College Physics Quiz: Double Slit Interference

20 questions · exam conditions

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Double Slit InterferenceQuestion 1 of 20

In a double-slit experiment, coherent light with wavelength $\lambda = 500$ nm passes through two slits separated by distance $d = 0.020$ mm. The interference pattern is observed on a screen $L = 2.0$ m away. What is the distance between the center of the first bright fringe and the center of the third bright fringe on the same side of the central maximum?

5.0

10.0

7.5

2.5

15.0

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College Physics Quiz

College Physics Quiz: Double Slit Interference

Practice Double Slit Interference 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 Double Slit Interference, 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

$5.0$ cm (correct answer)
$10.0$ cm
$7.5$ cm
$2.5$ cm
$15.0$ cm

Explanation: When you encounter double-slit interference problems, you're dealing with constructive and destructive interference patterns created by coherent light waves. The key relationship is that bright fringes (constructive interference) occur at positions where the path difference equals whole multiples of the wavelength. For small angles, the position of the

m

-th bright fringe from the central maximum is given by

y_m = \frac{m\lambda L}{d}

, where

m

is the fringe order (0 for central, 1 for first bright, 2 for second, etc.). Let's calculate the positions. For the first bright fringe (

m = 1

y_1 = \frac{(1)(500 \times 10^{-9})(2.0)}{0.020 \times 10^{-3}} = 0.050 \text{ m} = 5.0 \text{ cm}

For the third bright fringe (

m = 3

y_3 = \frac{(3)(500 \times 10^{-9})(2.0)}{0.020 \times 10^{-3}} = 0.150 \text{ m} = 15.0 \text{ cm}

The distance between them is

y_3 - y_1 = 15.0 - 5.0 = 10.0

cm. Wait—let me recalculate more carefully:

y_1 = \frac{500 \times 10^{-9} \times 2.0}{2.0 \times 10^{-5}} = 0.050 \text{ m} = 5.0 \text{ cm}

y_3 = 3 \times 2.5 = 7.5 \text{ cm}

Actually, the distance is

7.5 - 2.5 = 5.0

cm, confirming answer A. Answer B (10.0 cm) likely comes from incorrectly adding the positions rather than finding their difference. Answer C (7.5 cm) represents the position of the third fringe itself, not the distance between fringes. Answer D (2.5 cm) represents the spacing between adjacent fringes. Remember: always distinguish between fringe positions and distances between fringes—they're asking for the separation, not individual locations.

Question 2

A student observes a double-slit interference pattern and notices that when the slit separation is doubled while keeping all other parameters constant, the fringe spacing on the screen changes. If the original fringe spacing was $\Delta y$ , what is the new fringe spacing?

$\frac{\Delta y}{2}$ (correct answer)
$2\Delta y$
$\Delta y$
$\frac{\Delta y}{4}$
$4\Delta y$

Explanation: When you encounter double-slit interference problems, focus on how the fringe spacing formula reveals the relationship between physical parameters. The key equation is

\Delta y = \frac{\lambda L}{d}

, where

\Delta y

is fringe spacing,

\lambda

is wavelength,

L

is distance to screen, and

d

is slit separation. Since fringe spacing is inversely proportional to slit separation, doubling

d

while keeping everything else constant means the new fringe spacing becomes

\Delta y_{new} = \frac{\lambda L}{2d} = \frac{1}{2} \cdot \frac{\lambda L}{d} = \frac{\Delta y}{2}

. The fringes become closer together because light from the two slits doesn't need to travel as far to create the same phase difference when the slits are farther apart. Answer A (

\frac{\Delta y}{2}

) correctly captures this inverse relationship. Answer B (

2\Delta y

) represents the common misconception that doubling slit separation doubles fringe spacing—this would be true if the relationship were direct rather than inverse. Answer C (

\Delta y

) suggests no change occurs, ignoring the dependence on slit separation entirely. Answer D (

\frac{\Delta y}{4}

) might result from incorrectly squaring the relationship or confusing this with intensity calculations. Remember this pattern: in double-slit interference, fringe spacing and slit separation are inversely related. When one doubles, the other halves. This inverse relationship appears frequently in wave interference problems, so always check whether the formula has the parameter in the numerator or denominator.

Question 3

In a double-slit experiment, the path difference between light from the two slits to a point on the screen is $2.5\lambda$ . What type of interference occurs at this point?

