In double slit experiment, the distance between two slits is 0.6 mm and these are illuminated with light of wavelength 4800A0. The angular width of dark fringe on the screen at a distance 120 cm from slits will be

d=0.6mm=6×10−4m,λ=4800A0Angular fringe width βD=λd=4800×10−106×10−4=8×10−4rad

In double slit experiment, the distance between two slits is 0.6 mm and these are illuminated with light of wavelength 4800A0. The angular width of dark fringe on the screen at a distance 120 cm from slits will be

d=0.6mm=6×10−4m,λ=4800A0Angular fringe width βD=λd=4800×10−106×10−4=8×10−4rad

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In double slit experiment, the distance between two slits is 0.6 mm and these are illuminated with light of wavelength 4800A⁰. The angular width of dark fringe on the screen at a distance 120 cm from slits will be

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4 × 10^–4 raidan

16 × 10^–4 raidan

8 × 10^–4 raidan

6 × 10^–4 raidan

answer is C.

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Detailed Solution

$d = 0 .6 mm = 6 \times 10^{- 4} m, λ = 4800 A^{0}$
Angular fringe width $\frac{β}{D} = \frac{λ}{d}$
$= \frac{4800 \times 10^{- 10}}{6 \times 10^{- 4}} = 8 \times 10^{- 4} rad$

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In double slit experiment, the distance between two slits is 0.6 mm and these are illuminated with light of wavelength 4800A0. The angular width of dark fringe on the screen at a distance 120 cm from slits will be

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In double slit experiment, the distance between two slits is 0.6 mm and these are illuminated with light of wavelength 4800A⁰. The angular width of dark fringe on the screen at a distance 120 cm from slits will be