Question 9

Question: If the monochromatic source in Young’s double slit experiment is replaced by white light, then

Options: 

A. Interference pattern will disappear

B. There will be a central dark fringe surrounded by a few coloured fringes

C. There will be a central bright white fringe surrounded by a few coloured fringes

D. All bright fringes will be of equal width

Correct Answer: C

Year: NEET 2024

Solution: At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and other fringes depend on wavelength as $\beta = \lambda D / d$. Therefore, other fringes will be coloured.

Step Solution:

1.  Understand Path Difference: At the central point of the screen in YDSE, the path difference ($\Delta x$) is $0$ for all constituent wavelengths of white light.

2.  Identify Central Fringe Color: Since all wavelengths interfere constructively at the center, they recombine to form a white central bright fringe.

3.  Analyze Other Fringes: For any other point ($n \neq 0$), the fringe position is given by $y = n\lambda D/d$.

4.  Wavelength Dependency: Because different colors have different wavelengths ($\lambda$), their fringe positions and widths ($\beta = \lambda D/d$) differ.

5.  Conclusion: This wavelength dependency causes the colors to separate, making the non-central fringes appear coloured.

The Difficulty Level: Easy

The Concept Name: Young’s Double Slit Experiment (White Light)

Short cut solution: Zero path difference at the center for all $\lambda \rightarrow$ White. Other positions depend on $\lambda \rightarrow$ Coloured.

Question 14

Question: For Young's double slit experiment, two statements are given below:

Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant.

Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.

In the light of the above statements, choose the correct answer from the options given below:

Options: 

A. Both Statement I and Statement II are false.

B. Statement I is true but Statement II is false.

C. Statement I is false but Statement II is true.

D. Both Statement I and Statement II are true.

Correct Answer: B

Year: NEET 2023

Solution: For YDSE, angular fringe width is given by $\alpha = \frac{\lambda}{d}$. It does not depend on the distance of screen from the slit, so statement I is correct. Angular fringe width $\alpha \propto \lambda$. If $\lambda \uparrow$ angular separation of fringes increases. So, statement I is true and statement II is false.

Step Solution:

1.  Formula for Angular Separation: State the formula for angular fringe width: $\alpha = \lambda / d$, where $\lambda$ is wavelength and $d$ is slit separation.

2.  Evaluate Statement I: Observe that the formula for $\alpha$ contains no variable $D$ (distance to screen). Thus, moving the screen does not change $\alpha$. Statement I is true.

3.  Evaluate Statement II (Proportionality): From the formula, it is clear that $\alpha \propto \lambda$.

4.  Analyze Wavelength Change: If a larger wavelength is used ($\lambda \uparrow$), the angular separation $\alpha$ must also increase.

5.  Final Verdict: Statement II claims $\alpha$ decreases, which is incorrect. Therefore, Statement II is false.

The Difficulty Level: Medium

The Concept Name: Angular Fringe Width in YDSE

Short cut solution: $\alpha = \lambda / d$. It is independent of screen distance ($D$) and directly proportional to wavelength ($\lambda$).

Question 22

Question: If the screen is moved away from the plane of the slits in a Young's double slit experiment, then the :

Options: 

A. linear separation of the fringes decreases

B. angular separation of the fringes increases

C. angular separation of the fringes decreases

D. linear separation of the fringes increases

Correct Answer: D

Year: NEET Re-2022

Solution: We know fringe width $\beta = \frac{\lambda D}{d}$. As $D$ increases $\beta$ increases. i.e., Linear separation of fringes increases.

Step Solution:

1.  Linear Fringe Width Formula: State the formula for linear fringe width (separation): $\beta = \frac{\lambda D}{d}$.

2.  Identify Variables: $\lambda$ is wavelength, $d$ is slit separation, and $D$ is the distance between slits and screen.

3.  Analyze the Change: The problem states the screen is moved away, meaning $D$ increases.

4.  Relate Width to Distance: Since $\beta$ is directly proportional to $D$ ($\beta \propto D$), an increase in $D$ results in an increase in $\beta$.

5.  Conclusion: Therefore, the linear separation of the fringes increases.

The Difficulty Level: Easy

The Concept Name: Fringe Width in YDSE

Short cut solution: Linear width $\beta = \lambda D / d$. If $D$ (distance to screen) increases, $\beta$ must increase.

Question 25

Question: In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600nm wavelength is used. If the wavelength of light is changed to 400nm, then the number of fringes he would observe in the same region of the screen is

Options: 

A. 6

B. 8

C. 3

D. 12

Correct Answer: D

Year: NEET-2022

Solution: $\beta = \frac{\lambda D}{d}$. Let length of segment of screen $= 1 \Rightarrow I = 8 \beta_1 = \frac{8 \lambda_1 D}{d} \dots (1)$ and $I = n \beta_2 = \frac{n \lambda_2 D}{d} \dots (2)$. From (1) and (2) $8 \lambda_1 = n \lambda_2 \Rightarrow 8 (600 nm) = n (400 nm) \Rightarrow n = 12$.

Step Solution:

1.  Define Segment Length: The length of the screen segment ($L$) is the number of fringes ($n$) multiplied by the fringe width ($\beta$), so $L = n\beta$.

