A Young’s double slit experiment is conducted with slit separation 10mm, where the screen is 2m away from the slits. If wavelength of light used is 6000A⁰. If the wavelength is increased by 1000A⁰, and placed the whole apparatus in water of refractive index 4/3, the new fringe width in mm is

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0.210

0.315

0.420

0.105

answer is B.

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Detailed Solution

$\begin{array}{l} β' = \frac{β}{μ} = \frac{λ' D}{dμ}; β' = \frac{7000 \times 10^{- 10} \times 2}{\frac{4}{3} \times 10 \times 10^{- 3}} \\ = \frac{21}{2} \times 10^{- 5} m = 0 .105 \times 10^{- 3} m = 0 .105 mm \end{array}$

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