College Physics Quiz: Diffraction

20 questions · exam conditions

0:00

DiffractionQuestion 1 of 20

A diffraction grating with $N = 500$ lines per millimeter is illuminated by monochromatic light of wavelength $\lambda = 550 \text{ nm}$ . The number of complete orders of diffraction (including the central maximum) that can be observed is:

← Back to quizzes

College Physics Quiz

College Physics Quiz: Diffraction

Practice Diffraction in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Diffraction, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

3
4
5
7 (correct answer)
9

Explanation: When you encounter diffraction grating problems asking for the number of observable orders, you're determining how many bright fringes can physically exist given the grating spacing and wavelength constraints. The key insight is that diffraction orders are limited by the grating equation:

d \sin \theta = m\lambda

, where

d

is the line spacing,

m

is the order number, and

\theta

is the diffraction angle. Since

\sin \theta

cannot exceed 1, the maximum order occurs when

\sin \theta = 1

, giving us

m_{max} = \frac{d}{\lambda}

. First, find the grating spacing:

d = \frac{1}{500 \text{ lines/mm}} = 2.0 \times 10^{-6} \text{ m} = 2000 \text{ nm}

Now calculate the maximum order:

m_{max} = \frac{2000 \text{ nm}}{550 \text{ nm}} = 3.64

Since

m

must be an integer, the highest observable order is

m = 3

. However, diffraction produces orders on both sides of the central maximum (

m = 0

), so you observe:

m = -3, -2, -1, 0, +1, +2, +3

. That's 7 complete orders total. Choice A (3) counts only the positive orders, ignoring the central maximum and negative orders. Choice B (4) might represent

m_{max} + 1

but misses the symmetry. Choice C (5) could be

2m_{max} - 1

but uses incorrect reasoning. Choice D (7) correctly accounts for all observable orders:

2m_{max} + 1 = 2(3) + 1 = 7

. Remember: for diffraction gratings, always check both the maximum order calculation and the symmetry around the central maximum to count all observable orders.

Question 2

A single slit of width $a = 0.50 \text{ mm}$ is illuminated by coherent light of wavelength $\lambda = 600 \text{ nm}$ . On a screen located $2.0 \text{ m}$ from the slit, the distance from the center of the diffraction pattern to the first dark fringe is approximately:

$1.2 \text{ mm}$
$2.4 \text{ mm}$ (correct answer)
$3.6 \text{ mm}$
$4.8 \text{ mm}$
$6.0 \text{ mm}$

Explanation: When you encounter single-slit diffraction problems, you're dealing with the interference pattern created when coherent light passes through a narrow opening. The key relationship governs where dark fringes (minima) occur on a distant screen. For single-slit diffraction, the condition for the first dark fringe is

a \sin \theta = \lambda

, where

a

is the slit width,

\theta

is the angle from the central axis, and

\lambda

is the wavelength. For small angles (which applies here),

\sin \theta \approx \tan \theta = y/L

, where

y

is the distance from the center to the first dark fringe and

L

is the screen distance. Substituting:

a \cdot \frac{y}{L} = \lambda

Solving for

y

y = \frac{\lambda L}{a} = \frac{(600 \times 10^{-9} \text{ m})(2.0 \text{ m})}{0.50 \times 10^{-3} \text{ m}} = \frac{1.2 \times 10^{-6}}{0.50 \times 10^{-3}} = 2.4 \times 10^{-3} \text{ m} = 2.4 \text{ mm}

Choice A (1.2 mm) likely comes from forgetting to account for the screen distance properly or making an arithmetic error. Choice C (3.6 mm) might result from using the wrong multiple of the basic diffraction formula. Choice D (4.8 mm) could come from doubling the correct answer, perhaps confusing the distance to the first minimum with the width of the central maximum. Remember that single-slit diffraction problems follow the pattern

y = \frac{n\lambda L}{a}

for the

n

-th dark fringe. Always check that your small-angle approximation is valid (

\theta < 15°

) and keep track of units carefully.

Question 3

Sound waves of frequency $f = 1000 \text{ Hz}$ (speed $v = 340 \text{ m/s}$ ) pass through a doorway of width $w = 1.0 \text{ m}$ . A person standing $10 \text{ m}$ from the doorway and $6 \text{ m}$ to one side of the central axis will experience:

maximum intensity because the path difference is negligible compared to the wavelength
minimum intensity because they are positioned at the first diffraction minimum
reduced but significant intensity due to diffraction spreading the sound around the doorway (correct answer)
no sound because they are in the geometric shadow of the doorway opening
maximum intensity because sound diffracts completely around obstacles of this size

Explanation: When sound waves encounter an opening like a doorway, diffraction occurs - the waves bend and spread out beyond the geometric boundaries of the opening. The key is comparing the wavelength to the opening size and determining the observer's position relative to diffraction patterns. First, calculate the wavelength:

