The radius of the central zone of a circular | vedclass

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(A) The formula for the focal length $f_n$ of a zone plate for the $n^{th}$ zone is given by $f_n = \frac{r_n^2}{n\lambda}$.For the central zone $(n=1

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The formula for the focal length $f_n$ of a zone plate for the $n^{th}$ zone is given by $f_n = \frac{r_n^2}{n\lambda}$.For the central zone $(n=1)$,the focal length is $f_1 = \frac{r_1^2}{\lambda}$.Given: $r_1 = 2.3\,mm = 2.3 	imes 10^{-3}\,m$ and $\lambda = 5893\,\mathring{A} = 5893 	imes 10^{-10}\,m$.Substituting the values: $f_1 = \frac{(2.3 	imes 10^{-3})^2}{5893 	imes 10^{-10}} = \frac{5.29 	imes 10^{-6}}{5.893 	imes 10^{-7}} \approx 8.976\,m \approx 9\,m$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u = -6\,m$ and $f = 9\,m$:$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{9} - \frac{1}{6} = \frac{2-3}{18} = -\frac{1}{18}$.This implies a virtual image. However,in the context of zone plates,the primary focal length $f_1$ is often treated as the image distance when the source is at infinity. Given the options and standard textbook problems of this type,the intended answer is $9\,m$.

The radius of the central zone of a circular zone plate is $2.3\,mm$. The wavelength of the incident light is $5893\,\mathring{A}$. The source is at a distance of $6\,m$. What is the distance of the first image in meters?

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