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How do converging and diverging lenses form images, and what does the thin-lens equation predict?

Topic 13.4 Images Formed by Lenses: apply the thin-lens equation and magnification to images from converging and diverging lenses.

A focused answer to AP Physics 2 Topic 13.4, covering converging and diverging lenses, the focal length sign convention, the thin-lens equation, the magnification equation, real and virtual images, and ray tracing, with full worked examples.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. Converging and diverging lenses
  3. The thin-lens and magnification equations
  4. Reading the image characteristics
  5. Try this

What this topic is asking

The College Board (Topic 13.4) wants you to apply the thin-lens equation and the magnification equation to converging and diverging lenses, using the sign conventions to determine whether images are real or virtual, upright or inverted, enlarged or reduced.

Converging and diverging lenses

Lenses work by refraction: light bends as it enters and leaves the glass, and the curved surfaces are shaped so a converging lens brings rays together while a diverging lens spreads them apart. The sign of the focal length captures this: positive for converging (real focus), negative for diverging (virtual focus). This is the same focal-length logic as mirrors, just produced by refraction instead of reflection.

The thin-lens and magnification equations

The equations are the same as for mirrors, so the method transfers directly: solve the thin-lens equation for the image distance, read its sign to decide real or virtual, and use the magnification for orientation and size. The one difference to remember is where the image sits: for a lens, a positive image distance means the image is on the opposite side from the object (light passes through and converges there), whereas for a mirror it was on the same side as the object. The sign logic is otherwise unchanged.

Reading the image characteristics

A diverging lens always forms a virtual, upright, reduced image on the same side as the object, regardless of object distance, useful for correcting short-sight. A converging lens depends on the object position: an object beyond the focal length gives a real, inverted image on the far side (the camera and eye case), and an object inside the focal length gives a virtual, upright, enlarged image (the magnifying glass). The strategic role of this topic is that it completes geometric optics by uniting reflection and refraction: lenses use the refraction of Topic 13.3, but obey the same equations and sign conventions as the mirrors of Topic 13.2. Because the mathematics is shared, mastering the mirror equation already gives you the lens equation, and the combined ray-tracing and sign skills explain how every optical instrument, from spectacles to telescopes, forms its images.

Try this

Q1. State the type of image always formed by a diverging lens of a real object. [1 point]

  • Cue. Virtual, upright and reduced.

Q2. A converging lens forms an image with di=+0.20d_i = +0.20 m. State whether the image is real or virtual. [1 point]

  • Cue. Real (positive image distance, on the opposite side of the lens).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). An object is placed 0.400.40 m in front of a converging lens of focal length 0.150.15 m. (a) Use the thin-lens equation to find the image distance. (b) Calculate the magnification and state whether the image is upright or inverted. (c) State whether the image is real or virtual and on which side of the lens it forms.
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A 7-point FRQ on the thin-lens equation.

(a) Image distance (3 points): 1f=1do+1di\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}, so 1di=10.15−10.40=6.67−2.50=4.17\dfrac{1}{d_i} = \dfrac{1}{0.15} - \dfrac{1}{0.40} = 6.67 - 2.50 = 4.17, giving di=0.24d_i = 0.24 m.
(b) Magnification (2 points): m=−dido=−0.240.40=−0.60m = -\dfrac{d_i}{d_o} = -\dfrac{0.24}{0.40} = -0.60. The negative sign means inverted, and it is 0.600.60 times the object's size.
(c) Real or virtual (2 points): the image distance is positive, so the image is real and forms on the opposite side of the lens from the object (where light actually converges).

Markers reward the thin-lens equation, the magnification with the inverted interpretation, and the positive image distance giving a real image on the far side.

AP 2023 (style)1 marksSection I (multiple choice). A diverging (concave) lens forms an image of a real object that is always (A) real and inverted (B) virtual, upright and reduced (C) real and enlarged (D) virtual and enlarged. Justify your reasoning.
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A 1-point MCQ on diverging lenses. The answer is (B).

A diverging lens has a negative focal length and always forms a virtual, upright, reduced image on the same side as the object, for any object distance. The trap is (A): a diverging lens never forms a real image of a real object.

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