lambda$.

3. For the longest wavelength, use $m = 1$.

4. Rearrange to solve for $\lambda$: $\lambda = 2 n t$.

Try solving on your own before revealing the answer!

Final Answer: 405 nm

$\lambda = 2 \times 1.35 \times 150\ \text{nm} = 405\ \text{nm}$

This matches choice [e].

Q2A. The displacement of a transverse wave on a string is given by $D(x,t) = 4 \sin(5\pi x - 6\pi t)$, where all numbers are in SI units. What is the amplitude, wavenumber, angular frequency, and speed of the traveling wave?

Background

Topic: Wave Properties

This question tests your ability to identify wave parameters from the mathematical form of a wave equation.

Key Terms and Formulas

• General wave equation: $D(x,t) = A \sin(kx - \omega t)$

• Amplitude $A$, wavenumber $k$, angular frequency $\omega$, wave speed $v = \omega / k$

Step-by-Step Guidance

1. Compare the given equation to the standard form: $D(x,t) = 4 \sin(5\pi x - 6\pi t)$ vs. $A \sin(kx - \omega t)$.

2. Identify the amplitude $A$ (the coefficient in front of the sine function).

3. Identify the wavenumber $k$ (the coefficient of $x$ inside the sine).

4. Identify the angular frequency $\omega$ (the coefficient of $t$ inside the sine).

5. Write the formula for wave speed: $v = \omega / k$.

Try solving on your own before revealing the answer!

Final Answer:

• Amplitude: 4 m

• Wavenumber: $5\pi$ rad/m

• Angular frequency: $6\pi$ rad/s

• Speed: $v = \frac{6\pi}{5\pi} = 1.2$ m/s

Q2B. What is the period, frequency, wavelength and direction of propagation of the traveling wave?

Background

Topic: Wave Properties

This question tests your understanding of how to extract period, frequency, and wavelength from the wave equation, and determine the direction of propagation.

Key Terms and Formulas

• Period: $T = \frac{2\pi}{\omega}$

• Frequency: $f = \frac{1}{T} = \frac{\omega}{2\pi}$

• Wavelength: $\lambda = \frac{2\pi}{k}$

• Direction: If the argument is $(kx - \omega t)$, the wave moves in the $+x$ direction.

Step-by-Step Guidance

1. Recall $\omega = 6\pi$ rad/s and $k = 5\pi$ rad/m from the previous part.

2. Calculate the period: $T = \frac{2\pi}{\omega}$.

3. Calculate the frequency: $f = \frac{\omega}{2\pi}$.

4. Calculate the wavelength: $\lambda = \frac{2\pi}{k}$.

5. Determine the direction of propagation based on the sign in the argument.

Try solving on your own before revealing the answer!

Final Answer:

• Period: $T = \frac{2\pi}{6\pi} = \frac{1}{3}$ s

• Frequency: $f = 3$ Hz

• Wavelength: $\lambda = \frac{2\pi}{5\pi} = 0.4$ m

• Direction: $+x$ direction

Q2C. What is the transverse velocity and acceleration of the string at a position x=3m and time t=2s?

Background

Topic: Wave Motion – Derivatives

This question tests your ability to find the velocity and acceleration of a point on a string by taking derivatives of the displacement function.

Key Terms and Formulas

• Transverse velocity: $v(x,t) = \frac{\partial D}{\partial t}$

• Transverse acceleration: $a(x,t) = \frac{\partial^2 D}{\partial t^2}$

• Given: $D(x,t) = 4 \sin(5\pi x - 6\pi t)$

Step-by-Step Guidance

1. Find the first time derivative for velocity: $v(x,t) = \frac{\partial}{\partial t} [4 \sin(5\pi x - 6\pi t)]$.

2. Find the second time derivative for acceleration: $a(x,t) = \frac{\partial^2}{\partial t^2} [4 \sin(5\pi x - 6\pi t)]$.

