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 = A5λD1(A1+A2)\d(A3+A4)******X6=A6λD1(A1+A2)\(A3+A4)d
For s5, d=x5+x6=λD1[(A1+A2)(A5+A6)\(A3+A4)]\d X7=A7d(A3+A4)\(A1+A2)(A5+A6)
X8=A8d(A3+A4)\(A1+A2)(A5+A6)
For S6, “d”= d(A3+A4)(A7+A8)\(A1+A2)(A5+A6) X9=A9λD1(A1+A2)(A5+A6)\d(A3+A4)(A7+A8)
X10=A10λD1(A1+A2)(A5+A6)\(A3+A4)(A7+A8)
For S7 “d”=λD1\d{[(A1+A2)(A5+A6)(A9+A10)]\(A3+A4)(A7+A8)} X11= A11d (A3+A4)(A7+A8)
(A1+A2)(A5+A6)(A9+A10)
&X12= A12d (A3+A4)(A7+A8)
(A1+A2)(A5+A6)(A9+A10)
Since the fringe width is different for different “As” therefore we
Break our series in two parts:
1\2 TYPE “1” 3\4 TYPE “2”
5\6 7\8 have another
type of fringe width
9\10 have a particular 11\12
type of fringe width
now the general term of TYPE “1” series is {4m-3\4m-2} now if we have a no.
(n)such that m=n+3\4 or m=n+2\4 & both “m” & “n” obtained are integers then
we will say that (n) lies in type 1 now each element of type 1 has certain
common factor in it the uncommon factors are1\2=>1
5\6=>(A1+A2)\(A3+A4)
9\10=>(A1+A2)(A5+A6)\(A3+A4)(A7+A8)
now if we find general term of this series& generalize this situation then we will find general term of type “1” series BY OBSERVATION GENERAL TERM IS
(A1+A2)(A5+A7)(A9+A10)………………………………………………………..[Am+4 + An-4]
(A3+A7)(A7+A8)(A11+A12)…………………..…………………………………..[Am-2 +An-2]
Where m is the maxima no. just prior to n.
NOW FOR TYPE TWO SERIES, the variance is 3\4 =>1\A1+A2
7\8 =>A3+A4\(A1+A2)(A5+A6)
11\12=>(A3+A4)(A7+A8)
(A1+A2)(A5+A6)(A9+A10)
Again by simple observation the general series in this case is:
(A3+A4)(A7+A8)(A11+A12).............................................(Am-4+An-4)
(A1+A2)(A5+A6)(A9+A10)..............................................(Am-2+An-2)
FINALLY WE OBTAIN:
XN {for type 1} =ANλD1 (A1+A2)(A5+A6)(A9+A10).........(Am-4+An-4)
d (A3+A4)(A7+A8).............………(Am-2+An-2)
&
XN {for type 2} = And A3+A4)(A7+A8).....................(Am-4+An-4)
(A1+A2)(A5+A6)(A9+A10)...........(Am-2+An-2)
IN OUR RESULT BOTH Nr & Dr ARE CONSIDERED TILL [n].
THE CASE ΟF INCLINED SCREENS (SERIES)
Indeed, I must admit this was a very exhausting section, but I equally enjoyed it, this is a very general case in which one can calculate the fringe width of any maxima of a screen kept at angle “σ” now we try to generalize
For first maxima, First Max.
A
Central max.
d\2 B σ
C
d\2
L\2
Screen Fig.1 σ
D L\2cotσ X1
In this case we keep the screen at an angle σ with the horizontal, usual set up is made. Calculate the distance of first max. From “central max.”
Lets consider an arbitrary screen at a point were first max. is formed. Now for this screen => “D” =[D+L\2cotσ +X1] {LET L\2cotσ =y)
Thus AC= λ(D+L\2COTσ +X1)\d & hence for distance from central max.
Sinσ =λ(D+L\2cotσ+X1)\Dab =>ab=λ\d sinσ{D+Y+X1}
i.e.
A1=λ\d sinσ{D+Y+X1}……….(1)
Here we are defining x1,x2,x3…. Not just as a distance but a projection if the screen & as screen becomes a an angle 90* x1, x2 ,x3.=0 i.e. the projection becomes 0 . [CONSULT FORMULA “P” FOR IS MATHMATICAL PROOF]
SECOND MAX.
D
Second max.
A σ F
First max.
B C E
FIG.2
σ
D Y X1 X2
Again proceeding in the same way,
D for this screen = D+Y1 +X1+X2]
THUS DE= 2λ[D+Y1+X1+X2]
d
thus DF= DE-FE
ie DF= λ\d[2D +2Y +2X1 +2X2 –D-Y-X1] = λ\d[D+Y1+X1+2X2]
NOW sinσ = DF\AD = λ\DA2[D+Y+X1+2X2]
FINALLY: A2 = λ\d sinσ[D+Y1+X1+2X2]
3RD MAX.=>
G
D H
A F
B C E I
FIG.3
σ
D Y X1 X2 X3
Again for this Distance between slit & screen is = [D +Y +X1 +X2 +X3]
THUS GI = 3λ[D+Y+X1+X2+X3]\d
Thus GH = λ[3D+3Y +3X1 +3X2+3X3 –2D-2Y- 2X1-2X2]
d
thus GH= λ[D+Y1 +X+X2+3X3]
d
FINALLY: A3=λ[D+Y+X1+X2+3X3]
dsinσ
{calculation of x1,x2,x3,………}
consult fig. 1
tanσ = λ D + λY + λX1
dx1
X = λD + λY = λ{D+Y}
dtanσ−λ dtanσ−λ
Consult fig2 tanσ = λ{D+Y+X1+X2}\X2
=>X{dtanσ−2λ} = λ{D+Y+X1}
> X2= λ {D+Y+X1}
dtanσ−2λ
Consult fig.3 tanσ = λ {D+Y+X1+X2+3X3}
DX3
=>X3{dtanσ-3λ} = λ{D+Y+X1+X2}
THUS X3 = λ {D+Y+X1+X2}
dtanσ-nλ
GENERALISITION:
Xn = λ{D +Y+X1+X2+X3+X4+….+Xn-1}
dtanσ-nλ
We can surely say that f the second max. of screen at an angle will be the second max. of the arbitrary screen because between first & second max., since their is no other max. this light rays cant pass in that region & inferred to produce a maxima. This explanation is also true in case of first-second, second-third and third-forth & ay maxima.
Finally distance of a general max. From central maxima in a screen kept a an angle is:
An= λ{D+L\2(cotσ) +X1+X2+X3+……+nXn}
d sinσ
(A significant result.)
COLOURED FRINGES
If white light of wavelength {400-700} nm is sent through a ydse then we get colored fringes
For each of wavelength out of 400-700 we will obtain different fringes & hence different
Fringe width. The first maxima will be formed at different positions as shown below.
The central maxima will be white.
RED
ORANGE
YELLOW
GREEN
BLUE
INDIGO
VOILET
WHITE LIGHT
We may obtain some intensity due to λ also. This case may arise a possibility of MAXIMA overlapping between higher order maxima’s.
☺ANALYSIS ENDS ☺
Shubham Verma
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