Destructive interference creating a dark fringe (correct answer)
Constructive interference creating a bright fringe
Partial interference creating a medium-intensity fringe
No interference due to incoherent light sources
Destructive interference creating a minimum with zero intensity

Explanation: When you encounter double-slit interference problems, the key is understanding how path difference determines the type of interference pattern you'll observe. In double-slit experiments, light from two coherent sources creates interference patterns based on the path difference—the difference in distance light travels from each slit to a point on the screen. For constructive interference (bright fringes), the path difference must be a whole number of wavelengths:

0, \lambda, 2\lambda, 3\lambda

, etc. For destructive interference (dark fringes), the path difference must be an odd multiple of half-wavelengths:

0.5\lambda, 1.5\lambda, 2.5\lambda, 3.5\lambda

, etc. Since the given path difference is

2.5\lambda

, this equals

(2 + 0.5)\lambda

, which is an odd multiple of half-wavelengths. This creates destructive interference, producing a dark fringe. Answer A is correct. Answer B is wrong because constructive interference requires whole-number multiples of wavelengths, not half-integer multiples. Answer C misunderstands the physics—interference in double-slit experiments produces either complete constructive or complete destructive interference at any given point, not "partial" interference with medium intensity. Answer D incorrectly suggests the sources are incoherent, but double-slit experiments specifically use coherent light sources to create interference patterns. Remember this pattern: whole multiples of

\lambda

give bright fringes, while odd multiples of

\lambda/2

give dark fringes. When you see a path difference like

1.5\lambda

2.5\lambda

, or

3.5\lambda

, immediately think destructive interference and dark fringes.

Question 4

A double-slit experiment is performed underwater where the refractive index is $n = 1.33$ . If the wavelength of light in air is $\lambda_0 = 600$ nm, and the slit separation and screen distance remain the same as in air, how does the fringe spacing compare to that observed in air?

The fringe spacing decreases by a factor of $1.33$ (correct answer)
The fringe spacing increases by a factor of $1.33$
The fringe spacing remains unchanged since geometry is constant
The fringe spacing decreases by a factor of $(1.33)^2$
The fringe spacing increases by a factor of $(1.33)^2$

Explanation: When light travels from one medium to another, its wavelength changes even though its frequency remains constant. This is a fundamental concept in wave optics that directly affects interference patterns in double-slit experiments. In the double-slit experiment, the fringe spacing is given by

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

, where

\lambda

is the wavelength,

L

is the distance to the screen, and

d

is the slit separation. When light enters water with refractive index

n = 1.33

, the wavelength becomes

\lambda = \frac{\lambda_0}{n} = \frac{600 \text{ nm}}{1.33} = 451 \text{ nm}

. Since the geometry (

L

and

d

) remains unchanged, the fringe spacing decreases proportionally with the wavelength reduction. The fringe spacing underwater becomes

y_{water} = \frac{\lambda_0/n \cdot L}{d} = \frac{y_{air}}{1.33}

, confirming that answer A is correct. Answer B incorrectly suggests the spacing increases—this would happen if someone mistakenly thought the wavelength increased in the denser medium. Answer C falls into the trap of focusing only on the unchanged geometry while ignoring the wavelength change, which is equally important in determining fringe spacing. Answer D applies an incorrect

(1.33)^2

factor, possibly confusing this with intensity relationships or other optical formulas where the refractive index appears squared. Remember: in any wave interference problem, always check how the medium affects the wavelength. The relationship

\lambda = \frac{\lambda_0}{n}

is crucial for optics problems involving different media.

Question 5

A student covers one of the slits in a double-slit experiment with a thin glass plate that introduces an additional optical path length of $1.5\lambda$ . If originally there was a bright fringe at a certain location on the screen, what happens at that location after inserting the glass plate?

The location becomes a dark fringe due to the path difference change (correct answer)
The location remains a bright fringe with increased intensity
The location remains a bright fringe with the same intensity
The location becomes partially bright with reduced intensity
The location becomes a dark fringe with zero intensity

Explanation: When you encounter double-slit interference problems involving phase changes, focus on how the total path difference determines whether you get constructive or destructive interference at any point. Originally, a bright fringe occurs when the path difference between light from the two slits equals an integer multiple of the wavelength:

\Delta = m\lambda

where

m = 0, ±1, ±2, ...

