Question

What is the minimum distance between the object and images in a converging lens to form a real image?Tell in simple language.

Answer 1

Look at the figure. Here, O is point like object and I is it’s image.
u=object distance.
v= image distance.
f= focal length of the lens.
x= distance between object and image.
We have to find the minimum value of x for which we can get the real image.
We know that for real image we will have to have v positive.
The only weapon with us is the formula for a thin lens:
1/v-1/u=1/f………………(1)
Now, from the figure, u=x-v. We substitute this value of u in equation (1), with remembering that according to Cartesian system of sign convention u is negative, we get
1/v+1/(x-v)=1/f. Therefore,
(x-v+v)/[v(x-v)]=1/f. Therefore,
xf/(vx-v^2)=1
Or
xf=vx-v^2. Then,
v^2-vx+xf=0……………….(2).
We have to consider the condition for real root of this quadratic equation in v.
Now, the roots are
v=(1/2)[x+/-(x^2–4xf)^1/2].
For v to be real, x^2>4xf
OR x>4f……………(3).
Thus, for real image by converging lens the minimum distance between object and image should be 4f, where f is focal length of the lens

​    H⃢​    O⃢​    P⃢​    E⃢​     ​ 

   I⃢​    T⃢​     ​ 

   H⃢​    E⃢​    L⃢​    P⃢​    S⃢​     ​

    Y⃢​    O⃢​    U⃢​          ​