In Young's double-slit experiment, two coherent source are used. Intensity of one of the sources is I but for the other it is slightly different `I + dI`. Show that intensity at the minima is approximately `((delta I)^2)/(4 I)`.
Text Solution
Verified by Experts
From `I = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi` `I_(max) = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos 0^(@) = I + (I + delta I) + 2 sqrt(I(I + delta I))` As `deltaI lt lt I`, therefore, `I_(max) I + I 2 I = 4I` Again from `I = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi` `I_(min) = I + (I + deltaI) + 2sqrt(I(I + deltaI)) cos 180^(@) = 2I + delta I - 2 I(1 + (delta I)/(I))^(1//2)` `= 2 I + delta I - 2 I [1 + (1)/(2) (delta I)/(I) + ((1)/(2)((1)/(2) - 1))/(2!)((delta I)/(I))^(2)] = 2 I + delta I - 2 I - delta I + (1)/(4)I((delta I)/(I))^(2) = (delta I)^(2)/(4I)`
Topper's Solved these Questions
OPTICS
PRADEEP|Exercise Value based questions|3 Videos
OPTICS
PRADEEP|Exercise Exercise|582 Videos
OPTICS
PRADEEP|Exercise Long answer questions (NCERT Ch - 10)|2 Videos
In standard Young's double-slit experiment when sources of light are coherent, then intensity at the centre of the screen is found to be I_1 and when sources are incoherent, then intensity at the centre is found to be I_2 . Calculate I_1//I_2 . {:(0,1,2,3,4,5,6,7,8,9):}
In Young's double slit experiment, the ratio of maximum and minimum intensities in the fringe system is 9:1 the ratio of amplitudes of coherent sources is
In Young's double slit experiment, the intensity of central maximum is I . What will be the intensity at the same place if one slit is closed ?
Light waves from two coherent sources having intensities I and 2I cross each other at a point with a phase difference of 60^(@) . The intensity at the point will be
In Young's double slit experiment, the maximum intensity is I_(0) . What is the intensity at a point on the screen where the path difference between the interfering waves is lamda/4 ?
In Young's double slit experiment, the intensity of light at a point on the screen where path difference is lambda is I. If intensity at another point is I/4, then possible path differences at this point are
In young's double slit experiment if the seperation between coherent sources is halved and the distance of the screen from coherent sources is doubled, then the fringe width becomes:
In a Young's double-slit experiment, the intensity ratio of maxima and minima is infinite. The ratio of the amlitudes of two sources
In Young's double slit experiment, the constant phase difference between two source is (pi)/(2) . The intensity at a point equidistant from theslits in terms of maximum intensity I_(0) is