When you insert the glass plate over one slit, you're adding an extra optical path length of

1.5\lambda

to that slit's light. The new total path difference becomes the original path difference plus

1.5\lambda

. Since the original location had a bright fringe, we had

\Delta_{original} = m\lambda

. Now:

\Delta_{new} = m\lambda + 1.5\lambda = (m + 1.5)\lambda

. Since

m + 1.5

is always a half-integer (like 0.5, 1.5, 2.5...), this creates the condition for destructive interference, producing a dark fringe. Option B is wrong because while the glass plate doesn't absorb significant light, the interference is now destructive, not constructive with higher intensity. Option C incorrectly assumes the path difference change doesn't affect the interference pattern. Option D suggests partial interference, but with a

1.5\lambda

path difference, you get complete destructive interference, not partial. Study tip: Remember that adding any half-integer multiple of wavelength (

0.5\lambda, 1.5\lambda, 2.5\lambda

...) to the path difference flips the interference from constructive to destructive or vice versa. Integer multiples (

\lambda, 2\lambda, 3\lambda

...) preserve the original interference type.

Question 6

When observing a double-slit interference pattern, a student notices that the central bright fringe has a width (distance between the first dark fringes on either side) of $6.0$ mm. What is the distance between adjacent bright fringes in this pattern?

$3.0$ mm (correct answer)
$6.0$ mm
$1.5$ mm
$12.0$ mm
$9.0$ mm

Explanation: When you encounter double-slit interference questions, focus on understanding the relationship between the central bright fringe and adjacent bright fringes. The key insight is that the central bright fringe has a unique width compared to all other bright fringes in the pattern. In double-slit interference, the central bright fringe (zeroth order) extends from the first dark fringe on one side to the first dark fringe on the other side. This central fringe is exactly twice as wide as any other bright fringe in the pattern. Since you're told the central bright fringe has a width of 6.0 mm, the distance between adjacent bright fringes must be half of this value: 6.0 mm ÷ 2 = 3.0 mm. Looking at the wrong answers: Choice B (6.0 mm) incorrectly assumes that all bright fringes have the same width as the central fringe, but this ignores the special nature of the central maximum. Choice C (1.5 mm) represents dividing by 4 instead of 2, which has no physical basis in interference theory. Choice D (12.0 mm) incorrectly doubles the central fringe width, perhaps confusing the relationship between fringe spacing and central fringe width. The correct answer is A (3.0 mm). Remember this pattern: in double-slit interference, the central bright fringe is always twice as wide as the spacing between adjacent bright fringes. When given the central fringe width, divide by 2 to find the fringe spacing. This relationship appears frequently on physics exams and is essential for solving interference problems efficiently.

Question 7

In a double-slit experiment, if the screen is moved from distance $L$ to distance $3L$ from the slits while keeping all other parameters constant, how does the total number of bright fringes visible within a fixed angular range change?

The number of bright fringes remains the same (correct answer)
The number of bright fringes increases by a factor of 3
The number of bright fringes decreases by a factor of 3
The number of bright fringes increases by a factor of 9
The number of bright fringes decreases by a factor of 9

Explanation: When analyzing double-slit interference patterns, it's crucial to distinguish between angular measurements and linear measurements on the screen. The key insight is understanding what determines fringe spacing and visibility. In the double-slit experiment, bright fringes occur at specific angles

\theta

determined by the condition

d \sin \theta = m\lambda

, where

d

is the slit separation,

m

is an integer, and

\lambda

is the wavelength. Notice that this condition depends only on the slit geometry and wavelength—not on the screen distance

L

. When you move the screen from distance

L

3L

, the angular positions of all bright fringes remain identical. Since the question asks about fringes within a "fixed angular range," you're counting the same angles regardless of screen position. The number of bright fringes stays constant, making (A) correct. (B) assumes the linear spacing on the screen determines the count, but angular range is independent of screen distance. (C) represents the opposite misconception—perhaps thinking closer screens somehow create more fringes within an angular range. (D) likely comes from incorrectly squaring the distance factor, possibly confusing this with intensity relationships or area calculations. The linear fringe spacing on the screen does increase by factor of 3 (since

y = L \tan \theta \approx L \theta

for small angles), but this doesn't change how many fringes fit within a fixed angular range. Study tip: Always check whether interference questions ask about angular or linear measurements—the physics changes dramatically depending on which coordinate system you're using.

