In the following arrangement,y = 1.0 , mm,d = | vedclass

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(A) From the given figure,the distance between the central bright fringe and the first dark fringe is $y = \frac{\beta}{2}$,where $\beta$ is the fring

Vedclass · Accepted Answer

$0.9$

Vedclass · Answer

(A) From the given figure,the distance between the central bright fringe and the first dark fringe is $y = \frac{\beta}{2}$,where $\beta$ is the fringe width.Thus,$\beta = 2y = 2 	imes 1.0 \, mm = 2.0 \, mm = 2 	imes 10^{-3} \, m$.The fringe width is given by $\beta = \frac{\lambda D}{d}$.Therefore,$\lambda = \frac{\beta d}{D} = \frac{2 	imes 10^{-3} 	imes 0.24 	imes 10^{-3}}{1.2} = 0.4 	imes 10^{-6} \, m = 4000 \, 	ext{\AA}$.The energy of the incident photon is $E = \frac{12400}{\lambda (	ext{\AA})} \, eV = \frac{12400}{4000} \, eV = 3.1 \, eV$.Using Einstein's photoelectric equation,$K_{max} = E - W_0 = 3.1 \, eV - 2.2 \, eV = 0.9 \, eV$.The stopping potential $V_0$ is given by $K_{max} = eV_0$,so $V_0 = 0.9 \, V$.

In the following arrangement,$y = 1.0 \, mm$,$d = 0.24 \, mm$,and $D = 1.2 \, m$. The work function of the material of the emitter is $2.2 \, eV$. The stopping potential $V$ needed to stop the photocurrent will be....$V$

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