\lambda = v/f = 340/1000 = 0.34 \text{ m}

. Since the doorway width (1.0 m) is about 3 times the wavelength, significant diffraction will occur, but the opening isn't so small that sound spreads uniformly in all directions. For single-slit diffraction, the first minimum occurs at angle

\theta

where

\sin \theta = \lambda/w = 0.34/1.0 = 0.34

, giving

\theta \approx 20°

. The person's position creates an angle of

\tan^{-1}(6/10) \approx 31°

from the central axis - well beyond the first minimum but still within the diffracted sound field. Option A is wrong because path differences aren't negligible at this distance and angle. Option B incorrectly identifies this position as the first minimum - that occurs at a much smaller angle (20°). Option D applies geometric optics thinking, ignoring diffraction entirely; sound doesn't create sharp shadows like light because its wavelength is comparable to everyday openings. Option C correctly recognizes that while the person isn't in a minimum or maximum, they're in a region where diffracted sound reaches them with reduced but measurable intensity. Remember: when the opening size is a few wavelengths across, expect significant diffraction with alternating regions of high and low (but not zero) intensity extending well beyond the geometric shadow.

Question 4

A laser beam passes through a single slit and creates a diffraction pattern on a screen. If the slit is gradually made narrower while keeping all other parameters constant, which statement correctly describes the changes in the diffraction pattern?

The central maximum becomes narrower and brighter, while the secondary maxima become more pronounced
The central maximum becomes wider and dimmer, while the secondary maxima become less pronounced (correct answer)
The central maximum becomes wider but maintains the same brightness, while secondary maxima become more pronounced
The central maximum becomes narrower but maintains the same brightness, while the total pattern shifts toward the slit
The entire pattern scales proportionally larger while maintaining the same relative intensities throughout

Explanation: When you encounter single-slit diffraction problems, focus on the inverse relationship between slit width and diffraction effects. As the slit becomes narrower, diffraction becomes more pronounced because the opening approaches the wavelength of light. Here's what happens when you narrow the slit: The central maximum becomes wider because the first minima (dark fringes) move farther from the center. The angular position of the first minimum is given by

\sin \theta = \lambda/a

, where

a

is the slit width. As

a

decreases,

\theta

increases, spreading the central maximum. Additionally, the central maximum becomes dimmer because the same total light energy is now spread over a larger area. The secondary maxima also become less pronounced relative to the background because the overall intensity distribution becomes more spread out. Looking at the wrong answers: Choice A incorrectly suggests the central maximum narrows and brightens—this would happen if the slit got wider, not narrower. Choice C correctly identifies that the central maximum widens but wrongly claims brightness stays constant; energy conservation means spreading light over a larger area reduces brightness per unit area. Choice D incorrectly states the central maximum narrows and mentions a pattern shift, which doesn't occur when only slit width changes. Remember this key relationship: narrower slit equals wider, dimmer diffraction pattern. This inverse relationship appears frequently in wave optics problems, so whenever you see slit width changing, immediately think about how the diffraction angle and intensity distribution will respond oppositely.

Question 5

White light passes through a single slit of width $a = 2.0 \times 10^{-6} \text{ m}$ . On a screen $1.5 \text{ m}$ away, the first-order dark fringes for red light ( $\lambda = 700 \text{ nm}$ ) and violet light ( $\lambda = 400 \text{ nm}$ ) are separated by a distance of approximately:

$0.11 \text{ mm}$
$0.23 \text{ mm}$ (correct answer)
$0.34 \text{ mm}$
$0.45 \text{ mm}$
$0.56 \text{ mm}$

Explanation: When you encounter single-slit diffraction problems with multiple wavelengths, you're dealing with how different colors of light create interference patterns at different positions due to their wavelength differences. For single-slit diffraction, dark fringes occur when

a \sin \theta = m\lambda

, where

a

is the slit width,

m

is the order (1 for first-order), and

\lambda

is the wavelength. For small angles,

\sin \theta \approx \tan \theta = y/L

, so the position of dark fringes is

y = \frac{m\lambda L}{a}

. For the first-order dark fringe (

m = 1

), the positions are:

Red light: $y_{red} = \frac{(1)(700 \times 10^{-9})(1.5)}{2.0 \times 10^{-6}} = 0.525 \text{ mm}$
Violet light: $y_{violet} = \frac{(1)(400 \times 10^{-9})(1.5)}{2.0 \times 10^{-6}} = 0.300 \text{ mm}$

The separation is

0.525 - 0.300 = 0.225 \text{ mm} \approx 0.23 \text{ mm}

, confirming answer B. Answer A (0.11 mm) would result from incorrectly using half the wavelength difference. Answer C (0.34 mm) might come from calculation errors or using wrong values for the screen distance. Answer D (0.45 mm) could result from adding the positions instead of finding their difference, or doubling the correct answer. Study tip: In diffraction problems, longer wavelengths (red) always diffract more than shorter ones (violet), creating patterns farther from the center. Always subtract the smaller position from the larger when finding separations.