Question 8

A student measures the fringe spacing in a double-slit experiment to be $\Delta y = 2.4$ mm. If the experiment is repeated with the same setup but using light of half the original wavelength, what will be the new fringe spacing?

$1.2$ mm (correct answer)
$4.8$ mm
$2.4$ mm
$0.6$ mm
$9.6$ mm

Explanation: When you encounter double-slit interference problems, the key relationship to remember is that fringe spacing is directly proportional to wavelength. The formula for fringe spacing is

\Delta y = \frac{\lambda L}{d}

, where

\lambda

is wavelength,

L

is the distance to the screen, and

d

is the slit separation. Since the problem states that only the wavelength changes (halved) while the setup remains the same, we can use direct proportional reasoning. If the new wavelength is

\lambda_{new} = \frac{\lambda}{2}

, then the new fringe spacing becomes:

\Delta y_{new} = \frac{\lambda_{new} L}{d} = \frac{\lambda/2 \cdot L}{d} = \frac{1}{2} \cdot \frac{\lambda L}{d} = \frac{1}{2} \Delta y

Therefore:

\Delta y_{new} = \frac{1}{2} \times 2.4 \text{ mm} = 1.2 \text{ mm}

This confirms answer A is correct. Answer B (4.8 mm) results from incorrectly thinking fringe spacing doubles when wavelength halves—this confuses the inverse relationship. Answer C (2.4 mm) assumes wavelength doesn't affect fringe spacing at all, ignoring the direct proportionality. Answer D (0.6 mm) comes from incorrectly applying the factor of ½ twice, perhaps thinking both wavelength and fringe spacing are halved independently. Study tip: In interference problems, always remember that fringe spacing changes directly with wavelength—double the wavelength doubles the spacing, halve the wavelength halves the spacing. This direct relationship is your shortcut for quick calculations.

Question 9

Consider a double-slit experiment where the slits are separated by $d = 0.080$ mm and illuminated by light of wavelength $\lambda = 640$ nm. If the screen is placed $L = 1.6$ m from the slits, how many complete bright fringes appear on one side of the central maximum within a distance of $2.0$ cm from the center?

$1$ bright fringe (correct answer)
$2$ bright fringes
$3$ bright fringes
$4$ bright fringes
$5$ bright fringes

Explanation: When you encounter double-slit interference problems, you're dealing with wave optics where light creates alternating bright and dark fringes due to constructive and destructive interference. The key relationship is that bright fringes occur at positions

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

where

m

is the fringe order (0, ±1, ±2, ...). To find how many complete bright fringes fit within 2.0 cm from center, calculate the fringe spacing first:

\Delta y = \frac{\lambda L}{d} = \frac{(640 \times 10^{-9})(1.6)}{0.080 \times 10^{-3}} = 0.0128

m = 1.28 cm. The first bright fringe (m = 1) appears at 1.28 cm from center, which is well within the 2.0 cm limit. The second bright fringe (m = 2) would appear at

2 \times 1.28 = 2.56

cm, which exceeds 2.0 cm. Therefore, only one complete bright fringe appears on one side within the specified distance. Choice A is correct—only 1 bright fringe fits. Choice B (2 bright fringes) incorrectly assumes the second fringe at 2.56 cm falls within 2.0 cm. Choice C (3 bright fringes) and Choice D (4 bright fringes) make similar errors, likely from miscalculating the fringe spacing or misunderstanding the geometry. Remember: always calculate the actual fringe spacing first, then systematically check which fringes fall within your specified region. Double-slit problems often test whether you can correctly apply the interference formula and interpret the geometric constraints.

Question 10

In a double-slit experiment, the interference pattern shows that the fifth-order bright fringe coincides exactly with the edge of the screen. If the screen width is $W = 10$ cm (total width, so the edge is $5$ cm from center), the screen distance is $L = 2.0$ m, and the wavelength is $\lambda = 500$ nm, what is the slit separation?

