In a single slit diffraction pattern, a light of wavelength 6000A˚ is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen in placed 50 cm away from slits. The width of the slit is ______ ×10−4m.
Answer: 2
Solution
For the nth minimum of single-slit diffraction, bsinθ=nλ. For small angles sinθ≈tanθ=y/D, so the position of the nth minimum is
yn=bnλD
The separation between the first and third minima is
Δy=y3−y1=b3λD−bλD=b2λD
With λ=6000 Å =6×10−7 m, D=0.5 m, and Δy=3×10−3 m:
b=Δy2λD=3×10−32×6×10−7×0.5=2×10−4m
JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer
