Question

A converging lens forms a virtual image of a real object that is two times the objects size. The converging lens is replaced with a diverging lens having the same size focal length. What is the magnification of the image formed by the diverging lens?

A) –1 

B) –2/5 

C) 2/3 

D) 3/2 

E) 5/2

Answer 1

Option (c) is correct.

Explanation

The magnification of diverging lens for real object is always positive but less than one. The option (c) supports this.

It may be confirmed by calculation using lens formula as follows

Lens formula

I/image dist -1/obj dist= 1/focal length

For converging lens

Object distance =-u

Image distance =-2u as the size of the virtual image is double the size of the object

Hence applying lens formula we get

-1/2u+1/u=1/f, where f is the focal length.

=> f= 2u

Now for diverging lens

Object distance = -u

Image distance = v'

Focal length =-f (concave lens)

So 1/v'+1/u=-1/f=-1/(2u)

=>1/v'=-3/(2u)

=>v' = -(2u)/3

Hence magnification m = image distance/object distance

= (-2u/3)/-u=2/3

Answer 2

The correct option is: (C) 2/3 

Explanation:

The magnification is M = 2. Using M = – d_i / d_o … d_i = – 2d_o Lets assume a value of d_o = 10, then d_i = – 20, and from 1/f = 1/d_o + 1/d_i, the focal point is 20. Now redo the math with the focal point for the diverging lens being negative and the new d_i = –6.67, giving a new M = 0.67