Radiation Force and Pressure Questions in English

(A) For a non-reflecting (totally absorbing) surface,the radiation pressure $P$ is given by $P = \frac{F}{A} = \frac{I}{c}$,where $I$ is the intensity (energy flux) and $c$ is the speed of light.
Given:
Force $F = 10^{-6} \text{ N}$
Area $A = 15 \text{ cm}^2 = 15 \times 10^{-4} \text{ m}^2$
Speed of light $c = 3 \times 10^8 \text{ ms}^{-1}$
Rearranging the formula for intensity $I$:
$I = \frac{F \cdot c}{A}$
Substituting the values:
$I = \frac{10^{-6} \times 3 \times 10^8}{15 \times 10^{-4}}$
$I = \frac{3 \times 10^2}{15 \times 10^{-4}}$
$I = \frac{3}{15} \times 10^6 = 0.2 \times 10^6 \text{ Wm}^{-2} = 20 \times 10^4 \text{ Wm}^{-2}$.

(A) The intensity of radiation is given as $I = 0.5 \ Wm^{-2}$.
For a perfectly absorbing surface,the radiation pressure $p$ is given by the formula $p = \frac{I}{c}$,where $c$ is the speed of light in vacuum $(c \approx 3 \times 10^8 \ ms^{-1})$.
Substituting the values:
$p = \frac{0.5}{3 \times 10^8}$
$p = 0.166 \times 10^{-8} \ Nm^{-2}$.

(D) The radiation pressure force $F$ exerted by a beam of power $P$ on a perfectly absorbing surface is given by the formula $F = \frac{P}{c}$,where $c$ is the speed of light.
Given:
Power $P = 100 \,W$
Speed of light $c = 3 \times 10^8 \,m/s$
Substituting the values:
$F = \frac{100}{3 \times 10^8} \,N$
$F = 33.33 \times 10^{-8} \,N$
$F = 3.33 \times 10^{-7} \,N$
Thus,the force due to radiation pressure is approximately $3.3 \times 10^{-7} \,N$.

(A) The energy flux $I$ is $18 \,W \,cm^{-2}$.
The area $A$ is $20 \,cm^2$.
The total power $P$ incident on the surface is $P = I \times A = 18 \,W \,cm^{-2} \times 20 \,cm^2 = 360 \,W$.
For a non-reflecting surface, the radiation pressure exerts a force $F$ given by the formula $F = \frac{P}{c}$, where $c$ is the speed of light.
Substituting the values: $F = \frac{360 \,W}{3 \times 10^8 \,ms^{-1}} = 120 \times 10^{-8} \,N = 1.2 \times 10^{-6} \,N$.
Since the force is constant over the time interval, the average force is $1.2 \times 10^{-6} \,N$.

(C) The energy flux $I = 9 \ Wcm^{-2} = 9 \times 10^4 \ Wm^{-2}$.
The area $A = 20 \ cm^2 = 20 \times 10^{-4} \ m^2$.
The time $t = 1 \ hour = 3600 \ s$.
The speed of light $c = 3 \times 10^8 \ ms^{-1}$.
For a non-reflecting surface, the momentum $p$ delivered by radiation is given by $p = \frac{U}{c} = \frac{IAt}{c}$.
Substituting the values:
$p = \frac{(9 \times 10^4) \times (20 \times 10^{-4}) \times 3600}{3 \times 10^8}$.
$p = \frac{180 \times 3600}{3 \times 10^8} = \frac{648000}{3 \times 10^8} = 216000 \times 10^{-8} \ kgms^{-1}$.
$p = 2.16 \times 10^{-3} \ kgms^{-1}$.

(B) black surface is considered a perfectly absorbing surface. For a perfectly absorbing surface,the radiation pressure $P$ is given by the formula:
$P = \frac{I}{c}$
where $I$ is the intensity of the light and $c$ is the speed of light in vacuum $(c = 3 \times 10^8 m s^{-1})$.
Given:
$I = 12 W m^{-2}$
$c = 3 \times 10^8 m s^{-1}$
Substituting the values:
$P = \frac{12}{3 \times 10^8} Pa$
$P = 4 \times 10^{-8} Pa$

(C) The power of the light bulb is $P = 100 \ W$. The power converted into light is $P' = 0.66 \times 100 \ W = 66 \ W$.
The radius of the sphere is $r = 10 \ cm = 0.1 \ m$.
The intensity $I$ of the light at the surface of the sphere is given by $I = \frac{P'}{A} = \frac{P'}{4 \pi r^2}$.
Substituting the values: $I = \frac{66}{4 \times 3.14 \times (0.1)^2} = \frac{66}{4 \times 3.14 \times 0.01} = \frac{66}{0.1256} \approx 525.48 \ W/m^2$.
For a perfectly absorbing surface,the radiation pressure $P_r$ is given by $P_r = \frac{I}{c}$,where $c = 3 \times 10^8 \ m/s$ is the speed of light.
$P_r = \frac{525.48}{3 \times 10^8} \approx 1.75 \times 10^{-6} \ N \ m^{-2}$.

