An object is placed 22.4 cm in front of a converging lens with a focal length of 15 cm. What is the magnification of the image?

(Note: Your answer can be negative or positive)

Question

An object is placed 22.4 cm in front of a converging lens with a focal length of 15 cm. What is the magnification of the image?

(Note: Your answer can be negative or positive)

BlackTom AI · Accepted Answer

We start by restating the problem to keep the setup clear: an object is placed 22.4 cm in front of a converging lens with focal length 15 cm. We are asked for the magnification of the image (sign can be negative or positive).

First, apply the lens equation to find the image distance. Using the standard sign convention for lenses where distances are measured from the lens and taken as positive if they are to the right (and positive object distance is measured to the left, so the object distance u is treated as +22.4 cm in many common textbook conventions), the thin lens formula is 1/f = 1/v + 1/u. Substituting f = +15 cm and u = +22.4 cm gives:
1/v = 1/f − 1/u = (1/15) − (1/22.4).
Evaluating these fractions numerically: 1/15 ≈ 0.0666667 and 1/22.4 ≈ 0.0446429, so 1/v ≈ 0.0220238, which yields v ≈ 45.4 cm. This means the image forms 45.4 cm to the right of the lens, i.e., a real image.

Next, compute the magnification. The linear magnification m for a thin lens is m = − v / u, where u is the object distance and v is the image distance, both taken with the same sign convention as above. With u = 22.4 cm and v ≈ 45.4 cm, we get m ≈ −(45.4) / (22.4) ≈ −2.03.

Interpreting the result: the magnitude of the magnification is about 2.03, and the negative sign indicates the image is inverted relative to the object, which is expected for a real image formed by a converging lens when the object is placed beyond the focal length.

If you compare, using a slightly different but equivalent sign convention (taking u as negative for objects on the left) would yield the same physical outcome: a magnification with a negative sign, approximately −2.03, consistent with a real, inverted image.

Thus, the calculated magnification is approximately −2.03. The negative sign communicates inversion; the magnitude 2.03 tells you the image is about twice as large as the object.

Physics 1 Hon - Bowles - p06/p07 P10.2/10.22 Optics Formula Practice

An object is placed 22.4 cm in front of a converging lens with a focal length of 15 cm. What is the magnification of the image? (Note: Your answer can be negative or positive)

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