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JEE MAINS 2023
25.01.23 S1
Question
In Young's double slits experiment, the position of 5th bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is:
Select the correct option:
A
60 m
B
48 m
C
12 m
D
36 m
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. Given, $\mathrm{D}=1 \mathrm{~m}$ $$ \begin{array}{ll} & \lambda 600 \times 10^{-9} \mathrm{~m} \\ & n=5 \\ \text { As } & y_{n \mathrm{n}}=\frac{n \lambda \mathrm{D}}{\mathrm{~d}} \\ \Rightarrow & \frac{5 \times 600 \times 10^{-9} \times 1}{\mathrm{~d}}=5 \times 10^{-2} \\ \Rightarrow & d=\frac{5 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-2}}=60 \times 10^{-6} \mathrm{~m} \\ \Rightarrow & d=6 \mu \mathrm{~m} \end{array} $$
Question Tags
JEE Main
Physics
Easy
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