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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
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

or   ..............................(3)

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

........................... (4)

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

............................(5)

Distance between Object and screen is not changed . hence we have

| u | + v = | u' | + v'

( v x ) + v = [ v'  (4x / 3 ) ] + v'

...........................(6)

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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