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.
Let x_{o} denote the perpendicular distance of the object from the
centerline of the lens and let x_{o} be positive. Let x_{i}
denote the perpendicular distance of the image from the centerline of the lens. Then x_{i} can be found from the lens equation
1/x_{o }+ 1/x_{i }= 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 x_{o
}< f.
The focal length of a thin lens is related to the radii of curvature of its two
surfaces.
1/f = (n_{2 }- n_{1})(1/R_{1 }- 1/R_{2}).
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 R_{1} and R_{2} are positive. Therefore
1/f = (n_{2
}- n_{1})(1/R_{1 }- 1/R_{2})
= (1.5 - 1)(½ - ½.5) = 0.05. f = 20 cm.
The lens equation and the mirror equation
are written as
1/x_{o }+ 1/x_{i }= 1/f.
But the sign
conventions for x_{o}, x_{i}, 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.
Applying 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 x_{o}.
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/x_{o }+ 1/x_{i }= 1/f, x_{i }= fx_{o}/(x_{o
}- f)
(a) x_{i }= 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) x_{i }= 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.
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 = -x_{i}/x_{o }= 2, x_{i }= -2x_{o}.
1/x_{o}
- 1/(2x_{o}) = 1/(2x_{o}) = 1/f = 1/15.
x_{o
}= 7.5 cm.
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/x_{o
}+ 1/x_{i} = 1/f, 1/3 + 1/x_{i} = 1/4, x_{i}
= -12 cm.
M = -x_{i}/x_{o} = 12/3 = 4.
Assume two thin converging lenses with focal length f_{1} and f_{2},
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
f_{eff} from the lenses. Find f_{eff} in terms of f_{1}
and f_{2}.
Solution:
x_{o} = infinite, so for the first lens 1/x_{i
}= 1/f_{1},
x_{i} = f_{1}. 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 -f_{1}.
For the second lens we have -1/f_{1} + 1/x_{i} = 1/f_{2},
1/x_{i} = 1/f_{1} + 1/f_{2}.
But the image distance of the second lens is the focal length f_{eff}
of the whole system.
Therefore 1/f_{eff} = 1/f_{1} + 1/f_{2}.
Link:
Geometrical Optics (A Phet simulation) Try it!
Additional information:
The
Physics Classroom: Refraction and the Ray Model of Light Lesson 5