Interference MCQ Quiz - Objective Question with Answer for Interference - Download Free PDF

Concept:

Young’s Single Slit Experiment

• In the single-slit experiment, we use a long slit as the source of light in place of pin holes.
• The light coming out of the slit is intercepted on a screen placed parallel to the plane of the slits.
• The slit is illuminated by a parallel beam of monochromatic light.
• A series of dark and bright strips, called fringes, is observed on the screen.
• The nth-order minima are given by, 
  • d sinθ = nλ
• Where, d = width of the slit, λ = wavelength, θ = angle, n = nth fringe

​Calculation:

Given, wavelength λ = 600 nm, for first minimum angle θ = 30°, n = 1

Then width of the slit, d sinθ = nλ

⇒ d × sin30° = 1 × 600 

⇒ d = 600 × 2 = 1200 nm = 1.2 μm

The correct answer is Option 3: 0.3 mm.

Concept:

Fringe width (β): In Young’s double slit experiment, β = (λD)/d, where λ is wavelength, D is distance to screen, and d is slit separation.

Effect of changing parameters:

Fringe width is directly proportional to wavelength.

Fringe width is inversely proportional to slit separation.

Explanation:

Initial fringe width = 0.8 mm with λ = 640 nm.

New wavelength = 720 nm and slit separation becomes three times.

New fringe width = (720/640) × (1/3) × 0.8 mm.

New fringe width = (9/8) × (1/3) × 0.8 = 0.3 mm.

CONCEPT:

Young's double-slit experiment

• Young’s double-slit experiment helped in understanding the wave nature of light.
• The original Young’s double-slit experiment used diffracted light from a single monochromatic source of light.
• The light that comes from the monochromatic source is passed into two slits to be used as two coherent sources.
• At any point on the screen at a distance ‘y’ from the center, the waves travel distances l1 and l2 to create a path difference of Δl at that point.
• If there is a constructive interference on the point then the bright fringe occurs.
• If there is a destructive interference on the point then the dark fringe occurs.

• The distance of the nth bright fringe from the central fringe is given as,

\(⇒ y=\frac{nλ D}{d}\)

Where d = distance between slits, D = distance between slits and screen, and λ = wavelength

​

Explanation:

β (fringe width) = \(\frac{λ \mathrm{D}}{\mathrm{~d}}\)

In denser medium, λ ↓ ⇒ β ↓

⇒ fringe come closer 

Also, µ = \(\frac{c}{V} \) ⇒ V = \(\frac{c}{μ}\)

Frequency remains same, 

⇒ μ = \(\frac{\lambda_{\text {vac. }} \mathrm{f}}{\lambda_{\text {med }} \mathrm{f}}\) ⇒ \(\lambda_{\text {med }}\) = \(\frac{\lambda_{\mathrm{vac} .}}{\mu}\)

Calculation:

Let S be a source of sound and P the person or listener.

The waves from S reach point P directly following the path SMP and being reflected from the ceiling at point A following the path SAP.

M is mid-point of SP (i.e. SM = MP) and \(\angle SMA = 90^\circ\).

We have \(SAP = SA + AP = 2(SA)\)

\(= \sqrt{[(SM)^2+(MA)^2]} = 2\sqrt{(60^2+25^2)} = 130m\)

∴" id="MathJax-Element-262-Frame" role="presentation" style="position: relative;" tabindex="0">∴ ∴ $\therefore$ path difference \(= SAP - SMP = 130 - 120 = 10m\)

Path difference due to reflection from ceiling = \(\dfrac{\pi}{2}\)

∴" id="MathJax-Element-263-Frame" role="presentation" style="position: relative;" tabindex="0">∴ ∴ $\therefore$ Effective path difference \(\triangle x = 10 + \dfrac{\pi}{2}\)

For constructive interference

\(\triangle x = 10 + \dfrac{\lambda}{2} = \lambda \Rightarrow (2n - 1)\dfrac{\lambda}{2} = 10 (n = 1, 2, 3, ...)\)

∴" id="MathJax-Element-264-Frame" role="presentation" style="position: relative;" tabindex="0">∴ ∴ $\therefore$ Wavelength \(\lambda = \dfrac{2 \times 10}{(2n - 1)} = \dfrac{20}{2n - 1}\)

The possible wavelengths are \(\lambda = 20, \dfrac{20}{3}, \dfrac{20}{5}, \dfrac{20}{7}, \dfrac{20}{9}, ...\)

\(= 2m, 6.67m, 4m, 2.85m, 2.22m...\)

Concept:

When two light beams interfere, the intensity of maxima and minima is given by the following formulas:

• The intensity of maxima: I_max = (I₁ + I₂)
• The intensity of minima: I_min = |I₁ - I₂|

The ratio of the intensity of maxima and minima is given by:

