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Q)A point source of light is placed at the centre of curvature of a hemispherical surface. The radius of curvature is r and the inner surface is completely reflecting. Find the force on the hemisphere due to the light falling on it if the source emits a power W.

I tried attempting this question as follows:$$F=\frac{\Delta p}{\Delta t}$$ $$F= \frac{2E}{c\Delta t}$$ (since $p=\frac{h}{\lambda}$ and $E=\frac{hc}{\lambda}$) $$=\frac{2W\Delta t}{c\Delta t} $$ $$F=\frac{2W}{c}$$

However, this answer is wrong and is correctly attempted as follows:

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$$I=\frac{W}{4\pi r^2}$$ $$dE=\frac{I ds}{4\pi r^2}$$ $$dF=\frac{dE}{C}$$ By symmetry net force on hemisphere is along x, $$dF=\int\frac{2WdA cos\theta}{4\pi r^2c}$$ $$dF=\frac{2W \int dA cos \theta}{4\pi r^2c}$$ $$dF=\frac{W}{2c}$$ Which is different from what I attempted. Why is intensity taken in the official answer and why is my answer wrong?

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    $\begingroup$ First - you don't seem to have tossed out half the total power which doesn't hit the hemisphere. The other problem is that you didn't account for the vectorial nature of the total force, which requires integrating over "rings" of the hemisphere. The total force per $d\theta$ varies greatly from zero to 90 degrees. $\endgroup$ Feb 15, 2021 at 14:27
  • $\begingroup$ I got the vectorial part. But isn't power independent of area on which it falls and only dependent on energy and time. Energy is dependent on force and displacement. I can't see why I should toss half the power out. $\endgroup$ Feb 15, 2021 at 15:52
  • $\begingroup$ Half the light heads to the right and hits the hemisphere. The other half heads to the left and misses it. $\endgroup$
    – Ben51
    Feb 15, 2021 at 17:00
  • $\begingroup$ @Ben51 I do get that,however, power is independent of area(of the light it hits,according to formulas,please correct me if I am wrong),so why are we dividing it by 2? I know this is a really stupid question, but I can't find reasoning for it as I can't see power depending on area,even though it only falls on hemisphere and not completely on the sphere. $\endgroup$ Feb 15, 2021 at 17:03
  • $\begingroup$ I am sorry but I can’t fathom your confusion. Half the photons are caught by the hemisphere the other half aren’t. Why would you think that the half that aren’t would contribute to the power or the force? $\endgroup$
    – Ben51
    Feb 15, 2021 at 17:06

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(I don't have enough rep to comment)

The equation that you have used is derived for normal incidence (specifically for elastic collision assuming light as a particle).

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  • $\begingroup$ Since the point is at centre therefore wouldn't all the lines be perpendicular to the surface of the sphere? $\endgroup$ Feb 15, 2021 at 15:49
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    $\begingroup$ Yeah but what u hav calculated is the total force(as if the hemispher was a flat sheet).SInce u hav directly used the formula,forces aren't resolved into horizontal and vertical components. $\endgroup$
    – EVO
    Feb 15, 2021 at 17:07

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