Question 6

A student observes that when laser light passes through a very thin wire (treated as an opaque obstacle), a diffraction pattern appears on a screen behind the wire. This phenomenon is explained by:

Babinet's principle, which shows that the diffraction pattern from a thin wire is identical to that from a slit of the same width
Fresnel diffraction, where the wire creates a shadow with alternating bright and dark fringes due to path differences from edge waves (correct answer)
total internal reflection of light waves around the cylindrical surface of the wire creating interference patterns
Rayleigh scattering from the wire material dispersing different wavelengths at different angles behind the obstacle
geometric optics principles, where light rays bend around the wire due to the gravitational field of the wire material

Explanation: When you encounter diffraction problems involving obstacles like wires or edges, you're dealing with wave optics where light bends around barriers and creates interference patterns. The correct answer is B because Fresnel diffraction explains exactly what happens when light encounters a straight edge or thin obstacle. As laser light passes the wire, waves diffract around both edges of the wire, creating a shadow region behind it. These edge waves travel different path lengths to reach points on the screen, causing constructive and destructive interference that produces the alternating bright and dark fringes observed. This is classic Fresnel (near-field) diffraction behavior. Answer A incorrectly invokes Babinet's principle. While this principle does relate the diffraction patterns of complementary apertures (like a slit and a wire of the same width), it doesn't explain the physical mechanism creating the pattern – that's still Fresnel diffraction doing the work. Answer C misapplies total internal reflection, which occurs when light travels from a denser to less dense medium at angles greater than the critical angle. Light doesn't enter the opaque wire, so there's no interface for total internal reflection to occur. Answer D confuses the phenomenon with Rayleigh scattering, which involves tiny particles scattering different wavelengths by different amounts (like why the sky is blue). This has nothing to do with diffraction around macroscopic obstacles. Remember: when light encounters edges or obstacles comparable to its wavelength, think Fresnel diffraction. The key signature is interference patterns created by path differences between waves diffracting around the obstacle's edges.

Question 7

Two students debate whether diffraction is more pronounced for a narrow slit or a wide slit. Student A claims that narrow slits show more diffraction because "the waves have to squeeze through a smaller opening." Student B claims that wide slits show more diffraction because "more waves can fit through to interfere." Which student is correct and why?

Student A is correct; narrower slits cause greater angular spreading because diffraction angle is inversely proportional to slit width (correct answer)
Student B is correct; wider slits allow more wave fronts to contribute to the interference pattern, creating more pronounced diffraction effects
Both students are partially correct; narrow slits show more angular spreading while wide slits show more secondary maxima in the pattern
Neither student is correct; diffraction depends only on the wavelength of light, not on the slit width or number of contributing waves
Student A's reasoning is wrong but conclusion is correct; Student B's reasoning is correct but conclusion is wrong about the overall effect

Explanation: When analyzing diffraction phenomena, you need to understand how slit width affects the angular spreading of waves. Diffraction occurs when waves encounter an obstacle or aperture comparable to their wavelength, causing them to bend and spread out. The key relationship governing single-slit diffraction is that the angular width of the central maximum is inversely proportional to the slit width. Mathematically, the angle to the first minimum is given by

\sin \theta = \frac{\lambda}{a}

, where λ is the wavelength and a is the slit width. As the slit becomes narrower, θ increases, meaning the diffraction pattern spreads out more dramatically. Student A correctly identifies this inverse relationship. When waves pass through a narrow opening, they experience greater angular spreading because the constraining effect of the small aperture forces the waves to bend more significantly around the edges. Option B misunderstands the physics. While wider slits do allow more wave fronts to pass through, this actually reduces diffraction effects rather than enhancing them. More wave fronts mean the light behaves more like geometric optics with sharper shadows. Option C incorrectly suggests both students are right. Though narrow slits do show more angular spreading and wide slits can produce more secondary maxima, Student B's fundamental reasoning about "more pronounced diffraction" is wrong—wide slits show less overall diffraction. Option D is completely incorrect since diffraction clearly depends on both wavelength and slit width, as shown in the fundamental equation. Remember this inverse relationship: narrow slits create wide diffraction patterns, while wide slits create narrow patterns. This counterintuitive relationship appears frequently on physics exams.

