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: Start with the Sure Matches:
Closing a slit leaves one beam, so there is nothing to interfere with. B goes with IV. Lower frequency means longer wavelength, and fringe width $\beta = \lambda D/d$ grows with $\lambda$. So D goes with I.

Step 2: Remove Options:
Only options 3 and 4 have B-IV, so options 1 and 2 are out. Only option 3 has D-I, so option 4 is out too.

Step 3: Check the Remaining Pairs:
Option 3 says A-III and C-II. Wider slit A passes more light, so the peak intensity goes up: III. A wider source slit blurs the fringes: II. Both are physically correct.

Step 4: Result:
Option 3 is consistent for all four pairs.

Final Answer:
\[\boxed{\text{A-III, B-IV, C-II, D-I (option 3)}}\]