Answer to Question #200387 in Optics for Piolo

Question

A shopper standing 5.00 m from a convex security mirror sees his image with a magnification of 0.350.

(a) Where is his image? _____ m

(b) What is the focal length of the mirror? ______m

(c) What is its radius of curvature? ______ m

nearest 2 decimal places

Answer 1

It is given that the magnification of the mirror is positive, and since the object is real then this implies that the image is virtual, where according to the used sign convention the sign of the distance between the image and the mirror, is negative when the image is virtual.

also, it is given that the mirror is convex which also ensures that the image is virtual, where convex mirror only produces virtual images, also the mirror being convex implies that the focal length will be negative.

Knowing the magnification of the mirror and the distance d_o "distance between the mirror and the object", we can find the distance d_i "distance between the image and the mirror", using the following equation:-

M = - di/do [eq.1]

The sign convention which are use to associate the sign of each term in eq. 1 is as follows

f +ve for concave mirror

-ve for convex mirror

d_i +ve for real image

-ve for virtual image

d_o +ve for real objects

-ve for virtual objects

M +ve for upright image

-ve for inverted image

Given:-

It is given that, the mirror is convex and the magnification is positive, and the object is real this implies the following conditions..

f < 0 R < 0 d_i < 0

Where the image is virtual and upright.

Also it is given that the magnification is 0.350 and the distance between the object and the mirror is 5m, thus

PART [A

The distance between the image and the mirror d_i or the image is :-

substituting the givens in eq.1 to calculate the distance d_i we get

d_i = - M d_o

= - 0.350 x 500

= -175cm or -1.75m

PART [B]

The focal length of the mirror f:-

knowing each of the distance d_i and d_o , we can find the focal length of the mirror which is given by the mirror's equation:-

1/d_i + 1/d_o = 1/f [eq.2]

substituting the givens in eq.2, to find the focal length we get

1/f = 1/500 + 1/-175

1/f = - 13/3500

f = - 3500/13

f = -269.23 cm or -2.69m

Thus the focal length of the mirror is -269.23cm, which is negative as we expected thus the verify answer is true.

PART [C]

The radius of the curvature of the mirror R:

it is required to find the radius of the curvature of the mirror, which is easily can be calculated by knowing the focal length of the mirror which is given by the following:-

R = 2f [eq.3]

substituting the calculated focal length in eq.3 we get

R = 2 x (-269.23)

R = -538.46 cm or -5.38m

Thus, the magnitude of the radius of the curvature of the mirror is 538.46 cm, the negative sign implies that direction of the curvature of the mirror.

(a) Where is his image? _5.3846____ m.

(b) What is the focal length of the mirror? ____2.6923__m

(c) What is its radius of curvature? ___1.75___ m