Question 8

A student uses a laser pointer to create a diffraction pattern by shining the beam through a single human hair. The hair has a diameter of approximately $50 \text{ μm}$ and the laser wavelength is $650 \text{ nm}$ . If the screen is placed $1.0 \text{ m}$ from the hair, the spacing between the first dark fringes on either side of the central bright fringe is approximately:

$13 \text{ mm}$
$26 \text{ mm}$ (correct answer)
$39 \text{ mm}$
$52 \text{ mm}$
$65 \text{ mm}$

Explanation: When you encounter single-slit diffraction problems, you're dealing with the wave nature of light bending around obstacles. The key relationship is that smaller apertures create wider diffraction patterns. For single-slit diffraction, the angular position of the first dark fringe is given by

\sin \theta = \frac{\lambda}{a}

, where

\lambda

is the wavelength and

a

is the slit width (hair diameter). Since the screen is relatively far from the hair, we can use the small angle approximation:

\sin \theta \approx \tan \theta = \frac{y}{L}

, where

y

is the distance from the center to the first dark fringe and

L

is the screen distance. Combining these:

\frac{y}{L} = \frac{\lambda}{a}

, so

y = \frac{\lambda L}{a}

Substituting values:

y = \frac{(650 \times 10^{-9} \text{ m})(1.0 \text{ m})}{50 \times 10^{-6} \text{ m}} = 0.013 \text{ m} = 13 \text{ mm}

The spacing between the first dark fringes on either side is

2y = 26 \text{ mm}

, confirming answer B. Answer A (13 mm) represents just the distance from center to one dark fringe, not the spacing between both sides. Answer C (39 mm) might result from using the wrong diffraction formula or incorrect unit conversions. Answer D (52 mm) could come from doubling the correct answer again or using an inappropriate diffraction equation. Remember: in single-slit diffraction, always distinguish between the distance to one dark fringe versus the spacing between symmetric fringes on opposite sides of the central maximum.

Question 9

A diffraction grating is used to separate white light into its component colors. If the grating spacing is $d = 1.67 \times 10^{-6} \text{ m}$ and the screen is located $2.0 \text{ m}$ away, the linear separation on the screen between the first-order maxima of red light ( $\lambda = 700 \text{ nm}$ ) and blue light ( $\lambda = 450 \text{ nm}$ ) is approximately:

$0.15 \text{ m}$
$0.30 \text{ m}$
$0.37 \text{ m}$ (correct answer)
$0.45 \text{ m}$
$0.60 \text{ m}$

Explanation: When you encounter diffraction grating problems, you're dealing with the interference of light waves passing through multiple slits. The key relationship is the grating equation:

d \sin \theta = m\lambda

, where

d

is the grating spacing,

\theta

is the diffraction angle,

m

is the order, and

\lambda

is the wavelength. For first-order maxima (

m = 1

), you need to find the angle for each color, then calculate their positions on the screen. For red light:

\sin \theta_r = \frac{700 \times 10^{-9}}{1.67 \times 10^{-6}} = 0.419

, so

\theta_r = 24.7°

. For blue light:

\sin \theta_b = \frac{450 \times 10^{-9}}{1.67 \times 10^{-6}} = 0.269

, so

\theta_b = 15.6°

. The linear position on the screen is

y = L \tan \theta

, where

L = 2.0 \text{ m}

. For red:

y_r = 2.0 \times \tan(24.7°) = 0.92 \text{ m}

. For blue:

y_b = 2.0 \times \tan(15.6°) = 0.56 \text{ m}

. The separation is

0.92 - 0.56 = 0.36 \text{ m}

, which rounds to 0.37 m. Choice A (0.15 m) likely results from using the small angle approximation incorrectly. Choice B (0.30 m) might come from calculation errors or using sine instead of tangent for screen position. Choice D (0.45 m) could result from incorrect angle calculations or using the wrong grating spacing. Remember: for grating problems, use the grating equation to find angles, then

\tan \theta

(not

\sin \theta

) to find linear positions on distant screens.

Question 10

An adjustable slit is illuminated by coherent light of wavelength $\lambda = 500 \text{ nm}$ . When the slit width is $a_1 = 2.0 \times 10^{-6} \text{ m}$ , the angular position of the first diffraction minimum is $\theta_1$ . If the slit is adjusted so that the angular position of the first minimum becomes $\theta_2 = 2\theta_1$ , the new slit width $a_2$ is:

$1.0 \times 10^{-6} \text{ m}$ (correct answer)
$4.0 \times 10^{-6} \text{ m}$
$0.5 \times 10^{-6} \text{ m}$
$8.0 \times 10^{-6} \text{ m}$
$1.4 \times 10^{-6} \text{ m}$

