A monochromatic point source radiating wavelength `6000Å` with power `2` watt, an aperture A of diameter `0.1 m` and a large screen SC are placed as shown in fig, A photoemissive detector D of surface area `0.5 cm^(2)` is placed at the centre of the screen . The efficiency of the detector for the photoelectron generation per incident photon is `0.9`

(a) Calculate the photon flux at the centre of the screen and the photo current in the detector.
(b) If the concave lens L of focal length `0.6` m is inserted in the aperture as shown . find the new values of photon flux and photocurrent Assume a uniform average transmission of `80%` from the lens .
(c ) If the work function of the photoemissive surface is `1eV` . calculate the values of the stopping potential in the two cases (within and with the lens in the aperture).

Answer

(i) `1.1462 xx 10^(14) "Photons"//m^(2)-s`,
`2.063 xx 10^(-10)A`
(ii) `2.0574 xx 10^(13) "Photons"//m^(2)-s`,

Explanation

Power of source `= 21J//"sec"`
Energy of 1 photon `= (hc)/(lambda) = (12400)/(6000) = 2.067 eV`
No. of photons emiited /sec
`= (2 xx 10^(19))/(2.067 xx 16) = 6.0474 xx 10^(18)`
No. of photons striking on sphere of `0.6 m`
`= (6.0474 xx 10^(18))/(4pi(0.6)^(2))//m^(2)`
no. of photons passing through aperture
`= (6.0474 xx 10^(18))/(4pi(0.6)^(2)) xx pi ((0.1)/(2))^(2) = (4.2 xx 10^(6))/(2)` photons
No. of photons incident/area on screen (assuming aperature to act as a secondary point source)
`= (4.2 xx 10^(16))/(4 xx 4pi(5.4)^(2)) =` photon flux
`= 1.1462 xx 10^(14) "photons"//m^(2)` sec
No. of photon incident on detector /sec
`= 1.1462 xx 10^(14) xx (0.50)/(10000)`
Photo current `= etaoverset(.)(N)e = 2.063 xx 10^(-10)A`
When the lens of `f = - 0.6 m` is used,
`1/v - 1/u = 1/f rArr (1)/(v) = (-1)/(0.3) = v = -0.3 m`
Thus image will be at `0.3 m` from the lens in teh direction opposite to the screen.
Distance between screen & image `= 5.7 m`
No. of photons striking lens
`=` No. of photons striking the aperture
`= 1.05 xx 10^(16)` photons.
Photons transmitted through the lens
`= 0.8 xx 1.05 xx 10^(16) = 0.84 xx 10^(16) "photons"//s`
This new sitution, A point source emitting
`0.84 xx 10^(16) "photon"//"sec"` of `lambda = 6000 Å`
is kept at `5.7 m` away from the screen
`= 0.84 xx 10^(16)/(4pi(5.7)^(2)) = "photon"//"sec" = 2.0574 xx 10^(13)`
Electrons emitted `= 0.9 xx 2.0574 xx 10^(13) xx (0.5)/(10000)`
Current `= 1.4813 xx 10^(-10) A`
`E_(p) = 2.067 eV, phi = 1 eV`
`K.E._("max") = 1.067 eV,V_(S) = 1.067 V`