JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 9)
In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment,
a source of light of wavelengh 589 nm is used and the interference pattern is observed on a
screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :
4.9 mm
5.9 mm
6.9 mm
3.9 mm
Explanation
$$\beta = {{\lambda D} \over d} = {{589 \times {{10}^{ - 9}} \times 1.5} \over {0.15 \times {{10}^{ - 3}}}}$$ = 5.9 mm
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