In a Young's double slit experiment, the intensity at the center of the screen is I. If one of the slits is closed, the intensity at the center of the screen now will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{4} \)
At the centre of the screen the two waves travel equal paths and arrive in phase, so their amplitudes add, and intensity is proportional to the square of the amplitude. If each slit alone gives amplitude a and intensity \( I_0 \propto a^{2} \):
\( \displaystyle I \propto (a + a)^{2} = 4a^{2} \)
\( \displaystyle I = 4I_0 \)
With one slit closed, only one wave reaches the centre:
\( \displaystyle I_0 = \frac{I}{4} \)
Note: It seems that there is a mistake in the original NTA paper, which prints the options as 1/2, 1, 1/4 and 1/3 when they should read I/2, I, I/4 and I/3. The question can still be solved, because the options are clearly meant as fractions of the intensity I given in the question.