Read the following passage and answer the questions.
A student studies diffraction at a single slit. She uses a laser that gives light of wavelength 600 nm. The beam falls normally on a narrow slit. The width of the slit can be changed with a screw. A white screen is placed 1.0 m behind the slit. With the slit width set at 0.2 mm, she sees a broad bright band at the centre of the screen. On both sides of it there are dark and bright bands. She marks the centre of each band with a pencil. Later she changes the slit width, the wavelength and the screen distance, and notes how the bands move.
At home she tries a simpler version. She holds two razor blades with their edges parallel, so that a narrow slit is left between them. Through this slit she looks at a clear glass bulb with a straight filament. She keeps the slit parallel to the filament and close to her eye.
With the slit width at 0.2 mm, she measures the distance of the first secondary maximum from the centre of the screen. This maximum is the middle of the first bright band beside the central band. The distance is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4.5 mm
In single-slit diffraction, the secondary maxima are the weaker bright bands on either side of the central band. They lie at angles \( \theta\approx(n+\frac{1}{2})\lambda/a \). For the first one, \( n=1 \). Its distance from the centre of the screen is \( y=D\theta \).
Given:
\( \lambda=600\ \mathrm{nm}=6\times10^{-7}\ \mathrm{m} \)
\( a=0.2\ \mathrm{mm}=2\times10^{-4}\ \mathrm{m} \)
\( D=1.0\ \mathrm{m} \)
\( \begin{aligned} y &= \frac{3}{2}\cdot\frac{\lambda D}{a} \\ &= \frac{3}{2}\times\frac{6\times10^{-7}\times1.0}{2\times10^{-4}} \\ &= \frac{3}{2}\times3\times10^{-3}\ \mathrm{m} \\ &= 4.5\times10^{-3}\ \mathrm{m}=4.5\ \mathrm{mm} \end{aligned} \)
Hence, the first secondary maximum lies 4.5 mm from the centre of the screen.
NCERT (2026-27):
The single slit – “It is observed that the intensity has a central maximum at \( \theta=0 \) and other secondary maxima at \( \theta\approx(n+1/2)\lambda/a \), which go on becoming weaker and weaker with increasing n. …”