# Thin lens

A thin lens is an optical lens whose thickness is small compared to the radii of its refracting surfaces.

In paraxial optics , the thin lens is a concept according to which the finite thickness real lens is replaced by a plane. The two breaking processes of a light beam at the interfaces are combined into one breaking process at this level. If the real lens has a symmetrical cross-section (e.g. biconvex or biconcave ), its median plane becomes the refractive plane. A beam of light that hits the center of the plane passes through the lens with an unchanged direction and without a parallel offset.

The concept of the thin lens is an idealization of the finitely thick real lens and a good approximation for large radii of their interfaces and thus also a large focal length . The formal reduction to one plane does not mean that the lens has no refractive index or no curved interfaces, because its focal length depends on both. However, its focal length is sufficient to describe the behavior of the thin lens.

## Optical imaging using a thin lens

The two main planes and a general optical system coincide in the thin lens in their equivalent plane (central plane). The thin lens has only one main plane . Both focal lengths and are measured from the main plane . The ray- optical construction of the optical image is somewhat simpler and takes place as follows. ${\ displaystyle H_ {1}}$ ${\ displaystyle H_ {2}}$ ${\ displaystyle H}$ ${\ displaystyle {\ overline {OF}}}$ ${\ displaystyle {\ overline {OF '}}}$ ${\ displaystyle H}$ The image point B ' is found with the help of two of the three main rays parallel ray  (1), center ray  (3) or focal ray  (2) (in Figure 1 from top to bottom, numbering refers to Figure 2), which emanate from object point B. . The rays are refracted only once - namely at the middle / main plane (ray (3) remains unbroken). The parallel beam is refracted in such a way that it passes through the focal point F 'on the image side . The focal point beam goes through the object-side focal point F and is refracted at the center plane so that it becomes the parallel beam on the image side.

## Thick lens theory Figure 2: A biconvex lens with thickness d . With main planes and to which the focal lengths and relate. Rays coming from the left (solid red lines) are refracted twice, at each spherically curved interface. The main planes go through the intersection of the extension of the
parallel beam on one side and the extension of the focal beam on the other side (dashed lines 1 and 3).${\ displaystyle H_ {1}}$ ${\ displaystyle H_ {2},}$ ${\ displaystyle f_ {1}}$ ${\ displaystyle f_ {2}}$ In the following, the theory of the thick lens is presented and the lens equation is derived through a limit transition to a thin lens .

A (thick) lens consists of a transparent material with a refractive index . The lens forms two interfaces with its surroundings. Normally the environment is air and thus the refractive index . A beam of light coming from the left (see Figure 2) is refracted at each of the two interfaces according to Snellius' law of refraction . To calculate the focal length of a lens, the two refractions of the light beam are considered successively. ${\ displaystyle n_ {2}}$ ${\ displaystyle n_ {1} = 1}$ ### Refraction on a curved surface

First, the refraction of the light beam at the left interface is considered (Figure 3).

From a purely geometrical point of view

${\ displaystyle h = R \ sin \ delta = (f_ {2} -x) \ tan \ gamma '.}$ For rays close to the axis, the angles are small and it is approximately as well . Also applies . This results in ${\ displaystyle \ alpha '= \ delta, \ beta'}$ ${\ displaystyle x \ ll f_ {2}}$ ${\ displaystyle \ tan \ gamma '\ approx \ sin \ gamma'}$ ${\ displaystyle \ gamma '= \ alpha' - \ beta '}$ ${\ displaystyle f_ {2} \ approx \ left ({\ frac {\ sin \ alpha '} {\ sin (\ alpha' - \ beta ')}} \ right) R = f_ {2}.}$ With Snellius' law of refraction and results ${\ displaystyle n_ {1} \ sin \ alpha '= n_ {2} \ sin \ beta'}$ ${\ displaystyle \ sin (\ alpha '- \ beta') = \ sin \ alpha '\ cos \ beta' - \ cos \ alpha '\ sin \ beta' \ approx \ sin \ alpha '- \ sin \ beta'}$ ${\ displaystyle f_ {2} = R \ left ({\ frac {n_ {2}} {n_ {2} -n_ {1}}} \ right).}$ The image-side focal length of a curved interface for rays close to the axis therefore only depends on the refractive indices of the two materials and the curvature of the surface. If you now imagine that the light comes from the right, runs to the left and is then refracted, the above formula can also be used. The refractive indices in the formula must now be exchanged, as the light beam from the medium follows with it . ${\ displaystyle f_ {2}}$ ${\ displaystyle n_ {2}}$ ${\ displaystyle n_ {1}}$ ${\ displaystyle f_ {1} = R \ left ({\ frac {n_ {1}} {n_ {1} -n_ {2}}} \ right)}$ ${\ displaystyle f_ {1}}$ is called the object-side focal length .

