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In a YDSE experiment two slits S(1) and ...


In a YDSE experiment two slits `S_(1)` and `S_(2)` have separation of d=2mm the distance of the screen is `D=(8)/(5)` m source S starts moving from a very large distance towards `S_(2)` perpendicular to `S_(1)S_(2)` as shown in figure the wavelength of monochromatic light is 500 nm. The number of maximas observed on the screen at point P as the moves towards `S_(2)` is

A

4001

B

3999

C

3998

D

4000

Text Solution

Verified by Experts

The correct Answer is:
D
`S_(1)P-S_(2)P=(d^(2))/(2D)=(2xx10^(-3)xx2xx0^(-3))/(2xx(8)/(5))=(5)/(2)lamda" "(lamda=500nm)`
So when S is at `infty` there is `1^(st)` minima and when S is at `S_(2)` there is last minima because `d//lamda=4000`
So the number of minima's will be 4001 and number of maxima's will be 4000.
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Knowledge Check

  • In a Young's double slit experiment, let S_(1) and S_(2) be the two slits, and C be the centre of the screen. If angle S_(1)CS_(2)=theta and lambda is the wavelength, the fringe width will be

    A
    `(lambda)/(theta)`
    B
    `lambda theta`
    C
    `(2lambda)/(theta)`
    D
    `(lambda)/(2theta)`

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