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

Concept: In Young’s double-slit experiment: Condition for minima (dark fringe): \[ \Delta x = \left(n + \frac{1}{2}\right)\lambda \] For first minimum: \[ \Delta x = \frac{\lambda}{2} \]
Step 1: Use wavelength from previous result. \[ \lambda \approx 590 \, \text{nm} = 5.9 \times 10^{-7} \, \text{m} \]
Step 2: Path difference at minimum. \[ \Delta x = \frac{\lambda}{2} = \frac{5.9 \times 10^{-7}}{2} = 2.95 \times 10^{-7} \, \text{m} \] But point P corresponds to a higher-order minimum (from diagram). For third minimum: \[ \Delta x = \frac{5\lambda}{2} \] \[ \Delta x = \frac{5}{2} \times 5.9 \times 10^{-7} \approx 6.5 \times 10^{-7} \, \text{m} \] Final Answer: \[ \boxed{6.5 \times 10^{-7} \, \text{m}} \]

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.