Question

In Young's double slit experiment, the light emitted from source has $\lambda =6.5\times 10^{- 7} \, m$ and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth bright fringe will be

• A. 3.2 mm
• B. 1.63 mm
• C. 0.585 mm
• D. 2.31 mm

Answer 1

Answer: (B)

$x_{5}=n\frac{\lambda D}{d}=\frac{5 \times 6.5 \times 10^{- 7} \times 1}{10^{- 3}}=32.5\times 10^{- 4} \, m$
$\mathrm{x}_3=(2 \mathrm{n}-1) \frac{\lambda \mathrm{D}}{2 \mathrm{~d}}=\frac{5 \times 6.3 \times 10^{-7} \times 1}{2 \times 10^{-3}}=16.25 \times 10^{-4} \mathrm{~m}$
$\therefore x_{5}-x_{3}\approx 1.63 \, mm$