Radiation (General, Kirchoff's law, Black body, Prevost's Theory) Questions in English

(C) In the steady state,the energy absorbed by the middle plate is equal to the energy released by the middle plate.
Let the temperature of the middle plate be $T'$.
The radiation energy incident on the middle plate from the first plate (at $3T$) is $\sigma A (3T)^4$.
The radiation energy emitted by the middle plate towards the first plate is $\sigma A (T')^4$.
The net energy absorbed from the first plate is $\sigma A (3T)^4 - \sigma A (T')^4$.
The radiation energy incident on the middle plate from the third plate (at $2T$) is $\sigma A (2T)^4$.
The radiation energy emitted by the middle plate towards the third plate is $\sigma A (T')^4$.
The net energy released to the third plate is $\sigma A (T')^4 - \sigma A (2T)^4$.
Equating the net energy absorbed and released:
$\sigma A (3T)^4 - \sigma A (T')^4 = \sigma A (T')^4 - \sigma A (2T)^4$
$(3T)^4 - (T')^4 = (T')^4 - (2T)^4$
$81T^4 + 16T^4 = 2(T')^4$
$97T^4 = 2(T')^4$
$(T')^4 = \frac{97}{2} T^4$
$T' = \left(\frac{97}{2}\right)^{\frac{1}{4}} T$

(D) Given: Surface area $A = 64 \ mm^2 = 64 \times 10^{-6} \ m^2$,Temperature $T = 2500 \ K$,Distance $d = 100 \ m$,Pupil radius $R_e = 3 \ mm = 3 \times 10^{-3} \ m$.
$(A)$ Power radiated by the filament $P = \sigma A e T^4$. For a black body,emissivity $e = 1$.
$P = (5.67 \times 10^{-8}) \times (64 \times 10^{-6}) \times 1 \times (2500)^4 = 141.75 \ W$.
Thus,option $(A)$ is incorrect.
$(B)$ Power reaching the eye $P_{eye} = \frac{P}{4 \pi d^2} \times (\pi R_e^2) = \frac{141.75}{4 \times (100)^2} \times (3 \times 10^{-3})^2 = 3.189 \times 10^{-8} \ W$.
This is in the range $3.15 \times 10^{-8} \ W$ to $3.25 \times 10^{-8} \ W$. Thus,option $(B)$ is correct.
$(C)$ Using Wien's displacement law: $\lambda_m T = b$.
$\lambda_m = \frac{2.90 \times 10^{-3}}{2500} = 1.16 \times 10^{-6} \ m = 1160 \ nm$.
Thus,option $(C)$ is correct.
$(D)$ Power $P_{eye} = \dot{N} \frac{hc}{\lambda_{avg}}$,where $\dot{N}$ is the number of photons per second.
$\dot{N} = \frac{P_{eye} \lambda_{avg}}{hc} = \frac{3.189 \times 10^{-8} \times 1740 \times 10^{-9}}{6.63 \times 10^{-34} \times 3.00 \times 10^8} \approx 2.79 \times 10^{11}$.
This is in the range $2.75 \times 10^{11}$ to $2.85 \times 10^{11}$. Thus,option $(D)$ is correct.
Conclusion: Options $(B, C, D)$ are correct.

(C) Density is given by $\rho = \frac{M}{V}$. Since all objects are made of the same material,$\rho$ is the same. Given that the mass $M$ is also the same,the volume $V$ of all objects must be the same.
For a constant volume,the surface area $A$ of a thin circular plate is the maximum among a sphere,a cube,and a plate.
According to Stefan-Boltzmann law,the rate of heat loss is $\frac{dQ}{dt} \propto A(T^4 - T_0^4)$.
Since the surface area $A_{\text{plate}}$ is the largest,the rate of cooling is highest for the plate.
Therefore,the circular plate will reach room temperature the fastest.

(B) The total incident heat energy rate is $P_i = 1000 \ J \ min^{-1}$.
According to the law of conservation of energy for heat radiation,the sum of the coefficients of reflection $(r)$,absorption $(a)$,and transmission $(t)$ is equal to $1$,i.e.,$r + a + t = 1$.
Given $r = 0.1$ and $a = 0.8$,we can find the coefficient of transmission $(t)$:
$t = 1 - (r + a) = 1 - (0.1 + 0.8) = 1 - 0.9 = 0.1$.
The rate of transmitted heat energy is $P_t = t \times P_i = 0.1 \times 1000 \ J \ min^{-1} = 100 \ J \ min^{-1}$.
For a time duration of $5 \ minutes$,the total transmitted heat energy $(Q_t)$ is:
$Q_t = P_t \times \text{time} = 100 \ J \ min^{-1} \times 5 \ min = 500 \ J$.

