Geometrical OpticsHard
Question
A convex lens forms an image of an object on a screen . The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. Then high of this image is 4 cm. The distance between the object and the screen is 90cm.
Options
A.The distance between the two positions of the lens is 30 cm
B.The distance of the object from the its first position is 36 cm
C.The height of the object is 6 cm
D.The focal length of the lens 21.6 cm
Solution
Height of object
=
Height of object = 6 cm
m1 =
u2 ≡ v1 =
v1 ∴ X = 
u1 - u1 = 
= X
D = 90 = u2 + u1 =
u1 + u1
u1 = 36 ⇒ X =
= 18
Distance between two position of lens = X = 18
Focal length f =
= 21.6 cm
=
Height of object = 6 cm
m1 =
u2 ≡ v1 =
v1 ∴ X = u1 - u1 = = XD = 90 = u2 + u1 =
u1 + u1u1 = 36 ⇒ X =
= 18Distance between two position of lens = X = 18
Focal length f =
= 21.6 cmCreate a free account to view solution
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