2.  Substitute Fringe Width: Using $\beta = \frac{\lambda D}{d}$, the length remains constant: $L = \frac{n \lambda D}{d}$.

3.  Establish Proportionality: Since $D$, $d$, and $L$ are unchanged, the relation is $n_1 \lambda_1 = n_2 \lambda_2$.

4.  Insert Values: Substitute $n_1 = 8$, $\lambda_1 = 600 \text{ nm}$, and $\lambda_2 = 400 \text{ nm}$ into the equation: $8 \times 600 = n_2 \times 400$.

5.  Final Calculation: $n_2 = \frac{4800}{400} = \mathbf{12}$.

The difficulty level: Easy

The Concept Name: Fringe Width in YDSE

Short cut solution: Number of fringes is inversely proportional to wavelength ($n \propto 1/\lambda$). Thus, $n_2 = 8 \times (\frac{600}{400}) = 8 \times 1.5 = \mathbf{12}$.

Question 32

Question: In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :

Options: 

A. half

B. four times

C. one-fourth

D. double

Correct Answer: B

Year: 2020

Solution: Fringe width $\beta = \frac{\lambda D}{d}$. When, $d' = d/2$ and $D' = 2D$, New Fringe width, $\beta' = \frac{\lambda(2D)}{d/2} = \frac{4 \lambda D}{d} \Rightarrow \beta' = 4 \beta$. Fringe width becomes 4 times.

Step Solution:

1.  State Initial Formula: The standard fringe width is $\beta = \frac{\lambda D}{d}$.

2.  Identify Changes: According to the problem, the new distance $D' = 2D$ and new separation $d' = d/2$.

3.  Set up New Formula: Substitute the new values into the fringe width equation: $\beta' = \frac{\lambda D'}{d'}$.

4.  Substitute and Simplify: $\beta' = \frac{\lambda (2D)}{(d/2)} = \frac{2 \times 2 \times \lambda D}{d}$.

5.  Final Result: $\beta' = 4 \left( \frac{\lambda D}{d} \right) = \mathbf{4\beta}$.

The difficulty level: Easy

The Concept Name: Fringe Width in YDSE

Short cut solution: Since $\beta \propto \frac{D}{d}$, doubling the numerator and halving the denominator results in a factor of $2 \div 0.5 = \mathbf{4}$ times.

Question 40

Question: In a double slit experiment, when light of wavelength 400nm was used, the angular width of the first minima formed on a screen placed 1m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? ($\mu_{water} = 4/3$)

Options: 

A. 0.1°

B. 0.266°

C. 0.15°

D. 0.05°

Correct Answer: C

Year: NEET 2019

Solution: Angular width for first minima in Young's double slit experiment, $\theta = \frac{\lambda}{a}$. For given value of $a$, $\theta \propto \lambda$. $\frac{\theta}{\theta_w} = \frac{\lambda}{\lambda_w} = \frac{\lambda}{\lambda/\mu} = \mu \Rightarrow \theta_w = \frac{\theta}{\mu} = \frac{0.2^\circ}{4/3} = 0.15^\circ$.

Step Solution:

1.  Formula for Angular Width: The angular width ($\theta$) is given by $\theta = \frac{\lambda}{d}$.

2.  Effect of Medium: When immersed in water, the wavelength changes to $\lambda_w = \frac{\lambda_{air}}{\mu}$.

3.  Establish Relation: Since the slit separation $d$ is constant, $\theta$ is directly proportional to $\lambda$ ($\theta \propto \lambda$).

4.  Substitute Values: The new angular width is $\theta_w = \frac{\theta_{air}}{\mu}$.

5.  Final Calculation: $\theta_w = \frac{0.2^\circ}{4/3} = 0.2 \times \frac{3}{4} = \frac{0.6}{4} = \mathbf{0.15^\circ}$.

The difficulty level: Medium

The Concept Name: Angular Fringe Width in Media

Short cut solution: The angular width in a medium is simply the air width divided by the refractive index: $0.2^\circ / 1.33 = \mathbf{0.15^\circ}$.

Question 42

Question: In a Young' double slit experiment if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference.

Options: 

A. $20 \lambda / 2$

B. $10 \lambda / 2$

C. $9 \lambda / 2$

D. $11 \lambda / 2$

Correct Answer: C

Year: OD NEET 2019

Solution: Given, there is no initial phase difference. ∴ Initial phase $= \delta = 0$. Again, phase difference $= \frac{2\pi}{\lambda} \times$ path difference $\Rightarrow \delta' = \frac{2\pi}{\lambda} \times \Delta x \Rightarrow \Delta x = \frac{\lambda}{2\pi} \times \delta'$. Now, for the fifth minima we will consider $n = 4$ as initial phase difference is zero. ∴ For fifth minimum, $\delta = (8 + 1)\pi = 9\pi$. ∴ Path difference, $\Delta x = \frac{\lambda}{2\pi} \times 9\pi = \frac{9\lambda}{2}$.

Step Solution:

1.  State the condition for minima in Young's Double Slit Experiment (YDSE): path difference $\Delta x = (2n - 1) \frac{\lambda}{2}$, where $n$ is the order of the minimum.

