# In-Class Activities

By

Dr Lim Zhi Han

●

Published 2022-10-09

When light enters a material, it can be reflected, absorbed ot transmitted.
A blackbody can be imagined to be a perfect material that do not transmit nor reflect light. It absorbs all incoming light.
This does not mean that the blackbody itself does not emit light. In fact a blackbody will emit light as long as its temperature is not zero.
Blackbody emission is broadband. The intensity profile (over a range of wavelengths) is given by

\begin{equation}
I_\lambda = A \frac{2hc^2}{\lambda^5}\frac{1}{\exp(\frac{hc}{\lambda kT})-1}
\end{equation}

where h=6.63 \times 10^{-34} \text{ m}^2 \text{kgs}^{-1}, c = 3.0 \times 10^8 \text{ ms}^{-1}, k=1.38 \times 10^{-23} \text{ m}^2 \text{kgs}^{-2} \text{K}^{-1}, \lambda is the wavelength in meters (not nm), T is the temperature in Kelvins, and A is a geometrical scaling factor.

The total intensity of the radiation over all wavelength is given by the integral

I = \int^{\infin}_0 I_\lambda d\lambda

Show that

I \propto T^4

The peak of the blackbody emission profile can be found by finding its derivative and equate to 0, i.e. \frac{d I_\lambda}{d \lambda} =0. Show that

\lambda_{mode}T = 0.29 \text{ cm K}

Show that the maximum wavelength corresponds to photon energy of the order of kT.

Download the spectrum of an incandescent light bulb collected by an optical spectrometer in Canvas. Using A and T as your fitting parameter, fit the data with Eq.(8.22) What are the values of A and T from your fitting?

Also try fitting the data with

\begin{equation}
\tilde I_\lambda = A\frac{2hc^2}{\lambda^5}\exp \left(-\frac{hc}{\lambda kT}\right)
\end{equation}

Does it fit well?

Eq.(8.23) is known as the Wein's Tail of the distribution. The approximated expression works well for wavelengths much lower than the mode wavelength.

When E = \frac{hc}{\lambda} \ll kT,

\begin{equation}
I_\lambda \approx 2ckT\lambda^{-4}
\end{equation}

This is the approximated expression for the other end of the spectrum known as the Rayleigh-Jean side. Show Eq.(8.24)