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)

查看解析

登录即可查看完整答案

类似问题

Determine the magnification for an object placed 25.9 cm from a convex lens having a focal length of 12.2 cm.

In a consumer society, many adults channel creativity into buying things

Economic stress and unpredictable times have resulted in a booming industry for self-help products

People born without creativity never can develop it

A product has a selling price of $20, a contribution margin ratio of 40% and fixed cost of $120,000. To make a profit of $30,000. The number of units that must be sold is: Type the number without $ and a comma. Eg: 20000

Which of the following statement regarding cost is correct:

Under the assumptions used in cost-volume-profit analysis, as the activity increases:

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单