In the figure, two isotropic point sources of light ( S 1 and S 2 ) are separate
ID: 1376468 • Letter: I
Question
In the figure, two isotropic point sources of light (S1 and S2) are separated by distance 2.40 ?m along a y axis and emit in phase at wavelength 930 nm and at the same amplitude. A light detector is located at point P at coordinate xP on the xaxis. What is the greatest value of xP at which the detected light is minimum due to destructive interference?
In the figure, two isotropic point sources of light (S1 and S2) are separated by distance 2.40 ?m along a y axis and emit in phase at wavelength 930 nm and at the same amplitude. A light detector is located at point P at coordinate xP on the xaxis. What is the greatest value of xP at which the detected light is minimum due to destructive interference?Explanation / Answer
Destructive interference occurs when the phase difference is exactly (an odd multiple of) pi radians. This corresponds to length difference of half the wavelength. You need to find a point where the difference between the distance to S1 and S2 is exactly half the wavelength. The waves also destructively interfere at 3/2 the wavelenght, 5/2 the wavelength... but these occur closer to the origin.
The distance to S1 is simply x
The distance to S2 derives from the pythagorean theorem: sqrt(x^2+y^2). y is given.
sqrt(x^2+y^2)-x=w/2
sqrt(x^2+y^2)=w/2+x //isolate the square root
x^2+y^2=(w/2+x)^2 //undo the square root. This introduces no new solutions as both sides are positive.
x^2+y^2=w^2/4+2wx/2+x^2 //perform the multiplication
y^2=w^2/4+wx //subtract x^2 and simplify
wx=y^2-w^2/4 //isolate terms with x
x=(y^2-(w/2)^2)/w //isolate x
x=(2400^2-465^2)/930 //substitute values in nm. 1um=1000nm.
x~=5961
5961nm=5.961um
The distance is roughly 5.961 micrometers.
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