2.  For the fifth minimum, set the value of $n = 5$.

3.  Substitute $n$ into the formula: $\Delta x = (2 \times 5 - 1) \frac{\lambda}{2}$.

4.  Perform the arithmetic calculation: $\Delta x = (10 - 1) \frac{\lambda}{2}$.

5.  Result: $\Delta x = \mathbf{\frac{9\lambda}{2}}$.

The difficulty level: Medium

The Concept Name: Condition for Minima in YDSE

Short cut solution: The path difference for the $n^{th}$ minimum is always $(n - 0.5)\lambda$. For $n=5$, $\Delta x = (5 - 0.5)\lambda = 4.5\lambda = \mathbf{9\lambda/2}$.

Question 46

Question: In Young's double slit experiment the separation $d$ between the slits is 2mm the wavelength $\lambda$ of the light used is 5896 Å and distance $D$ between the screen and slits is 100cm. It is found that the angular width of the fringes is $0.20^\circ$. To increase the fringe angular width to $0.21^\circ$ (with same $\lambda$ and $D$) the separation between the slits needs to be changed to

Options: 

A. 1.8mm

B. 1.9mm

C. 2.1mm

D. 1.7mm

Correct Answer: B

Year: NEET 2018

Solution: Angular width $= \frac{\lambda}{d}$. $0.20^\circ = \frac{\lambda}{2\text{ mm}}$ and $0.21^\circ = \frac{\lambda}{d}$. Dividing we get, $\frac{0.20}{0.21} = \frac{d}{2\text{ mm}}$. $\therefore d = 1.9\text{ mm}$.

Step Solution:

1.  Identify the formula for angular fringe width ($\theta$): $\theta = \frac{\lambda}{d}$, where $d$ is the slit separation.

2.  Since $\lambda$ remains constant, the relationship between angular widths and separations is $\theta_1 d_1 = \theta_2 d_2$.

3.  Substitute the given values: $0.20^\circ \times 2\text{ mm} = 0.21^\circ \times d_{new}$.

4.  Rearrange to solve for the new separation: $d_{new} = \frac{0.20 \times 2}{0.21}$.

5.  Calculate the result: $d_{new} = \frac{0.40}{0.21} \approx \mathbf{1.9\text{ mm}}$.

The difficulty level: Easy

The Concept Name: Angular Fringe Width in YDSE

Short cut solution: Use the inverse proportionality $\theta \propto \frac{1}{d}$. If the width increases from $0.20$ to $0.21$, the separation must decrease: $d_{new} = 2 \times \frac{0.20}{0.21} = \mathbf{1.9\text{ mm}}$.

Question 49

Question: Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly

Options: 

A. 1.59

B. 1.69

C. 1.78

D. 1.25

Correct Answer: C

Year: 2017 NEET

Solution: Position of 8th bright fringe in medium, $x = \frac{8 \lambda_m D}{d}$. Position of 5th dark fringe in air, $x' = \frac{(5 - \frac{1}{2}) \lambda_{air} D}{d} = \frac{4.5 \lambda_{air} D}{d}$. Given $x = x'$. $\therefore \frac{8 \lambda_m D}{d} = \frac{4.5 \lambda_{air} D}{d}$. $\mu_m = \frac{\lambda_{air}}{\lambda_m} = \frac{8}{4.5} \approx 1.78$.

Step Solution:

1.  Write the position of the $8^{th}$ bright fringe in a medium ($\lambda_m = \lambda/\mu$): $y_8 = \frac{8 \lambda D}{\mu d}$.

2.  Write the position of the $5^{th}$ dark fringe in air: $y'_5 = \frac{(5 - 0.5) \lambda D}{d} = \frac{4.5 \lambda D}{d}$.

3.  Equate the two positions as per the problem statement: $\frac{8 \lambda D}{\mu d} = \frac{4.5 \lambda D}{d}$.

4.  Cancel the common terms ($\lambda, D, d$) to simplify the equation: $\frac{8}{\mu} = 4.5$.

5.  Solve for the refractive index ($\mu$): $\mu = \frac{8}{4.5} = \frac{16}{9} \approx \mathbf{1.78}$.

The difficulty level: Medium

The Concept Name: Fringe Position and Refractive Index in YDSE

Short cut solution: Use the equality of optical paths: $n_{bright} \times \frac{\lambda}{\mu} = (m_{dark} - 0.5) \lambda$. For $n=8$ and $m=5$, $\frac{8}{\mu} = 4.5$, so $\mu = \frac{8}{4.5} = \mathbf{1.78}$.

Question 56

Question: The intensity at the maximum in a Young’s double slit experiment is $I_0$. Distance between two slits is $d = 5\lambda$, where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

Options: 

A. $\frac{3}{4}I_0$

B. $\frac{I_0}{2}$

C. $I_0$

D. $\frac{I_0}{4}$

Correct Answer: B

Year: 2016 NEET Phase-I

Solution: Here, $d = 5\lambda$, $D = 10d$, $y = \frac{d}{2}$. Resultant Intensity at $y = \frac{d}{2}, I_y = ?$ The path difference between two waves at $y = \frac{d}{2}$ is $\Delta x = d \tan \theta = d \times \frac{y}{D} = \frac{d \times \frac{d}{2}}{10d} = \frac{d}{20} = \frac{5\lambda}{20} = \frac{\lambda}{4}$. Corresponding phase difference, $\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{\pi}{2}$. Now, maximum intensity $I_{max} = I_0$. Required intensity $I = I_{max} \cos^2(\frac{\phi}{2}) = I_0 \cos^2(\frac{\pi}{4}) = \frac{I_0}{2}$.

