Question

A 1.8 -m-tall girl stands $2.4 \mathrm{m}$ from a store's security mirror. Her image appears to be $0.36 \mathrm{m}$ tall. What is the focal length of the mirror?

   A 1.8 -m-tall girl stands $2.4 \mathrm{m}$ from a store's security mirror. Her image appears to be $0.36 \mathrm{m}$ tall. What is the focal length of the mirror?

Physics Principle and Problems

Physics Principle and Problems

Paul W. Zitzewitz,… 1st Edition

Chapter 17, Problem 21

Instant Answer

Solved by Expert Mayukh Banik

05/25/2020

Step 1

The magnification is given by the ratio of the image height to the object height. In this case, the image height is 0.36 m and the object height is 1.8 m. So, the magnification is given by: \[m = \frac{h_i}{h_o} = \frac{0.36}{1.8} = 0.2\] Show more…

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A 1.8 -m-tall girl stands $2.4 \mathrm{m}$ from a store's security mirror. Her image appears to be $0.36 \mathrm{m}$ tall. What is the focal length of the mirror?

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Key Concepts

Mirror Equation

The mirror equation, given by 1/f = 1/d? + 1/d?, relates the focal length of a mirror to the distances of the object and its image from the mirror. This fundamental formula is used in geometrical optics to solve for unknown quantities when two are known and is applicable to both concave and convex mirrors under appropriate sign conventions.

Magnification

Magnification in mirror optics is defined as the ratio of the image height to the object height and is also related to the distances by m = -d?/d?. This concept helps in determining whether the image is magnified or reduced, as well as whether it is upright or inverted relative to the object.

Sign Conventions

Sign conventions in mirror optics are crucial for correctly applying equations like the mirror equation and the magnification formula. They determine the positive and negative directions for distances, focal lengths, and magnification, and are essential for distinguishing between real and virtual images as well as between different types of mirrors (concave or convex).

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