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

Explanation

Given : Fringe width
` y = 1.0 mm xx 2 = 2.0 mm `
` d = 0.24 mm`,
`W_0 = 2.2 ev`
D = 1.2 m `
` y = lambda D / d `
` or `lambda = yd / D `
` = 2 xx 10^(-3) xx 0.24 xx 10^(-3) / 1.2 `
` = 4 xx 10^(-7) m`
` E = hc / lambda `
` 4.14 xx 10^(-15) xx 3 xx 10^(8) / 4 xx 10^(-7) `
` = 3.105 ev`
Stopping potential
` eV_0 = 3.105 - 2.2 = 0.905 eV`
`V_0 = 0.905 xx / 1.6 xx 10^(-19) xx 1.6 xx 10^(-19) V `
` = 0.905 V ` .