Question #202104

 A beam of parallel light of wavelength 6000 Å is incident normally on a 1 rupee  coin of diameter 3 cm. If an observation screen is placed at a distance of 1 m from  the coin, how many Fresnel zones would be obstructed by it (the coin)? For what  separation between the coin and the screen, only 5 zones would be cut off?


Expert's answer

1)

m=d24λa=0.0324⋅6⋅10−7⋅1=375,m=\frac{d^2}{4\lambda a }=\frac{0.03^2}{4\cdot 6\cdot 10^{-7}\cdot1}=375,

2)

a′=d24λm′=0.0324⋅6⋅10−7⋅5=75 m.a'=\frac{d^2}{4\lambda m'}=\frac{0.03^2}{4\cdot 6\cdot 10^{-7}\cdot 5}=75~\text m.


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