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

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answer is 2.

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Detailed Solution

For n^thminima
$bsin θ = n λ$
$(λ is small so \sin θ is small, hence \sin θ ≃ \tan θ)$
$btan θ = n λ$
$b \frac{y}{D} = n λ$
$\Rightarrow y_{n} = \frac{n λ D}{b} (Position of n^{th} minima)$
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$B \to 1^{st} minima, A \to 3^{rd} minima$
$\begin{aligned} y_{3} = \frac{3 λ D}{b}, y_{1} = \frac{λ D}{b} \\ Δy = y_{3} - y_{1} = \frac{2 λ D}{b} \\ 3 \times 10^{- 3} = \frac{2 \times 6000 \times 10^{- 10} \times 0.5}{b} \\ b = \frac{2 \times 6000 \times 10^{- 10} \times 0.5}{3 \times 10^{- 3}} \\ b = 2 \times 10^{- 4} m \\ x = 2 \end{aligned}$