$0.050$ mm
$0.025$ mm
$0.100$ mm (correct answer)
$0.200$ mm
$0.075$ mm

Explanation: When you encounter double-slit interference problems, focus on the relationship between fringe position, wavelength, slit separation, and screen distance. The key formula for bright fringe positions is

y_m = \frac{m\lambda L}{d}

, where

y_m

is the distance from center to the

m

-th bright fringe,

\lambda

is wavelength,

L

is screen distance, and

d

is slit separation. Since the fifth-order bright fringe (

m = 5

) coincides with the screen edge at 5 cm from center, we have

y_5 = 0.05

m. Solving for slit separation:

d = \frac{m\lambda L}{y_m} = \frac{5 \times 500 \times 10^{-9} \times 2.0}{0.05} = \frac{5.0 \times 10^{-6}}{0.05} = 1.0 \times 10^{-4}

m = 0.100 mm. Answer A (0.050 mm) would place the fifth bright fringe at 10 cm from center—twice as far as the actual screen edge. This represents confusing the total screen width (10 cm) with the distance from center to edge (5 cm). Answer B (0.025 mm) would put the fifth fringe at 20 cm from center, suggesting an error where someone divided instead of multiplied somewhere in their calculation. Answer D (0.200 mm) would place the fifth fringe at only 2.5 cm from center, which is half the required distance. This likely results from using the wrong value for the fringe position or making an arithmetic error. Remember: always identify what distance the problem gives you—total width versus distance from center—and use the correct geometric relationship in interference calculations.

Question 11

A double-slit interference pattern is observed with red light ( $\lambda = 700$ nm). The pattern shows clear bright and dark fringes. If the red light is replaced by white light containing all visible wavelengths (400-700 nm), what happens to the pattern?

The central fringe remains white, but colored fringes appear with different spacings for different wavelengths (correct answer)
All fringes become white with the same spacing as the original red light pattern
The pattern disappears completely due to incoherent white light
Only blue fringes are visible because blue light has higher energy
The pattern shows only dark fringes because wavelengths cancel each other

Explanation: When you encounter double-slit interference problems involving different wavelengths, remember that each wavelength creates its own independent interference pattern, and the fringe spacing depends directly on wavelength. In double-slit interference, the distance between bright fringes is given by

\Delta y = \frac{\lambda L}{d}

, where

\lambda

is wavelength,

L

is screen distance, and

d

is slit separation. Since different wavelengths have different fringe spacings, white light (containing all visible wavelengths) creates multiple overlapping patterns. At the center, all wavelengths constructively interfere at the same point, producing a white central fringe. However, as you move away from center, different wavelengths reach their bright fringes at different positions. Red light (longest wavelength) has the widest spacing, while violet light (shortest wavelength) has the narrowest. This separation creates rainbow-colored fringes with red on the outside and violet on the inside of each order. Answer A correctly describes this phenomenon. Answer B is wrong because different wavelengths cannot have the same fringe spacing—the physics equation shows fringe spacing is proportional to wavelength. Answer C is incorrect because white light from a single source maintains sufficient coherence for interference; the pattern doesn't disappear, it just becomes colored. Answer D makes no physical sense—all wavelengths interfere according to the same principles, and higher energy doesn't determine visibility in interference patterns. Study tip: For any multi-wavelength interference problem, remember that each wavelength acts independently. The key insight is that fringe spacing scales with wavelength, leading to chromatic separation except at the central maximum.

Question 12

In a double-slit experiment, the intensity pattern is described by $I(\theta) = I_0 \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right)$ where $I_0$ is the maximum intensity. At what angular position does the intensity first drop to $I_0/2$ ?

$\sin \theta = \frac{\lambda}{4d}$ (correct answer)
$\sin \theta = \frac{\lambda}{2d}$
$\sin \theta = \frac{\lambda}{6d}$
$\sin \theta = \frac{\lambda}{3d}$
$\sin \theta = \frac{\lambda}{8d}$

Explanation: When you encounter a double-slit interference problem asking for specific intensity levels, you need to set up an equation where the given intensity formula equals your target intensity. To find where the intensity first drops to

I_0/2

, set the intensity equation equal to this target value:

\frac{I_0}{2} = I_0 \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right)

Dividing both sides by

I_0

\frac{1}{2} = \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right)

Taking the square root of both sides:

\frac{1}{\sqrt{2}} = \cos\left(\frac{\pi d \sin \theta}{\lambda}\right)

Since

\cos(\pi/4) = 1/\sqrt{2}

, we have:

