Question 133 of 191

In a single slit diffraction pattern, a light of wavelength $6000A^{o}$ is used. The distance between the first and third minima in the diffraction pattern is found to be $3mm$ when the screen is placed $50cm$ away from slits. The width of the slit is ____$\times {10}^{-4}m$.

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Answer: 2

For ${n}^{th}$ minima:
$\Rightarrow b\sin \theta =n\lambda$
($\lambda$ is very small, so $\sin \theta$ is very small, hence $\sin \theta ≃\tan \theta$.
$\Rightarrow b\tan \theta =n\lambda$
$\Rightarrow b\dfrac{y}{D}=n\lambda$
$\Rightarrow {y}_{n}=\dfrac{n\lambda D}{b}$ (Position of ${n}^{th}$ minima)

$B\rightarrow {1}^{\text{st}}$ minima and $A\rightarrow {3}^{\text{rd}}$:
${y}_{3}=\dfrac{3\lambda D}{b},{y}_{1}=\dfrac{\lambda D}{b}$
$\Rightarrow \Delta y={y}_{3}-{y}_{1}=\dfrac{2\lambda D}{b}$
$\Rightarrow 3\times {10}^{-3}=\dfrac{2\times 6000\times {10}^{-10}\times 0.5}{b}$
$\Rightarrow b=\dfrac{2\times 6000\times {10}^{-10}\times 0.5}{3\times {10}^{-3}}$
$\Rightarrow b=2\times {10}^{-4}m$