Optics 6 Question 28
30. 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 $2 m$ is
(1994, 1M)
(a) $1.2 cm$
(b) $1.2 mm$
(c) $2.4 cm$
(d) $2.4 mm$
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Solution:
- For first dark fringe on either side $d \sin \theta=\lambda$
$$ \begin{array}{rlrl} \text { or } & & \frac{d y}{D} & =\lambda \\ \therefore & y & =\frac{\lambda D}{d} \end{array} $$
Therefore, distance between two dark fringes on either side
$$ =2 y=\frac{2 \lambda D}{d} $$
Substituting the values, we have
$$ \text { Distance }=\frac{2\left(600 \times 10^{-6} mm\right)\left(2 \times 10^{3} mm\right)}{(1.0 mm)}=2.4 mm $$
