- #1
nordmoon
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- 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/
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/