In a Young’ double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. The wavelength of light used in the experiment

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600 nm

6 nm

60 nm

0.6 nm

answer is B.

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Detailed Solution

In YDSE we know that distance of nth bright fringe from central bright fringe is given as $x_{nB} = \frac{nλD}{d}$
Here D = 1.4 m
$\begin{array}{l} \Rightarrow d = 0 .28 \times 10^{- 3} m \\ \Rightarrow x_{nB} = 1 .2 \times 10^{- 2} m \\ For n = 4 \\ \Rightarrow 1 .2 \times 10^{- 2} = \frac{4 \times λ \times 1 .4}{0 .28 \times 10^{- 3}} \\ \Rightarrow λ = \frac{1 .2 \times 10^{- 2} \times 0 .28 \times 10^{- 3}}{4 \times 1 .4} \\ = 6 \times 10^{- 7} m = 600 nm \end{array}$

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