# Should the thickness of a plano-convex lens affect its focal length and intensity of light at the image?

Using Lens maker formula for thick lens,

$\frac{1}{f}=(n-1)(\frac{1}{R_{1}}-\frac{1}{R_{2}}+\frac{(n-1)d}{n{R_{1}}{R_{2}}})$

where f is the focal length

R1 and R2 are the radii of the two sides

d is the thickness

n is the index of refraction

For a plano-convex lens, R2=∞. So the equation reduces to

$1/f=(n-1)(\frac{1}{R_{1}})$

Given d is not found in the final equation does that mean it has no effect on the focal point of the lens, or is it because of the approximation of R2 = ∞ that makes it have no effect?

Also another problem is I cannot find any relationships between the thickness of a lens and the intensity of light at the image.

• Welcome to physics.SE. Please mark up your math properly using mathjax (which is basically latex). – Ben Crowell Mar 30 '18 at 16:07

Let us assume we are talking about a positive plano-convex lens, where the 1st surface is convex, and the 2nd surface is plano. Your suspicion is correct - Since $R2=\infty$ , the second surface does not contribute any power to the lens, and the efl is determined solely by the first surface. The focal length of lens depends on its construction, but where the image is located also depends on the construction of the lens. At this point, it is important to note which surfaces we are using for a reference point. Since the lens powers is all on the first surface, the thickness of the lens serves only to displace the image. As the lens thickness increases, the image will displace further from the first surface. A plane parallel plate will displace an image by an amount $\Delta z=\frac{N-1}{N}$ where $N$ is the refractive index of the plate (assuming the plate is in air).
So say your lens has $R1=50$ mm, $N=1.5$. The effective focal length will be 100. If the lens has no thickness, the distance from the R1 or R2 surface to the image (object at infinity) would be 100 mm. If you make the lens 10 mm thick, the image is displaced backwards an additional 3.3 mm. Since you put in a 10 mm thick plate, the image, relative to the R2 surfaces, is now $BFL=100-10+3.3=93.3$ mm. Note the importance of which surface you are using for measuring the distance - people commonly use the distance from the R2 surface to the image, even though in our example all the optical power is at R1. For a single lens, with the object at infinity, this distance is called the BFL, or back focal length.