In a single-slit experiment, the slit width is 0.3 mm and the screen is 1.5 m away. The first secondary maxima on the two sides of the central maximum are 9 mm apart. The wavelength of the light is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 600 nm
The first secondary maximum is at angle \( \frac{3}{2}\lambda/a \) on each side of the centre. On the screen it is at \( y=\frac{3}{2}\lambda D/a \). The two first secondary maxima are therefore \( 3\lambda D/a \) apart.
\( a=0.3\ \mathrm{mm}=3\times10^{-4}\ \mathrm{m} \)
\( D=1.5\ \mathrm{m} \)
\( s=9\ \mathrm{mm}=9\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} s &= 2\times\frac{3}{2}\frac{\lambda D}{a}=\frac{3\lambda D}{a} \\ \lambda &= \frac{sa}{3D} \\ &= \frac{9\times10^{-3}\times3\times10^{-4}}{3\times1.5} \\ &= 6\times10^{-7}\ \mathrm{m}=600\ \mathrm{nm} \end{aligned} \)
Hence, the wavelength of the light is 600 nm.
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”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)