Step Solution:

1.  Determine Point Position: In front of one slit, the distance from the center is $y = d/2$.

2.  Calculate Path Difference: $\Delta x = \frac{yd}{D} = \frac{(d/2) \cdot d}{10d} = \frac{d}{20}$.

3.  Substitute $d = 5\lambda$: $\Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4}$.

4.  Find Phase Difference: $\phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$ radians.

5.  Calculate Intensity: $I = I_{max} \cos^2(\frac{\phi}{2}) = I_0 \cos^2(\frac{\pi}{4}) = I_0 \cdot (\frac{1}{\sqrt{2}})^2 = \mathbf{\frac{I_0}{2}}$.

The difficulty level: Hard

The Concept Name: Intensity in Young’s Double Slit Experiment

Short cut solution: Path difference at $y=d/2$ is $d^2/2D$. Given $D=10d$, $\Delta x = d/20 = 5\lambda/20 = \lambda/4$. A path difference of $\lambda/4$ corresponds to a phase of $\pi/2$, which always yields half of the maximum intensity.

Question 61

Question: The interference pattern is obtained with two coherent light sources of intensity ratio $n$. In the interference pattern the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ will be

Options: 

A. $\frac{\sqrt{n}}{n+1}$

B. $\frac{2\sqrt{n}}{n+1}$

C. $\frac{\sqrt{n}}{(n+1)^2}$

D. $\frac{2\sqrt{n}}{(n+1)^2}$

Correct Answer: B

Year: 2016 NEET Phase-II

Solution: Here, $\frac{I_1}{I_2} = n$. $\frac{I_{max}}{I_{min}} = (\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}})^2 = (\frac{\sqrt{n} + 1}{\sqrt{n} - 1})^2$. $\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{(\frac{\sqrt{n} + 1}{\sqrt{n} - 1})^2 - 1}{(\frac{\sqrt{n} + 1}{\sqrt{n} - 1})^2 + 1} = \frac{(\sqrt{n} + 1)^2 - (\sqrt{n} - 1)^2}{(\sqrt{n} + 1)^2 + (\sqrt{n} - 1)^2} = \frac{4\sqrt{n}}{2(n+1)} = \frac{2\sqrt{n}}{n+1}$.

Step Solution:

1.  Intensity Ratio: Let $I_1 = nI$ and $I_2 = I$.

2.  Define Max/Min: $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.

3.  Find $I_{max} - I_{min}$: $(\sqrt{nI} + \sqrt{I})^2 - (\sqrt{nI} - \sqrt{I})^2 = 4\sqrt{n}I$.

4.  Find $I_{max} + I_{min}$: $(\sqrt{nI} + \sqrt{I})^2 + (\sqrt{nI} - \sqrt{I})^2 = 2(nI + I) = 2I(n+1)$.

5.  Calculate Ratio: $\frac{4\sqrt{n}I}{2I(n+1)} = \mathbf{\frac{2\sqrt{n}}{n+1}}$.

The difficulty level: Medium

The Concept Name: Fringe Visibility in Interference

Short cut solution: The requested ratio is the definition of fringe visibility ($V$). For two sources, $V = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$. Dividing numerator and denominator by $I_2$, we get $\frac{2\sqrt{I_1/I_2}}{(I_1/I_2) + 1} = \mathbf{\frac{2\sqrt{n}}{n+1}}$.

Question 66

Question: In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

Options: 

A. 0.5 mm

B. 0.02 mm

C. 0.2 mm

D. 0.1 mm

Correct Answer: C

Year: 2015 Cancelled

Solution: For double slit experiment, $d = 1\text{ mm} = 10^{-3}\text{ m}, D = 1\text{ m}, \lambda = 500 \times 10^{-9}\text{ m}$. Fringe width $\beta = \frac{D\lambda}{d}$. Width of central maxima in a single slit $= \frac{2\lambda D}{a}$. As per question, width of central maxima of single slit pattern = width of 10 maxima of double slit pattern. $\frac{2\lambda D}{a} = 10(\frac{\lambda D}{d}) \Rightarrow a = \frac{2d}{10} = \frac{2 \times 10^{-3}}{10} = 0.2\text{ mm}$.

Step Solution:

1.  State Central Maxima Width: The angular width of the central maxima in diffraction is $\frac{2\lambda}{a}$.

2.  State Interference Maxima Width: The angular width of 10 interference maxima is $10 \cdot \frac{\lambda}{d}$.

3.  Equate Widths: $\frac{2\lambda}{a} = \frac{10\lambda}{d}$.

4.  Simplify for Slit Width ($a$): $a = \frac{2d}{10} = \frac{d}{5}$.

5.  Calculate Value: $a = \frac{1\text{ mm}}{5} = \mathbf{0.2\text{ mm}}$.

The difficulty level: Medium

The Concept Name: Superposition of Diffraction and Interference

Short cut solution: Use the relation $a = \frac{2d}{n}$, where $n$ is the number of interference maxima within the central diffraction envelope. For $n=10$, $a = \frac{2 \times 1\text{ mm}}{10} = \mathbf{0.2\text{ mm}}$.

