Doubts in Photoelectric effect !!

1) A point source of light is placed at the centre of curvature of a hemispherical surface.the radius if 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?

2)A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture.If the beam carries an intensity I,find the force exerted by the beam on the sphere???

12 Answers

1
rajatjain_ix ·

1) w/c
2) 4/(3c)
r dese correct?

1
abcd ·

dude the answer is
W/2c
and
pi r^2 I/c

1
Jesh ·

@rajat ur both ans are wrong dude!!

1
Jesh ·

hey ishan u r right! but i want solutions !!

29
govind ·

Jesh in the first one completely reflecting means the change in momentum = 2c
in the second I.Ï€r2 = F.v ..
so F comes out to be I πr2 / c

1
Jesh ·

anyway i got the first two probs can u also try this!!

3) In a photo cell the plates P and Q have a separation of 10cm, which are connected through a galvanometer without any cell.Bichromatic light of wavelengths 4000 Ao and 6000 A0 are incident on plate Q whose work function is 2.39eV. If a uniform magnetic field B exists parallel to the plates, find the minimum value of B for which the galvanometer shows zero deflection.

13
Avik ·

Engy withrespective lights ::
4000A0 ,E=3.09 eV
6000A0 ,E=2.06eV

So, only the 4000A0 wavelength waali light wud be responsible fr ejecting the electrons.

MAx, KE of ejected e- = 3.09-2.39 = 0.7eV.

29
govind ·

The current will stop..use the relation r = mv/qB..
so wen the radius is less than 10cm the partcles will get deflected and will not reach the opposite plate..
btw the answer i am getting is B ≥ 2.8 * 10-5 T

13
Avik ·

Yeh, i was taking the mag. field perpendicular to the plates instead of parallel... :P, thnx.

1
radhika roy ·

A totally reflecting , small plane mirror placed horizontally faces a parallel beam of light. The mass of the mirror is 20g. Assume that there is no absorption in the lens and that 30% of the light emitted by the source goes through the lens. Find the power of the source needed to support The weight of the mirror. Take g=10m/s2

11
SANDIPAN CHAKRABORTY ·

Since the mirror is totally reflecting...

Prad = 2I/c (where I =intensity = power /unit area )

so 2 Ic x 30100 = mg

put all the values....

you get I = 108

1
radhika roy ·

OH !!!! THAT WAS EASY. . THANKS SANDIPAN . .

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