The radius of the fifth zone of a zone plate is 2 mm. Considering the zone plate as a converging lens, calculate its focal length for light of wavelength 4800 Å
Answer
Given data in question
r5=0.002mr_5=0.002mr5=0.002m
n=5
λ=4800×10−10m\lambda=4800\times10^{-10}mλ=4800×10−10m
Thus, the focal length of the zone plate,
f=rn2nλ=(0.002)25×4800×10−10=166.67mf =\frac{ r_n^2}{n\lambda}\\= \frac{ (0.002)^2}{5\times 4800\times10^{-10} } \\=166.67mf=nλrn2=5×4800×10−10(0.002)2=166.67m
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