Question 71

Question: Two slits in Youngs experiment have widths in the ratio 1:25. The ratio of intensity at the maxima and minima in the interference pattern, $I_{max} / I_{min}$ is

Options: 

A. 49/121

B. 4/9

C. 9/4

D. 121/9

Correct Answer: C

Year: 2015

Solution: As, intensity $I \propto$ width of slit $W$. Also, intensity $I \propto$ square of amplitude $A$. $\therefore \frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{A_1^2}{A_2^2}$. But $\frac{W_1}{W_2} = \frac{1}{25}$ (given). $\therefore \frac{A_1^2}{A_2^2} = \frac{1}{25}$ or $\frac{A_1}{A_2} = \sqrt{\frac{1}{25}} = \frac{1}{5}$. $\therefore \frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{(\frac{A_1}{A_2} + 1)^2}{(\frac{A_1}{A_2} - 1)^2} = \frac{(\frac{1}{5} + 1)^2}{(\frac{1}{5} - 1)^2} = \frac{(\frac{6}{5})^2}{(- \frac{4}{5})^2} = \frac{36}{16} = \frac{9}{4}$.

Step Solution:

1.  Relate Slit Width to Intensity: Given slit width ratio $W_1/W_2 = 1/25$. Since intensity $I \propto W$, then $I_1/I_2 = 1/25$.

2.  Determine Amplitude Ratio: Intensity is proportional to the square of amplitude ($I \propto A^2$), so $\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{1}{25}} = \frac{1}{5}$.

3.  State Max/Min Intensity Formula: The ratio of intensities is given by $\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2}$.

4.  Substitute Values: Assume $A_1 = 1$ and $A_2 = 5$. Thus, $\frac{I_{max}}{I_{min}} = \frac{(1 + 5)^2}{(1 - 5)^2} = \frac{6^2}{(-4)^2}$.

5.  Calculate Final Ratio: $\frac{36}{16} = \mathbf{\frac{9}{4}}$.

The difficulty level: Medium

The Concept Name: Intensity and Slit Width in Young's Double Slit Experiment

Short cut solution: Use $\frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{W_2} + \sqrt{W_1}}{\sqrt{W_2} - \sqrt{W_1}} \right)^2 = \left( \frac{5 + 1}{5 - 1} \right)^2 = \left( \frac{6}{4} \right)^2 = \mathbf{\frac{9}{4}}$.

Question 73

Question: In the Young’s double slit experiment, the intensity of light at a point on the screen where the path difference $\lambda$ is K, ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\frac{\lambda}{4}$ will be

Options: 

A. K

B. K/4

C. K/2

D. zero

Correct Answer: C

Year: 2014

Solution: Intensity at any point on the screen is $I = 4I_0 \cos^2 \frac{\phi}{2}$ where $I_0$ is the intensity of either wave and $\phi$ is the phase difference between two waves. Phase difference, $\phi = \frac{2\pi}{\lambda} \times$ Path difference. When path difference is $\lambda$, then $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$. $\therefore I = 4I_0 \cos^2 (\frac{2\pi}{2}) = 4I_0 \cos^2 (\pi) = 4I_0 = K$. When path difference is $\frac{\lambda}{4}$, then $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$. $\therefore I' = 4I_0 \cos^2 (\frac{\pi}{4}) = 2I_0 = \frac{K}{2}$.

Step Solution:

1.  Calculate Initial Phase: For path difference $\Delta x = \lambda$, the phase difference $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.

2.  Determine Maximum Intensity: Using $I = I_{max} \cos^2(\phi/2)$, we get $K = I_{max} \cos^2(\pi) = I_{max}$. Thus, $I_{max} = K$.

3.  Calculate New Phase: For path difference $\Delta x = \lambda/4$, the new phase difference $\phi' = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.

4.  Substitute into Intensity Formula: New intensity $I' = K \cos^2(\frac{\pi/2}{2}) = K \cos^2(\frac{\pi}{4})$.

5.  Calculate Final Intensity: $I' = K \times (\frac{1}{\sqrt{2}})^2 = \mathbf{\frac{K}{2}}$.

The difficulty level: Medium

The Concept Name: Intensity Distribution in YDSE / Phase-Path Relationship

Short cut solution: A path difference of $\lambda$ corresponds to constructive interference (Intensity $K$). A path difference of $\lambda/4$ corresponds to a $90^\circ (\pi/2)$ phase difference. The intensity at $90^\circ$ phase difference is always $I_{max} / 2 = K/2$.

Question 77

Question: In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths $\lambda_1 = 12000 \text{ \AA}$ and $\lambda_2 = 10000 \text{ \AA}$. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Options: 

A. 4 mm

B. 3 m

C. 8 mm

D. 6 mm

Correct Answer: D

Year: 2013 NEET

Solution: Let $n_1$ bright fringe of $\lambda_1$ coincides with $n_2$ bright fringe of $\lambda_2$. Then $\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$ or $n_1 \lambda_1 = n_2 \lambda_2$. $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{10000}{12000} = \frac{5}{6}$. Let $x$ be given distance. $\therefore X = \frac{n_1 \lambda_1 D}{d}$. Here, $n_1 = 5, D = 2 m, d = 2 mm = 2 \times 10^{-3} m$. $\lambda_1 = 12000 \text{ \AA} = 12 \times 10^{-7} m$. $x = \frac{5 \times 12 \times 10^{-7} m \times 2 m}{2 \times 10^{-3} m} = 6 \times 10^{-3} m = 6 mm$.