A relationship between the image distance and the object distance can now also be derived. ${\ displaystyle b}$ ${\ displaystyle a}$ The following angle relationships apply: as well as . Therefore, in the small-angle approximation for Snellius' law of refraction: ${\ displaystyle \ alpha = \ delta + \ varepsilon}$ ${\ displaystyle \ beta = \ delta - \ gamma}$ ${\ displaystyle n_ {1} (\ delta + \ varepsilon) \ approx n_ {2} (\ delta - \ gamma)}$ ${\ displaystyle h = (a + x) \ tan \ varepsilon \ approx a \ varepsilon}$ ${\ displaystyle h = (bx) \ tan \ gamma \ approx b \ gamma}$ ${\ displaystyle h = R \ sin \ delta \ approx R \ delta}$ Insertion into the small-angle approximation of Snell's law of refraction, whereby all approximations are treated as equations, results

${\ displaystyle {\ frac {n_ {1}} {a}} + {\ frac {n_ {2}} {b}} = {\ frac {n_ {2} -n_ {1}} {R}}. }$ The right side can now be expressed using the focal length derived above:

${\ displaystyle {\ frac {n_ {1}} {a}} + {\ frac {n_ {2}} {b}} = {\ frac {n_ {2}} {f_ {2}}} = - { \ frac {n_ {1}} {f_ {1}}}}$ ### Refraction on a thick lens

The refraction of near-axis rays on a lens corresponds to two consecutive refractions at curved interfaces. It is assumed that the light is incident from the left. If it is a convex lens (shown in the upper figure), then the radius of curvature of the 1st interface is positive, while the radius of curvature of the 2nd interface is negative. In the case of a diverging lens (concave lens) the situation is exactly the opposite. ${\ displaystyle R_ {2}}$ For the second refraction, the image of the first refraction is used as the object. It is the object distance of the 1st refraction and the object distance of the 2nd refraction. The same agreement applies to the image widths and the radii of curvature . ${\ displaystyle a_ {1}}$ ${\ displaystyle a_ {2}}$ ${\ displaystyle b_ {1}, b_ {2}}$ ${\ displaystyle R_ {1}> 0, R_ {2} <0}$ The following applies to the image distance after the first refraction: ${\ displaystyle b_ {1}}$ ${\ displaystyle {\ frac {n_ {1}} {a_ {1}}} + {\ frac {n_ {2}} {b_ {1}}} = {\ frac {n_ {2} -n_ {1} } {R_ {1}}}}$ Now you insert the image distance as the object distance in the formula for the 2nd refraction: with the thickness of the lens (as indicated in the 1st illustration). The shift by has to be done because the derivation above assumed that the curved surface goes through the origin. Of course the refractive indices have to be reversed again. The following applies to the 2nd refraction: ${\ displaystyle b_ {1}}$ ${\ displaystyle a_ {2}}$ ${\ displaystyle a_ {2} = - (b_ {1} -d)}$ ${\ displaystyle d}$ ${\ displaystyle d}$ ${\ displaystyle {\ frac {n_ {2}} {a_ {2}}} + {\ frac {n_ {1}} {b_ {2}}} = {\ frac {n_ {1} -n_ {2} } {R_ {2}}}}$ Insertion of results ${\ displaystyle a_ {2} = - (b_ {1} -d)}$ ${\ displaystyle - {\ frac {n_ {2}} {b_ {1} -d}} + {\ frac {n_ {1}} {b_ {2}}} = {\ frac {n_ {1} -n_ {2}} {R_ {2}}}.}$ ${\ displaystyle {\ frac {n_ {1}} {a_ {1}}} + {\ frac {n_ {2}} {b_ {1}}} - {\ frac {n_ {2}} {b_ {1 } -d}} + {\ frac {n_ {1}} {b_ {2}}} = {\ frac {n_ {2} -n_ {1}} {R_ {1}}} + {\ frac {n_ {1} -n_ {2}} {R_ {2}}}}$ ${\ displaystyle \ Leftrightarrow {\ frac {n_ {1}} {a_ {1}}} + {\ frac {n_ {1}} {b_ {2}}} = (n_ {2} -n_ {1}) \ left ({\ frac {1} {R_ {1}}} - {\ frac {1} {R_ {2}}} \ right) + {\ frac {n_ {2} d} {b_ {1} ( b_ {1} -d)}}}$ The left side of the lens and the right side of the lens are now counted. ${\ displaystyle a_ {1}}$ ${\ displaystyle b_ {2}}$ ### Thin lens approximation

You can now introduce the object distance and the image distance for the entire lens: , . A lens is said to be thin if and applies. This simplifies the above formula to ${\ displaystyle a}$ ${\ displaystyle b}$ ${\ displaystyle a = a_ {1} + d / 2}$ ${\ displaystyle b = b_ {2} + d / 2}$ ${\ displaystyle d \ ll b_ {2}}$ ${\ displaystyle d \ ll a_ {1}}$ ${\ displaystyle {\ frac {1} {a}} + {\ frac {1} {b}} = (n_ {2} -n_ {1}) \ left ({\ frac {1} {R_ {1} }} - {\ frac {1} {R_ {2}}} \ right).}$ This is the well-known lens equation and describes the imaging of on-axis rays with a thin lens. For axially parallel rays applies (the object lies at infinity). On the image side, the rays must now pass through the focal point. Therefore, for the focal length of a thin lens: ${\ displaystyle a = \ infty}$ ${\ displaystyle f = b}$ ${\ displaystyle f = {\ frac {1} {n_ {2} -n_ {1}}} \ left ({\ frac {R_ {1} R_ {2}} {R_ {2} -R_ {1}} } \ right)}$ For a biconvex lens with applies ${\ displaystyle R_ {1} = R = -R_ {2}}$ ${\ displaystyle f = {\ frac {R} {2 (n_ {2} -n_ {1})}}.}$ Insertion into the imaging equation results in the imaging equation for thin lenses :

${\ displaystyle {\ frac {1} {a}} + {\ frac {1} {b}} = {\ frac {1} {f}}}$ In the end, the lens grinding formula was also derived:

${\ displaystyle {\ frac {1} {f}} = (n_ {2} -n_ {1}) \ left ({\ frac {1} {R_ {1}}} - {\ frac {1} {R_ {2}}} \ right)}$ 