0
$\begingroup$

I need to focus a laser at a certain distance. But everytime I perform the experiment, the distance between the laser and the surface will be slightly different. Therefore I will fix the lens with the short focal length and move the lens with the high focal length. For the effective focal length, the lenses can be interchanged without difference. But the rays will be different when interchanged. For the combination of the lenses, the thin lens formula can not be used to describe the 2-lens system as one effective lens, as is proven below. Which sequence is better, first the strong lens or the other way around?

Using x, the distance x of the point from the optical axis, and θ the angle of the ray with the horizontal; (x,θ) for the incoming ray can be transformed in (x′,θ′) for the outgoing ray:

$\begin{pmatrix}x'\\\theta'\end{pmatrix}=\begin{pmatrix}1-\dfrac{d}{f_1}&d\\-\left(\dfrac{1}{f_1}+\dfrac{1}{f_2}-\dfrac{d}{f_1f_2}\right)&1-\dfrac{d}{f_2}\end{pmatrix}\begin{pmatrix}x\\\theta\end{pmatrix}$

$\endgroup$
1
$\begingroup$

If you start with a collimated beam and first place the strong lens the beams will become diverging after some distance and in that case the weak lens is not strong enough to make theses rays converging. A virtual image will be formed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.