Lenses and mirrors: predict image location, size, and orientation

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Use one sign convention consistently to predict image location, orientation, and size, then connect focal length to optical power and simple vision correction.

About 17 minutes, including converging-lens examples and checks.

Use the thin-lens equation with a consistent sign convention

For a thin lens,

1/f = 1/d_o + 1/d_i.

In the common real-is-positive convention used here, a converging lens has positive focal length and a diverging lens has negative focal length. A real object placed on the incoming-light side has positive object distance. A positive image distance describes a real image on the outgoing-light side; a negative image distance describes a virtual image on the same side as the object.

Other sign conventions exist. The physics is unchanged, but mixing conventions within one calculation produces contradictions. Use the convention supplied by the passage or state your own before calculating.

Concept: magnification adds orientation and size

Linear magnification is

m = h_i/h_o = −d_i/d_o.

A negative magnification indicates an inverted image in this convention; a positive magnification indicates an upright image. The magnitude |m| reports relative size. A value of −0.50 means inverted and half as tall as the object. A value of +2 means upright and twice as tall.

Worked application: move the object across the focal point

An original converging-lens model has focal length f = +10 cm.

Case A: object at 30 cm.

1/10 = 1/30 + 1/d_i, so 1/d_i = 2/30 and d_i = +15 cm. Magnification is m = −15/30 = −0.50. The image is real, inverted, and half-size.

Case B: object at 5 cm.

1/10 = 1/5 + 1/d_i, so 1/d_i = −1/10 and d_i = −10 cm. Magnification is m = −(−10)/5 = +2. The image is virtual, upright, and magnified.

A ray sketch should agree with the algebra. For a converging lens, an object inside the focal length produces diverging outgoing rays whose backward extensions meet at a virtual image.

Pattern: lens power is the inverse of focal length

Optical power in diopters is

P = 1/f

when f is measured in meters. A +0.50 m converging lens has power +2.0 D. A −0.25 m diverging lens has power −4.0 D.

A simplified eye model uses the same idea. Myopia focuses distant objects in front of the retina and can be corrected with a diverging lens. Hyperopia places the unaccommodated focus behind the retina and can be corrected with a converging lens. A passage may supply a far point or near point and ask for the lens power required to shift the effective object or image location.

Rule: a second lens uses the first image as its object

For a two-lens system, solve the first lens first. Its image becomes the object for the second lens, with the second object distance determined by the image position relative to lens 2. Do not reuse the original object distance for both lenses.

Total magnification is the product of the individual magnifications when the intermediate image is handled consistently. A passage can therefore separate image-location reasoning from overall size change.

Check your reasoning

Choose an answer before opening its feedback.

Continue learning

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• Interference and diffraction
• Browse Chemistry & Physics MCAT lessons

Sources

1. AAMC. Content Category 4D.
2. OpenStax Physics. Lenses.