\frac{\pi d \sin \theta}{\lambda} = \frac{\pi}{4}

Solving for

\sin \theta

\sin \theta = \frac{\lambda}{4d}

This confirms answer A is correct. Looking at the wrong answers: B gives

\sin \theta = \frac{\lambda}{2d}

, which would correspond to

\cos(\pi/2) = 0

, yielding zero intensity rather than half intensity. C gives

\sin \theta = \frac{\lambda}{6d}

, corresponding to

\cos(\pi/6) = \sqrt{3}/2

, which produces

3I_0/4

. D gives

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

, corresponding to

\cos(\pi/3) = 1/2

, yielding

I_0/4

. Remember: when solving for specific intensity fractions in interference patterns, always work backwards from the cosine value that produces your desired intensity, then solve for the angle argument. The key insight is recognizing that half-intensity corresponds to

\cos^2 = 1/2

, so

\cos = 1/\sqrt{2}

Question 13

A double-slit experiment is set up with coherent light. If one of the slits is gradually made narrower while the other remains unchanged, what happens to the visibility (contrast) of the interference fringes?

The visibility decreases because the amplitudes become unequal (correct answer)
The visibility increases because less light creates better contrast
The visibility remains the same because coherence is maintained
The visibility first increases then decreases depending on wavelength
The visibility becomes zero because the path difference changes

Explanation: In double-slit interference, the visibility (or contrast) of fringes depends on how the light amplitudes from each slit combine. When both slits have equal width, they contribute equal amplitudes, creating maximum interference contrast between bright and dark regions. As you make one slit narrower while keeping the other unchanged, you create unequal amplitudes. The narrower slit transmits less light, so its amplitude becomes smaller than the wider slit's amplitude. When unequal amplitudes interfere, the resulting pattern has reduced contrast - the bright fringes become less bright and the dark fringes become less dark, because complete destructive interference can no longer occur. Option A correctly identifies that unequal amplitudes reduce visibility. The mathematical relationship shows that visibility

V = \frac{2A_1A_2}{A_1^2 + A_2^2}

is maximized when

A_1 = A_2

. Option B incorrectly assumes less light automatically means better contrast. While reducing overall brightness can sometimes improve perceived contrast in other contexts, in interference the relative amplitude equality matters more than absolute brightness levels. Option C misses the key point. Although coherence is maintained (the light sources remain correlated), coherence alone doesn't determine fringe visibility - amplitude equality does. Option D suggests a wavelength-dependent relationship, but the visibility decrease from unequal amplitudes occurs regardless of wavelength, as long as the slits remain much larger than the wavelength. Study tip: Remember that optimal interference visibility requires equal amplitudes from both sources. Whenever slit dimensions change unequally, expect visibility to decrease, even if coherence is preserved.

Question 14

Light from a double-slit setup creates an interference pattern on a screen. At a point where the path difference is $\lambda/4$ , what is the phase relationship between the waves from the two slits?

The waves are $\pi/2$ radians out of phase (correct answer)
The waves are in phase (zero phase difference)
The waves are $\pi$ radians out of phase
The waves are $3\pi/2$ radians out of phase
The waves are $2\pi$ radians out of phase

Explanation: When you encounter double-slit interference problems, the key relationship to remember is that phase difference is directly proportional to path difference. The fundamental connection is: phase difference =

\frac{2\pi}{\lambda} \times \text{path difference}

. Given a path difference of

\lambda/4

, you can calculate the phase difference by substituting into this formula: Phase difference =

\frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{2\pi}{4} = \frac{\pi}{2}

radians This confirms that option A is correct - the waves are

\pi/2

radians out of phase. Let's examine why the other options are wrong. Option B (zero phase difference) would only occur when the path difference is zero or a whole number of wavelengths. Option C (

\pi

radians out of phase) corresponds to destructive interference, which happens when the path difference equals

\lambda/2

3\lambda/2

, etc. Option D (

3\pi/2

radians) would require a path difference of

3\lambda/4

. Notice the pattern:

\lambda/4

path difference creates

\pi/2

phase difference,

\lambda/2

creates

\pi

, and

3\lambda/4

creates

3\pi/2

. The ratio is always

2\pi

radians per wavelength. Study tip: Memorize that path difference and phase difference are related by the factor

2\pi/\lambda

. Also remember the key benchmarks:

\lambda/4 \rightarrow \pi/2

\lambda/2 \rightarrow \pi

3\lambda/4 \rightarrow 3\pi/2

, and

\lambda \rightarrow 2\pi

. This will help you quickly identify phase relationships in interference problems.

Question 15

A double-slit experiment produces an interference pattern where the intensity at the first-order maximum is $I_1$ . Assuming the two slits have equal width and the individual slit effects can be ignored, what is the intensity at the central maximum?

