Sign Conventions & Ray-Diagram Discipline
- Apply the sign conventions for focal length, object and image distance
- Distinguish a real image from a virtual image on a diagram and in the arithmetic
- Draw the three principal rays for a lens or mirror
One equation, and the signs do the work
The thin-lens and mirror equation is the same in both cases: 1/f = 1/d_o + 1/d_i. The physics that distinguishes a converging lens from a diverging one, or a real image from a virtual one, lives entirely in the signs. Get the conventions right and the equation handles every case; get them wrong and the arithmetic is impeccable and the answer is meaningless.
Real versus virtual, physically
A real image is formed where light rays actually converge, so it can be caught on a screen — a projector or a camera sensor relies on this. A virtual image is where the rays only appear to come from when extrapolated backward; nothing converges there, and a screen placed at that location shows nothing. Your reflection in a flat mirror is virtual, which is why you cannot project it. The arithmetic signals which you have: d_i negative means virtual, with no exceptions.
The three principal rays
For a converging lens: a ray parallel to the axis refracts through the far focal point; a ray through the near focal point emerges parallel; and a ray through the center goes straight on. Any two locate the image. For a concave mirror the analogues are: parallel in, reflects through the focal point; through the focal point, reflects out parallel; and through the center of curvature, reflects straight back. Drawing two rays correctly earns the diagram marks even without arithmetic.
What each device can and cannot do
A diverging lens or convex mirror always produces a virtual, upright, reduced image, whatever the object distance — no exceptions, which makes those questions quick. A converging lens or concave mirror depends on where the object sits: beyond f gives a real inverted image, inside f gives a virtual upright enlarged one (this is a magnifying glass), and at f gives no image at all, since the rays emerge parallel.
An object sits 15 cm from a converging lens of focal length 10 cm. Find the image distance and magnification, and describe the image.
- 1.1/d_i = 1/f − 1/d_o = 1/10 − 1/15.
- 2.Common denominator: 3/30 − 2/30 = 1/30, so d_i = +30 cm.
- 3.Positive d_i means the image is REAL, on the opposite side of the lens.
- 4.Magnification: m = −d_i/d_o = −30/15 = −2.0.
- 5.Negative m means inverted; magnitude 2.0 means twice the object size.
When solving 1/f = 1/d_o + 1/d_i, remember the final reciprocal. Computing 1/d_i = 1/30 and reporting d_i = 1/30 cm rather than 30 cm is a common slip, and the implausible magnitude is the clue.
A calculation gives an image distance of −12 cm. The image is:
A diverging lens forms an image of a real object. The image is always:
An object is placed exactly at the focal point of a converging lens. The result is:
State the sign convention you are using at the top of an optics free response. Rubrics accept either standard convention consistently applied, but they cannot award marks for a d_i whose sign meaning is unstated.
Answer the 3 checkpoints as you read.
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