Show that 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

General
1

show that 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.

2

Show that 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

Answer:
In optics, when dealing with convex lenses, there is a relationship between the object distance (u), image distance (v), and the focal length (f) of the lens which is defined by the lens formula:

1/f = 1/v + 1/u

We need to prove that the least possible distance between the object and its real image in a convex lens is 4f. And to find this, we start by considering the case when the object is placed at a distance beyond 2f from the lens.

Let’s assume the object distance (u) is taken as 2f since we are looking for the minimum distance where the virtual image becomes real. Substituting this into the lens formula gives:

1/f = 1/v + 1/(2f)

By simplifying the equation, we get:

1/v = 1/f - 1/(2f)
1/v = 2/(2f) - 1/(2f)
1/v = 1/(2f)

From this calculation, we see that the image distance (v) is equal to 2f. This case refers to the situation when the object is placed at 2f, and the image is formed at 2f on the other side of the lens.

Now, for the least possible distance between the object and its real image, the image formed is at the same distance as the focal length away from the lens on the opposite side. Therefore, the image distance (v) is 2f away from the lens. And considering the total distance the light travels from the object to the image and back, we have 2f + 2f = 4f.

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