(B) The momentum $p$ carried by a photon of energy $E$ is given by $p = E/c$,where $c$ is the speed of light.
When a photon strikes a perfectly reflecting surface,it is reflected back in the opposite direction with the same energy $E$ and thus the same magnitude of momentum $p = E/c$.
The initial momentum of the photon is $p_i = E/c$ (taking the direction towards the surface as positive).
The final momentum of the photon after reflection is $p_f = -E/c$ (since it moves in the opposite direction).
The change in momentum of the photon is $\Delta p = p_f - p_i = -E/c - E/c = -2E/c$.
According to the law of conservation of momentum,the momentum transferred to the surface is equal in magnitude to the change in momentum of the photon.
Therefore,the momentum transferred to the surface is $|\Delta p| = 2E/c$.

(A) The radiation pressure $P$ exerted by electromagnetic radiation on a perfectly absorbing (black) surface is given by the formula:
$P = \frac{I}{c}$
where $I$ is the intensity of the radiation and $c$ is the speed of light in vacuum.
Given:
Intensity $I = 0.6 \ W/m^2$
Speed of light $c = 3 \times 10^8 \ m/s$
Substituting the values:
$P = \frac{0.6}{3 \times 10^8}$
$P = 0.2 \times 10^{-8} \ N/m^2$
$P = 2 \times 10^{-9} \ N/m^2$

(C) The energy of the incident radiation is $E$.
The momentum of the incident radiation is given by $p = E/c$.
Since the surface is perfectly reflecting,the radiation reflects back with the same energy $E$.
The momentum of the reflected radiation is $p' = -E/c$ (the negative sign indicates the opposite direction).
The total momentum transferred to the surface is the change in momentum: $\Delta p = p - p'$.
$\Delta p = E/c - (-E/c) = 2E/c$.

(C) When a light beam is incident at an angle of incidence $\theta$,the effective intensity on the surface is $I \cos \theta$ and the effective area is $A \cos \theta$. The momentum transferred per unit time per unit area by the incident light is $\frac{I \cos \theta}{c}$. The normal component of this momentum transfer is $\frac{I \cos \theta}{c} \times \cos \theta = \frac{I \cos^2 \theta}{c}$.
Since the surface is perfectly reflecting,the reflected light also exerts an equal pressure due to the change in momentum. The total radiation pressure $p_{\text{net}}$ is the sum of the pressure due to incident and reflected light:
$p_{\text{net}} = \frac{I \cos^2 \theta}{c} + \frac{I \cos^2 \theta}{c} = \frac{2I \cos^2 \theta}{c}$.
The force $F$ exerted on the surface of area $A$ is given by $F = p_{\text{net}} \times A$.
Therefore,$F = \frac{2IA \cos^2 \theta}{c}$.

(C) Given, intensity of flashlight, $I = 9 \, W/cm^2 = 9 \times 10^4 \, W/m^2$.
Area, $A = 300 \, cm^2 = 3 \times 10^{-2} \, m^2$.
For a perfectly reflective surface, the radiation pressure $p$ is given by $p = \frac{2I}{c}$, where $c = 3.0 \times 10^8 \, m/s$ is the speed of light.
Substituting the values: $p = \frac{2 \times 9 \times 10^4}{3 \times 10^8} = 6 \times 10^{-4} \, N/m^2$.
The average force $F$ exerted on the surface is $F = p \times A$.
$F = (6 \times 10^{-4} \, N/m^2) \times (3 \times 10^{-2} \, m^2) = 18 \times 10^{-6} \, N = 18 \, \mu N$.

(D) The force $F$ exerted by electromagnetic waves on a non-reflecting (perfectly absorbing) surface is given by the formula $F = \frac{P}{c}$,where $P$ is the power of the waves and $c$ is the speed of light in vacuum.
Given,power $P = 600 \ W$.
The speed of light $c = 3 \times 10^8 \ m/s$.
Substituting the values,we get $F = \frac{600}{3 \times 10^8} \ N$.
$F = 200 \times 10^{-8} \ N$.
$F = 2 \times 10^{-6} \ N$.
Therefore,the correct option is $D$.

(B) The force exerted by a beam of light on a surface is given by the rate of change of momentum.
For a beam of power $P$ incident normally:
$1$. Force due to absorbed part $(70 \%)$: $F_{abs} = \frac{P_{abs}}{c} = \frac{0.7 P}{c}$
$2$. Force due to reflected part $(30 \%)$: $F_{ref} = \frac{2 P_{ref}}{c} = \frac{2 \times 0.3 P}{c} = \frac{0.6 P}{c}$
Total force $F = F_{abs} + F_{ref} = \frac{0.7 P + 0.6 P}{c} = \frac{1.3 P}{c}$
Given $P = 10 \ W$ and $c = 3 \times 10^8 \ m/s$:
$F = \frac{1.3 \times 10}{3 \times 10^8} = \frac{13}{3} \times 10^{-8} \ N \approx 4.33 \times 10^{-8} \ N$.

(B) Electromagnetic waves carry both energy and momentum,and they exert radiation pressure on surfaces they strike. Assertion $(A)$ is true.
Reason $(R)$ states there is no mass associated with electromagnetic waves,which is true because photons are massless.
However,the reason for radiation pressure is the transfer of momentum,not the presence of mass.
Thus,$(A)$ and $(R)$ are true,but $(R)$ is not the correct explanation for $(A)$.