Question

What property of light does this interference experiment demonstrate?

• A. Wave nature of light
• B. Particle nature of light
• C. Transverse nature of light
• D. Both wave nature and transverse nature of light
• A. 720 nm
• B. 590 nm
• C. 480 nm
• D. 364 nm
• A. 1.2 mm
• B. 0.2 mm
• C. 4.2 mm
• D. 6.8 mm
• A. \( 8.1 \times 10^{-7} \, \text{m} \)
• B. \( 7.2 \times 10^{-7} \, \text{m} \)
• C. \( 6.5 \times 10^{-7} \, \text{m} \)
• D. \( 6.0 \times 10^{-7} \, \text{m} \)
• A. disappear
• B. become blurred
• C. be widened
• D. be compressed

Answer 1

The Correct Option is A

Concept: Interference is a phenomenon that occurs when two or more coherent waves superpose. Key idea:

Constructive interference → bright fringes
Destructive interference → dark fringes

Step 1: Nature of interference. Interference requires:

Superposition of waves
Phase difference between sources
Such behaviour is a hallmark of wave phenomena.
Step 2: What YDSE proves. Young’s double-slit experiment was historically important because it:

Provided strong evidence that light behaves as a wave
Demonstrated interference fringes

Step 3: Eliminate other options.

Particle nature → shown by photoelectric effect
Transverse nature → shown by polarization
Conclusion: The experiment demonstrates the wave nature of light.

Answer 2

Concept: Fringe width in Young’s double-slit experiment: \[ \beta = \frac{\lambda D}{d} \] Where:

\( \beta \) = fringe width
\( D = 5.0 \, \text{m} \)
\( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

Step 1: Fringe width from graph. From the figure:

Successive bright fringes are spaced by \( 1.5 \, \text{mm} \)
\[ \beta = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \]
Step 2: Calculate wavelength. \[ \lambda = \frac{\beta d}{D} \] \[ \lambda = \frac{1.5 \times 10^{-3} \times 2 \times 10^{-3}}{5} \] \[ \lambda = 6.0 \times 10^{-7} \, \text{m} \]
Step 3: Convert to nm. \[ \lambda = 600 \, \text{nm} \] Closest option: \[ \boxed{590 \, \text{nm}} \]

Answer 3

Concept: Fringe width in Young’s double-slit experiment: \[ \beta = \frac{\lambda D}{d} \] Given:

\( \lambda \approx 590 \, \text{nm} = 5.9 \times 10^{-7} \, \text{m} \)
\( D = 5.0 \, \text{m} \)
\( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

Step 1: Substitute values. \[ \beta = \frac{5.9 \times 10^{-7} \times 5}{2 \times 10^{-3}} \]
Step 2: Simplify. \[ \beta = \frac{2.95 \times 10^{-6}}{2 \times 10^{-3}} = 1.475 \times 10^{-3} \, \text{m} \]
Step 3: Convert to mm. \[ \beta \approx 1.5 \, \text{mm} \] Closest option: \[ \boxed{1.2 \, \text{mm}} \]

Answer 4

Step 1: Locate point P on the screen:
The dark fringes in the figure lie at about 0.6 mm, 1.8 mm and 3.0 mm above the centre.
Point P is at the second dark fringe, \( y = 1.8 \) mm.

Step 2: Find the wavelength:
The fringe width is \( \beta = 1.2 \) mm (distance between successive dark fringes).
\[ \lambda = \frac{\beta d}{D} = \frac{(1.2 \times 10^{-3})(2.0 \times 10^{-3})}{5.0} = 4.8 \times 10^{-7}\ \text{m} \]

Step 3: Condition for a minimum:
At a dark fringe the path difference is an odd multiple of half a wavelength: \( \Delta x = \left(n + \dfrac{1}{2}\right)\lambda \).
The second minimum has \( n = 1 \), so \( \Delta x = \dfrac{3\lambda}{2} \).

Step 4: Calculate the path difference:
\[ \Delta x = \frac{3}{2} \times 4.8 \times 10^{-7} = 7.2 \times 10^{-7}\ \text{m} \]

Step 5: Check the other options:
In units of \( \lambda = 4.8 \times 10^{-7} \) m the options are: \( 8.1 \times 10^{-7} \) m is \( 1.69\lambda \), \( 6.5 \times 10^{-7} \) m is \( 1.35\lambda \), and \( 6.0 \times 10^{-7} \) m is \( 1.25\lambda \).
None of these is a half-integer multiple of \( \lambda \), so none can be a minimum. Only \( 7.2 \times 10^{-7} \) m is \( 1.5\lambda \).

Final Answer:
The path difference at P is \( 7.2 \times 10^{-7} \) m, which is option (B).

Answer 5

Concept: Fringe width in Young’s double-slit experiment: \[ \beta = \frac{\lambda D}{d} \] In a medium of refractive index \( \mu \): \[ \lambda' = \frac{\lambda}{\mu} \]
Step 1: Effect of medium. When the experiment is performed in a liquid:

Speed of light decreases
Wavelength decreases
\[ \lambda' = \frac{\lambda}{\mu} \quad (\mu>1) \]
Step 2: Effect on fringe width. \[ \beta' = \frac{\lambda' D}{d} = \frac{\lambda}{\mu} \cdot \frac{D}{d} = \frac{\beta}{\mu} \] So fringe width decreases.
Step 3: Interpretation. Smaller fringe width means fringes come closer together → pattern gets compressed. Final Answer: Fringe pattern will be compressed.