Question

1. An object of size 9 cm is placed at 20 cm in front of a concave mirror of focal length
15 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed
image can be obtained? Find the size and the nature of the image.​

Answer 1

Answer :

Position of object = 9cm

Distance of object = 20cm

Focal length = 15cm

Type of mirror : concave

We have to find position of object and height of image$.$

★ Focal length of concave mirror is taken negative and that of convex mirror is taken positive.

Position of image can be measured by using mirror formula.

• 1/u + 1/v = 1/f

u denotes position of object

v denotes position of image

f denotes focal length

By substituting the values, we get

➝ 1/(-20) + 1/v = 1/(-15)

➝ 1/v = -1/15 + 1/20

➝ 1/v = (-4 + 3)/60

➝ v = -60 cm

There for screen should be placed at 60cm in front of mirror in order to get sharp focussed image.

Formula of magnification is given by

➝ m = -v/u = h'/h

• h' = height of image
• h = height of object

➝ -(-60)/(-15) = h'/9

➝ -4 = h'/9

➝ h' = -36 cm

Negative sign indicates that image is real and inverted.

Nature : Real, inverted and enlarged

Answer 2

$\fbox{♠Given}$

• Size= 9 cm

• Distance of object = 28 cm

• Focal length= 15cm

• Mirror: concave mirror

_____________________________________

$\red{focal \: lenght \: of \: concave \: mirror \: is \: taken \: as \: negative}$

Mirror Formula

• 1/u+1/v= 1/f

$\to \: \frac{1}{ - 20} + \frac{1}{v} = \frac{1}{ - 15} \\ \\ \to \frac{1}{v} = \frac{ - 4 + 3}{60} \\ \\ \to \: v = - 60 \: cm$

♠️screen should be placed at 60 CM in front of the mirror to get a sharp focused image.

magnification= -v/u = h'/h

• h'= height of the image
• h= height of the object

$m = \frac{ - ( - 60)}{ - 15} = \frac{h'}{9} \\ h' = - 36 \: cm$

Image:

$\pink{Real , inverted \: and \: enlarged}$