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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