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In Young's double slits experiment, the position of 5th bright fringe  from the central maximum is 5cm.The distance between slits and screen is 1m and wavelength of used monochromatic light is 600 nm.The separation between the slits is:

Option: 1

48 \mu \mathrm{m}


Option: 2

36 \mu \mathrm{m}


Option: 3

12 \mu \mathrm{m}


Option: 4

60 \mu \mathrm{m}


Answers (1)

\begin{aligned} & 5 \beta=5 \mathrm{~cm} \\ & \Rightarrow \beta=1 \mathrm{~cm} \\ & \frac{\lambda \mathrm{D}}{\mathrm{d}}=1 \mathrm{~cm}=\frac{1}{100} \mathrm{~m} \\ & \Rightarrow \mathrm{d}=600 \times 10^{-9} \times 100 \times 1 \\ & =60 \times 10^{-6} \mathrm{~m} \\ & =60 \mu \mathrm{m} \end{aligned}

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Ramraj Saini

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