Question

How can I derive the equation of young double slit experiment y=λD/a where y is the distance between maxima, D is the distance from screen and a is the split spacing and λ the wavelength?

Answer 1

as shown in above figure.... path differance between waves reaching P from S1 & S2 is (S2P-S1P)
suppose, S1M & S2N are perpendicular on screen
thn,
from fig,
PM=X-d & PN=X+d

in triangle S2NP,
(S2P)^2=(S2N)^2+(PN)^2
therfore,
(S2P)^2=D^2+(X+d)^2
in triangle S1MP,
(S1P)^2=(S1M)^2+(S1M)^2
therfor,
(S1P)^2=D^2+(X-d)^2
for path diff as i told u,
(S2P)^2-(S1p)^2=D^2+(X-d)^2-D^2-(X-d)^2
equals to, S2P-S1p=2Xd/(S2P-S1P)

Answer 2

but practically, the dist X & d are very small as compaired to D .to a first approximation we can write
S1P=S2P=D or S2P+S1P=2D
hence,the path difference between two wave is given by,
S2P-S1P=2Xd/2D=Xd/D

Answer 3

now the intensity at P will be max or min according to the path difference is an odd multiple of\[\lambda \div2\]

Answer 4

i.e S2P- S1P =Xd/D=2n(\[\lambda \div2\])

Answer 5

where n=0,1,2,3.........................
or
x=\[n lambdaD divd\]

Answer 6

the pt p will be dark if path diff is an odd integral multiple of \[\lambda \div2\]

Answer 7

let Xn & Xn+1 denotes the distance of nth & (n+1)th bright band on the same side of central bright band, from equation of
X=n\[\lambda D \div d\]

Answer 8

Xn+1=(n+1)\[\lambda D \div d\]
when u ll substract above two equation u ll get the equation\[X =\lambda D \div d\]
where X=Xn+1-Xn is the band width or fringe width.

Answer 9

simillarly for dark band is \[Xm=(2m-1)\lambda D \div 2d\]
and X\[Xm+1=(2(m+1)-1)\lambda D \div2d\]
substracting above equations u will get
\[X'=\lambda D \div d\] here X' =Xm+1-Xm

Answer 10

thus the distance between two adj bright or dark bands is equal i.e.the band width of bright and dark band is same

Answer 11

I lost you at about ...."Now the intensity at P....". I will have to come back and take another look when I am fresher. That first part helped me to answer another question, so thanks

Answer 12

tell me wht doent u understand,............i ll hlp u to sort out it..

Answer 13

Ok. When I actually put pencil to paper I'm stuck higher up. I understand up to here:
(S2P)^2=D^2+(X+d)^2 and (S1P)^2=D^2+(X-d)^2

but for (S2P)^2-(S1p)^2 I get
(S2P)^2-(S1p)^2= [D^2+(X+d)^2] - [D^2+(X-d)^2]

= [D^2 + X^2 + 2dX + d^2] - [D^2 + X^2 - 2dX + d^2]

(S2P)^2-(S1p)^2= 4Xd
This is where I'm stuck. How did you get S2P-S1p=2Xd/(S2P-S1P) from this?

Answer 14

sry ...for ur inconvience....

Answer 15

take a look at attachments....i hope it will be more clear to u...

Answer 16

here...i hav changed notations of distances.....instead of using '2d' i m using 'd' coz its very small distance....& so remaing changes..
ask me if u ll have any prob.....

Answer 17

ok. Thanks. I'll have a look at it and let u know how it goes.

Answer 18

Ok. Thanks much for this. I get it now.

Answer 19

yup....wlcm...:)