Q. A narrow slit of width 1 mm is illuminated by monochromatic light of wavelength 600 nm. The distance between the first minima on either side of a screen at a distance of 2m is
IIT JEEIIT JEE 1994
Solution:
For first dark fringe on either side,
$
\begin{array}{l}
d \sin \theta=\lambda \\
\Rightarrow \frac{d y}{D}=\lambda [\sin \theta \simeq \tan \theta] \\
\therefore y=\frac{\lambda D}{d}
\end{array}
$
There fore, dis $\tan$ ce between two dark fringes on either side $=2 y=\frac{2 \lambda D}{d}$
Substituting the values, we have dis $\tan c e=\frac{2\left(600 \times 10^{-6} mm \right)\left(2 \times 10^3 mm \right)}{(1.0 mm )}=2.4 mm$