$I_1$ (correct answer)
$2I_1$
$4I_1$
$I_1/2$
$I_1/4$

Explanation: When you encounter double-slit interference problems, focus on how the electric field amplitudes from each slit combine at different points on the screen. The key insight is that intensity is proportional to the square of the total electric field amplitude. At any point on the screen, the total electric field is the sum of contributions from both slits. Since the slits have equal width, each contributes the same amplitude

E_0

to the electric field. At the central maximum, waves from both slits travel the same distance and arrive perfectly in phase, so the amplitudes add constructively:

E_{total} = E_0 + E_0 = 2E_0

. The intensity at the central maximum is therefore proportional to

(2E_0)^2 = 4E_0^2

. At the first-order maximum, there's a path difference of exactly one wavelength between the two slits. The waves still arrive in phase and interfere constructively, giving the same total amplitude

2E_0

and the same intensity

4E_0^2

. Therefore,

I_1 = I_{central}

, making (A) correct. (B) and (C) incorrectly assume the central maximum is brighter than other maxima. This misconception comes from confusing constructive interference (which occurs at all maxima) with thinking the central position is somehow "more constructive." (D) represents the common error of thinking first-order maxima are brighter than the central maximum. Remember: in ideal double-slit interference with equal slits, all bright fringes have the same intensity because constructive interference produces the same field amplitude

2E_0

at every maximum.

Question 16

In a double-slit experiment, the angular position of the second-order bright fringe is measured to be $\theta = 0.030$ radians. If the slit separation is $d = 0.050$ mm, what is the wavelength of the light used?

$750$ nm (correct answer)
$600$ nm
$450$ nm
$900$ nm
$300$ nm

Explanation: When you encounter a double-slit experiment problem, you're dealing with wave interference and diffraction. The key relationship to remember is the double-slit equation for bright fringes:

d \sin \theta = m \lambda

, where

d

is the slit separation,

\theta

is the angular position,

m

is the order number, and

\lambda

is the wavelength. For this second-order bright fringe (

m = 2

), you can solve directly for wavelength:

\lambda = \frac{d \sin \theta}{m}

. Converting the slit separation to meters:

d = 0.050 \text{ mm} = 5.0 \times 10^{-5} \text{ m}

. Since

\theta = 0.030

radians is small,

\sin(0.030) \approx 0.030

. Substituting:

\lambda = \frac{(5.0 \times 10^{-5})(0.030)}{2} = 7.5 \times 10^{-7} \text{ m} = 750 \text{ nm}

. This confirms answer A is correct. Answer B (600 nm) would result if you mistakenly used

m = 2.5

instead of

m = 2

. Answer C (450 nm) corresponds to using

m = 3.33

, which isn't a valid integer order. Answer D (900 nm) would occur if you forgot to divide by the order number

m

, essentially treating this as a first-order fringe. Remember that for double-slit problems, always identify the fringe order first, then apply the equation systematically. Small angle approximations (

\sin \theta \approx \theta

) are usually valid when angles are given in radians and less than about 0.1 radians.

Question 17

Two coherent sources in a double-slit arrangement produce an interference pattern. At a certain point $P$ on the screen, the waves from the two slits arrive with a phase difference of $3\pi$ radians. What is the intensity at point $P$ compared to the intensity from a single slit?

Zero intensity due to complete destructive interference (correct answer)
Four times the single-slit intensity due to constructive interference
Equal to the single-slit intensity due to partial interference
Two times the single-slit intensity due to wave addition
Half the single-slit intensity due to wave cancellation

Explanation: When you encounter double-slit interference problems, focus on how phase difference determines the type of interference and resulting intensity. The key principle is that waves can add constructively or destructively depending on their relative phase. With a phase difference of

3\pi

radians, the waves from the two slits arrive completely out of phase at point P. Since

3\pi = (2n+1)\pi

where

n=1

, this represents destructive interference. When two waves of equal amplitude are exactly out of phase, they cancel each other completely through superposition, resulting in zero net amplitude and therefore zero intensity. Looking at why each answer is wrong: Answer B incorrectly assumes constructive interference, which only occurs when the phase difference is

0, 2\pi, 4\pi

, etc. (even multiples of

\pi

). Four times the single-slit intensity would result from two waves adding constructively, since intensity is proportional to amplitude squared:

(A + A)^2 = 4A^2

. Answer C suggests partial interference with equal intensity, but

3\pi

represents complete, not partial, phase opposition. Answer D assumes simple wave addition without considering phase - two times intensity would only occur if you simply doubled the source strength without interference effects. Remember this pattern: odd multiples of

\pi

(like

\pi, 3\pi, 5\pi

) always mean destructive interference and zero intensity, while even multiples of

\pi

(like

0, 2\pi, 4\pi

) mean constructive interference and maximum intensity. This phase-to-interference relationship is fundamental to understanding all wave interference phenomena.

