dsinθ=nλ
∴λ=ndsinθ=10−4×401×n1
∴λ=n2.5×10−6=n2500×10−9=n2500(nm)
Putting n=1,λ1=2500nm
n=2,λ2=1250nm
n=3,λ3=833nm
n=4,λ4=625nm
n=5,λ5=500nm
n=6,λ6=357nm
JEE Main 2019 — Physics Optics
In a Young's double slit experiment slit separation 0.1mm, one observes a bright fringe at angle 401rad by using light of wavelength λ1. When the light of wavelength λ2 is used a bright fringe is seen at the same angle in the same set up. Given that λ1 and λ2 are in visible range (380nmto740nm), their values are:
Held on 10 Jan 2019 · Verified 6 Jul 2026.
400nm,500nm
380nm,525nm
625nm,500nm
380nm,500nm
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Work through every JEE Main Optics PYQ, year by year.