Question

A concave mirror forms on a screen a real image of thrice the linear dimensions of the object. Object and screen are moved until the image is twice the size of the object. If the shift of the object is 6 cm then find the shift of the screen and the focal length of the mirror. 

a)36 cm, 54cm, 

b)54 cm, 36 cm, 

c)36 cm, 36 cm, 

d)None

Answer 1

Initial magnification = 3 

i.e. v₁/u₁ = 3 

v₁ = 3u₁ 

f = u₁v₁/(u₁+v₁) 

f= u₁(3u₁)/(u₁+3u₁) = (3/4)u₁ ....................(1) 

In the second case, m = 2 

i.e. v₂/u₂ = 2 

v₂ = 2u₂ 

f = u₂v₂/(u₂+v₂) 

f= u₂(2u₂)/(u₂+2u₂) = (2/3)u₂ ..................(2) 

From eq. (1) and (2) focal length is same 

(3/4)u₁ = (2/3)u₂ 

or u₂ = (9/8)u₁ ...............(3) 

It is given that the shift of the object = 6 cm= u₂- u₁ 

∴ [(9/8)u₁]–u₁ = 6 

[(9/8)–1]u₁ = 6 (1/8) 

u₁ = 6 u₁ = 6*8 = 48 cm 

and v₁ = 3 u₁ =3*48 = 144 cm 

substituting the value of u₁ and v₁ in equation (1) 

f= (3/4)u₁ 

f= (3/4)48 = 36 cm 

From equation (3) 

u₂= (9/8)u₁ 

u₂= (9/8)48 = 54 cm 

v₂= 2u₂ = 2*54 = 108 cm 

Thus, shift of the screen = v₁-v₂ 

v₁-v₂ = 144-108 = 36 cm