AS Daily A Level Physics question
In a double‑slit experiment with monochromatic light, the central point O on the screen is initially the brightest spot when both identical slits are unobstructed. Step 1: a filter over one slit reduces the amplitude from that slit at the screen to half, without adding any extra phase change. Step 2: a thin slide over the same slit increases its path by half a wavelength relative to the other slit. After both steps, what is the intensity at O relative to the original central maximum, and is O a maximum or a minimum?
Answer
The correct answer is C.
Correct: C — About 6% of the original and a minimum. With equal slits, the original central intensity ∝ (A + A)^2 = 4A^2. After Step 1 the amplitudes are A and 0.5A; after Step 2 the extra half‑wavelength makes them out of phase at O, so the resultant amplitude is |A − 0.5A| = 0.5A and intensity ∝ (0.5A)^2 = 0.25A^2. Relative to 4A^2 this is 1/16 ≈ 6%, and O is a minimum (not zero because amplitudes are unequal). A treats intensity as proportional to amplitude, using 1.5A/2A = 0.75, and also ignores the phase reversal from the half‑wavelength path change. B matches only the situation after Step 1: (A + 0.5A)^2/ (2A)^2 = 2.25/4 ≈ 56%, but it wrongly assumes O stays a maximum when the slide adds a half‑wavelength (confusing path difference effects with just a lateral fringe shift). C is correct as shown above: subtract amplitudes due to the π phase shift, then square and compare to the original. D assumes perfect cancellation for a half‑wavelength shift regardless of amplitudes; complete darkness would be the limiting case only if the two amplitudes were equal.