In Young’s double slit experiment the distance between the two slits is 0.1 mm and th wavelength of light used is 4 x 10-7m. If the width of the fringe on the screen is 4mm. The distance between screen and slits isa) 0.1 mmb) 1 cmc) 0.1 cmd) 1 m
Radhika Batra , 11 Years ago
Grade 11
2 Answers
Arun
Last Activity: 4 Years ago
The position of n
th
dark fringe. So position of first dark fringe in x
1
= λD/2d.
d = 20 cm, D = 0.1mm, λ=5460
A
o
, x
1
= 0.16
Vikas TU
Last Activity: 4 Years ago
PE = qV where q =1.6 x 10^-19 C and V = 1000 V
Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:
PE* = 0, KE* = 1/(2mv^2)
Now since PE +KE = Total energy =PE* + KE*
qV + 0 = 0 + 1/2mv^2
Or: KE = qV = 1.6 10^-15 J
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