β=dλD=0.15×10−3589×10−9×1.5=5.9mm
In a Young’s double slit experiment, the separation between the slits is 0.15mm . In the experiment, a source of light of wavelength 589nm is used…
JEE Main 2020 — Physics Optics
In a Young’s double slit experiment, the separation between the slits is 0.15mm . In the experiment, a source of light of wavelength 589nm is used and the interference pattern is observed on a screen kept 1.5m away. The separation between the successive bright fringes on the screen is:
Held on 7 Jan 2020 · Verified 6 Jul 2026.
Options
- A
6.9mm
- B
3.9mm
- C
5.9mm
- D
4.9mm
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