Step Solution:

1.  Equate Fringe Positions: For coincidence, $n_1 \lambda_1 = n_2 \lambda_2$.

2.  Find Order of Coincidence: $\frac{n_1}{n_2} = \frac{10000}{12000} = \frac{5}{6}$. The first coincidence occurs at $n_1 = 5$ and $n_2 = 6$.

3.  State Position Formula: $x = \frac{n_1 \lambda_1 D}{d}$.

4.  Substitute Parameters: $x = \frac{5 \times (12000 \times 10^{-10}) \times 2}{2 \times 10^{-3}}$.

5.  Perform Final Calculation: $x = \frac{5 \times 12 \times 10^{-7} \times 2}{2 \times 10^{-3}} = \mathbf{6 \times 10^{-3} \text{ m} = 6 \text{ mm}}$.

The difficulty level: Hard

The Concept Name: Coinciding Fringes in Young's Double Slit Experiment

Short cut solution: Use $n_1 \lambda_1 = n_2 \lambda_2$ to find $5 \times 1.2 = 6 \times 1.0$. The distance is $5 \times \beta_1 = 5 \times \frac{(1.2 \times 10^{-3} \text{ mm}) \times 2}{2 \times 10^{-3}} = \mathbf{6 \text{ mm}}$.

Question 120

Question: Colours appear on a thin soap film and on soap bubbles due to the phenomenon of

Options:

A. interference

B. dispersion

C. refraction

D. diffraction

Correct Answer: A

Year: 1999

Solution: (Source provides the answer "interference" but does not detail the mechanism; however, the correct mechanism is the superposition of light waves reflected from the top and bottom surfaces of the thin film).

Step Solution:

1.  Light incident on a thin soap film reflects from both the upper and lower surfaces.

2.  The two reflected waves are coherent as they originate from the same source.

3.  A path difference is created between these two waves depending on the film's thickness and the refractive index.

4.  Superposition occurs between these waves, leading to constructive or destructive interference.

5.  Since white light contains various wavelengths, different colors satisfy the condition for constructive interference at different points, resulting in the appearance of colors.

The difficulty level: Easy

The Concept Name: Interference in Thin Films

Short cut solution: The colorful appearance of thin layers (oil on water, soap bubbles) is a classic example of interference.

Question 134

Question: In a Fresnel biprism experiment, the two positions of lens give separation between the slits as 16 cm and 9 cm respectively. What is the actual distance of separation?

Options:

A. 13 cm

B. 14 cm

C. 12.5 cm

D. 12 cm

Correct Answer: D

Year: 1995

Solution: Separations between the slits $(d_1) = 16 \text{ cm}$ and $(d_2) = 9 \text{ cm}$. Actual distance of separation $(d) = \sqrt{d_1 d_2} = \sqrt{16 \times 9} = 12 \text{ cm}$.

Step Solution:

1.  Identify the given values for the two virtual images of the slits: $d_1 = 16 \text{ cm}$ and $d_2 = 9 \text{ cm}$.

2.  Use the displacement method principle for a biprism, which states that the actual separation $d$ is the geometric mean of the separations at two positions.

3.  State the formula: $d = \sqrt{d_1 \times d_2}$.

4.  Substitute the values: $d = \sqrt{16 \times 9}$.

5.  Calculate the result: $\sqrt{144} = \mathbf{12 \text{ cm}}$.

The difficulty level: Easy

The Concept Name: Fresnel Biprism / Displacement Method

Short cut solution: Use the geometric mean: $d = \sqrt{16 \times 9} = 4 \times 3 = 12 \text{ cm}$.

Question 140

Question: Interference was observed in interference chamber where air was present, now the chamber is evacuated, and if the same light is used, a careful observer will see

Options:

A. no interference

B. interference with brighter bands

C. interference with dark bands

D. interference with larger width

Correct Answer: D

Year: 1993

Solution: In vacuum, $\lambda$ increases very slightly compared to that in air. As $\beta \propto \lambda$, therefore, width of interference fringe increases slightly.

Step Solution:

1.  The wavelength of light in a medium is given by $\lambda_m = \lambda_0 / \mu$, where $\mu$ is the refractive index.

2.  For air, $\mu > 1$ (slightly), and for vacuum (evacuated chamber), $\mu = 1$.

3.  Therefore, the wavelength in vacuum ($\lambda_{vac}$) is slightly larger than the wavelength in air ($\lambda_{air}$).

4.  The fringe width ($\beta$) in interference is given by $\beta = \lambda D / d$.

5.  Since $\beta$ is directly proportional to $\lambda$, the increase in wavelength results in a larger fringe width.

The difficulty level: Medium

The Concept Name: Fringe Width Dependency on Medium

Short cut solution: Evacuating air reduces the refractive index to its minimum (1), which increases the wavelength, and since $\beta \propto \lambda$, the fringe width increases.

