Electromagnetic Waves - Result Question 6
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6. Light with an energy flux of $25 \times 10^{4} Wm^{-2}$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15 cm^{2}$, the average force exerted on the surface is :[2014]
======= ####6. Light with an energy flux of $25 \times 10^{4} Wm^{-2}$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15 cm^{2}$, the average force exerted on the surface is :[2014]
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/electromagnetic-waves/electromagnetic-waves—result-question-6.md (a) $1.25 \times 10^{-6} N$
(c) $1.20 \times 10^{-6} N$
(b) $2.50 \times 10^{-6} N$
(d) $3.0 \times 10^{-6} N$
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Answer:
Correct Answer: 6. (b)
Solution:
- (b) Average force $F _{av}=\frac{\Delta p}{\Delta t}=\frac{2 IA}{c}$
$ \begin{aligned} & =\frac{2 \times 25 \times 10^{4} \times 15 \times 10^{-4}}{3 \times 10^{8}} \\ & =2.50 \times 10^{-6} N \end{aligned} $
Pressure $=\frac{\text{ Force }}{\text{ Area }}=\frac{F}{A}$
Force is the rate of change of momentum $F=\frac{d p}{d t}$
But energy, $U=p c$
$ \therefore \text{ Pressure }=\frac{1}{A} \frac{U}{c \cdot d t}=\frac{I}{c} $
$ [\because \text{ Intensity }=\frac{U}{A d t}] $
For perfectly reflecting surface, $P_r=\frac{2 I}{c}$
$\therefore$ For perfectly reflecting surface, Force $=\frac{2 I A}{c}$
