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In the gure below, two isotropic point sources of light (S1 and S2) are separated by distance 2.70 mm along a y-axis and emit in phase at wavelength 900 nm and at the same amplitude. A light detector is located at point P at coordinate xP on the x-axis. What is the greatest value of xP at which the detected light is minimum due to destructive interference

Respuesta :

The figure is missing, so i have attached it.

The distance between S1 and S2 is 2.70μm

Answer:

7.88 μm

Explanation:

Let the distance in question be x.

The path difference between rays starting from S1 and S2 up to points on the x axis when x is greater than zero is given as;

(√(d² + x²)) - x = (m + ½)λ

Making x the subject of the formula, we have;

x = (d²/(2m + 1)λ) - (2m + 1)λ/4

m is an integer e.g, 0,1,2..

The greatest value of xP will be at m = 0

Thus, putting m = 0,we have;

x = (d²/λ) - (λ/4)

We are given;

d = 2.70 μm = 2700 nm

λ = 900 nm

Thus;

x = (2700²/900) - (900/4)

x = 7875 nm = 7.875 μm ≈ 7.88 μm

Ver imagen AFOKE88