CBSE Class 10 Answered
image 1 cm long of object is formed on a screen by convex lens keeping the object and the screen is the lens is moved until a s image is formed on the screen if the image of 0.75 cm long what is the lenght of the object
Asked by udaypratapyadav0001 | 14 Oct, 2021, 20:57: PM
We have lens equation : - ( 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 of convex lens .
magnification m of image, m = (v / u ) = I / O ....................................(2)
where I is image size and O be object size
In above equations, we need to follow the cartesian sign convention . As per cartesian sign convention v is positive , u is negative and f is positive .
Since we get inverted image I is negative and O is positive
Let x be the object size and we are given that initially image size is 1 cm .
m = ( v / u ) = -1/x , hence we get u = - ( v x )
If we use the above substitution u = - ( v x ) , we write eqn.(1) as


By moving the lens , image-to-lens distance iand object-to-lens distance are changed as v' and u' respectively .
New image size is (3/4) cm
Lens equation after changing the distances is given as
( 1 / v' ) - ( 1 / u' ) = ( 1 / f ) ......................(4)
New magnification m = ( v' / u' ) = [ -3 / (4x) ] , hence u' = (-4/3) v' x
If we substitute u' = (-4/3) v' x in eqn.(4) , we get after simplification as

Since same lens is used to get both images , i.e. focal length f is same in eqn.(3) and (4) .
Hence by equating eqn.(3) and eqn.(4) , we get

Distance between Object and screen is not changed . hence we have
| u | + v = | u' | + v'
( v x ) + v = [ v' (4x / 3 ) ] + v'

from eqn.(5) and eqn.(6) , we get

Above expression is equation in one variable x . By solving above equation we get , x = 1.15 cm
hence size of object is 1.15 cm
Answered by Thiyagarajan K | 15 Oct, 2021, 09:16: AM
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