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In Fraunhofer diffraction pattern, slit...

In Fraunhofer diffraction pattern, slit width is `0.2 mm` and screen is at 2 m away from the lens. If wavelength of light used is `5000 Å`, then the distance between the first minimum on either side of the central maximum is (`theta` is small and measured in radian)

A

`10^(-1)m`

B

`10^(-2)m`

C

`2xx10^(-2)m`

D

`2xx 10^(-1)m`

Text Solution

Verified by Experts

The correct Answer is:
B
`therefore a=0.2 xx 10^(-3) m, D=2m`
`lambda=5 xx 10^(-7)m`
`x =(lamda D)/(a)=(5xx10^(-7)xx2)/(0.2 xx 10^(-3))`
`therefore x=(5xx 10^(-7))/(10^(-4))=5xx10^(-7+4)m`
`therefore x=5xx10^(-3)m`
Distance between 1st minima on either side `=5xx10^(-3) +5 xx 10^(-3)`
`=10 xx 10^(-3)=10^(-2)m`
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Knowledge Check

  • In YDSE, the distance between the slits is 1 m m and screen is 25cm away from the slits . If the wavelength of light is 6000 Å , the fringe width on the secreen is

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