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. `1/3`sec
B. `2/3`sec
C. `1/2`sec
D. `1/4`sec

Answer

A, C, D

Explanation

`2acos(pit)=aimplies cos(pit_(1))=+(1)/(2)implies pit_(1)=(pi)/(3), t_(1)=(1)/(3)sec`
Detector will receive signal in `(0 to (1)/(3) sec)`
And `cos(pit_(2))=-(1)/(2) implies pit_(2) =(2pi)/(3) implies t_(2)=(2)/(3)sec`.
It again receive signal between `(t_(2)=(2)/(3)sec implies 1sec)`
The time interval for which detector remains idle is between `(1)/(3)` sec to `(2)/(3)`sec i.e., `Deltat=(1)/(3)sec`.