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Question on Young's double slit EXPERIMENT
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In YDSE EXPERIMENT THE DISTANCE BETWEEN THE SLITS AND SCREEN IS a &DISTANCE BETWEEN TWO SLITS IS b.WAVELENGTH OF LIGHT USED IS LEMBDA. THE DISTANCE BETWEEN THE THIRD MAXIMA ON ONE SIDE AND FOURTH MINIMA ON OPPOSITE SIDE OF THE SCREEN IS :?:
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3 Comments sorted by
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hi
ans is 13(LAMBDA)a/2b.
if correct thn ask for solution.
if no thn give me correct ans i'll gt it solved -
u gt the answer rite ... good work
I share the solution:
Maxima occurs under the condition :
n(LAMBDA) = Xb/a
Minima occurs under the condition:
(n+1/2)(LAMBDA) = Xb/a
Thus,
For the 3rd Maxima :
Xmax = 3(LAMBDA)a/b
For 4th Minima :
Xmin = 7(LAMBDA)a/2b
Since they are on the opposite sides ,
Required distance :
X = Xmax + Xmin
= 13(LAMBDA)a/2b
add anything to the solution if u wish !! -
mylie said:
In YDSE EXPERIMENT THE DISTANCE BETWEEN THE SLITS AND SCREEN IS a &DISTANCE BETWEEN TWO SLITS IS b.WAVELENGTH OF LIGHT USED IS LEMBDA. THE DISTANCE BETWEEN THE THIRD MAXIMA ON ONE SIDE AND FOURTH MINIMA ON OPPOSITE SIDE OF THE SCREEN IS :?:
answer is 13(LAMBDA)a/2b.
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- Friction December 2008
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