Thin lenses

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Converging lens

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Diverging lens

A single spherical interface between two transparent media can lead to image formation.  But viewing these images is not always possible.  Consider an air-glass interface.  When the object is on the air side, the image can only be viewed from inside the glass.  Giving the glass a finite thickness and adding a second spherical interface between glass and air will produce an image that can be viewed by in air.  A piece of glass of finite thickness with two spherical boundaries is a lens.  If the thickness of the lens is much smaller than the diameter of the lens, we call it a thin lens.

Thin lenses can be converging or diverging.  Converging lenses are thicker in the middle than near the edges, and diverging lenses are thicker near the edges than in the middle.  A thin lens has two focal points, located on the optical axis, a distance f from the center of the lens on either side of the lens.  Rays parallel to the optic axis passing through a thin converging lens come together at the focus f on the opposite side of the lens, and rays parallel to the optic axis passing through a diverging lens diverge as if they were leaving the focus on the incident side on a straight line path.

A plane through a focus perpendicular to the optic axis is called a focal plane.  Parallel rays converge in the focal plane or diverge from the focal plane, independent of the angle the make with the optic axis.


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Real image

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Virtual image

Thin lenses can form real and virtual image.

Let xo denote the perpendicular distance of the object from the centerline of the lens and let xo be positive.  Let xi denote the perpendicular distance of the image from the centerline of the lens.  Then xi can be found from the lens equation

1/xo + 1/xi = 1/f,

provided we use the following sign convention.

We can find the position and size of the image geometrically.  Only two rays must be drawn.

Diverging lenses only form virtual images.
Converging lenses form real inverted images if x
o> f and virtual upright images if xo < f.

The focal length of a thin lens is related to the radii of curvature of its two surfaces.
1/f = (n2 - n1)(1/R1 - 1/R2).

imageProblem:

A contact lens is made of plastic with an index of refraction of 1.5.  The lens has an outer radius of curvature of 2 cm and an inner radius of curvature of 2.5 cm. What is the focal length of the lens?

Solution:
Both R1 and R2 are positive. Therefore
1/f = (n2 - n1)(1/R1 - 1/R2) = (1.5 - 1)(½ - ½.5) = 0.05.  f = 20 cm.


The lens equation and the mirror equation are written as 1/xo + 1/xi = 1/f.
But the sign conventions for xo, xi, and f are different for lenses and mirrors.

Here are some things that always go together for a thin lens.

The lens and mirror equations are valid for paraxial rays.  If all the light rays involved in image formation make small angles with the optic axis, so that sinθ ≈ tanθ ≈ θ, then near perfect images form at the locations predicted by these equations.  For off-axis or skew rays, the image of a point becomes blurry and we observe aberrations.

imageApplying the lens equation to thin lenses, we can safely assume the light bends at the center of the lens, and neglect the thickness of the lens.  For thick lenses we have to be more careful.
One must calculate the location of principal planes, from which to measure distances to the object, the image, and the focal points.


Many optical instruments have more than one optical element.  A combination of mirrors and lenses can be analyzed by treating the image of the first element as the object of the second element, and so on.  If the image of the first element falls behind the second element we can still use the lens or mirror equation, but we then must use a negative object distance xo.

Problem:

A thin lens has a focal length of 25 cm.  Locate the image when the object is placed
(a) 26 cm
(b) 24 cm
in front of the lens.  Describe the image in each case.

Solution:
1/xo + 1/xi = 1/f,  xi = fxo/(xo - f)

(a) xi = 25*26/(26 - 25) = 650.  The image a real image 650 cm behind the lens. 
Its magnification is M = -650/26 = -25.
The image is inverted and enlarged.

(b) xi = 25*24/(24 - 25) = -600.
The image a virtual image 600 cm in front of the lens.
Its magnification is M = 600/24 = 25.
The image is upright and enlarged.

Problem:

A magnifying glass is a converging lens of focal length 15 cm.  At what distance from a postage stamp should you hold this lens to get a magnification of +2?

Solution:
M = -xi/xo = 2,  xi = -2xo.
1/xo - 1/(2xo) = 1/(2xo) = 1/f = 1/15.
xo = 7.5 cm.

Problem:

A converging lens of focal length 4 cm is used as a magnifier.  If an object is placed 3 cm from the lens, what is the magnification?

Solution;
1/xo + 1/xi = 1/f,  1/3 + 1/xi = 1/4, xi = -12 cm.
M = -xi/xo = 12/3 = 4.

Problem:

Assume two thin converging lenses with focal length f1 and f2, respectively, are very close to each other, so we can neglect the distance between them. They focus the light of a distant star at a focal point a distance feff from the lenses.  Find feff in terms of f1 and f2.

Solution:
xo = infinite, so for the first lens 1/xi = 1/f1, xi = f1.  The image of the first lens becomes the object of the second lens.  It lies behind the second lens, and so the object distance for the second lens is -f1.
For the second lens we have -1/f1 + 1/xi = 1/f2,   1/xi = 1/f1 + 1/f2.
But the image distance of the second lens is the focal length feff of the whole system.
Therefore 1/feff = 1/f1 + 1/f2.


Link:  Geometrical Optics  (A Phet simulation)  Try it!

Additional information:  The Physics Classroom: Refraction and the Ray Model of Light  Lesson 5