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. The height of this image is 4 cm. The distance between the object and the screen is 90 cm.

• A. The distance between the two positon of the lens is 30 cm
• B. The distance of the object from the lens in its first position is 36 cm
• C. The height of the object is 6 cm
• D. The focal length of the lens is 21.6 cm

Answer 1

Answer: (B), (C), (D)

The distance of the object from the lens in its first position is 36 cm
The height of the object is 6 cm
The focal length of the lens is 21.6 cm

$u+v=90$

$h_{object}^2=h_{image1}\times h_{image2}=9\times 6=36$

$h_{object}=6$

for the first image $\frac{v}{u}=\frac{9}{6}=1.5$

$v=1.5u$

$u+v=90$

$2.5u=90$

$u=36$ ; $v=54$

distance of the object from the lens in its first position is 36 cm

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{54}+\frac{1}{36}=\frac{1}{f}$

$f=+21.6$

option $B,C,D$ are correct