Question: I'm a zoology minor and we are doing protein estimation by colorimetric method. I have stumbled upon a term 'Optical density'. I don't understand the term well. Is it a measure of the extent of light that can pass through a particular object?

I've checked a [related question](https://physics.stackexchange.com/questions/132094/why-does-the-optical-media-have-different-refractive-indices) of this community and it doesn't solve my question completely.

Answer: 
You're a little confused probably because there are *two* usages of the words "optical density". 

The first usage is as a synonym for *refractive index*, as described in the answers to the related question you cite. This is the commoner usage in physics.

The second usage is the total attenuation afforded by a protective screen, neutral density filter, laser goggles or the like. $ODx\; \lambda=y$ or even $ODx\; y$ means that the filter, goggles etc afford a power attenuation factor of $10^x$ at a light wavelength of $y$ or light wavelength range $y$. That is, the power transmitted through the filter is $10^{-x}$ of the incident power when the wavelength is as stated.

For example, laser goggles marked $OD7\;488{\rm nm}$ means that the goggles will reduce incident power at 488nm by a factor of $10^7$. Goggles marked with a lone wavelength rather than a wavelength range are always meant for use with a particular kind of laser. For example, the $OD7\;488{\rm nm}$ goggles are meant for use with an argon ion laser. You cannot rely on them using another source of wavelength 485nm, for example. 

For generic use, a wavelength range must be specified. So, for example, one often sees $OD7\;450{\rm nm} - 510{\rm nm}$, meaning, pretty obviously, goggles that will give you seven orders of magnitude of attenuation over the whole range $450{\rm nm} \leq \lambda \leq 510{\rm nm}$.
Optical density is a mathematical way of describing the extent of attenuation of visual lightwaves.

Density is the log of opacity.

Opacity is the reciprocal of transmittance.

So a transmittance of 50% of the light (half of the light) would have an opacity of 2. The log of 2 = .3

So a density of .3 blocks half of the light energy.