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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 -7 m. If the width of the fringe on the screen is 4mm. The distance between screen and slits is a) 0.1 mm b) 1 cm c) 0.1 cm d) 1 m

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 is


a)      0.1 mm


b)      1 cm


c)      0.1 cm


 


d)      1 m

Grade:11

2 Answers

Arun
25768 Points
10 months 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
14148 Points
9 months 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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