Show that the fringe pattern on the screen is actually a superposition of single slit diffraction from each slit?
Answers
Answer:
Volume of cube = Volume of wire ⇒1cm
3
=πr
2
h
⇒
7
22
×(0.05)
2
×h=(1×10)
3
mm
3
=1000mm
3
⇒h=
22×0.0025
1000×7
=12727.73mm
⇒h=
1000
127272.73
m[∵1mm=
1000
1
m]
⇒h=127.3m, ∴ Length of wire =127.3m
The intensity variation in the fringe pattern obtained on a screen in a Young’s double slit experiment corresponds to both single slit diffraction and double slit interference because the two sources are slits of finite width in the double slit experiment.
The intensity variation in the fringe pattern obtained on a screen in a Young’s double slit experiment corresponds to both single slit diffraction and double slit interference because the two sources are slits of finite width in the double slit experiment.If a lens is placed in front of the double slits and when one of the slit S1 is closed and the other kept open, a single slit diffraction is formed on the screen. A similar diffraction pattern is obtained on the screen if the slit S1 is kept open and S2 is closed. Both diffraction patterns form on the same position on the screen in the focal plane of the lens. When both slits open simultaneously, the resulting total intensity pattern on the screen is actually the superposition of the single slit diffraction pattern formed by waves from various point sources of each slit and a double slit interference pattern as shown. The actual double slit intensity pattern consists of the interference pattern (solid lines) formed within the diffraction pattern (dotted lines).