Explanation: When you encounter single-slit diffraction problems, you're dealing with the relationship between slit width and the angular positions of diffraction minima. The key equation is

a \sin \theta = m\lambda

, where

a

is the slit width,

\theta

is the angular position of the

m

th minimum, and

\lambda

is the wavelength. For the first minimum (

m = 1

), we have

a_1 \sin \theta_1 = \lambda

and

a_2 \sin \theta_2 = \lambda

. Since both equal

\lambda

, we can set them equal:

a_1 \sin \theta_1 = a_2 \sin \theta_2

. Given that

\theta_2 = 2\theta_1

, we substitute:

a_1 \sin \theta_1 = a_2 \sin(2\theta_1)

. Using the double-angle identity,

\sin(2\theta_1) = 2\sin \theta_1 \cos \theta_1

, so:

a_1 \sin \theta_1 = a_2 \cdot 2\sin \theta_1 \cos \theta_1

Dividing both sides by

\sin \theta_1

a_1 = 2a_2 \cos \theta_1

Therefore:

a_2 = \frac{a_1}{2\cos \theta_1}

From the original condition,

\sin \theta_1 = \frac{\lambda}{a_1} = \frac{500 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.25

This gives

\cos \theta_1 = \sqrt{1 - \sin^2 \theta_1} = \sqrt{1 - 0.0625} = 0.968

a_2 = \frac{2.0 \times 10^{-6}}{2(0.968)} = 1.0 \times 10^{-6} \text{ m}

Answer A is correct. Answer B (

4.0 \times 10^{-6}

) would result if you incorrectly assumed the slit width doubles when the angle doubles. Answers C and D represent other incorrect proportional relationships. Remember: in diffraction, smaller slits produce wider diffraction patterns, so doubling the angle requires a smaller slit width.

Question 11

A student observes a single-slit diffraction pattern and measures the distance between the first minima on either side of the central maximum to be $W = 4.0 \text{ cm}$ . The screen is $L = 2.0 \text{ m}$ from the slit, and the light wavelength is $\lambda = 632 \text{ nm}$ . The width of the slit is approximately:

$32 \text{ μm}$
$63 \text{ μm}$ (correct answer)
$95 \text{ μm}$
$126 \text{ μm}$
$158 \text{ μm}$

Explanation: When you encounter single-slit diffraction problems, you're dealing with wave interference patterns created when light passes through a narrow opening. The key relationship connects the slit width to the positions of dark fringes (minima). For single-slit diffraction, the first minimum occurs at an angle θ where

a \sin θ = λ

, where

a

is the slit width and

λ

is the wavelength. Since the screen distance

L

is much larger than the fringe spacing, we can use the small angle approximation:

\sin θ ≈ \tan θ = y/L

, where

y

is the distance from the central maximum to the first minimum. The problem gives you

W = 4.0

cm as the distance between the first minima on either side of the central maximum. This means each first minimum is at

y = W/2 = 2.0

cm from the center. Substituting into our equation:

a \cdot \frac{y}{L} = λ

Solving for slit width:

a = \frac{λL}{y} = \frac{632 × 10^{-9} \text{ m} × 2.0 \text{ m}}{0.020 \text{ m}} = 63.2 × 10^{-6} \text{ m} = 63 \text{ μm}

This matches answer B. Answer A (32 μm) would result if you incorrectly used the full width

W

instead of

W/2

. Answer C (95 μm) might come from calculation errors with unit conversions. Answer D (126 μm) would result from using

W

and making an additional factor-of-2 error. Always remember: in diffraction problems, carefully identify whether given measurements represent distances from center or total widths between symmetric features.

Question 12

Two identical single slits are placed side by side with their centers separated by a distance $D$ . Each slit has width $a$ where $a < D$ . When illuminated by coherent light of wavelength $\lambda$ , the resulting pattern on a distant screen shows both single-slit diffraction effects and double-slit interference effects. The condition for the first diffraction minimum to occur at the same angle as the third interference maximum is:

$D = 3a$ (correct answer)
$D = a/3$
$D = 3\lambda$
$a = 3\lambda$
$a = \lambda/3$

Explanation: When you encounter problems involving both single-slit diffraction and double-slit interference, you need to apply the conditions for each effect separately and then find where they coincide. For single-slit diffraction, the first minimum occurs when

a \sin \theta = \lambda

, where

a

is the slit width. For double-slit interference, the

m

-th maximum occurs when

D \sin \theta = m\lambda

, where

D

is the separation between slit centers. The third interference maximum corresponds to

m = 3

, so

D \sin \theta = 3\lambda

. Since both effects occur at the same angle, we can set

\sin \theta

equal in both equations. From the diffraction condition:

\sin \theta = \lambda/a

. From the interference condition:

\sin \theta = 3\lambda/D

. Setting these equal:

\frac{\lambda}{a} = \frac{3\lambda}{D}

. Canceling

\lambda

and solving gives us

D = 3a

. Looking at the wrong answers: Choice B (

D = a/3

) would make the slits impossibly close since we're told

a < D

. Choice C (

D = 3\lambda

) relates the slit separation to wavelength rather than slit width, missing the diffraction relationship entirely. Choice D (

a = 3\lambda

) incorrectly relates slit width to wavelength and ignores the interference condition. The correct answer is A:

D = 3a

. Study tip: In combined diffraction-interference problems, always write both conditions separately first, then use the fact that they occur at the same angle to eliminate

\sin \theta

and solve for the relationship between parameters.

