Question

Match List-I with List-II in the context of Young's Double Slit Experiment.

Choose the correct answer from the options given below:

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

Hint:

Remember these three YDSE relations: \[ \beta=\frac{\lambda D}{d} \] \[ \text{Bright Fringe: }\Delta=n\lambda \] \[ \text{Dark Fringe: }\Delta=\frac{(2n+1)\lambda}{2} \] Central maximum occurs at \(\Delta=0\).

Answer 1

The Correct Option is D

Concept: In Young's Double Slit Experiment (YDSE), Fringe width: \[ \beta=\frac{\lambda D}{d} \] Condition for bright fringe (constructive interference): \[ \Delta=n\lambda \] Condition for dark fringe (destructive interference): \[ \Delta=\frac{(2n+1)\lambda}{2} \] For central maximum, \[ \Delta=0 \]

Step 1:Match fringe width. \[ \beta=\frac{\lambda D}{d} \] Hence, \[ (A)\rightarrow(III) \]

Step 2: Match bright fringe condition. \[ \Delta=n\lambda \] Hence, \[ (B)\rightarrow(II) \]

Step 3: Match dark fringe condition. \[ \Delta=\frac{(2n+1)\lambda}{2} \] Hence, \[ (C)\rightarrow(I) \]

Step 4: Match central maximum condition. \[ \Delta=0 \] Hence, \[ (D)\rightarrow(IV) \]

Step 5: Write the final matching. \[ (A)\rightarrow(III) \] \[ (B)\rightarrow(II) \] \[ (C)\rightarrow(I) \] \[ (D)\rightarrow(IV) \] \[ { \begin{array}{c} (A)-(III),\ (B)-(II), (C)-(I),\ (D)-(IV) \end{array} } \] Hence, the correct option is \[ {(D)} \]