Question 143

Question: If yellow light emitted by sodium lamp in Young's double slit experiment is replaced by monochromatic blue light of the same intensity

Options: 

A. fringe width will decrease

B. fringe width will increase

C. fringe width will remain unchanged

D. fringes will becomes less intense

Correct Answer: A

Year: 1992

Solution: As $\beta = \frac{\lambda D}{d}$ and $\lambda_b < \lambda_y$, $\therefore$ Fringe width $\beta$ will decrease.

Step Solution:

1.  State the Formula: The fringe width ($\beta$) in Young’s double slit experiment is given by $\beta = \frac{\lambda D}{d}$.

2.  Identify Proportionality: Note that the fringe width is directly proportional to the wavelength ($\beta \propto \lambda$).

3.  Compare Wavelengths: Blue light has a shorter wavelength than yellow light ($\lambda_{blue} < \lambda_{yellow}$).

4.  Analyze the Change: Since the source is switched to a shorter wavelength, the value of $\beta$ must decrease.

The difficulty level: Easy

The Concept Name: Fringe Width in YDSE

Short cut solution: Since $\beta \propto \lambda$ and blue light has the shorter wavelength in the visible spectrum compared to yellow, the fringes must become narrower (decrease).

Question 144

Question: In Young's double slit experiment carried out with light of wavelength $\lambda = 5000\text{ \AA}$, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at $X = 0$. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to

Options: 

A. 1.67 cm

B. 1.5 cm

C. 0.5 cm

D. 5.0 cm

Correct Answer: B

Year: 1992

Solution: $x = (n) \lambda \frac{D}{d} = 3 \times 5000 \times 10^{-10} \times \frac{2}{0.2 \times 10^{-3}} = 1.5 \times 10^{-2} \text{ m} = 1.5 \text{ cm}$.

Step Solution:

1.  List Given Values: $n = 3$, $\lambda = 5000 \times 10^{-10} \text{ m}$, $D = 2 \text{ m}$, and $d = 0.2 \times 10^{-3} \text{ m}$.

2.  State the Position Formula: The distance of the $n^{th}$ bright fringe from the center is $x = \frac{n\lambda D}{d}$.

3.  Substitute Values: $x = \frac{3 \times (5 \times 10^{-7}) \times 2}{2 \times 10^{-4}}$.

4.  Perform Calculation: $x = \frac{30 \times 10^{-7}}{2 \times 10^{-4}} = 15 \times 10^{-3} \text{ m}$.

5.  Convert to cm: $15 \times 10^{-3} \text{ m} = \mathbf{1.5 \text{ cm}}$.

The difficulty level: Medium

The Concept Name: Position of Maxima in YDSE

Short cut solution: Calculate fringe width $\beta = \frac{\lambda D}{d} = \frac{5 \times 10^{-7} \times 2}{2 \times 10^{-4}} = 0.5 \text{ cm}$. For the $3^{rd}$ maximum, $x = 3\beta = 3 \times 0.5 = \mathbf{1.5 \text{ cm}}$.

Question 147

Question: Ratio of intensities of two waves are given by 4 : 1. Then ratio of the amplitudes of the two waves is

Options: 

A. 2: 1

B. 1: 2

C. 4: 1

D. 1: 4

Correct Answer: A

Year: 1991

Solution: $\frac{I_1}{I_2} = \frac{a^2}{b^2} = \frac{4}{1} \therefore \frac{a}{b} = \frac{2}{1}$.

Step Solution:

1.  State the Relation: Intensity ($I$) of a wave is directly proportional to the square of its amplitude ($a$), expressed as $I \propto a^2$.

2.  Set up the Ratio: $\frac{I_1}{I_2} = \frac{a_1^2}{a_2^2}$.

3.  Substitute Given Values: $\frac{4}{1} = \frac{a_1^2}{a_2^2}$.

4.  Solve for Amplitude Ratio: Take the square root of both sides: $\sqrt{\frac{4}{1}} = \frac{a_1}{a_2}$.

5.  Calculate Result: $\frac{a_1}{a_2} = \mathbf{\frac{2}{1}}$.

The difficulty level: Easy

The Concept Name: Intensity and Amplitude Relationship

Short cut solution: The amplitude ratio is simply the square root of the intensity ratio: $\sqrt{4} : \sqrt{1} = \mathbf{2 : 1}$.

Question 148

Question: In Young's experiment, two coherent sources are placed 0.90 mm apart and fringes are observed one metre away. If it produces second dark fringe at a distance of 1 mm from central fringe, the wavelength of monochromatic light is used would be

Options: 

A. $60 \times 10^{-4}$ cm

B. $10 \times 10^{-4}$ cm

C. $.10 \times 10^{-5}$ cm

D. $6 \times 10^{-5}$ cm

Correct Answer: D

Year: 1991

Solution: (as Given in the Source) For dark fringe, $x = (2n - 1) \frac{\lambda D}{2d} \Rightarrow \lambda = \frac{2xd}{(2n - 1)D} = \frac{2 \times 10^{-3} \times 0.9 \times 10^{-3}}{(2 \times 2 - 1) \times 1} \Rightarrow \lambda = 0.6 \times 10^{-6} \text{ m} = 6 \times 10^{-5} \text{ cm}$.

