Light of wavelength 500 nm falls normally on a slit of width 0.25 mm. The central maximum on the screen is 8 mm wide. The distance of the screen from the slit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.0 m
The central maximum lies between the first minima on the two sides. These minima are at angles \( \pm\lambda/a \). So the width of the central maximum on a screen at distance \( D \) is \( w=2\lambda D/a \).
\( \lambda=500\ \mathrm{nm}=5\times10^{-7}\ \mathrm{m} \)
\( a=0.25\ \mathrm{mm}=2.5\times10^{-4}\ \mathrm{m} \)
\( w=8\ \mathrm{mm}=8\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} w &= \frac{2\lambda D}{a} \\ D &= \frac{wa}{2\lambda} \\ &= \frac{8\times10^{-3}\times2.5\times10^{-4}}{2\times5\times10^{-7}} \\ &= 2.0\ \mathrm{m} \end{aligned} \)
Hence, the screen is 2.0 m from the slit.
NCERT (2026-27):
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”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)