Question

Light of wavelength 600 nm is incident upon a single slit with width $4\times {10}^{–4}m$. The figure shows the pattern observed on a screen positioned 2 m from the slits. Determine the distance s.

• A. 0.002 m
• B. 0.003 m
• C. 0.004 m
• D. 0.006 m

Answer 1

The correct option is D 0.006 m

We can see that it is diffraction pattern.
We can see, second minima of diffraction at distance 'S'.
$d\sin \theta =n\lambda ,$ (here n = 2)
$(4\times {10}^{-4}m)\sin \theta =\left(2\right)(600\times {10}^{-9}m)$
$\Rightarrow sin\theta =0.003$
As $\theta$ is small, so $\sin \theta \approx \theta \approx tan\theta$
$\tan \theta =\frac{S}{D}=0.003$
$\Rightarrow S=\left(2\right)\left(0.003\right)$
$\Rightarrow S=6\text{mm}$
$\Rightarrow S=0.006m$