Modern Physics 7 Question 74
4. Light is incident normally on a completely absorbing surface with an energy flux of $25 W cm^{-2}$. If the surface has an area of $25 cm^{2}$, the momentum transferred to the surface in $40 min$ time duration will be
(a) $3.5 \times 10^{-6} N-s$
(b) $6.3 \times 10^{-4} N-s$
(c) $1.4 \times 10^{-6} N-s$
(d) $5.0 \times 10^{-3} N-s$
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Solution:
- Radiation pressure over an absorbing surface is, $p=\frac{I}{c}$ where, $I$ =intensity or energy flux
and $c=$ speed of light.
If $A=$ area of surface, then force due to radiation on the surface is
$$ F=p \times A=\frac{I A}{c} $$
If force $F$ acts for a duration of $\Delta t$ seconds, then momentum transferred to the surface is
$$ \Delta p=F \times \Delta t=\frac{I A}{c} \times \Delta t $$
Here, $I=25 W cm^{-2}, A=25 cm^{2}$,
$c=3 \times 10^{8} ms^{-1}, \Delta t=40 min=2400 s$
So, momentum transferred to the surface,
$$ \Delta p=\frac{25 \times 25 \times 2400}{3 \times 10^{8}}=5 \times 10^{-3} N-s $$
