In the following arrangement y = 1.0 mm,d = 0 | vedclass

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(A) In Young's double-slit experiment,the fringe width $\beta$ is the distance between two consecutive bright or dark fringes,given by $\beta = \frac{

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$0.9$

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(A) In Young's double-slit experiment,the fringe width $\beta$ is the distance between two consecutive bright or dark fringes,given by $\beta = \frac{\lambda D}{d}$.From the figure,the distance between a central bright fringe and the first dark fringe is $y = 1.0\ mm$. Since the distance between a bright and the next dark fringe is $\frac{\beta}{2}$,we have $y = \frac{\beta}{2}$,which implies $\beta = 2y = 2.0\ mm = 2.0 	imes 10^{-3}\ m$.Using the fringe width formula: $\lambda = \frac{\beta d}{D} = \frac{2.0 	imes 10^{-3} 	imes 0.24 	imes 10^{-3}}{1.2} = 4 	imes 10^{-7}\ m = 4000\ \mathring{A}$.The energy of the incident photon is $E = \frac{12400}{\lambda(\mathring{A})} \approx \frac{12400}{4000} = 3.1\ eV$.According to Einstein's photoelectric equation,$E = W_0 + eV_0$,where $W_0 = 2.2\ eV$ is the work function.$eV_0 = E - W_0 = 3.1\ eV - 2.2\ eV = 0.9\ eV$.Therefore,the stopping potential $V_0 = 0.9\ V$.

Column $I$	Column $II$
$(A)$ Transition between two atomic energy levels	$(p)$ Characteristic $X$-rays
$(B)$ Electron emission from a material	$(q)$ Photoelectric effect
$(C)$ Mosley's law	$(r)$ Hydrogen spectrum
$(D)$ Change of photon energy into kinetic energy of electrons	$(s)$ $\beta$-decay

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 $V_0$ needed to stop the photocurrent will be .............. $V$.

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