lambda_{1} and lambda_{2} are used to illumin | vedclass

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Question

(A) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the waveleng

Vedclass · Accepted Answer

$\beta_{1} < \beta_{2}$

Vedclass · Answer

(A) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit separation.This implies that $\beta \propto \lambda$.According to the photoelectric effect,the energy of a photon is $E = \frac{hc}{\lambda}$. For the photoelectric effect to occur,the energy of the incident photon must be greater than or equal to the work function of the metal $(\Phi)$,i.e.,$\frac{hc}{\lambda} \geq \Phi$.This means that a smaller wavelength corresponds to higher energy.Since $\lambda_{1}$ can produce the photoelectric effect and $\lambda_{2}$ cannot,it implies that $\lambda_{1}$ has higher energy than $\lambda_{2}$,which means $\lambda_{1} Since $\beta \propto \lambda$,it follows that $\beta_{1} < \beta_{2}$.

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