The distance between the first and the sixth minima in the diffraction pattern of a single slit is 0.5mm. The screen is 0.5m away from the slit. If wavelength of the light used is 5000 A⁰, then the slit width will be

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2.5 mm

5 mm

1.25 mm

1.0 mm

answer is B.

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Detailed Solution

$\begin{array}{l} a \frac{y}{D} = nλ \Rightarrow \frac{(y_{2} - y_{1}) a}{D} = λ (n_{2} - n_{1}) \\ \frac{(a) [0 .5 \times 10^{- 3}]}{0 .5} = 5000 \times 10^{- 10} [6 - 1] \\ a = 25 \times 10^{- 4} m = 2 .5 mm \end{array}$

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