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Karnataka Board PUCPUC Science Class 11

A Totally Reflecting, Small Plane Mirror Placed Horizontally Faces a Parallel Beam of Light, as Shown in the Figure. the Mass of the Mirror is 20 G.

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Question

A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light, as shown in the figure. The mass of the mirror is 20 g. 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.

(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)

Sum
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Solution

Given :-

Mass of the mirror, m = 20 g = 20 × 10−3 kg

The weight of the mirror will be balanced if the force exerted by the photons will be equal to the weight of the mirror.

Now,

Relation between wavelength `(λ)` and momentum (p) :-

`p = h/λ`

On divinding both sides by t , we get 

`P/t = h/(λt)...............(1)`

Energy,

`E = (hc)/λ`

⇒ `E/t = (hc)/(λt)`

Let P be the power. Then,

`P = E/t = (hc)/(λt)`

`P = (pc)/t ............ ["Using equation (1)"]`

⇒ `P/c = p/t`

Force , 

`F = P/t = P/c   ..............("Since F" = ("Momentum")/("Time"))`

Thus, rate of change of momentum = Power/c

As the light gets reflected normally,

Force exerted = 2 (Rate of change of momentum) = 2 × Power/c

`30% "of" ((2 xx "Power")/c) = "mg"`

`⇒ "Power" = (20 xx 10^-3 xx 10 xx 3 xx 10^8 xx 10)/(2 xx 3)= 100" MW"`

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Chapter 42: Photoelectric Effect and Wave-Particle Duality - Exercises [Page 365]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 42 Photoelectric Effect and Wave-Particle Duality
Exercises | Q 8 | Page 365

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