Question

Hi guys
I'm preparing for an exam. It will be very helpful if you could explain some questions to me. I don't have solutions to these(but i have answers):
1) Four charges are placed at the four corners of a square of side a as shown in the figure. What will be the electric dipole moment (in vector form) of this configuration?
2) In Young's double slit exp. , the distance b/w slits is 1mm, wavelength = 6000 Angstrom, distance b/w slits and screen D is 1m. The slits produce same intensity on the screen. The minimum distance b/w two points on the screen having 75% intensity of the maximum is ?

Answer 1

this is the configuration for the 1st question.

Answer 2

@iambatman

Answer 3

@Vincent-Lyon.Fr

Answer 4

Dipole moment definition about O is:
\(\vec P=\sum q_i \vec {OA_i}\)

It should lead to \(\vec P = 3qa \,\hat x+3qa \,\hat y\)

Answer 5

Is the answer to the second one 0.40 mm ?

Answer 6

Answer to the first question is :
\[-qa (i) -qa(j)\]

Answer 7

Then, there must be some minus sign missing in your sketch. (could be -2q in the top-right corner).

Answer 8

and btw answer to 2nd question is 0.20mm

Answer 9

It's correct. I had made the wrong sketch ;-)
Do you know how to do it now?

Answer 10

no i don't
plz tell me

Answer 11

Narrowest Δx is as shown.


If a is the distance between slits and D is the slit-screen distance, then
intensity is I(x) = Imax/2 (1 + cos (2πax/λD))

Solving for I = 0.75 Imax, you will get Δx = λD/3a