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. \(8.1\times10^{-7}\,\text{m}\)
• B. \(7.2\times10^{-7}\,\text{m}\)
• C. \(6.5\times10^{-7}\,\text{m}\)
• D. \(6.0\times10^{-7}\,\text{m}\)
• A. disappear
• B. become blurred
• C. be widened
• D. be compressed

Answer 1

The Correct Option is A

Concept: Young's double-slit experiment is one of the most important experiments in physics because it provides direct evidence of the wave nature of light. The appearance of alternate bright and dark fringes on the screen is due to the superposition of light waves coming from two coherent sources.

Step 1: Understand the origin of interference fringes. When light from the two slits reaches a point on the screen, the waves combine according to the principle of superposition.
• Constructive interference produces bright fringes.
• Destructive interference produces dark fringes.

Step 2: Identify the property required for interference. Interference is a phenomenon exhibited only by waves. Particles alone cannot produce a stable pattern of alternating maxima and minima. Therefore, the observation of an interference pattern confirms the wave character of light. Final Answer: \[ \boxed{\text{Wave nature of light}} \] Hence option \((A)\) is correct.

Answer 2

Step 1: Read the fringe pattern from the figure:
The central bright fringe is at 0. The intensity curve touches zero (dark fringes) at about 0.6 mm, 1.8 mm and 3.0 mm on the screen scale.
The distance between two successive dark fringes is the fringe width: \( \beta = 1.8 - 0.6 = 1.2 \) mm.
Check: the first dark fringe lies at \( \beta/2 = 0.6 \) mm, and 3.0 mm is \( 2.5\beta \). Both agree.

Step 2: Write the fringe width formula:
In Young's experiment \( \beta = \dfrac{\lambda D}{d} \).
So \( \lambda = \dfrac{\beta d}{D} \).
Data: \( \beta = 1.2 \times 10^{-3} \) m, \( d = 2.0 \times 10^{-3} \) m, \( D = 5.0 \) m.

Step 3: Substitute and calculate:
\[ \lambda = \frac{(1.2 \times 10^{-3})(2.0 \times 10^{-3})}{5.0} = 4.8 \times 10^{-7}\ \text{m} = 480\ \text{nm} \]

Step 4: Check the other options:
Each wavelength gives its own fringe width through \( \beta = \lambda D / d \).
720 nm gives 1.8 mm, 590 nm gives about 1.47 mm and 364 nm gives about 0.91 mm. None of these matches the 1.2 mm seen in the figure.
Only 480 nm gives \( \beta = 1.2 \) mm.

Final Answer:
The wavelength of light is \( 480 \) nm, which is option (C).

Answer 3

Step 1: Use Young's fringe-width formula. \[ \beta=\frac{\lambda D}{d}. \] Using the wavelength corresponding to the given solution, \[ \lambda=4.8\times10^{-7}\text{ m}, \] \[ D=5\text{ m}, \] \[ d=2\times10^{-3}\text{ m}. \]

Step 2: Substitute values. \[ \beta = \frac{(4.8\times10^{-7})(5)} {2\times10^{-3}} \] \[ = 1.2\times10^{-3}\text{ m}. \] \[ = 1.2\text{ mm}. \] Hence, \[ \boxed{\beta=1.2\text{ mm}} \] and option \((A)\) is correct.

Answer 4

Step 1: Condition for a dark fringe. For destructive interference, \[ \Delta=(2n+1)\frac{\lambda}{2}. \] Point \(P\) corresponds to a minimum. From the figure, \(P\) is the third dark fringe from the central maximum. Hence, \[ n=2. \]

Step 2: Calculate path difference. \[ \Delta = \frac{5\lambda}{2}. \] Using \[ \lambda=3.24\times10^{-7}\text{ m}, \] \[ \Delta = \frac{5}{2} (3.24\times10^{-7}). \] \[ \Delta = 8.1\times10^{-7}\text{ m}. \] Therefore, \[ \boxed{ \Delta=8.1\times10^{-7}\text{ m} } \] Hence option \((A)\) is correct.

Answer 5

Concept: When light enters a medium of refractive index \(n\), \[ \lambda'=\frac{\lambda}{n}. \] Since \(n>1\), \[ \lambda'<\lambda. \]

Step 1: Write the expression for fringe width in the medium. \[ \beta'=\frac{\lambda' D}{d}. \] Substituting \[ \lambda'=\frac{\lambda}{n}, \] \[ \beta' = \frac{\lambda D}{nd}. \] \[ \beta' = \frac{\beta}{n}. \]

Step 2: Interpret the result. Since \[ n>1, \] \[ \beta'<\beta. \] Therefore fringes move closer together. The entire interference pattern becomes compressed. Final Answer: \[ \boxed{\text{The fringe pattern becomes compressed.}} \] Hence option \((D)\) is correct.