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In Young's double slit experiment, light of wavelength \mathrm{6000 \, \AA} is used to get an interference pattern on a screen. The fringe width changes by \mathrm{1.5 \mathrm{~mm},} when the screen is brought towards the double slit by 50 \mathrm{~cm}. The distance between the two slits is:

Option: 1

0.2 mm


Option: 2

0.4 mm


Option: 3

0.6 mm


Option: 4

0.8 mm


Answers (1)

best_answer

Fringe width, \mathrm{\beta=\frac{D \lambda}{d}}

As  \mathrm{\lambda \, and \, \mathrm{d}} are constant, so the change in fringe width is given by

\mathrm{ \Delta \beta=\frac{\lambda \Delta \mathrm{D}}{\mathrm{d}} }
\mathrm{\therefore \quad} Distance between the two slits,

\mathrm{ \mathrm{d}=\frac{\lambda \Delta \mathrm{D}}{\Delta \beta}=\frac{6000 \times 10^{-10} \times 50 \times 10^{-2}}{1.5 \times 10^{-3}}=0.2 \times 10^{-3} \mathrm{~m}=0.2 \mathrm{~mm} }

Posted by

Gautam harsolia

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