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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-IV, C-II, D-I
In Young's experiment the pattern comes from the superposition of two coherent waves;
(A) The width of one slit is slightly increased. — this matches (III) Maximum intensity increases. More light passes through the wider slit, so the intensity at a maximum, \( (a_1 + a_2)^{2} \), increases.
(B) One slit is closed. — this matches (IV) Interference pattern disappears. Only one wave reaches the screen, so there is nothing for it to interfere with.
(C) The width of the source slit is increased. — this matches (II) Interference pattern becomes less sharp. A wider source acts as many point sources whose slightly shifted fringe systems blur one another.
(D) Light of smaller frequency is used. — this matches (I) The fringe width increases. A smaller frequency means a longer wavelength, and the fringe width is proportional to it:
\( \displaystyle \beta = \frac{\lambda D}{d} \)
Note: Item (A) also makes the pattern less sharp, entry (II), because with unequal slit widths the minima are no longer completely dark. The question can still be solved: items (B) and (D) have only one possible match each, (IV) and (I), and Option (3) is the only option that contains both B-IV and D-I.