Image position in an optical system?

[nordmoon]

Homework Statement

I have an plano-convex lens with focal length 75 mm. The object is 325mm from the lens.
a) Where is the location of the image after the lens?
b) Where is the location of the image if an beamsplitter is placed after the lens?

Relevant Equations

lens formula

I have an plano-convex lens with focal length 75 mm. The object is 325mm from the lens.
a) Where is the location of the image after the lens?
b) Where is the location of the image if an beamsplitter is placed after the lens?

My solution, questions:
a)
a = 325 mm
b = ? mm
f = 75 mmm, plano-concave, for example https://www.edmundoptics.com/p/25mm-dia-x-75mm-fl-nir-ii-coated-achromatic-lens/6253/

lens formula is 1/a + 1/b = 1/f

Thus, the image is then
b = 1/(1/f-1/a) = 97.5 mm after the lens, the plane side.

Exactly from where relative the lens is the image located? Is it 97.5 mm from the back-end of the lens?
The effective focal length (EFL) is not really easy to figure out, but I can see where the back side of the lens is, as a reference point.
Can this be determined or calculated?
What is BFL and EFL?

b)
What happens if you also place an beamsplitter right after the lens?

The image is split 50/50 into two directions, straight forward and side-ways of the beamsplitter.

Is the image located at the same location (straight forward) as in example a)?
Or does it move to another location as a result of the beam splitter?
How do you compute this new location with the lens formula?

The beam splitter is a non-polarizing beamsplitter, 38x38 mm 50/50 cube. For example https://www.edmundoptics.com/p/nir-coated-c-mounted-non-polarizing-cube-beamsplitter/10188/

 

[Cutter Ketch]

The lens equation is an approximation that assumes the lens is thin compared to the object, image, and focal distances. It does not tell you from where in the few mm thickness of the lens to measure these distances. Generally, since it is an approximation anyway, you just pick a place. This is an easier decision with a symmetric lens, as it seems logical to pick the middle. There is no really good reason for this, particularly when the image and object distances are not symmetric, but it still feels good. With a Plano-convex, there isn’t an obvious choice. You might say the plano surface, but you’d probably be closer to right to pick the middle, ie halfway between the front and back face. In any case, it doesn’t matter much because you won’t get the true focus exactly right in any case. An exact distance measured from one surface or the other, and depending on which way you turn the lens, can be calculated by using Snell’s law at each surface. Unfortunately the place where the rays meet will vary with the distance out from the center of the lens (particularly if you use the lens in the worst orientation). This is spherical aberration. There will also be errors based on the field angle (coma). However for large distances compared to the thickness and the diameter of the lens (“paraxial”) the lens equation will get you pretty close.

 

[Cutter Ketch]

What happens with the beam splitter? Forget about a cube for a second and imagine it’s a very thin membrane. Part of the light goes through, part reflects like a flat mirror. What happens to the image given a flat mirror?

 

[Tom.G]

The lens equation is an approximation that assumes the lens is thin compared to the object, image, and focal distances. It does not tell you from where in the few mm thickness of the lens to measure these distances.

Still an approximation, but the measurement point, actually a plane, is called the Principal Plane of a lens.

For a thin, symmetrical, lens it is generally a plane that slices the lens into two symmetrical pieces. For instance a double-convex 'thin' lens is considered to be two plano-convex lenses mated at their flat surfaces.

A 'thick' double-convex lens, or a lens with different curvatures on its two sides, will have two principal planes, one for each surface.

For a nice explanation that starts out with what you are after, then gets deep, see:
http://www.physics.purdue.edu/~jones105/phys42200_Spring2013/notes/Phys42200_Lecture32.pdf

Cheers,
Tom

 

[nordmoon]

What happens with the beam splitter? Forget about a cube for a second and imagine it’s a very thin membrane. Part of the light goes through, part reflects like a flat mirror. What happens to the image given a flat mirror?

Okay, but for the light that goes through the beam splitter? Will it change the position of the image (assuming there is the same lens in front of the beam splitter), with and without the cube beam splitter? I always thought it was unchanged but I have noticed a difference of the location of my image.

 

[Tom.G]

Okay, but for the light that goes through the beam splitter? Will it change the position of the image

By a tiny amount.
A plain, optically flat, glass plate shows these effects:

• With a light beam perpendicular to the surface, the plate acts as a very weak Negative lens.
• With a light beam at an angle to the surface, the plate additionally acts as a very weak wedge (i.e. a prism with almost-parallel entry and exit sides)

Most people either don't notice or ignore the phenomena.

If you have a Laser pointer and a sheet of glass you may be able to see the wedge effect if you project across the room. The experiment is easier if you can mount the Laser and glass on an optical bench so the light beam passes thru the center-of-rotation of the glass. That avoids manyany spatial variations in the glass.

Cheers,
Tom