Question

Match the LIST-I with LIST-II (In the context of Young's double slit experiment)

LIST-I	LIST-II
A. The width of one slit is slightly increased.	I. The fringe width increases.
B. One slit is closed.	II. Interference pattern becomes less sharp.
C. The width of the source slit is increased.	III. Maximum intensity increases
D. Light of smaller frequency is used.	IV. Interference pattern disappears

Choose the correct answer from the options given below:

• A. A-I, B-II, C-III, D-IV
• B. A-IV, B-III, C-II, D-I
• C. A-III, B-IV, C-II, D-I
• D. A-III, B-IV, C-I, D-II

Hint:

Closed slit means no interference. Lower frequency means larger wavelength and larger fringe width.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We match each change in Young's double slit setup with its effect on the pattern.

Step 2: Match A:
When one slit is made a little wider, it lets more light through. The total light reaching the screen rises, so the maximum intensity increases. So A matches III.

Step 3: Match B:
If one slit is closed, there is only one source of light on the screen. There are no two waves to interfere. So the interference pattern disappears. B matches IV.

Step 4: Match C:
A wider source slit acts like many nearby sources. Each one makes its own fringe system, and these are slightly shifted. They overlap and wash out the dark and bright bands. So the pattern becomes less sharp. C matches II.

Step 5: Match D:
Fringe width is \(\beta = \lambda D / d\). Smaller frequency means larger wavelength, since \(\lambda = c/f\). So \(\beta\) increases. D matches I.

Step 6: Choose the Option:
We have A-III, B-IV, C-II, D-I. This is the third printed option. The other options break at least one of these matches.

Final Answer:
The matching is A-III, B-IV, C-II, D-I, so option 3 is correct. \[ \boxed{\text{A-III, B-IV, C-II, D-I}} \]