Question

A spherical, concave shaving mirror has a radius of curvature of 0.983 m. What is the magnification of a person's face when it is 0.155 m from the vertex of the mirror (answer sign and magnitude)

Answer 1

Answer:

The magnification is 1.5.

Explanation:

radius of curvature, R = - 0.983 m

distance of object, u = - 0.155 m

Let the distance of image is v.

focal length, f = R/2 = - 0.492 m

Use the mirror equation

$\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m$

The magnification is given by

m = - v/u

m = 0.226/0.155

m = 1.5

Answer 2

Answer:

Magnification = 1

Explanation:

given data

radius of curvature r = - 0.983 m

image distance u = - 0.155

solution

we get here first focal length that is

Focal length, f = R/2     ...................1

f = -0.4915 m

we use here formula that is

$\frac{1}{v} + \frac{1}{u} + \frac{1}{f}$      .................2

put here value and we get

$\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}$  

v = 0.155 m

so

Magnification will be here as

m = $- \frac{v}{u}$

m =  $\frac{0.155}{0.155}$

m = 1