1.3 Basic Geometrical Optics (E-III-4)

D. Lens formulas for thin lenses

As with mirrors, convenient formulas can be used to locate the image mathematically. The derivation of such formulas—as was carried out for spherical mirrors in the previous section—can be found in most texts on geometrical optics. The derivation essentially traces an arbitrary ray geometrically and mathematically from an object point through the two surfaces of a thin lens to the corresponding image point. Snell’s law is applied for the ray at each spherical refracting surface. The details of the derivation involve the geometry of triangles and the approximations mentioned earlier—sin j @ f, tan j @ f, and cos j @ 1—to simplify the final results. Figure 3-27 shows the essential elements that show up in the final equations, relating object distance p to image distance q, for a lens of focal length f with radii of curvature r₁ and r₂ and refractive index n_g. For generality, the lens is shown situated in an arbitrary medium of refractive index n. If the medium is air, then, of course, n = 1.

Figure 3-27 Defining quantities for image formation with a thin lens

1. Equations for thin lens calculations. The thin lens equation is given by Equation 3-10.

(3-10)

where p is the object distance (from object to lens vertex V )

q is the image distance (from image to lens vertex V )

and f is the focal length (from either focal point F or F¢ to the lens vertex V )

For a lens of refractive index n_g situated in a medium of refractive index n, the relationship between the parameters n, n_g, r₁, r₂ and the focal length f is given by the lensmaker’s equation in Equation 3-11.

(3-11)

where n is the index of refraction of the surrounding medium

n_g is the index of refraction of the lens materials

r₁ is the radius of curvature of the front face of the lens

r₂ is the radius of curvature of the rear face of the lens

The magnification m produced by a thin lens is given in Equation 3-12.

(3-12)

where m is the magnification (ratio of image size to object size)

h_i is the transverse size of the image

h_o is the transverse size of the object

p and q are object and image distance respectively

2. Sign convention for thin lens formulas. Just as for mirrors, we must agree on a sign convention to be used in the application of Equations 3-10, 3-11, and 3-12. It is:

Light travels initially from left to right toward the lens.
Object distance p is positive for real objects located to the left of the lens and negative for virtual objects located to the right of the lens.
Image distance q is positive for real images formed to the right of the lens and negative for virtual images formed to the left of the lens.
The focal length f is positive for a converging lens, negative for a diverging lens.
The radius of curvature r is positive for a convex surface, negative for a concave surface.
Transverse distances (h_o and h_i) are positive above the optical axis, negative below.

Now let’s apply Equations 3-10, 3-11, and 3-12 in several examples, where the use of the sign convention is illustrated and where the size, orientation, and location of a final image are determined.

Example 8

A double-convex thin lens such as that shown in Figure 3-21 can be used as a simple “magnifier.” It has a front surface with a radius of curvature of 20 cm and a rear surface with a radius of curvature of 15 cm. The lens material has a refractive index of 1.52. Answer the following questions to learn more about this simple magnifying lens.

(a)    What is its focal length in air?
(b)    What is its focal length in water (n = 1.33)?
(c)    Does it matter which lens face is turned toward the light?
(d)    How far would you hold an index card from this lens to form a sharp image of the sun on the card?

Solution:

(a) Use the lensmaker’s equation. With the sign convention given, we have n_g = 1.52, n = 1.00, r₁ = +20 cm, and r₂ = – 15 cm. Then

So f = +16.5 cm (a converging lens, so the sign is positive, as it should be)

(b)

f = 60 cm (converging but less so than in air)

(c) No, the magnifying lens behaves the same, having the same focal length, no matter which surface faces the light. You can prove this by reversing the lens and repeating the calculation with Equation 3-11. Results are the same. But note carefully, reversing a thick lens changes its effect on the light passing through it. The two orientations are not equivalent.

(d) Since the sun is very far wary, its light is collimated (parallel rays) as it strikes the lens and will come to a focus at the lens focal point. Thus, one should hold the lens about 16.5 cm from the index card to form a sharp image on the card.

Example 9

A two-lens system is made up of a converging lens followed by a diverging lens, each of focal length 15 cm. The system is used to form an image of a short nail, 1.5 cm high, standing erect, 25 cm from the first lens. The two lenses are separated by a distance of 60 cm. See accompanying diagram. (Refer to Figure 3-26 for a ray-trace diagram of what’s going on in this problem.)

Locate the final image, determine its size, and state whether it is real or virtual, erect or inverted.

Solution: We apply the thin lens equations to each lens in turn, making use of the correct sign convention at each step.

Lens L₁:

(f₁ is + since lens L₁ is converging.)

q₁ = +37.5 cm (Since the sign is positive, the image is real and located 37.5 cm to the right of lens L₁.

Lens L₂:

where p₂ = (60 – 37.5) = 22.5 cm

Since the first image, a distance q₁ from L₁, serves as the object for the lens L₂, this object is to the left of lens L₂, and thus its distance p₂ is positive. The focal length for L₂ is negative since it is a diverging lens. So, the thin lens equation becomes

giving q₂ = –9cm

Since q₂ is negative, it locates a virtual image, 9 cm to the left of lens L₂. (See Figure 3-26.)

The overall magnification for the two-lens system is given by the combined magnification of the lenses. Then

Thus, the final image is inverted (since overall magnification is negative) and is of final size
(0.6 × 1.5 cm) = 0.9 cm.