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Question: If light of wavelength 4000Å incidents on a slit of width 0.5 mm and diffraction pattern is obtained...

If light of wavelength 4000Å incidents on a slit of width 0.5 mm and diffraction pattern is obtained on a screen 2m from slit then distance of first bright fringe from central bright ring

A

1.2 mm

B

0.8 mm

C

0.4 mm

D

None

Answer

0.8 mm

Explanation

Solution

3Dλ2a\frac{3D\lambda}{2a} – Dλa\frac{D\lambda}{a} = Dλ2a\frac{D\lambda}{2a} = 2×4000×10–102×0.5×10–3\frac{2 \times 4000 \times 10^{–10}}{2 \times 0.5 \times 10^{–3}}

= 0.8 × 10–3 m = 0.8 mm