Request a call back

A screen bearing a real image of magnification m1,formed by a convex lens,is moved by a distance x.The object is then moved until a new image of magnification m2 is formed on screen.The focal lenght of lens is: (1) . x/m2-m1 (2) . m2-m1/x (3) . x/m1-m2 (4) . m1-m2/x
Asked by dfojajjar | 16 Feb, 2020, 01:35: PM
We have lens formula :-  (1/v) - (1/u) = 1/f  ..............(1)

where v is lens-to-image distance, u is lens-to-object distance and f is focal length

By multiplying eqn.(1) with v, we get ,  1 - (v/u) = (v/f)

since magnification m of the image is given by, m1 = (v/u),

then we get from above equation, 1 - m1 = (v/f)  ...................(2)

if image-to-lens distance is changed to v+x, then magnification becomes m2

Equation (2) with new magnification m2 is written as ,  1 - m2 = (v+x)/f  = (v/f) + (x/f)  ...........(3)

Substituting for (v/f) from eqn.(2) in eqn.(3), we get,  1- m2 = 1 - m1 + (x/f)

Hence we get ,  (x/f) = [ m1 - m2 ]  ,   hence  focal length f = x / [ m1 - m2 ]

Answered by Thiyagarajan K | 16 Feb, 2020, 05:06: PM

Concept Videos

CBSE 10 - Physics
Asked by ruchipandey20071978 | 11 May, 2023, 06:53: AM
CBSE 10 - Physics
Asked by priyanshdhiman2008 | 17 Mar, 2023, 09:17: AM
CBSE 10 - Physics
Asked by piustoppo9thb | 24 Jun, 2022, 09:58: PM
CBSE 10 - Physics
Asked by priyankakiran2017 | 22 Jun, 2022, 07:07: PM
CBSE 10 - Physics
Asked by topperlearningforcontent | 02 Feb, 2022, 03:32: PM
CBSE 10 - Physics
Asked by bibektripathy17 | 15 Jun, 2021, 09:13: AM