Step Solution:

1. Identify given values: Slit separation $d = 0.90 \text{ mm} = 0.9 \times 10^{-3} \text{ m}$, screen distance $D = 1 \text{ m}$, fringe position $x = 1 \text{ mm} = 10^{-3} \text{ m}$, and order $n = 2$ for the second dark fringe.

2. State the condition for dark fringes: The distance from the central fringe is given by the formula $x = (2n - 1) \frac{\lambda D}{2d}$.

3. Rearrange for wavelength: $\lambda = \frac{2xd}{(2n - 1)D}$.

4. Substitute the values: $\lambda = \frac{2 \times (10^{-3} \text{ m}) \times (0.9 \times 10^{-3} \text{ m})}{(2 \times 2 - 1) \times 1 \text{ m}} = \frac{1.8 \times 10^{-6}}{3} = 0.6 \times 10^{-6} \text{ m}$.

5. Convert units: $0.6 \times 10^{-6} \text{ m} = 6 \times 10^{-7} \text{ m} = \mathbf{6 \times 10^{-5} \text{ cm}}$.

The difficulty level: Medium

The Concept Name: Condition for Dark Fringes in YDSE

Short cut solution: Use $\lambda = \frac{2xd}{3D}$ for $n=2$. $\lambda = \frac{2 \times 1 \times 0.9}{3 \times 1000} = 0.6 \times 10^{-3} \text{ mm} = \mathbf{6 \times 10^{-5} \text{ cm}}$.

Question 149

Question: In Young's double slit experiment, the fringes width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index $4/3$, without disturbing the geometrical arrangement, the new fringe width will be

Options: 

A. 0.30 mm

B. 0.40 mm

C. 0.53 mm

D. 450 microns

Correct Answer: A

Year: 1990

Solution: (as Given in the Source) $\beta' = \frac{\beta}{\mu} = \frac{0.4}{4/3} = 0.3 \text{ mm}$.

Step Solution:

1. Identify parameters: Initial fringe width $\beta = 0.4 \text{ mm}$ and refractive index of the medium $\mu = 4/3$.

2. State the fringe width formula: $\beta = \frac{\lambda D}{d}$, which shows that fringe width is directly proportional to wavelength ($\beta \propto \lambda$).

3. Determine wavelength in medium: The wavelength in water is $\lambda' = \lambda / \mu$.

4. Relate new fringe width: Since the geometry ($D$ and $d$) is unchanged, the new width is $\beta' = \beta / \mu$.

5. Calculate result: $\beta' = \frac{0.4 \text{ mm}}{4/3} = 0.4 \times \frac{3}{4} = \mathbf{0.3 \text{ mm}}$.

The difficulty level: Easy

The Concept Name: Effect of Medium on Fringe Width

Short cut solution: Divide the air fringe width by the refractive index: $0.4 \text{ mm} \div 1.33 = \mathbf{0.3 \text{ mm}}$.

Question 150

Question: The Young's double slit experiment is performed with blue and with green light (wavelengths 4360 Å and 5460 Å respectively). If x is the distance of 4th maxima from the central one, then

Options:

A. x (blue) = x (green)

B. x (blue) > x (green)

C. x (blue) < x (green)

D. x (blue) / x (green) = 5460 / 4360

Correct Answer: C

Year: 1990

Solution: (as Given in the Source) Distance of $n^{th}$ maxima $x = n \lambda \frac{D}{d} \propto \lambda$. As $\lambda_b < \lambda_g$, $\therefore$ x(blue) < x(green).

Step Solution:

1.  State the formula for the position of the $n^{th}$ bright fringe (maxima) from the center: $x = \frac{n \lambda D}{d}$.

2.  Identify the constants in this experiment: the order $n = 4$, the screen distance $D$, and the slit separation $d$.

3.  Establish the proportionality: Since $n, D,$ and $d$ are constant, the distance is directly proportional to the wavelength ($x \propto \lambda$).

4.  Compare the given wavelengths: $\lambda_{\text{blue}} = 4360 \text{ \AA}$ and $\lambda_{\text{green}} = 5460 \text{ \AA}$.

5.  Determine the relationship: Since $\lambda_{\text{blue}} < \lambda_{\text{green}}$, it follows that $x(\text{blue}) < x(\text{green})$.

The difficulty level: Easy

The Concept Name: Position of Maxima in YDSE

Short cut solution: In Young's Double Slit Experiment, the fringe pattern scales with wavelength ($x \propto \lambda$). Because blue light has a shorter wavelength than green light, its 4th maximum will be closer to the center.

Question 151

Question: Interference is possible in

Options:

A. light waves only

B. sound waves only

C. both light and sound waves

D. neither light nor sound waves.

Correct Answer: C

Year: 1989

Solution: (as Given in the Source) Interference is a wave phenomenon shown by both the light waves and sound waves.

Step Solution:

1.  Define interference as a phenomenon of superposition that occurs when two or more waves overlap.

2.  Recognize that interference is a fundamental property of all wave types.

3.  Identify that light is a transverse electromagnetic wave.

4.  Identify that sound is a longitudinal mechanical wave.

5.  Conclude that since both light and sound behave as waves, interference is possible in both.

The difficulty level: Easy

The Concept Name: Wave Nature of Interference

Short cut solution: Interference is a universal wave phenomenon. It is not restricted by the type of wave (longitudinal or transverse), thus it occurs in both sound and light.