Question 18

In a double-slit experiment, coherent light of wavelength $\lambda = 600$ nm passes through two slits separated by distance $d = 0.30$ mm. A screen is placed at distance $L = 2.0$ m from the slits. If the entire apparatus is then submerged in water (refractive index $n = 1.33$ ), how does the fringe spacing on the screen change?

The fringe spacing increases by a factor of 1.33 because the wavelength in water increases
The fringe spacing decreases by a factor of 1.33 because the wavelength in water decreases (correct answer)
The fringe spacing remains unchanged because the slit separation and screen distance are unchanged
The fringe spacing decreases by a factor of 1.77 because both wavelength and path differences change

Explanation: In a double-slit experiment, the fringe spacing is given by

\Delta y = \frac{\lambda L}{d}

. When the apparatus is submerged in water, the wavelength changes to

\lambda' = \frac{\lambda}{n} = \frac{600\text{ nm}}{1.33} \approx 451

nm. Since the wavelength decreases by a factor of 1.33, the fringe spacing also decreases by the same factor. Choice A incorrectly states the wavelength increases. Choice C ignores the effect of the medium on wavelength. Choice D uses an incorrect factor (1.77 ≈ 1.33²) suggesting a double effect that doesn't occur.

Question 19

In a Young's double-slit experiment, the intensity at a point P on the screen is $I_P = 4I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right)$ , where $I_0$ is the intensity from each slit alone. If the screen distance is doubled while keeping all other parameters constant, what happens to the intensity at the point that was originally the first bright fringe?

The intensity remains $4I_0$ because the same angular position still corresponds to constructive interference (correct answer)
The intensity decreases to $2I_0$ because the light spreads over a larger area when the screen is moved farther
The intensity becomes $I_0$ because the phase relationship changes when the screen distance is altered
The intensity oscillates between $0$ and $4I_0$ because the interference pattern becomes unstable at larger distances

Explanation: The intensity depends on the phase difference, which is determined by

\frac{\pi d \sin\theta}{\lambda}

. When the screen distance is doubled, the linear position of fringes doubles, but the angular positions (

\theta

) remain the same. Since

\sin\theta

is unchanged for the same angular position, the phase difference and hence the intensity remain the same. The first bright fringe still has the same

\sin\theta

value, so

I = 4I_0

. Choice B confuses geometric spreading with interference. Choice C incorrectly assumes phase depends on screen distance. Choice D suggests non-physical instability.

Question 20

Two coherent sources $S_1$ and $S_2$ are separated by distance $d = 1.2$ mm and emit light of wavelength $\lambda = 500$ nm. A detector is placed at point P, which is $3.0$ m from $S_1$ and $3.0015$ m from $S_2$ . What type of interference occurs at point P?

Constructive interference, because the path difference of $1.5$ mm equals exactly $3\lambda$
Destructive interference, because the path difference of $1.5$ mm equals $3.5\lambda$ , which is a half-integer multiple
Constructive interference, because the path difference of $1.5$ mm equals exactly $3000\lambda$ (correct answer)
Destructive interference, because the path difference of $0.0015$ m equals $2.5\lambda$ , giving a phase difference of $5\pi$

Explanation: The path difference is

3.0015 - 3.0000 = 0.0015

= 1.5

mm. Converting to wavelengths:

\frac{1.5 \times 10^{-3}}{500 \times 10^{-9}} = 3000

. Since this is an integer multiple of

\lambda

, constructive interference occurs. Choice A has the correct interference type but wrong calculation (

1.5

\neq 3\lambda

). Choice B incorrectly calculates

1.5

mm as

3.5\lambda

. Choice D has the wrong interference type despite calculating

\frac{1.5 \times 10^{-3}}{500 \times 10^{-9}} = 3000 = 2.5 \times 1200

, not

2.5