Question

Young's double slit experiment is performed with sodium (Yellow) light of wavelength $589.3\, nm$ and the interference pattern is observed on a screen $100 \,cm$ away. The 10 th bright fringe has its centre at a distance of $12\, nm$ from the central maximum. The separation between the slits is

• A. 0.49 mm
• B. 0.6 mm
• C. 0.7 mm
• D. 0.53 mm

Answer 1

Answer: (A)

Position of $10th$ bright fringe $=\frac{10 \lambda D}{d}$
Also, $\frac{10 \lambda D}{d}=12$
$\Rightarrow d=\frac{10 \lambda D}{12}$
The separation between the slits
$=\frac{10 \times 589.3 \times 10^{-9} \times 1}{12 \times 10^{-3}}$
$=4.9 \times 10^{-4} m =0.49\, mm$