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, 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
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
Application Videos
-
Power and nature of a lens
This video explains the relation between the focal length and power of lens... -
Combination of lenses
This video explains the concept of finding the focal length of combination ... -
Power and Nature of Lens
This video explains the power of lens, its SI unit and shows how to iden...
Concept Videos
-
Refraction of Light
Define convex and concave lenses, optical centre, principle axis, centre o... -
Refraction of Light
Define convex and concave lenses, optical centre, principle axis, centre o... -
Refraction of Light
Define convex and concave lenses, optical centre, principle axis, centre o... -
Spherical lens and its power
This video describes the way to solve problems based on lens formula to cal...
CBSE 10 - Physics
Asked by udaypratapyadav0001 | 14 Oct, 2021, 08:57: PM
ANSWERED BY EXPERTCBSE 10 - Physics
Asked by akshata.shukla1910 | 13 Aug, 2021, 12:43: PM
ANSWERED BY EXPERTCBSE 10 - Physics
why are telescopes launched in space ? explain in brife about the hubble telescope launched in space
Asked by yashsingh2443 | 01 Aug, 2021, 06:53: PM
ANSWERED BY EXPERTCBSE 10 - Physics
Asked by chandel9807 | 10 Mar, 2021, 11:09: PM
ANSWERED BY EXPERTCBSE 10 - Physics
Asked by shivendralodhi194194 | 26 Jan, 2021, 01:45: PM
ANSWERED BY EXPERTCBSE 10 - Physics
Asked by rathodveeru41 | 19 Dec, 2020, 10:33: AM
ANSWERED BY EXPERTCBSE 10 - Physics
Asked by rathodveeru41 | 18 Dec, 2020, 10:02: PM
ANSWERED BY EXPERTCBSE 10 - Physics
Asked by abdibnzubair | 06 Oct, 2020, 12:52: PM
ANSWERED BY EXPERTCBSE 10 - Physics
Asked by Omkar | 17 Sep, 2020, 06:56: PM
ANSWERED BY EXPERTCBSE 10 - Physics
Asked by adipadmakarri | 19 Aug, 2020, 09:01: AM
ANSWERED BY EXPERT
