A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20cm. If the distance of the object is 30 cm from the lens. Find the position, nature, and size of the image and also find its magnification.
Answers
Mirror formula :
1/u + 1/v = 1/f
1/20 - 1/30 = 1/60
Position of the image is 60cm.
Magnification = image distance ÷ object distance
M = 60/30 = 2
Magnification = image height ÷ height of the object
2 × 4 = 8cm
Height of the object = 8cm
Answer:
Given - f=+24cm,u=−16cm ,
As ∣u∣(16cm)<∣f∣(24cm) ,it means object lies between F and C , in this position of object the rays from object cannot meet at any point on the other side of lens ,which is also clear from the position of image found below ,
From lens equation v=
u+f
uf
=
−16+24
−16×24
=−48cm ,
-ive sign shows that image will be formed on the same side of lens , where the object is placed .
Now linear magnification m=I/O=v/u=−48/−16=+3 ,
or I=3×4=12cm (given O=4cm) ,
since m=+ive and m>1 ,therefore image will be virtual ,erect and magnified .