In a Young's double slit experiment `lamda= 500nm, d=1.0 mm andD=1.0m`. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

Options

A. `2 xx 10^(-4) m`
B. `1.25 xx 10^(-4) m`
C. `4 xx 10^(-4) m`
D. `2.5 xx 10^(-4) m`

Answer

b

Explanation

Intensity of central maxima, `I_(u) = (2 A_(0))^(2) = 4 k A_(0)^(2) = 4 I_(0)` ltbgt Intensity at distance x from the central maxima is halft of the maximum intensity if
`I = 4I_(0) cos ^(2) ((phi)/(2)) = (4I_(0))/(2) implies cos^(2) ((phi)/(2)) = (1)/(2)`
`:. x = 1.25 xx 10^(-4) m`
`(lambda D)/(4d) = (500 xx 10^(-9) xx 1)/(4 xx 10^(-3))`