In a double-slit experiment, the pattern on the screen shows interference fringes within the central maximum of the single-slit diffraction pattern. If the separation between the slits is doubled, keeping the width of each slit, the wavelength and the slit-screen distance unchanged, then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → the fringe width is halved and the number of interference fringes within the central maximum is doubled
The pattern of a double slit is the interference pattern of the two slits superposed on the diffraction pattern of each slit.
The fringe width \( \beta=\lambda D/d \) is the spacing of the equally spaced fringes. It depends on the slit separation \( d \).
The central maximum of the diffraction pattern spreads between the first minima at \( \theta=\pm\lambda/a \). So its angular width \( 2\lambda/a \) depends only on the slit width \( a \).
The number of fringes inside the central maximum is about \( (2\lambda/a)/(\lambda/d)=2d/a \).
When \( d \) is doubled, \( \beta \) becomes half. The central maximum does not change, so the number of fringes in it, \( 2d/a \), becomes twice as large.
NCERT (2026-27):
Interference of Light Waves and Young’s Experiment – “We will have constructive interference resulting in a bright region when \( \displaystyle\frac{xd}{D}=n\lambda \). That is,
\( \displaystyle x=x_n=\frac{n\lambda D}{d} \)\( ;\ n=0,\pm1,\pm2,\ldots \qquad (10.13) \)”
The Single Slit – “In the double-slit experiment, we must note that the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole, and the double-slit interference pattern.”
Summary – “5. A single slit of width a gives a diffraction pattern with a central maximum. The intensity falls to zero at angles of ±λ/a, ±2λ/a, etc., with successively weaker secondary maxima in between.”