Question

In Young's double slit experiment, two slits are separated by $1$ m. The slits are illuminated by a light of wavelength $650nm$. The source of light is placed symmetrically with respect to the two slits. An interference pattern is observed on a screen at a distance of $1$ m from the slits. The distance between the third dark fringe and the fifth bright fringe from the center of the pattern will be

• A. $1.62mm$
• B. $2.62mm$
• C. $5.62mm$
• D. $3.62mm$

Answer 1

Answer: (A)

$1.62mm$

Given : $D=1m$, $\lambda =650\times {10}^{-9}m$, $d=1mm={10}^{-3}m$

Distance of third dark fringe from centre of pattern
$x_1=(3-\frac{1}{2})\frac{D\lambda }{d}=\frac{5}{2}\times \frac{1\times 650\times {10}^{-9}}{1\times {10}^{-3}}$
$=1.62mm$
Distance of fifth bright fringe from centre of pattern
$x_2=5\frac{D\lambda }{d}=3.24mm$
$\therefore$ Required distance $\Delta x=x_2-x_1$
$\Delta x=3.24-1.62=1.62mm$