# Is $I(\lambda)=I(\tilde{\nu})$ true in general in Fourier Optics?

Say the spectrum of a light bulb is given by $$I(\lambda)=I_0exp\left(-\frac{(\lambda-\lambda_0)^2}{2(\delta \lambda)^2}\right)$$ (i.e. a gaussian)

if I want the intensity in terms of the wavenumber $$\tilde{\nu}=\frac{1}{\lambda}$$, wouldn't it just be $$I(\tilde{\nu})=I_0exp\left(-\frac{\left(\frac{1}{\tilde{\nu}}-\lambda_0\right)^2}{2(\delta \lambda)^2} \right)$$ ?? which is not a gaussian in general

Or do I have to carry out a Fourier Transform of some kind $$I(\lambda) \to I (\tilde{\nu})$$ but this doesn't make much sense to me, why can I not just substitude $$\tilde{\nu}=\frac{1}{\lambda}$$ to obtain $$I(\tilde{\nu})$$?

I am asking this because in Michealson Inteferometry one is often provided with a light source spectrum $$I_S(\lambda)$$ and is asked to predict the pattern on the inteferogram $$I(\Delta)$$ with $$\Delta$$ being the retardation. But relation $$I_(\Delta)=\frac{1}{2}\int_{-\infty}^{\infty}I_S(\tilde{\nu})e^{-i2\pi\tilde{\nu}\Delta}d \tilde{\nu}$$ seems to hold for $$I_S(\tilde{\nu})$$ only which means I cannot apply $$I(\Delta) \neq \frac{1}{2}\int_{-\infty}^{\infty}I_S(\lambda)e^{-i2\pi \lambda \Delta}d\lambda$$ because $$\tilde{\nu}\neq \lambda, but =1/\lambda$$ ?...

I think the substitution $$\tilde{\nu} = 1/\lambda$$ would work over the formula for $$I(\lambda)$$, to get $$I(\tilde{\nu})$$.

Because essentially $$I(\lambda) = I(1/\tilde{\nu})$$ has to be what we're looking for right?

• @Thanks. I was confused because my lab manual suggests that a Gaussian light source in terms of $\lambda$ will produce a Gaussian profile with beats interferogram in a Michealson Inteferometer. But I thought this would only be true if $I_S=I_S(\tilde{\nu})=gaussian$ since if $I_S(\lambda)=gaussian$ then $I_S(\tilde{\nu})$ will not be a gaussian so the inteferogram cannot be a gaussian. Is there something wrong with the lab script? Commented Nov 1, 2021 at 0:17
• I am sorry, but I am unaware on anything about interferograms. I don't think fourier transforms work since wavelength and wavenumber are essentially the same thing. On the other hand if it were between $\omega$ and $t$ (which we normally encounter), the relation between them are pretty obscure right? Since the intuition we use to do fourier transformations is to see the signal represented in a different way - between a light pulse (say for example) in an interval of time, and its constituent frequencies that make such a light pulse. I hope I haven't confused you. Commented Nov 1, 2021 at 0:24

if I want the intensity in terms of the wavenumber $$\tilde{\nu}=\frac{1}{\lambda}$$, wouldn't it just be $$I(\tilde{\nu})=I_0exp\left(-\frac{(\frac{1}{\tilde{\nu}}-\lambda_0)^2}{2(\delta \lambda)^2} \right)$$ ??

Unfortunately it's not that simple.

why can I not just substitude $$\tilde{\nu}=\frac{1}{\lambda}$$ to obtain $$I(\tilde{\nu})$$?

The issue is that when we talk about the spectrum of a source, we're talking about the spectral intensity. If we take the spectrum in terms of wavelength, then this is the power output of the source per unit solid angle and per unit wavelength. But if we take the spectrum in terms of wavenumber, then we should give the power output per unit solid angle and per unit wavenumber instead.

If you simply replace $$\lambda$$ in the formula with $$\frac{1}{\tilde{\nu}}$$, you will have expressed the spectrum in terms of wavenumber, but you will still be giving the density per unit wavelength rather than per unit wavenumber.

Considering the chain rule from derivative calculus, you should get (if I haven't flubbed the math) $$I(\tilde{\nu})=I_0\tilde{\nu}^2exp\left(-\frac{(\frac{1}{\tilde{\nu}}-\lambda_0)^2}{2(\delta \lambda)^2} \right)$$

If the spectral width of the source isn't very wide, this probably won't make much of a difference, but if the spectrum is broad (as it would typically be for an incandescent light bulb) then it will change your result substantially.

Also be aware (if you are working on a practical problem and not a classroom exercise) that a Gaussian is probably not a very good approximation of an incandescent bulb's spectrum, whether in wavelength terms or wavenumber terms.