Show that the least possible distance between an object and its real image in aconvex lens is 4f, where f is the focal length of the lens

Sure, here is the step by step solution:

1. The lens formula for a convex lens is given by 1/v - 1/u = 1/f, where v is the image distance, u is the object distance, and f is the focal length of the lens.

2. We are looking for the minimum value of the distance between the object and its real image, which is |v - u|.

3. To find the minimum value of a function, we can take its derivative and set it equal to zero. However, in this case, it's easier to complete the square.

4. Rewrite the lens formula as 1/v = 1/f + 1/u.

5. Then, rewrite it again as v = 1/(1/f + 1/u).

6. Now, express u in terms of v and f: u = fv/(v - f).

7. Substitute u back into the expression for |v - u| to get |v - fv/(v - f)|.

8. Simplify this to get |(v^2 - 2fv + f^2)/(v - f)| = |(v - f)^2/(v - f)| = |v - f|.

9. The minimum value of |v - u| = |v - f| occurs when v = 2f.

10. Substitute v = 2f back into the lens formula to get 1/2f - 1/u = 1/f, which simplifies to u = 2f.

11. Therefore, the minimum distance between the object and its real image is |2f - 2f| = 4f.

So, the least possible distance between an object and its real image in a convex lens is 4f, where f is the focal length of the lens.