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In a young's double slit experiment, slits are illuminated by a monochromatic source of wavelength 6000 \AA and fringes are obtained. If screen is moved by a distance of 5 \mathrm{~cm} towards slits, change in fringe width is 3 \times 10^{-5} \mathrm{~m}. Then separation between the slits will be

Option: 1

1 \mathrm{~mm}


Option: 2

1.2 \mathrm{~mm}


Option: 3

1.5 \mathrm{~mm}


Option: 4

1.63 \mathrm{~mm}


Answers (1)

best_answer

\mathrm{\beta=\frac{\lambda D}{d} \text { and } \beta^{\prime}=\frac{\lambda D^{\prime}}{d}}
\mathrm{\therefore \beta-\beta^{\prime}=\frac{\lambda\left(D-D^{\prime}\right)}{d}}

or
\mathrm{d=\frac{\lambda\left(D-D^{\prime}\right)}{\left(\beta-\beta^{\prime}\right)}}
\mathrm{=\frac{6000 \times 10^{-10} \times 5 \times 10^{-2}}{3 \times 10^{-5}} \mathrm{~m}=10^{-3} \mathrm{~m}=1 \mathrm{~mm}}

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mansi

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