Lenses

Lenses are found in a huge array of optical instruments, ranging from a simple magnifying glass to the eye to a camera’s zoom lens. Law of refraction is used to explore the properties of lenses and how they form images.

The Main Idea

Convex

Concave

Index of refraction depends on the wavelength. Thus, light of different wavelengths is bent, or deflected, by different amounts as it passes through a lens. The shape of a lens, either concave or convex, also plays a role in the deflection pattern of light. The images above show that how these two shapes determines the behavior of the light rays. A lens where the middle is thicker than the two ends is called a "convex" lens, through which incoming light rays converge towards the center axis of the lens. A lens where the middle is thinner than the two ends is called a "concave" lens the prisms represent a "diverging" lens, through which incoming light rays diverge away from the center axis. The angle at which light rays converge or diverge is called the deflection angle. Deflection angles for thin lenses will be modeled mathematically in the following section. Thin lenses are lenses where the y position of a light ray does not change very much as the light ray travels through it. In other words, the lens is thick enough to refract light rays, but does not allow dispersion or aberrations.

A Mathematical Model

Law of Refraction

Refraction occurs when light travels through an area of space that has a changing index of refraction. The simplest case of refraction involves a uniform medium with index of refraction [math]\displaystyle{ n_1 }[/math] and another medium with index of refraction [math]\displaystyle{ n_2 }[/math]. The following equation describes the resulting deflection of the light ray:

[math]\displaystyle{ n_1\sin\theta_1 = n_2\sin\theta_2\ }[/math]

Thin Lens Equation and Magnification

A ray tracing diagram for a converging lens.

Thin lenses produce focal points on either side that can be modeled using the lensmaker's equation. Thin lenses follow a simple equation that determines the location of the images given a particular focal length ([math]\displaystyle{ f }[/math]) and object distance ([math]\displaystyle{ S_1 }[/math]):

[math]\displaystyle{ \frac{1}{S_1} + \frac{1}{S_2} = \frac{1}{f} }[/math]

The magnification of a lens is given by

[math]\displaystyle{ M = - \frac{S_2}{S_1} = \frac{f}{f - S_1} }[/math]

A Computational Model

Lens Simulation

Examples

Determine the Index of Refraction from Refraction Data

Figure1

Find the index of refraction for medium 2 in Figure1 (a), assuming medium 1 is air and given the incident angle is 30.0º and the angle of refraction is 22.0º.

Strategy

The index of refraction for air is taken to be 1 in most cases (and up to four significant figures, it is 1.000). Thus [math]\displaystyle{ n_1 = 1.00 }[/math] here. From the given information, [math]\displaystyle{ \theta_1 = 30.0º }[/math] and [math]\displaystyle{ \theta_2 = 22.0º }[/math]. With this information, the only unknown in the lens equation is [math]\displaystyle{ n_2 }[/math], so that it can be used to find this unknown.

Solution

[math]\displaystyle{ n_1\sin\theta_1 = n_2\sin\theta_2\ }[/math]

Entering known values,[math]\displaystyle{ n_2 = 1.33 }[/math].

Fining the Image of a Light Bulb Filament by Ray Tracing and by the Thin Lens Equations

Figure2

A clear glass light bulb is placed 0.750 m from a convex lens having a 0.500 m focal length, as shown in Figure2. Use ray tracing to get an approximate location for the image. Then use the thin lens equations to calculate (a) the location of the image and (b) its magnification. Verify that ray tracing and the thin lens equations produce consistent results.

Solution

The ray tracing to scale shows two rays from a point on the bulb’s filament crossing about 1.50 m on the farside of the lens. Thus the image [math]\displaystyle{ d_i }[/math] distance is about 1.50 m. Similarly, the image height based on ray tracing is greater than the object height by about a factor of 2, and the image is inverted. Thus [math]\displaystyle{ m }[/math] is about –2. The minus sign indicates that the image is inverted.

[math]\displaystyle{ \frac{1}{S_1} + \frac{1}{S_2} = \frac{1}{f} }[/math]

Solving for [math]\displaystyle{ d_i }[/math] from the given known values gives [math]\displaystyle{ d_i = 1.50m }[/math].

The magnification of a lens is given by

[math]\displaystyle{ M = - \frac{S_2}{S_1} = \frac{f}{f - S_1} }[/math]

Solving for [math]\displaystyle{ m }[/math] gives [math]\displaystyle{ m = -2.00 }[/math]

Image Produced by a Concave Lens

Suppose an object such as a book page is held 7.50 cm from a concave lens of focal length –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced?

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