Magnification & Two-Element Systems

You’ll be able to

Compute magnification from distances and from image and object heights
Trace an image through two lenses in sequence
Compute the overall magnification of a compound system

Two expressions for magnification

Magnification can be written as m = −d_i/d_o or as m = h_i/h_o. Both are always valid, so equating them lets you find a height from distances or a distance from heights. The sign carries the orientation: negative m means the image is inverted, positive means upright. And |m| > 1 means enlarged while |m| < 1 means reduced, independently of the sign.

Magnification

m = −d_i/d_o = h_i/h_o · sign gives orientation, magnitude gives size · overall for two elements: m_total = m₁ × m₂

Multiply, do not add. Two inversions give an upright final image, since two negatives multiply to a positive.

The two-lens procedure

Solve the first lens completely, then use its image as the object for the second lens. The object distance for the second lens is the separation minus the first image distance. If that arithmetic comes out negative, the first image lies beyond the second lens, which makes it a virtual object — perfectly legitimate, and the equation handles it if you keep the sign. Overall magnification is the product of the individual magnifications.

Telescope and microscope, briefly

A refracting telescope places two converging lenses so the objective's image sits at the eyepiece's focal point, giving angular magnification f_objective/f_eyepiece — so a long objective and a short eyepiece magnify most. A compound microscope puts the specimen just beyond the objective's focal point to form a large real image, which the eyepiece then magnifies again as a virtual image. Both are two-element systems solved by the same sequential procedure.

Worked example

An object 4.0 cm tall sits 30 cm from a converging lens of focal length 20 cm. A second converging lens of focal length 10 cm sits 90 cm beyond the first. Find the final image position and overall magnification.

1.First lens: 1/d_i = 1/20 − 1/30 = 3/60 − 2/60 = 1/60, so d_i₁ = +60 cm. m₁ = −60/30 = −2.0.
2.The first image is 60 cm past lens 1, so it is 90 − 60 = 30 cm in front of lens 2. That is the second object distance.
3.Second lens: 1/d_i = 1/10 − 1/30 = 3/30 − 1/30 = 2/30, so d_i₂ = +15 cm. m₂ = −15/30 = −0.50.
4.Overall: m = m₁ × m₂ = (−2.0)(−0.50) = +1.0.
5.Final image height = (+1.0)(4.0) = 4.0 cm, upright.

Answer: The final image is real, 15 cm past the second lens, upright and 4.0 cm tall — the same size as the object. The two inversions cancel, which is why the product of two negative magnifications is positive.

Watch out

Multiply the magnifications; never add them. And take the sign of each seriously — two inverting elements produce an upright final image, which is the result students most often get wrong by tracking only magnitudes.

Checkpoint

An image has magnification −0.25. The image is:

Checkpoint

Two lenses each produce a magnification of −3.0. The overall magnification of the system is:

Checkpoint

In a two-lens system, the object distance for the second lens is found by:

On the exam

Solve two-element problems in strict sequence and write each intermediate d_i and m down. Rubrics award the first lens and the second lens separately, so a correct first stage earns credit even if the second goes wrong.

Answer the 3 checkpoints as you read.