In Young’s double slit experiment with monochromatic light of wavelength 600 mm, the distance between slits is $10^{- 3}$ m. For changing fringe width by $3 imes 10^{- 5}$ m

Question

In Young’s double slit experiment with monochromatic light of wavelength 600 mm, the distance between slits is $10^{- 3}$ m. For changing fringe width by $3 	imes 10^{- 5}$ m

Dr. Abubacker · Accepted Answer

(d) $\beta = \frac{D}{d}\lambda$
$\mathrm{\Delta}\beta = \frac{\mathrm{\Delta}D}{d}\lambda$
$\Delta D = \frac{d\Delta\beta}{\lambda}$
$= \frac{10^{- 3} 	imes 3 	imes 10^{- 5}}{600 	imes 10^{- 9}}$
$= 5\ cm\ $away or towards the slits

Q. In Young’s double slit experiment with monochromatic light of wavelength 600 mm, the distance between slits is $10^{- 3}$ m. For changing fringe width by $3 \times 10^{- 5}$ m

Explanation

Reviewed by Dr. Abubacker

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