Question

If a lense draw a reflection on a curtain of the same size of a substance.Then show that the distance between the lense and the curtain is f(M+1)

Answer 1

Answer:

Explanation: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.