Question

In Young?? double slit experiment, the slits are $3 \,mm$ apart. The wavelength of light used is $5000\, ??$ and the distance between the slits and the screen is $90 \,cm$ . The fringe width in $mm$ is :

• A. $1.5$
• B. $0.015$
• C. $2.0$
• D. $0.15$

Answer 1

Answer: (D)

$0.15$

Answer 2

Let $?$ be wavelength of monochromatic light, used to illuminate the slit s, and d be the
distance between coherent sources, then width of slits is given by

$W=\frac{D\,\lambda}{d}$
where $D$ is distance between screen and source.
Given, $d=3\, mm , \lambda=5000 \, ??= =5 \times 10^{-7} \,m$
$=5 \times 10^{-4}\, mm$
$D=90\, cm =900 \,mm$
$\therefore \, W=\frac{5 \times 10^{-4} \times 900}{3}=15 \times 10^{-2}\, mm$
$=0.15 \,mm$