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. For AB, particle is ` " "` Moving in +ve X-direction with an increasing speed
B. For BC, particle is ` " "` Moving in +ve X-direction with a decreeing speed
C. For CD, particle is `" "` Moving in –ve X-direction with an increasing speed
D. For DE, particle is` " "` Moving in –ve X-direction with a decreasing speed

Answer

A, B, C, D

Explanation

`(A-P,B-P,C-Q,D-R) `
For AB and BC : a = slope of tangent = positive and v is positive ` rArr ` Motion in `+X -direction with increasing speed .
For CD : v is positive ` rArr ` Motion is in ` +X-direction.
a= slope of tangent= negative
` rArr overset- a ` is in -X direction .`rArr ` speed decreases . For DE : v is negative `rArr ` motion is in-X direction.
a= slope of tangent ` lt 0 " "rArr overset-a` is also in -X direction ` rArr speed increases.