Question

An object 3 cm high is placed at a distance of 10 cm in front of a concave mirror of focal length 20 cm Find the
position, nature and size of the image formed.​

Answer 1

use the mirror formula

1/v + 1/u = 1/f

=> 1/v= 1/-20 - ( 1/-10)

=>1/ v = 1/20

=> V = 20cm

position of image is 20 cm ..

nature = virtual and erect...

therefore magnification = -(v)/u = -20/-10 = 2

hence also magnification = size of image / size of objects = h1/h2 = 2 => h1/3 =2

h1 = 6cm

so size of image is 6cm

$<marquee>....I hope it will help you...$

Answer 2

ANSWER:

CHARACTERISTICS OF IMAGE FORMED:

• VIRTUAL
• ERECT
• MAGNIFIED

HEIGHT OF IMAGE

• 6 cm

GIVEN:

Height of object = 3 cm

Object distance = (-10) cm

Focal length = (-20) cm

TO FIND:

position, nature and size of the image formed.

SOLUTION:

By using Mirror formula:

$\implies \: \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \implies \: \frac{1}{v} = \frac{ - 1}{20} + \frac{1}{10} \\ \implies \frac{1}{v} = \frac{ - 1 + 2}{20} \\ \implies \frac{1}{v} = \frac{1}{20} \: cm \\ \implies \: v = 20 \: cm$

Now finding Magnification:

$m = \frac{ - v}{u} \\ m = \frac{ - 20}{ - 10} \\ m = 2$

Now height of image

$\implies \frac{h(i)}{h(o)} = m \\ \\ \implies \: h(i) = 2 \times 3 \\ \implies \: h(i) = 6 \: cm$

CHARACTERISTICS OF IMAGE FORMED:

• VIRTUAL
• ERECT
• MAGNIFIED

HEIGHT OF IMAGE

• 6 cm