Question

0). An object 3cm high is placed at a distance of 8cm from
a concave mirror Which produces a virtual image 4.5 cm
high:
1)What is the focal length of the mirror?
(2)what is the position of image?
1) What is the positions of image?​

Answer 1

1. Question⇒  An object 3cm high is placed at a distance of 8cm from  a concave mirror Which produces a virtual image 4.5 cm high:  1)What is the focal length of the mirror? (2)what is the position of image?  1) What is the positions of image?​

Explanation:

• Hey!
• ___________________________
• Height of the object (h) = 3 cm
• Distance of object from concave mirror (u) = -8 cm
• Height of the virtual image formed (h') = 4.5
• Let's find 'v' first =
• - v/u = h'/h
• v = - u × h'/h
• v = - ( -8) × 4.5 / 3
• v = 12 cm
• ________
• (i) Focal length of mirror = ?
• 1/f = 1/v + 1/u
• 1/f = 1/12 + 1/ -8
• 1/f = -1/24
• 1 × 24 = -f
• 24 = -f
• f = -24 cm
• _________
• (ii) Position of image (v) = 12 cm
• _________
• (iii) Ray diagram - Refer pic
• (Image is vertical , thus erect)
• ___________________________
• Hope it helps...!!!

Answer 2

Answer:(1)-24cm

(2)+12

(3)virtual and erect

Explanation:As

Magnification of the object is 1.5

M=-v/u =H/h

Also u=8

So v=12

Then using mirror formula

1/v+1/u=1/f

Please be remember the sign convection 'u' is negative

Vis positive as given virtual image in question