Case of Doubled screen Maths Investigation

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This is a portion of the main project taken from between.

Case of Doubled screen(diagram only)

        First max.

        First Max.

                                                                               

GENERALIZATION   FOR   N – SCREENS.

First of all let us consider a simple case in which we find the fringe width on nth screen of first maxima.

The diagram may be

                 S1                S2                S3        S4

N => odd

λ                ……………….

                                                                                                            ……………….

………………..        ………………

………………..        …………………

                           D        D        D        D

Again if n is odd then we would obtain the fringe width d/2 & thus the separation between two maximas

= d.

Therefore, if n is odd then fringe width of first maxima = d /2.

And if n is even, then we would get fringe width greater then that when n is odd and it would be λD/d.

Note: Here we are considering diagram type A, ie n=>odd suppose we considered B, then also we would get same result.

THIS IS A VERY GENERAL CASE IN WHICH WE CAN CALCULATE “ FRINGE WIDTH”

FOR ANY SCREEN NUMBER .THE MAIN PART OF THIS IS THAT WE HAVE

CONSIDERED THIS FOR FIRST MAXIMA.

A MORE GENERAL CASE IS DISSUSED BELOW: 

        

        …………………………….

        …………………………….

 

   d/2         X1    X3    X5       X7     X9          Xn-1

        

   d/2        X2     X4  X6        X8     X10        Xn

        ………………………….

        ………………………….

The size of First two slits is “d/2” & distance between the slits is “D’. Now a monochromatic source of light is emitted having wavelength λ . Obviously the light rays from the slits would interfere & From “MAXIMAS” & “MINIMAS” .Now we choose A1th and A2th  “MAXIMA” on screen &, make holes on the  places it was formed so that “LIGHT RAYS” can pass through that points  we carry on in this way till the nth “MAXIMA” . NOW FIND THE FRINGE WIDTH OF nth “MAXIMA”.

Join now!

Here we go a little more mathematical & less linguistic

X1 is the distance of A1th “MAXIMA”

X2        A2

X3                                   A3

X4                                   A4

Xn                      An

        (keep smilling)

For S2  ………….X1=AIλD1\d*************X2=A2 D1\d1

For S3, d = X1 +X2=λD1(A1+A2)\d ****** thus x3=dA3\A1+A2      &x4=dA4\A1+A2

For S4, d=X3+X4 thus X5 = ...

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