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Chapter 2: Problem 36

A shopper standing \(3.00 \mathrm{m}\) from a convex security mirror sees his image with a magnification of 0.250 . (a) Where is his image? (b) What is the focal length of the mirror? (c) What is its radius of curvature?

Short Answer

Expert verified
(a) The image distance (v) is -0.750 m. (b) The focal length (f) of the mirror is -0.333 m. (c) The radius of curvature (R) of the mirror is -0.666 m.

Step by step solution

01

Find the Image Distance (v) Using the Magnification Formula

We are given that the magnification is 0.250, and the object distance (u) is 3.00 m. We can plug in the given values into the magnification formula: \( M = -\frac{v}{u} \) \(0.250 = -\frac{v}{3.00}\) Now, we can solve for the image distance (v).
02

Solve for Image Distance (v)

To solve for v, we can multiply both sides of the equation by -3.00: \(-0.750 = v\) So, the image distance (v) is -0.750 m (negative sign indicates that the image is virtual and on the opposite side of the mirror).
03

Use the Mirror Formula to Find the Focal Length (f)

Now we have both u and v, and we can use the mirror formula to find the focal length (f): \(1/f = 1/u + 1/v\) Plug in the values of u and v we found: \(1/f = 1/3.00 + 1/(-0.750)\)
04

Solve for Focal Length (f)

To find f, we first calculate the sum of the fractions on the right side of the equation, then take the reciprocal of that number: \(1/f = \frac{1}{3.00} - \frac{1}{0.750} = \frac{1}{3.00} - \frac{4}{3.00} = -\frac{3}{3.00}\) Now, take the reciprocal to find f: \(f = \frac{-1}{3}\) So, the focal length of the mirror is -0.333 m (negative sign indicates that the mirror is convex and the focus is behind the mirror).
05

Find the Radius of Curvature (R)

For a convex mirror, the radius of curvature R is related to the focal length by the equation: \(R = 2f\) Now, plug in the value of f we found: \(R = 2(-0.333)\) R = -0.666 m So, the radius of curvature of the mirror is -0.666 m (negative sign indicates that the mirror is convex). In summary, we found the following values: (a) Image distance (v) = -0.750 m (b) Focal length (f) = -0.333 m (c) Radius of curvature (R) = -0.666 m

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Most popular questions from this chapter

If the comea is to be reshaped (this can be done surgically or with contact lenses) to correct myopia, should its curvature be made greater or smaller? Explain.

Unless otherwise stated, the lens-to-retina distance is 2.00 Calculate the power of the eye when viewing an object \(3.00 \mathrm{m}\) away.

The power for normal distant vision is 50.0 D. A severely myopic patient has a far point of \(5.00 \mathrm{cm}\). By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him?

An object is located in water \(15 \mathrm{cm}\) from the vertex of a concave surface made of glass with a radius of curvature 10 \(\mathrm{cm} .\) Where does the image by refraction form and what is its magnification? Use \(n_{\text {water }}=4 / 3\) and \(n_{\text {glaxs }}=1.5\)

Derive the following equation for a convex mirror: $$\frac{1}{V O}-\frac{1}{V I}=-\frac{1}{V F}$$ $$\frac{1}{V O}-\frac{1}{V I}=-\frac{1}{V F}$$
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