Question

A light beam of intensity 20W/cm2 is incident normally on a perfectly reflecting surface of sides 25cm×15cm. The momentum imparted to the surface by the light per second is
A. 2×10−5kgms−1
B. 1×10−5kgms−1
C. 5×10−5kgms−1
D. 1.2×10−5kgms−1

Answer 1

Answer:

I think this should be your answer

Explanation:

C)5×10-5kgms-1

Answer 2

Given:

A light beam of intensity 20W/cm2 is incident normally on a perfectly reflecting surface of sides 25cm×15cm.

To find:

The momentum imparted to the surface by the light per second.

Calculation:

Energy of the light beam be E ;

$\therefore \: \rm{Intensity = \dfrac{E}{(area)t} }$

$= > \rm{20 = \dfrac{E}{(25 \times 15)1} }$

$= > \rm{E = 25 \times 15 \times 20}$

$= > \rm{E = 7500 \: joules}$

Now , this energy can be represented in terms of wavelength:

$\rm{ \therefore \: E = \dfrac{hc}{ \lambda} }$

Again , the wavelength can be represented in form of Momentum (as per De-Broglie wavelength)

$\rm{ = > \: E = \dfrac{hc}{ (\frac{h}{p}) } }$

$\rm{ = > \: E = p \times c }$

$\rm{ = > \: 7500 = p \times (3 \times {10}^{8} )}$

$\rm{ = > \: p = 2.5 \times {10}^{ - 5} \: kgm {s}^{ - 1} }$

This is the momentum of incident ray.

So, the momentum imparted to the surface be P_{net}:

$|P_{net}| = 2 \times 2.5 \times {10}^{-5}$

$=>|P_{net}| = 5\times {10}^{-5}\: kgm{s}^{-1}$

So, final answer is:

$\boxed{\bf{|P_{net}| = 5\times {10}^{-5}\: kgm{s}^{-1}}}$