Question 13

A parallel beam of microwaves with wavelength $\lambda = 3.0 \text{ cm}$ is incident on a metal sheet with a rectangular aperture measuring $6.0 \text{ cm} \times 12.0 \text{ cm}$ . At a distance of $2.0 \text{ m}$ from the aperture, the angular half-width of the central diffraction maximum in the direction parallel to the $6.0 \text{ cm}$ side is approximately:

$14°$
$25°$
$30°$ (correct answer)
$45°$
$60°$

Explanation: When you encounter diffraction through a rectangular aperture, you're dealing with single-slit diffraction theory applied to each dimension independently. The key insight is that diffraction effects are most pronounced when the aperture dimension is small relative to the wavelength. For a rectangular aperture, the angular half-width of the central maximum is determined by the smaller dimension (which creates the strongest diffraction). The formula is

\sin \theta = \frac{\lambda}{a}

, where

\theta

is the angular half-width and

a

is the aperture width in the direction of interest. Since we want the diffraction pattern parallel to the 6.0 cm side, we use the 6.0 cm dimension:

\sin \theta = \frac{3.0 \text{ cm}}{6.0 \text{ cm}} = 0.5

. Therefore,

\theta = \arcsin(0.5) = 30°

. Looking at the wrong answers: Choice (A) 14° would result from incorrectly using the larger 12.0 cm dimension or making calculation errors. Choice (B) 25° doesn't correspond to any standard diffraction formula application with these values. Choice (D) 45° might come from assuming

\sin \theta = \frac{\lambda}{a} = 0.707

(which would require

\lambda = 4.24

cm), suggesting confusion about the given wavelength. The distance of 2.0 m is irrelevant to calculating the angular width—it's included as a distractor since many students expect all given information to be necessary. Study tip: In diffraction problems, always identify which dimension controls the diffraction pattern you're analyzing, and remember that the 2.0 m distance doesn't affect angular measurements, only linear measurements of the pattern size.

Question 14

A circular aperture of diameter $D$ produces a diffraction pattern when illuminated by plane waves of wavelength $\lambda$ . The angular radius of the first dark ring (from the optical axis to the first minimum) is given by $\theta_1 \approx 1.22\lambda/D$ for small angles. If the aperture diameter is reduced by a factor of 3, the linear radius of the first dark ring on a screen at distance $L$ will:

decrease by a factor of 3
remain unchanged
increase by a factor of 3 (correct answer)
increase by a factor of 9
decrease by a factor of 9

Explanation: When you encounter circular aperture diffraction problems, focus on how the aperture size affects the spreading of light. Circular apertures create characteristic diffraction patterns with concentric dark and bright rings, where smaller apertures cause more spreading. The angular radius of the first dark ring is given by

\theta_1 \approx 1.22\lambda/D

. When the diameter

D

is reduced by a factor of 3, the new angular radius becomes

\theta_1' = 1.22\lambda/(D/3) = 3 \times 1.22\lambda/D = 3\theta_1

. So the angular radius triples. The linear radius on a screen at distance

L

is related to the angular radius by

r = L\tan\theta \approx L\theta

for small angles. Since the angular radius increases by a factor of 3, the linear radius also increases by a factor of 3. This confirms answer C is correct. Answer A incorrectly assumes the linear radius changes inversely with aperture diameter, perhaps confusing this with how aperture size affects light gathering. Answer B suggests no change, which would only be true if angular radius were independent of aperture size—clearly wrong from the given formula. Answer D applies the factor of 3 twice (

3^2 = 9

), possibly from incorrectly thinking both angular radius and the geometric projection each contribute a factor of 3. Remember: smaller apertures create larger diffraction patterns. The relationship is inversely proportional—when aperture diameter decreases by a factor, diffraction angles increase by that same factor, and linear dimensions scale accordingly.

Question 15

In a single-slit Fraunhofer diffraction experiment, the intensity pattern is given by $I(\theta) = I_0 \left[\frac{\sin(\beta)}{\beta}\right]^2$ where $\beta = \frac{\pi a \sin \theta}{\lambda}$ . At what value of $\beta$ does the first minimum occur?

