The minimum distance between an object and it's real image

[eep]

Maybe it's just late, but I'm having an extremely difficult time proving that the minimum distance between an object and it's real image (geometric optics, thin lense equation) is 4f. I can see that it is true, however I'm unsure how to go about proving it mathematically.

 

[gnpatterson]

write an equation for the total distance d1+d2 in term of just d1 then differentiate it wrt d1 find value of d1 for which dif = zero. then find d1+d2

 

[quicknote]

The minimum distance between an object and its real image is indeed 4f, where f is the focal length of the lens. This is known as the thin lens equation, and it can be mathematically derived from the principles of geometric optics.

To prove this, we can use the following steps:

1. Start with the basic equation for a thin lens: 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

2. Since we are looking for the minimum distance between the object and its real image, we can assume that the object is placed at the focal point of the lens, so do = f.

3. This simplifies the equation to 1/f = 1/f + 1/di.

4. Rearranging the equation, we get 1/di = 1/f - 1/f = 0.

5. Solving for di, we get di = 0. This means that the image is formed at the same point as the object, which is known as the focal point.

6. However, in order to have a real image, the object must be placed at a distance greater than the focal length of the lens. Therefore, the minimum distance between the object and its real image is 4f, since 4f - f = 3f is the minimum distance that satisfies the condition of having a real image.

In conclusion, the minimum distance between an object and its real image can be mathematically derived from the thin lens equation and is equal to 4 times the focal length of the lens. This is an important concept in understanding the behavior of light and lenses in geometric optics.