In Young’s double–slit experiment, the slits are 0.5 mm apart and the interference is observed on a screen at a distance of 100 cm from the slits. It is found that the ninth bright fringe is at a distance of 7.5 mm from the second dark fringe from the center of the fringe pattern. The wavelength of the light used is

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$5000 \overset{o}{A}$

$\frac{5000}{7} \overset{o}{A}$

$2500 \overset{o}{A}$

$\frac{2500}{7} \overset{o}{A}$

answer is A.

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Detailed Solution

$y_{9} =$ position of 9th bright fringe $= 9 (\frac{λ D}{d})$

$y_{2} =$ position of 2nd dark fringe = $(2 - \frac{1}{2}) \frac{λ D}{d} = \frac{3}{2} \frac{λ D}{d}$

$y_{9} - y_{2} = 7.5 m m \Rightarrow \frac{λ D}{d} (9 - \frac{3}{2}) = 7.5 \times 10^{- 3}$

$∴ λ = (7.5 \times 10^{- 3}) (\frac{2}{15}) (\frac{0.5 \times 10^{- 3}}{100 \times 10^{- 2}})$

$= (75) (\frac{2}{15}) (5) (10^{- 4 - 4}) = 50 \times 10^{- 8} m$

$= 5000 \overset{o}{A}$

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