(C) For any surface,the sum of the coefficient of absorption $(a)$,the coefficient of reflection $(r)$,and the coefficient of transmission $(t)$ is equal to $1$.
$a + r + t = 1$
Given: $a = 0.77$,$r = 0.17$.
Substituting these values: $0.77 + 0.17 + t = 1$
$0.94 + t = 1$
$t = 1 - 0.94 = 0.06$
The quantity of heat transmitted is given by the product of the incident heat $(Q)$ and the coefficient of transmission $(t)$.
$Q_{\text{transmitted}} = Q \times t$
$Q_{\text{transmitted}} = 250 \ kcal \times 0.06$
$Q_{\text{transmitted}} = 15 \ kcal$
Therefore,the correct option is $C$.

(B) perfectly black body is defined as an object that absorbs all incident radiation of any wavelength. According to Kirchhoff's law of thermal radiation,for an arbitrary body in thermal equilibrium with its surroundings,the emissivity is equal to its absorptivity. Since a perfectly black body has an absorptivity of $1$,its emissivity (coefficient of emission) must also be $1$,which is referred to as unity.

(C) According to Stefan-Boltzmann law,the rate of cooling $R$ of a body is given by $R \propto (T^4 - T_0^4)$,where $T$ is the temperature of the body and $T_0$ is the temperature of the surroundings.
For the first case: $R_0 = k(400^4 - 300^4) = k(256 \times 10^8 - 81 \times 10^8) = k(175 \times 10^8)$.
For the second case: $R' = k(900^4 - 300^4) = k(6561 \times 10^8 - 81 \times 10^8) = k(6480 \times 10^8)$.
Taking the ratio: $\frac{R'}{R_0} = \frac{6480}{175} \approx 37.02$.
Comparing this with the given options,the closest value is $36 R_0$.

(C) Let the total heat incident on the body be $Q = x$ joule.
Let the heat reflected be $Q_r$ and the heat transmitted be $Q_t$.
Given that the sum of reflected and transmitted heat is $Q_r + Q_t = y$ joule.
The total incident heat is the sum of absorbed heat $(Q_a)$,reflected heat $(Q_r)$,and transmitted heat $(Q_t)$:
$Q = Q_a + Q_r + Q_t$
Substituting the given values:
$x = Q_a + y$
Therefore,the absorbed heat is $Q_a = x - y$.
The absorption coefficient $(a)$ is defined as the ratio of absorbed heat to the total incident heat:
$a = \frac{Q_a}{Q} = \frac{x - y}{x}$.

(A) According to Stefan-Boltzmann Law,the rate of heat loss by radiation is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$,where $e$ is the emissivity of the surface.
For a given temperature and surface area,the rate of cooling is directly proportional to the emissivity $(e)$ of the surface.
Black surfaces have the highest emissivity (close to $1$),followed by dark colors like red,while white or polished surfaces have the lowest emissivity.
Therefore,the order of cooling rate from maximum to minimum is: black > red > white.

(A) black body is an ideal body that absorbs and emits all incident electromagnetic radiation. According to Planck's Law of black body radiation,the intensity of radiation emitted by a black body depends on the wavelength and temperature.
$1$. The intensity is not the same for all wavelengths; it follows a specific distribution curve (Planck's curve) that peaks at a particular wavelength depending on the temperature.
$2$. Therefore,the statement 'For all wavelengths,intensity is same' is incorrect.
$3$. As temperature increases,the peak intensity shifts towards shorter wavelengths,but the intensity varies continuously across the spectrum.

(A) For any body,the sum of the coefficients of absorption $(a)$,reflection $(r)$,and transmission $(t)$ is always equal to $1$,i.e.,$a + r + t = 1$.
An opaque body is defined as a body that does not allow any radiation to pass through it.
Therefore,the transmission coefficient $t$ for an opaque body is $0$.
Substituting $t=0$ into the equation $a + r + t = 1$,we get $a + r = 1$.
Thus,for an opaque body,$t=0$ and $a+r=1$.

(D) black body is an idealized physical body that absorbs all incident electromagnetic radiation.
According to Planck's law of black body radiation,the intensity of radiation emitted by a black body varies with wavelength.
The emission spectrum is continuous,meaning it emits radiation across all wavelengths.
However,the intensity is not constant for all wavelengths; it follows a specific distribution curve (Planck's distribution) that depends on the temperature of the body.
Therefore,the statement that intensity is the same for all wavelengths is incorrect.

(A) black body is an ideal body that emits and absorbs all frequencies of radiation. According to Planck's law of black body radiation,the intensity of radiation emitted by a black body depends on the wavelength and temperature. The intensity distribution is not uniform for all wavelengths; it follows a specific curve (Planck's curve) where intensity varies with wavelength. Therefore,the statement that 'intensity is same for all wavelengths' is incorrect. Additionally,the intensity does not strictly increase or decrease monotonically for all ranges; it peaks at a specific wavelength determined by Wien's displacement law. However,among the given options,the statement that intensity is the same for all wavelengths is fundamentally wrong.

(B) The coefficient of transmission $(T)$ is defined as the ratio of the radiant energy transmitted through a substance to the total radiant energy incident upon it.
An athermanous substance is a material that does not allow thermal radiation to pass through it.
Since no radiant energy is transmitted through an athermanous substance,the transmitted energy is $0$.
Therefore,the coefficient of transmission $T = 0$.

