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 | 14th Oct, 2021, 08:57: PM

Expert Answer:

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
 
begin mathsize 14px style 1 over v space plus space fraction numerator 1 over denominator v space x end fraction space equals space 1 over f space end style  or   begin mathsize 14px style 1 over v space open parentheses 1 space plus space 1 over x close parentheses space equals space 1 over f end style..............................(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
 
begin mathsize 14px style fraction numerator 1 over denominator v apostrophe end fraction space open parentheses 1 space plus space fraction numerator 3 over denominator 4 x end fraction close parentheses space equals space 1 over f end style ........................... (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
 
begin mathsize 14px style fraction numerator v apostrophe over denominator v end fraction space equals space fraction numerator 1 plus begin display style fraction numerator 3 over denominator 4 x end fraction end style over denominator 1 plus begin display style 1 over x end style end fraction end style ............................(5)

Distance between Object and screen is not changed . hence we have
 
| u | + v = | u' | + v'
 
( v x ) + v = [ v'  (4x / 3 ) ] + v'
 
begin mathsize 14px style fraction numerator v apostrophe over denominator v end fraction space equals space fraction numerator left parenthesis 1 plus x right parenthesis over denominator open parentheses 1 plus 4 over 3 x close parentheses end fraction end style   ...........................(6)
 
from eqn.(5) and eqn.(6) , we get

begin mathsize 14px style space fraction numerator 1 plus begin display style fraction numerator 3 over denominator 4 x end fraction end style over denominator 1 plus begin display style 1 over x end style end fraction space equals space fraction numerator 1 plus x over denominator 1 plus begin display style 4 over 3 end style x end fraction end style
 
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 | 15th Oct, 2021, 09:16: AM