Question

A convex mirror of focal length 33 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror.If the height of the image is 7.0 cm,where is the object located,and how tall is it? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.

Answer 1

Answer:

Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42

The image is a virtual and upright image.

Explanation:

The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.

The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.

Given f = -33.0cm, v = -19.0cm

using thr mirror formula to get the object distance u, we have;

$\frac{1}{f}=\frac{1}{u} + \frac{1}{v}\\ \frac{1}{u}=\frac{1}{f} - \frac{1}{v}\\\frac{1}{u}=\frac{1}{-33} - \frac{1}{-19}\\\frac{1}{u}=\frac{-19+33}{627} \\\frac{1}{u}=\frac{14}{627} \\u=\frac{627}{14} \\u = 44.8cm$

To calculate the image height, we will use the magnification formula

M = $\frac{image\ height}{object\ height}=\frac{image\ distance}{object\ distance}$

M = $\frac{Hi}{HI}=\frac{v}{u}$

Given Hi = 7.0cm

v = 19.0cm

u = 44.8cm

HI = 7*44.8/19

HI = 16.5cm

The object height is 16.5cm

Magnification = v/u = 19.0/44.8 = 0.42

SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image