Q86 / 90 · JEE Main 2024 (29 Jan Shift 2) · Physics

In a single slit diffraction pattern, a light of wavelength 6000 Ao is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from slits. The width of the slit is ____×10-4 m.

Correct answer: 2

Solution

For nth minima:

⇒bsinθ=nλ

(λ is very small, so sinθ is very small, hence sinθ≃tanθ.

⇒btanθ=nλ

⇒byD=nλ

⇒yn=nλDb (Position of nth minima)

Figure

B→1st  minima and A→3rd :

y3=3λDb,y1=λDb

⇒Δy=y3-y1=2λDb

⇒3×10-3=2×6000×10-10×0.5b

⇒b=2×6000×10-10×0.53×10-3

⇒b=2×10-4 m

Hence, the correct answer is 2.

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