In a Young 's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m . A parallel beam of light of wavelength 600 nm is incident on the slits at angle α as shown

Question

In a Young 's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m . A parallel beam of light of wavelength 600 nm is incident on the slits at angle $α$ as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 nm . The minimum value of

$α ($ in $1 0^{- 3} rad)$ so that point P is a maxima is :

TestHub · Accepted Answer

Δ x = d sin α + D d y ​ d α = Δ x − D d y ​ ⇒ α = d Δ x ​ − D y ​ ​ α = 3 × 1 0 − 4 n × 6 × 1 0 − 7 ​ − 1 11 × 1 0 − 3 ​ = 2 n × 1 0 − 3 − 11 × 1 0 − 3 = 1 0 − 3 [ 2 n − 11 ] ​ ​ 2 n − 11 > 0 n > 5.5 × 1 0 − 3 n = 6 × 1 0 − 3 rad

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