PHY222 Midterm #3 Study Guidance: Diffraction, Optics, and Polarization

Q1A. What is the angular separation (in degrees) between the first two minima (m=1 and m=2) of the diffraction pattern for a single slit 0.5 mm wide illuminated by 600 nm light?

Background

Topic: Single-Slit Diffraction

This question tests your understanding of how light diffracts through a single slit and how to calculate the angular positions of minima in the diffraction pattern.

Key formula:

$a \sin\theta = m\lambda$

Where:

• $a$ = slit width (in meters)

• $\theta$ = angle from the center of the slit

• $m$ = order of the minimum (integer, $m=1,2,...$)

• $\lambda$ = wavelength of light (in meters)

Step-by-Step Guidance

1. Convert the slit width and wavelength to meters: $a = 0.5 \text{ mm} = 5 \times 10^{-4} \text{ m}$, $\lambda = 600 \text{ nm} = 6 \times 10^{-7} \text{ m}$.

2. Use the formula $a \sin\theta = m\lambda$ to find $\theta_1$ for $m=1$ and $\theta_2$ for $m=2$.

3. Set up the equations: $\sin\theta_1 = \frac{1 \times \lambda}{a}$ and $\sin\theta_2 = \frac{2 \times \lambda}{a}$.

4. Calculate $\theta_1$ and $\theta_2$ using the inverse sine function (but do not compute the final values yet).

Try solving on your own before revealing the answer!

Final Answer: 0.069°

$\theta_1 = \arcsin\left(\frac{6 \times 10^{-7}}{5 \times 10^{-4}}\right)$, $\theta_2 = \arcsin\left(\frac{2 \times 6 \times 10^{-7}}{5 \times 10^{-4}}\right)$, and the angular separation is $\theta_2 - \theta_1 \approx 0.069^\circ$.

This uses the single-slit diffraction formula and the correct conversion of units.

Q1B. What is the intensity of light at an angle of 0.5° from the center of two slits (each 0.1 mm wide, separated by 0.7 mm, illuminated by 400 nm light), in terms of the peak intensity $I_0$?

Background

Topic: Double-Slit Diffraction and Interference

This question tests your ability to combine the effects of diffraction (from each slit) and interference (from the separation between slits) to find the intensity at a specific angle.

Key formulas:

$I(\theta) = I_0 \left[ \frac{\sin(\beta)}{\beta} \right]^2 \cos^2(\alpha)$

• $\beta = \frac{\pi a \sin\theta}{\lambda}$ (diffraction envelope)

• $\alpha = \frac{\pi d \sin\theta}{\lambda}$ (interference term)

• $a$ = slit width, $d$ = slit separation, $\lambda$ = wavelength

Step-by-Step Guidance

1. Convert all values to meters: $a = 1 \times 10^{-4} \text{ m}$, $d = 7 \times 10^{-4} \text{ m}$, $\lambda = 4 \times 10^{-7} \text{ m}$.

2. Calculate $\sin\theta$ for $\theta = 0.5^\circ$.

3. Compute $\beta$ and $\alpha$ using the formulas above.

4. Set up the intensity formula $I(\theta) = I_0 \left[ \frac{\sin(\beta)}{\beta} \right]^2 \cos^2(\alpha)$, but do not plug in the final numbers yet.

Try solving on your own before revealing the answer!

Final Answer: 0.0062I0

After plugging in the values and calculating, the intensity at $0.5^\circ$ is $0.0062I_0$.

This combines both the diffraction envelope and interference pattern.

Q1C. How far apart must two shiny pennies be on the ground to be resolved by the JWST (6.5 m diameter mirror, 407 km altitude, 600 nm wavelength)?

Background

Topic: Angular Resolution and Rayleigh Criterion

This question tests your understanding of the resolving power of a telescope and how to use the Rayleigh criterion to determine the minimum separation of two objects that can be distinguished.

Key formula:

$\theta_{min} = 1.22 \frac{\lambda}{D}$

• $\lambda$ = wavelength (in meters)

• $D$ = diameter of the mirror (in meters)

• $\theta_{min}$ = minimum resolvable angle (in radians)

Step-by-Step Guidance

1. Convert $\lambda = 600 \text{ nm} = 6 \times 10^{-7} \text{ m}$, $D = 6.5 \text{ m}$.

2. Calculate $\theta_{min}$ using $\theta_{min} = 1.22 \frac{\lambda}{D}$.

3. Find the minimum separation $s$ using $s = h \cdot \theta_{min}$, where $h = 407,000 \text{ m}$ (altitude).

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

Try solving on your own before revealing the answer!

Final Answer: 4.6 cm

Using the Rayleigh criterion and the altitude, the minimum separation is $s = h \cdot \theta_{min} \approx 4.6 \text{ cm}$.

This shows the resolving power of the JWST for ground targets.

Q1D. What is the angular separation (in degrees) between the red and green principal maxima for the first order (m=1) in a diffraction grating with 10,000 lines/cm, using 700 nm and 500 nm lasers?

Background

Topic: Diffraction Grating

This question tests your ability to use the diffraction grating equation to find the angular positions of maxima for different wavelengths and calculate their separation.

Key formula:

$d \sin\theta = m\lambda$

• $d$ = grating spacing (in meters)

• $\theta$ = angle of principal maxima

• $m$ = order (here, $m=1$)

• $\lambda$ = wavelength (in meters)

Step-by-Step Guidance

1. Calculate $d = \frac{1}{10,000 \text{ lines/cm}} = 1 \times 10^{-4} \text{ cm} = 1 \times 10^{-6} \text{ m}$.

2. Set up $\sin\theta_{red} = \frac{700 \times 10^{-9}}{d}$ and $\sin\theta_{green} = \frac{500 \times 10^{-9}}{d}$ for $m=1$.

3. Calculate $\theta_{red}$ and $\theta_{green}$ using the inverse sine function (but do not compute the final values yet).

4. Find the angular separation $\Delta\theta = \theta_{red} - \theta_{green}$.

Try solving on your own before revealing the answer!

Final Answer: 14.4°

The angular separation between the red and green maxima is $\Delta\theta \approx 14.4^\circ$.

This uses the grating equation and correct unit conversions.

Q1E. What is the minimum number of polarizers needed to reduce the output intensity to 10% of the input intensity, if each polarizer is rotated 10° from the previous?

Background

Topic: Polarization and Malus's Law

This question tests your understanding of how the intensity of polarized light changes as it passes through multiple polarizers, each rotated by a fixed angle.

Key formula:

$I = I_0 \cos^2\theta$ (Malus's Law for one polarizer)

For $n$ polarizers, each rotated by $\Delta\theta$, $I = I_0 \left( \cos^2\Delta\theta \right)^n$

• $I_0$ = initial intensity

• $\Delta\theta$ = angle between each polarizer

• $n$ = number of polarizers

Step-by-Step Guidance

1. Set up the equation: $I = I_0 \left( \cos^2 10^\circ \right)^n$

2. Set $I = 0.1 I_0$ and solve for $n$.

3. Take the logarithm of both sides to isolate $n$.

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

Try solving on your own before revealing the answer!

Final Answer: 53

Solving $0.1 = (\cos^2 10^\circ)^n$ gives $n \approx 53$ polarizers.

This uses Malus's Law and logarithms to find the minimum number needed.