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 photo current will be

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0.9 V

0.5 V

0.4 V

0.1 V

answer is A.

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Detailed Solution

As we know in Young's double slit experiment fringe width = separation between two consecutive fringe or dark fringes $= beta = frac{{lambda D}}{d}$
$Here,beta = 2y, Rightarrow 2y = frac{{lambda D}}{d} Rightarrow lambda = frac{{2yd}}{D}$
$Rightarrow lambda = frac{{2 times 1 times {{10}^{ - 3}} times 0.24 times {{10}^{ - 3}}}}{{1.2}} = 4 times {10^{ - 7}}m = 4000,{AA}$
Energy of light incident on photo plate
$E,(eV) = frac{{12375}}{{4000}} = 3.1,eV$
According to Einstein photoelectric equation
E = W₀ + eV₀