Ratio = I_max / I_min

Given that the ratio of intensities of two light beams is 9:4, we have I₁ = 9k and I₂ = 4k, where k is a constant. Let’s calculate the ratio of the intensity of maxima and minima:

First, the formula for intensity of maxima becomes:

I_max = (I₁ + I₂) = 9k + 4k = 13k

Next, the formula for intensity of minima is:

I_min = |I₁ - I₂| = |9k - 4k| = 5k

Now, the ratio of intensity of maxima and minima is:

Ratio = I_max / I_min = (13k) / (5k) = 13 / 5

However, to get the correct ratio, we need to square the terms and consider the square root factors, leading to the correct result:

√(I₁ / I₂) = √(9 / 4) = 3 / 2

Then, using the relation between the electric field intensities:

(√(I₁) + √(I₂))² / (√(I₁) - √(I₂))² = 5² = 25

∴ The correct answer is option 4: 25 : 1.

Concept:

Young’s Single Slit Experiment

• In the single-slit experiment, we use a long slit as the source of light in place of pin holes.
• The light coming out of the slit is intercepted on a screen placed parallel to the plane of the slits.
• The slit is illuminated by a parallel beam of monochromatic light.
• A series of dark and bright strips, called fringes, is observed on the screen.
• The nth-order minima are given by, 
  • d sinθ = nλ
• Where, d = width of the slit, λ = wavelength, θ = angle, n = nth fringe

​Calculation:

Given, wavelength λ = 600 nm, for first minimum angle θ = 30°, n = 1

Then width of the slit, d sinθ = nλ

⇒ d × sin30° = 1 × 600 

⇒ d = 600 × 2 = 1200 nm = 1.2 μm

Concept:

Young’s Single Slit Experiment

• In the single-slit experiment, we use a long slit as the source of light in place of pin holes.
• The light coming out of the slit is intercepted on a screen placed parallel to the plane of the slits.
• The slit is illuminated by a parallel beam of monochromatic light.
• A series of dark and bright strips, called fringes, is observed on the screen.
• The nth-order minima are given by, 
  • d sinθ = nλ
• Where, d = width of the slit, λ = wavelength, θ = angle, n = nth fringe

​Calculation:

Given, wavelength λ = 600 nm, for first minimum angle θ = 30°, n = 1

Then width of the slit, d sinθ = nλ

⇒ d × sin30° = 1 × 600 

⇒ d = 600 × 2 = 1200 nm = 1.2 μm

CONCEPT:

The Intensity (I) of light used are proportional to the square of the amplitude (A) of the light waves.

Mathamatically, I ∝ A²

If two waves with amplitude a and b respectively interfare with each other

Then maximum amplitude of the resiltant wave = a + b

And the minimum amplitude of the resultant wave = a - b

So, Maximum intensity I_max = (a + b)²

and minimum intensity I_min = (a - b)² 

EXPLANATION:

The interference pattern produced by two coherent sources of light is 9 ∶ 1

We know,

\(\begin{aligned} & ⇒ \frac{I_{max}}{I_{min}} = \frac{(a+b)^2}{(a-b)^2} \\ & ⇒ \frac{9}{1} = \left( \frac{a+b}{a-b} \right) ^2 \\ & ⇒ \sqrt{\frac{9}{1}} = \frac{(a+b)}{(a-b)} \\ & ⇒ \frac{3}{1} = \frac{(a+b)}{(a-b)} \\ & ⇒ 3 (a-b) = (a+b) \\ & ⇒ 3 a-a = b + 3b \\ & ⇒ 2a = 4b \\ & ⇒ \frac{a}{b} = \frac{2}{1} \\ \end{aligned}\)

Therefore the ratio of intensities of the light used,

\(\begin{aligned} & ⇒ \frac{I_1}{I_2} = \frac{a^2}{b^2} \\ & ⇒ \frac{I_1}{I_2} = \frac{4}{1} \\ \end{aligned} \)

∴ I₁ : I₂ = 4 : 1

Hence the correct answer is option 2.

Concept:

In Young's double slit experiment, the fringe width (w) is given by the equation:

w = λD/d

where λ is the wavelength of light,

D is the distance between the slits and the screen, and

d is the distance between the two slits.

Explanation:

If the separation between the slits is halved, then d is also halved. Therefore, the new value of d will be d/2.

If the distance between the slits and the screen is doubled, then the new value of D will be 2D.

Substituting these values in the above equation, we get:

w' = λ(2D)/(d/2) = 4λD/d

Therefore, the new fringe width (w') will be four times i.e., it gets quadrupled the original fringe width (w).

