In young's experiment the distance of the screen from the two slits is 3.5 m. When light of wavelength $λ$ is allowed to fall on the slits, the width of the fringes obtained on the screen is 3.4 mm. The distance between the two slits is 0.05cm. The wavelength $λ$ is ....... (nm)

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485

600

625

515

answer is D.

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Detailed Solution

$\begin{array}{l} β = \frac{λD}{d}; 3 .4 \times 10^{- 3} = \frac{λ (3 .5)}{0 .05 \times 10^{- 2}} \\ λ = \frac{34 \times 5 \times 10^{- 8}}{3 .5} = 48 .5 \times 10^{8} = 485 \times 10^{- 9} m \end{array}$

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