Young's double slit experiment two |WaveOpticsJEEMain2021

JEE MAIN 2021

26-02-2021 S1

Question

In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be:

Select the correct option:

0.25 mm

0.50 mm

0.75 mm

1 mm

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \beta=\frac{\lambda D}{d}=\frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}} \\ & \beta=\frac{5}{2} \times 10^{-4} \mathrm{~m}=2.5 \times 10^{-1} \mathrm{~mm} \\ & b=0.25 \mathrm{~mm}\end{aligned}$

Question Tags

JEE Main

Physics

Easy

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