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Light of intensity I falls along the axi...

Light of intensity I falls along the axis on a perfectly refleccting right circular cone having semi-vertical angle `theta` and base radius R. If E is the energy of one photon and c is the speed oh light, then find
Q. the number of photons hitting the cone per second

A

`(piR^2I)/(2E)`

B

`(2piR^2I)/(E)`

C

`(piR^2I)/(4E)`

D

`(piR^2I)/(E)`

Text Solution

Verified by Experts

The correct Answer is:
D

Power of light received by the cone `=(piR^2)`
Let number of photons hitting the cone per second is n.
Then, `nE=IpiR^2impliesn=(piR^2I)/(E)`
By symmetry, the net force on the cone will be vertically downward.
Force due to one photon:
`f=2(h)/(lamda)sintheta`
This force is perpendicular to the surface of cone. Hence, net force on the cone will be
`F=nfsintheta=n(2(h)/(lamda)sintheta)sintheta`
`=(piR^2I)/(c)(1-cos2theta)`
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Knowledge Check

  • Light of intensity I falls along the axis on a perfectly refleccting right circular cone having semi-vertical angle theta and base radius R. If E is the energy of one photon and c is the speed oh light, then find Q. The net force of the cone.

    A
    `((piR^2I))/(c(1-cos2theta))`
    B
    `((piR^2I))/(2c(1-cos2theta))`
    C
    `(2(piR^2I))/(c(1-cos2theta))`
    D
    `((piR^2I))/2c(1+cos2theta)`

  • A light of intensity 16 mW and energy of each photon 10 eV incident on a metal plate of work function 5 eV and area 10^(-4)m^(2) then find the maximum kinetic energy of emitted electrons and the number of photoelectrons emitted per second if photon efficiency is 10% .

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    5 eV, `10^(11)`
    B
    10 eV, `10^(12)`
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    B
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    C
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