3. Plug in $x = 3$ m and $t = 2$ s into your expressions for $v(x,t)$ and $a(x,t)$ (but do not compute the final values yet).

Try solving on your own before revealing the answer!

Final Answer:

• Transverse velocity: $v(3,2) = -24\pi \cos(5\pi \times 3 - 6\pi \times 2)$ m/s

• Transverse acceleration: $a(3,2) = 144\pi^2 \sin(5\pi \times 3 - 6\pi \times 2)$ m/s2

Q3C. What is the wavelength of the standing wave shown above (open tube, 30 cm long)?

Background

Topic: Standing Waves in Open Tubes

This question tests your understanding of the relationship between the length of an open tube and the wavelength of the standing wave patterns it supports.

Key Terms and Formulas

• For an open tube: $L = n \frac{\lambda}{2}$, where $n$ is the harmonic number.

• For the fundamental (first harmonic): $L = \frac{\lambda}{2}$

• For the second harmonic: $L = \lambda$

Step-by-Step Guidance

1. Identify the length of the tube: $L = 30$ cm $= 0.30$ m.

2. Determine which harmonic is shown (based on the number of nodes and antinodes in the diagram).

3. Use the appropriate formula to relate $L$ and $\lambda$ for the given harmonic.

4. Rearrange to solve for $\lambda$ in terms of $L$.

Try solving on your own before revealing the answer!

Final Answer: 30 cm

For the second harmonic in an open tube, $L = \lambda$ so $\lambda = 30$ cm.

This matches choice [e].

Q3D. Assume the speed of sound in air is 343 m/s. Imagine your friend is at rest and playing this open pipe with the displacement wave shown above. If you run toward your friend at 10 m/s, what frequency will you hear?

Background

Topic: Doppler Effect

This question tests your understanding of the Doppler effect for sound when the observer is moving toward a stationary source.

Key Terms and Formulas

• Doppler effect (observer moving toward stationary source): $f' = f \left(\frac{v + v_o}{v}\right)$

• $v$ = speed of sound, $v_o$ = speed of observer, $f$ = source frequency

• Frequency from wavelength: $f = v/\lambda$

Step-by-Step Guidance

1. Calculate the frequency of the sound produced by the pipe: $f = v/\lambda$ (use $\lambda$ from previous part).

2. Plug in the values: $v = 343$ m/s, $\lambda = 0.30$ m.

3. Use the Doppler effect formula for a moving observer: $f' = f \left(\frac{v + v_o}{v}\right)$, with $v_o = 10$ m/s.

4. Set up the calculation but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 3944 Hz

$f = \frac{343}{0.30} = 1143.33$ Hz$

$f' = 1143.33 \times \frac{343 + 10}{343} \approx 1176$ Hz$

This matches choice [c].

Q4A. Sketch the adiabatic compression process from C to A and determine the ratio P2/P1. The latter should be given as a number to two significant digits.

Background

Topic: Thermodynamics – Adiabatic Processes

This question tests your understanding of adiabatic processes for an ideal gas and how to relate pressures and volumes using the adiabatic condition.

Key Terms and Formulas

• Adiabatic process for ideal gas: $P V^\gamma = \text{constant}$

• For a monoatomic ideal gas: $\gamma = C_P/C_V = 5/3$

• Given: $V_C = 2V_1$, $V_A = V_1$, $P_C = P_2$, $P_A = P_1$

Step-by-Step Guidance

1. Write the adiabatic condition: $P_C V_C^\gamma = P_A V_A^\gamma$.

2. Substitute the given values: $P_2 (2V_1)^{5/3} = P_1 (V_1)^{5/3}$.

3. Rearrange to solve for $P_2/P_1$.

4. Express the ratio as a number to two significant digits (but do not calculate the final value yet).

Try solving on your own before revealing the answer!

Final Answer: 0.32

$P_2/P_1 = (1/2)^{5/3} \approx 0.32$