(D) Every body at all times,at all temperatures,emits electromagnetic radiation.
For a human body at a normal temperature of approximately $37^{\circ}C$ $(310 \ K)$,the peak wavelength of the emitted radiation falls within the infrared region of the electromagnetic spectrum.
Therefore,the radiation emitted by the human body is in the infrared region.

(A) An opaque body does not transmit any radiation.
By definition,the coefficient of transmission $(t)$ is the ratio of the transmitted energy to the incident energy.
Since no radiation passes through an opaque body,the transmitted energy is $0$.
Therefore,the coefficient of transmission for an opaque body is $0$.

(B) When thermal radiation $(Q)$ falls on a body,it is partly reflected $(Q_{r})$,partly absorbed $(Q_{a})$,and partly transmitted $(Q_{t})$.
The total energy incident is given by $Q = Q_{a} + Q_{r} + Q_{t}$.
Dividing by $Q$,we get the relation: $a + r + t = 1$,where $a$ is the coefficient of absorption,$r$ is the coefficient of reflection,and $t$ is the coefficient of transmission.
Given: $Q = 150 \,J$,$Q_{r} = 15 \,J$,and $a = 0.6$.
The coefficient of reflection $r = \frac{Q_{r}}{Q} = \frac{15}{150} = 0.1$.
Using the relation $a + r + t = 1$:
$0.6 + 0.1 + t = 1$
$0.7 + t = 1$
$t = 0.3$.
The energy transmitted $Q_{t} = t \times Q = 0.3 \times 150 \,J = 45 \,J$.

(C) An open window acts as a perfectly black body because any radiation entering it is absorbed by the interior of the room and does not escape.
For a perfectly black body,the absorption coefficient is defined as $1$.

(B) The Earth's atmosphere acts as a blanket that traps heat through the greenhouse effect,primarily due to gases like $CO_2$ and water vapor.
Without an atmosphere,the heat received from the Sun during the day would radiate back into space at night without being trapped.
Consequently,the average surface temperature of the Earth would be significantly lower than it is currently.
Therefore,the correct option is $B$.

(B) Thermal radiation consists of electromagnetic waves,which behave similarly to light.
When electromagnetic radiation travels from one medium to another,its speed and wavelength change due to the change in the refractive index of the medium.
However,the frequency of the radiation is a property of the source and remains constant regardless of the medium through which it travels.
Therefore,the statement that frequency changes is incorrect.

(D) perfect black body is defined as an ideal body that absorbs all the incident radiation falling on it,regardless of the frequency or angle of incidence.
By definition,the absorption coefficient $\alpha$ is the ratio of the energy absorbed to the total energy incident on the body.
Since a perfect black body absorbs all incident radiation,the energy absorbed equals the total incident energy.
Therefore,the absorption coefficient $\alpha = 1$.

(A) perfectly black body absorbs $100 \%$ of the radiation incident on it. Therefore,the absorption coefficient $(a_r)$ is defined as the ratio of the amount of absorbed radiation $(Q_a)$ to the amount of incident radiation $(Q_i)$.
Since for a perfect black body,$Q_a = Q_i$,we have:
$a_r = \frac{Q_a}{Q_i} = \frac{Q_a}{Q_a} = 1$.

(D) thermopile is a device that converts thermal energy into electrical energy. It is based on the thermoelectric effect (Seebeck effect),where a voltage is generated due to a temperature difference between two dissimilar metals. Because it is highly sensitive to heat,it is primarily used to detect infrared radiation.

(D) Thermal radiation is the process of heat transfer in the form of electromagnetic waves.
These waves travel in straight lines at the speed of light $(c \approx 3 \times 10^8 \ m/s)$ and do not require a material medium for propagation.

(D) The energy density $u$ of blackbody radiation at temperature $T$ is given by $u = aT^4$,where $a$ is the radiation constant.
The total energy $E$ in a container of volume $V$ is $E = uV = aVT^4$.
The heat required to increase the temperature from $T_1$ to $T_2$ is $\Delta E = aV(T_2^4 - T_1^4)$.
Since the volume $V$ of the container is not provided in the problem statement,it is impossible to calculate the exact numerical value of the heat required.
Therefore,the information provided is insufficient.

(A) The energy density $u$ of black body radiation in a cavity at temperature $T$ is given by $u = aT^4$,where $a$ is the radiation constant.
The total energy $E$ contained in the box is the product of energy density $u$ and the volume $V$ of the box.
$E = u \times V = aT^4 \times V$.
Initially,the side length is $L$,so the volume $V_1 = L^3$. The energy is $E_1 = aT^4 L^3$.
Finally,the side length is doubled,so $L' = 2L$,and the new volume $V_2 = (2L)^3 = 8L^3$.
The temperature is halved,so $T' = T/2$.
The new total energy $E_2 = a(T')^4 V_2 = a(T/2)^4 (8L^3)$.
$E_2 = a(T^4 / 16) (8L^3) = (8/16) aT^4 L^3 = (1/2) aT^4 L^3$.
$E_2 = E_1 / 2$.
Therefore,the total energy halves.