$\beta = 0$
$\beta = \pi/2$
$\beta = \pi$ (correct answer)
$\beta = 3\pi/2$
$\beta = 2\pi$

Explanation: When analyzing single-slit diffraction patterns, you need to understand that minima occur when the intensity drops to zero, which happens when the numerator of the intensity function equals zero while the denominator remains finite. The intensity function

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

reaches zero when

\sin(\beta) = 0

. This occurs when

\beta

equals integer multiples of

\pi

\beta = n\pi

where

n = ±1, ±2, ±3, ...

Note that

n = 0

is excluded because at

\beta = 0

, we have the central maximum where

\lim_{\beta \to 0} \frac{\sin(\beta)}{\beta} = 1

. The first minimum occurs at the smallest positive value, which is

\beta = \pi

, making C correct. Let's examine why the other options are wrong: A)

\beta = 0

represents the central maximum, not a minimum. At this point, the intensity is at its peak value

I_0

. B)

\beta = \pi/2

gives

\sin(\pi/2) = 1

, so the intensity is non-zero:

I = I_0 \left[\frac{1}{\pi/2}\right]^2

. D)

\beta = 3\pi/2

does produce a minimum since

\sin(3\pi/2) = -1

, but this is the second minimum, not the first. Remember this key pattern: for single-slit diffraction, minima always occur at

\beta = n\pi

where

n

is any non-zero integer. The first minimum is always at

\beta = \pi

, regardless of the specific values of slit width, wavelength, or angle.

Question 16

Two identical slits, each of width $a$ , are separated by a distance $d$ where $d > a$ . When illuminated by monochromatic light, both single-slit diffraction and double-slit interference patterns are observed. If the third-order interference maximum coincides with the first diffraction minimum, which relationship must be satisfied?

$d = 3a$ (correct answer)
$d = a/3$
$d = 3a/2$
$d = 2a/3$
$d = a/2$

Explanation: When you encounter problems combining single-slit diffraction and double-slit interference, you need to apply the conditions for both phenomena simultaneously. The interference pattern comes from light passing through two slits, while diffraction occurs at each individual slit. For double-slit interference, the condition for the mth-order maximum is

d \sin \theta = m\lambda

. The third-order maximum occurs when

d \sin \theta = 3\lambda

. For single-slit diffraction, the condition for the first minimum is

a \sin \theta = \lambda

, where the angle

\theta

is measured from the central axis. Since these two conditions must occur at the same angle (they "coincide"), you can set the sine values equal:

\sin \theta = \frac{3\lambda}{d} = \frac{\lambda}{a}

. Solving this equation:

\frac{3\lambda}{d} = \frac{\lambda}{a}

. The wavelength

\lambda

cancels out, giving you

\frac{3}{d} = \frac{1}{a}

, which simplifies to

d = 3a

. Answer A (

d = 3a

) is correct based on this derivation. Answer B (

d = a/3

) would result from incorrectly inverting the relationship. Answer C (

d = 3a/2

) might come from confusing the factor of 3 with a factor of 3/2, perhaps mixing up different diffraction orders. Answer D (

d = 2a/3

) could result from algebraic errors when manipulating the relationship between the two conditions. Remember: when optical phenomena coincide, set their angular conditions equal and solve algebraically. Always check that your final relationship makes physical sense given the constraint

d > a

Question 17

Two identical circular apertures, each of diameter $D$ , are placed side by side with their centers separated by distance $2D$ . When coherent light of wavelength $\lambda$ illuminates both apertures simultaneously, the resulting pattern on a distant screen shows both diffraction and interference effects. What determines the envelope function that modulates the interference fringes?

The diffraction pattern from a single aperture of diameter $D$ provides the envelope function (correct answer)
The interference between waves from the two aperture centers creates the envelope modulation pattern
The combined effective aperture of size $2D$ determines the envelope through its diffraction pattern
The envelope is determined by the diffraction from an aperture of width $2D$ minus the blocked central region

Explanation: When you encounter problems involving multiple apertures, you're dealing with a combination of single-aperture diffraction and multi-aperture interference. The key insight is understanding how these two phenomena interact to create the final pattern. In this two-aperture system, each individual aperture of diameter

D

creates its own diffraction pattern. Since the apertures are identical, they produce identical diffraction envelopes. The interference between waves from the two apertures creates closely-spaced fringes, but these fringes don't have uniform intensity across the screen. Instead, they're modulated by the diffraction envelope from each single aperture. Option A is correct because the envelope function that modulates the interference fringes is determined by the diffraction pattern of a single aperture of diameter

D

. This envelope sets the overall intensity distribution, with the interference creating fine structure within it. Option B confuses cause and effect – the interference between aperture centers creates the fine fringe structure, not the envelope modulation. Option C incorrectly suggests the envelope comes from a

2D

aperture, but the physical apertures are each diameter

D

, not

2D

. Option D attempts to account for the separation between apertures, but the envelope modulation doesn't depend on the blocked region between them – it depends on the diffraction from each individual aperture. Remember this principle: in multiple-aperture systems, single-aperture diffraction provides the envelope, while multi-aperture interference creates the detailed fringe pattern within that envelope. The envelope always comes from the individual aperture characteristics, not the overall geometry.

