# Determining the light intensity in a double convex lens

1. Feb 22, 2016

### 01baftb

Attached is an illustration of an image being demagnified using a double convex lens.

I want to determine the intensity of the light (mW/cm2) if I were to put a lux meter at various locations after the convex lens given that the object is at a fixed location. The object is a monochromatic light projected by a projection system which goes through demagnification due to the double convex lens.

My assumption would be that be that as you move closer to the focal plane, the intensity will increase. I know that the intensity is inversely related to the the magnification (M = -di/do). Thus, if M = -1/2, then intensity at the focal plane is I2 = I0/M = 2*I0. But what if I put a lux meter before and after the focal plane? How can I determine the intensity of the light which is out of focus?

How can I derive the relationship between the light intensity of the object (I0) and the intensity of the image before the focal plane (I1), at the focal plane (I2), and after the focal plane (I3)?

#### Attached Files:

• ###### demag-light-intensity.png
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Last edited: Feb 22, 2016
2. Feb 23, 2016

### Simon Bridge

All the light from the object, that passes through the lens, will pass through each plane[*]. But you will notice that the light from the point of the arrow is more spread out through some planes than others. What does that suggest about the intensity?

Notice - it is not the light that is out of focus - "focus" is a word used to describe how sharp the picture formed on a screen would be.
In this context, the picture on the screen is not the same as the "image". When the closer the screen position is to the image position, the sharper the picture. The sharpest picture is said to be "in focus".

[*] ... some light from the object will be scattered, reflected, or absorbed, when it is intercepted by the lens.

3. Feb 23, 2016

### Andy Resnick

In radiometry and photometry, W/cm2 is irradiance, not intensity. Intensity is W/sr. The irradiance will increase to a maximum at the focal plane, but the intensity will not (AFAIK). Lux meters measure irradiance, and the output should vary accordingly. If you want to calculate the irradiance at arbitrary locations, you need to know the f/# because that tells you how much the light 'diverges' away from focus. Alternatively, if you know the Rayleigh length, that would work as well.

4. Feb 26, 2016

### Simon Bridge

5. Feb 26, 2016

### lightarrow

6. Feb 26, 2016

### Andy Resnick

7. Feb 26, 2016

### Andy Resnick

Yeah, it can be confusing- there's a steady trickle of folks who post here asking the same basic questions that basically just involve conversion between the various quantities. Apostilbs! skots!

8. Feb 28, 2016

### Simon Bridge

Granted - however, what is not wrong with the problem statement is the choice of units for intensity, whatever units any data may be collected in.

Taking a look at the problem statement:
... now I read that as saying "I want to get intensity in these specific units using a lux meter" ... this would presumably involve a unit conversion.
But there's more:
- and the context is "general physics".

Looks like homework to me so I won't go further without feedback from OP.