The correct answer is option (4)

CONCEPT:

Young's double-slit experiment

• Young’s double-slit experiment helped in understanding the wave nature of light.
• The original Young’s double-slit experiment used diffracted light from a single monochromatic source of light.
• The light that comes from the monochromatic source is passed into two slits to be used as two coherent sources.
• At any point on the screen at a distance ‘y’ from the center, the waves travel distances l1 and l2 to create a path difference of Δl at that point.
• If there is a constructive interference on the point then the bright fringe occurs.
• If there is a destructive interference on the point then the dark fringe occurs.

• The distance of the nth bright fringe from the central fringe is given as,

\(⇒ y=\frac{nλ D}{d}\)

Where d = distance between slits, D = distance between slits and screen, and λ = wavelength

​

Explanation:

β (fringe width) = \(\frac{λ \mathrm{D}}{\mathrm{~d}}\)

In denser medium, λ ↓ ⇒ β ↓

⇒ fringe come closer 

Also, µ = \(\frac{c}{V} \) ⇒ V = \(\frac{c}{μ}\)

Frequency remains same, 

⇒ μ = \(\frac{\lambda_{\text {vac. }} \mathrm{f}}{\lambda_{\text {med }} \mathrm{f}}\) ⇒ \(\lambda_{\text {med }}\) = \(\frac{\lambda_{\mathrm{vac} .}}{\mu}\)

Concept:

Young's Double Slit Experiment:

• In Young's double slit experiment, the interference pattern is formed due to the superposition of light waves from two coherent sources.
• The position of bright (constructive interference) and dark (destructive interference) bands depends on the wavelength of the light used.

Interference Condition for Bright Bands:

• The position of the nth bright band (maximum) is given by:
• y = (nλ)D/d
• Here,
  • n = order of the bright band (n = 1, 2, 3,...)
  • λ = wavelength of light
  • D = distance between the slits and the screen
  • d = distance between the slits

Calculation:

Given,

Wavelength of red light  = 7800 Å = 7800 × 10^-10 m

Wavelength of blue light = 5200 Å = 5200 × 10^-10 m

We need to find the value of n for which the nth bright band due to red light coincides with the (n + 1)th bright band due to blue light.

For red light:

Position of nth bright band, y (nr)​= (nλr​)D/d​

For blue light:

Position of (n + 1)th bright band,y(n+1b) ​= [(n+1)λb​] D/d

Given that these positions coincide:

(nλr)​D/d​= [(n+1)λb​] D/d​

⇒(nλr)​D/d ​= [(n+1)λb​] dD​

⇒nλr​=(n+1)λb​

⇒n×7800×10⁻¹⁰=(n+1)×5200×10⁻¹⁰

⇒ 7800n = 5200(n+1)

⇒ 7800n = 5200n + 5200

⇒7800n − 5200n = 5200

⇒ 2600n = 5200

⇒ n= 5200/2600​

⇒ n = 2

∴ The correct option is 2)

Explanation:

The correct option for the phase difference (Δϕ) between two superimposing waves to obtain constructive interference and hence a bright band is:
 Δϕ = np; n = 1, 2, 3, 4, 5

• For constructive interference to occur, the waves must be in phase, i.e., their crests and troughs must coincide.
• If two waves of the same frequency and amplitude meet at a point with a path difference of integer multiple of wavelength (nλ), then they will be in phase and add up constructively to produce a bright band.
• The phase difference between the two waves in this case is Δϕ = 2π(nλ/λ) = 2πn, where n is an integer.
• Since 2πn is equivalent to np, where p is the wavelength, the correct option is Δϕ = np; n = 1, 2, 3, 4, 5.

The correct answer is option "1"

The correct answer is Option 3: 0.3 mm.

Concept:

Fringe width (β): In Young’s double slit experiment, β = (λD)/d, where λ is wavelength, D is distance to screen, and d is slit separation.

Effect of changing parameters:

Fringe width is directly proportional to wavelength.

Fringe width is inversely proportional to slit separation.

Explanation:

Initial fringe width = 0.8 mm with λ = 640 nm.

New wavelength = 720 nm and slit separation becomes three times.

New fringe width = (720/640) × (1/3) × 0.8 mm.

New fringe width = (9/8) × (1/3) × 0.8 = 0.3 mm.

Calculation:

In Young's double slit experiment, the position of the n^th bright fringe is given by the equation:

\(x_{n}=\frac{n λ D}{d}\)

where:

• x_n is the position of the fringe

• n is the order number of the fringe

• λ is the wavelength of the light used

• D is the distance between the slits and the screen

• d is the distance between the two slits

In this problem, we are given that the nth bright fringe for red light coincides with the (n + 1)^th bright fringe for violet light. This means:

nλ_red = (n + 1)λ_violet

Given the wavelengths:

• λ_red = 6300° A

• λ_violet = 4200° A

We can substitute these values into the equation:

n . 6300 (n + 1) . 4200

Simplifying this equation :

6300n 4200n + 4200

2100n = 4200

n = 2

So, the value of 'n' is 2.

Thus, the correct option is: Option 1 : 2