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A point source of radiation of power P i...

A point source of radiation of power `P` is placed on the axis of completely absorbing disc. The distance between the source and the disc is `2 "times"` the radius of the disc. The force that light exerts on the disc is `(Px)/(40c)`. Then the value of `x`

A

`1`

B

`2`

C

`3`

D

`4`

Text Solution

Verified by Experts

The correct Answer is:
B
Number of photons striking per second `N=(IA)/(hv)`
Area `A` here is the area perpendicular to the direction of intensity or direction of energy flow.
Consider a ring of radius `x` and width `dx` on the disc. Intensity `I` on the ring due to source is `I=(P)/(4pir^(2))`.

now `dA_(_|_)=dAcostheta` and `dA=2pix dx`
`:. dA_(_|_)=(2pix dx)costheta`
A photon will exert force `F` as shown, only the `Fcostheta` component will remain and `F sintheta` will cancel out as we integrate on the ring `(F)=(Nh)/(lambda)` [as `N` photons strike per second and this leads to a loss of `(Nh)/(lambda)` momentum per second].
as only `Fcostheta` component of force remains
`dF=((IdA_(_|_))/(hv))xx(h)/(lambda)costheta`
`F=int_(0)^(R)dFint_(0)^(R)[(P)/(4pi(4R^(2)+x^(2)))]xx(2pidx cos theta)/((hC)/(lambda))xx(h)/(lambda)costheta`
`[:' r=sqrt(4R^(2)+x^(2))]`. Solving we get `F=(P)/(20c)`
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