Question

Light of wavelength $6.5\times {10}^{-7}$ meters is made incident on two slits, $1mm$ apart. The distance between third dark fringes and fifth bright fringe on a screen distant $1m$ from the slits will be:

• A. $0.35mm$
• B. $0.65mm$
• C. $1.63mm$
• D. $3.25mm$

Answer 1

Answer: (C)

$1.63mm$

Distance of $3^{rd}$ dark fringe from centre:
$x_3=(2n-1)\frac{\lambda D}{2d}=(2\times 3-1)\frac{\lambda D}{2d}=\frac{5\lambda D}{2d}$
Distance of $5^{th}$ bright fringe from centre:
$x_5=\frac{n\lambda D}{d}=\frac{5\lambda D}{d}$
$\therefore x_5-x_3=\frac{5\lambda D}{d}-\frac{5\lambda D}{2d}=\frac{5\lambda D}{2d}=\frac{5\times 6.5\times {10}^{-7}\times 1}{2\times (1\times {10}^{-3})}$,
which gives distance = 1.63 mm