I have always understood intensity to be defined as power divided by area. However, I have read (in the Wikipedia article on intensity) that intensity in optics can mean something slightly different. For example, depending on the context, it can be defined as power per unit solid angle. I am not to sure what a solid angle is, but I would think that this would indicate that a larger laser beam, even if it had the same power per area as a smaller beam, would be more intense.

In particular, I am wondering if this applies to intensity in the context of interference. I have that intensity is directly proportional to the slit width, so I am essentially wondering if this is a valid statement.

  • $\begingroup$ A solid angle is the area of a part of a sphere divided by the radius of the sphere. The solid angle of the full sphere is therefor $4\pi$, that of one of the hemispheres is $2\pi$ etc.. The quantity that you are talking about is called "radiance" and it has nothing to do with slits. $\endgroup$
    – CuriousOne
    Mar 9 '16 at 5:21
  • $\begingroup$ Read this: Radiant Intensity $\endgroup$
    – hxri
    Mar 9 '16 at 5:23
  • $\begingroup$ So does increasing slit width increase the intensity of the light, considering the slit as a source? $\endgroup$ Mar 9 '16 at 5:32
  • $\begingroup$ W/m${}^2$ is "officially" a unit of irradiance, but it is commonly called intensity. The W/sr is "officially" a unit of intensity. You have to tell by context which is meant. See also Table 2 in [this Wikipedia article}(en.wikipedia.org/wiki/Photometry_(optics)). I believe that in other fields, e.g. acoustics, W/m${}^2$ is a unit of intensity (but I'm not an expert in acoustics). I can't answer your question about slits, because you don't tell me where your source is and where your slit is and where you are measuring. $\endgroup$
    – garyp
    Mar 9 '16 at 14:59
  • $\begingroup$ My question at this point is as follows: imagine you have a doubles slit experiment in which one slit has a width two times the width of the other. Now, all things being equal, will the intensity from the larger slit be twice the intensity of the smaller slit? $\endgroup$ Mar 10 '16 at 2:14

What is a solid angle?

The solid angle $\Omega$ describes the size of the cone of intensity radiated by the source.

Consider a sphere with radius $R$ and a cone which intersects the sphere surface to form a circle with a radius $r$. For the case of a narrow cone and small cone area on the sphere surface, the following approximation holds. The area of the spot is $A_{circle}$ and the surface area of the sphere is $A_{sphere}$, as shown in the figure below.

There are $4\pi$ steradians over the entire surface of a sphere. So the ratio $\frac{A_{circle}}{A_{sphere}}$ is the fraction of the total $4\pi[sr]$ of the sphere which is encompassed by the cone.

enter image description here

The power $P [W]$ from a source that is directed into a particular direction along the center of a cone encompassing a solid angle $[steradians]$ or $[sr]$ is called the Radiant Intensity, $I$ $[W/sr]$.

enter image description here

Source: http://omlc.org/classroom/ece532/class1/intensity.html

Does increasing slit width increase the intensity of the light, considering the slit as a source?

This is a different kind of question.

The width of a slit in interference is directly proportional to the intensity of light from the slit(Considering slit as a source of light).


Where ${W_1}$ and ${W_2}$ are the widths of the slits and ${I_1}$ and ${I_2}$ are the intensities of light from the respective slits.

So increasing slit width increases the intensity of the light.

  • $\begingroup$ My question at this point is in fact whether intensity of light from a slit is proportional to the slit width. If I am correct, you are quoting a previous post, but I am not so sure that post is correct. If you are using another source, however, please let me know $\endgroup$ Mar 9 '16 at 6:06
  • $\begingroup$ Yes that's true! and you know it : physics.stackexchange.com/questions/102134/… $\endgroup$
    – hxri
    Mar 9 '16 at 6:45
  • $\begingroup$ But my question is how do we know that intensity is directly proportional to slit width because it would appear to me that only power is proportional to slit width. $\endgroup$ Mar 9 '16 at 14:27

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