In an interference experiment,let D be the di | vedclass

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Question

(C) The fringe width $Z$ in an interference experiment is given by the formula $Z = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D

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$\left(\frac{d_{B}}{d_{A}}\right)$

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(C) The fringe width $Z$ in an interference experiment is given by the formula $Z = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the source and the screen (or eyepiece),and $d$ is the distance between the two virtual sources.Since $\lambda$ and $D$ are constant,the fringe width is inversely proportional to the distance between the sources,i.e.,$Z \propto \frac{1}{d}$.Therefore,we can write the ratio as $\frac{Z_{A}}{Z_{B}} = \frac{d_{B}}{d_{A}}$.Thus,the correct option is $C$.

In an interference experiment,let $D$ be the distance between the slit and the eyepiece. When the distance between two virtual sources is changed from $d_{A}$ to $d_{B}$,the fringe width changes from $Z_{A}$ to $Z_{B}$. The ratio $Z_{A} / Z_{B}$ is:

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