An object of 5 mm height is placed at a distance of 15 cm from a convex lens of focal 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the object. Find
(a) the position of the final image.
(b) its nature and (c) its size.

Explanation

Give: `h_(1) = 5 mm = 0.5 cm, u_(1) = - 15 cm f_(1) = 10 cm f_(2) = 5 cm, d = 40 cm`
For the first lens, the lens equation is , `(1)/(v_(1))-(1)/(u_(1))=(1)/(f_(1))`
Substituting the values, `((1)/(v_(1))-(1)/15)=(1)/(10):(1)/(v_(1))+(1)/(15)=(1)/(10)`
`(1)/(v_(1))=(1)/(10)-(1)/(15)=(15-10)/(150)=(5)/(150)=(1)/(30)`
`v_(1) = 30 cm`
First lens from image 30 cm to the right of first lens.
Let us find the heigh of this image.
Equation for magnification is, ` m = (h_(2)) = (v)/(u)`
Substituting the values, `(h_(2))/(0.5)=(30)/(-15)`
`h_(2)=0.5xx(30)/(-15)=-1cm`
As the height fo the lens is negative, the image is inverted, real image.
Object is at 10 cm to the left of the second lens (40 - 30 = 10), Hence, `u_(2) = - 10 cm`
For the second, the lens equation is, `(1)/(v_(2))-(1)/(u_(2))=(1)/(f_(2))`
Substituting the values, `(1)/(v_(2))-(1)/(-10)=(1)/(5),(1)/(v_(2))+(1)/(10)=(1)/(5)`
`(1)/(v_(2))=(1)/(5)-(1)/(10)=(10-5)/(50)=(5)/(50)=(1)/(10)`
`v_(2) = 10 cm`
The image is formed 10 cm to the right of the second lens.
Let us find the height of the final image. Assume, the final heigh of the image formed by the second lens is `h_(2)` and the height of the object for the second lens `h_(1)` is the image height of the first lens `h_(1) = h_(2)`
Equation for magnification is, `m = (h_(2))/(h_(1))=(v_(2))/(u_(2))`
Substituting the values, `(h_(2))/(-1) = (10)/(-10)`
`h_(2)=(-1)((10)/(-10))=1cm=10cm`
As the heigh of the image is positive, the image is erect, and it is real.