A slit of width 0.2 mm is lit first by red light of wavelength 700 nm and then by blue light of wavelength 450 nm. The screen is 1 m from the slit. The central maximum for red light is wider than that for blue light by
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.5 mm
The width of the central maximum is \( 2\lambda D/a \). It grows with the wavelength. So red light gives a wider central maximum than blue light.
\( a=0.2\ \mathrm{mm}=2\times10^{-4}\ \mathrm{m} \)
\( D=1\ \mathrm{m} \)
\( \lambda_r=700\ \mathrm{nm} \)
\( \lambda_b=450\ \mathrm{nm} \)
\( \begin{aligned} w_r-w_b &= \frac{2D(\lambda_r-\lambda_b)}{a} \\ &= \frac{2\times1\times(700-450)\times10^{-9}}{2\times10^{-4}} \\ &= 2.5\times10^{-3}\ \mathrm{m}=2.5\ \mathrm{mm} \end{aligned} \)
Hence, the central maximum for red light is wider by 2.5 mm.
NCERT (2026-27):
Seeing the single slit diffraction pattern – “… Using a filter for red or blue will make the fringes clearer. With both filters available, the wider fringes for red compared to blue can be seen.”
The single slit – “It is observed that the intensity has a central maximum at θ = 0 and other secondary maxima at”
\( \displaystyle \theta \approx \left(n+\frac{1}{2}\right)\frac{\lambda}{a} \)
“which go on becoming weaker and weaker with increasing n. The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)