Question

A student conducts an activity using a flask of height 15 cm and a concave mirror. Hefinds that the image formed is 45 cm in height. What is the magnification of the image? ​

Answer 1

Given: Height of the flask $=15cm$

Height of the image that is formed $=45cm$

To find: The magnification of the image

Solution:

The height of the flask = $15cm$ = The height of the object

The height of the image of the flask that is formed $=45cm$

Now, the question says that the student conducts the activity using a concave mirror.

For a concave mirror, the image height will be positive hence, the overall magnification will be negative.

The formula of magnification is given by,

$m=-\frac{h_i}{h_o}$

where, $h_i$ is the height of the image and $h_o$ is the height of the object.

Putting the values,

$m=-\frac{45}{13}$

⇒ $m=-3$

Since magnification has no unit, therefore, it is written as only $-3$

Final answer:

The magnification of the image will be -3.

Answer 2

Answer:

given

v=-15cm

u=45cm

m=-v/u

m=-45/-15

m=3