Question 18

Sound waves of frequency $f = 1000$ Hz are incident on a doorway of width $w = 1.0$ m. A person standing $10$ m away from the doorway at an angle of $30°$ from the forward direction can clearly hear the sound. Using the speed of sound $v = 340$ m/s, what would happen to the sound intensity at this same position if the frequency were increased to $3000$ Hz?

The intensity would increase because higher frequency waves carry more energy per photon
The intensity would remain approximately the same since the position is still within the main diffraction lobe
The intensity would decrease significantly because the position now falls near a diffraction minimum (correct answer)
The intensity would decrease slightly due to increased atmospheric absorption of higher frequency sound

Explanation: For the first frequency:

\lambda_1 = v/f_1 = 340/1000 = 0.34

m. The diffraction condition is

\sin \theta = n\lambda/w

. At

\theta = 30°

\sin 30° = 0.5 = n(0.34)/1.0

, giving

n = 1.47

, so the position is between first and second minima. For

f = 3000

Hz:

\lambda_2 = 340/3000 = 0.113

m. Now

n = 0.5 \times 1.0/0.113 = 4.4

, placing the position very close to the fourth minimum where intensity is much lower. Choice A incorrectly applies photon concepts to sound. Choice B misses the shift in diffraction pattern. Choice D mentions a real but minor effect compared to diffraction.

Question 19

A plane wave is incident on an opaque disk of diameter $D$ . According to Fresnel diffraction theory, there is a bright spot (Poisson spot) at the center of the shadow on a screen placed behind the disk. If the distance from the disk to the screen is doubled while keeping the wavelength and disk diameter constant, how does the intensity of the Poisson spot change?

The intensity remains constant because it depends only on the interference of edge-diffracted waves (correct answer)
The intensity doubles because the spot size increases with distance, collecting more diffracted light
The intensity decreases by a factor of 2 because the Fresnel number changes with distance
The intensity decreases by a factor of 4 following an inverse square law with screen distance

Explanation: When you encounter Fresnel diffraction problems, focus on understanding what creates the interference pattern rather than assuming simple geometric optics applies. The Poisson spot forms because light waves diffract around the disk's edge and interfere constructively at the center of the geometric shadow. This bright spot exists because all the diffracted waves from points around the disk's circumference travel nearly the same distance to reach the center point, arriving in phase regardless of the screen distance. The key insight is that the Poisson spot's intensity depends only on the interference of these edge-diffracted waves, not on geometric factors like distance. When you double the screen distance, the diffracted waves still maintain their phase relationships at the center point, so the intensity remains constant. Option A correctly identifies that intensity depends solely on edge-wave interference. Option B incorrectly suggests the spot collects more light as it grows - while the spot does get slightly larger, the intensity per unit area (what we measure) stays the same. Option C mentions the Fresnel number

F = D^2/(4\lambda z)

, which does change with distance

z

, but this affects the overall diffraction pattern shape, not specifically the central spot intensity. Option D applies an inverse square law, which would be relevant for a point source but doesn't apply here since we're dealing with wave interference, not simple intensity spreading. Remember: In Fresnel diffraction, always consider the wave nature of light and phase relationships. The Poisson spot is a pure interference effect, making it independent of screen distance.

Question 20

Two students observe diffraction of water waves in a ripple tank. Student A uses a straight barrier with a $2$ cm gap, while Student B uses a circular obstacle of $2$ cm diameter. Both observe the wave patterns on a screen placed $50$ cm away. If the water waves have wavelength $\lambda = 1.5$ cm, which statement correctly compares their observations?

Both patterns will be identical because the characteristic dimension is the same in both cases
Student A will see a linear diffraction pattern while Student B will observe concentric circular fringes
Student B will see a bright spot in the center of the shadow while Student A will see darkness (correct answer)
The angular spreads will be identical, but Student A will see higher contrast due to linear geometry

Explanation: Student A observes single-slit diffraction through a gap, which creates a dark region directly behind the center of the gap. Student B observes diffraction around a circular obstacle, which creates a Poisson spot (bright spot) in the center of the geometric shadow due to constructive interference of waves diffracted around the obstacle's edge. This is the key qualitative difference between aperture diffraction (gap) and obstacle diffraction (disk). Choice A ignores the fundamental difference between aperture and obstacle diffraction. Choice B incorrectly describes the geometries. Choice D incorrectly focuses on contrast rather than the central bright spot phenomenon.