Chapter 23, Problem 22Q

To determine

To Explain:

If the diverging lens form a real image under any circumstances

Answer to Problem 22Q

Solution:

Diverging lens, if obtains a virtual object by another lens or lens system, can create a real image.

Explanation of Solution

Diverging lenses always produce an upright virtual image for any real object, no matter where the object is.

A lens combination with the first one being a converging lens and the second being a diverging lens can be used to obtain the virtual image. The combination is chosen such that the image formed by the conversing lens is virtual and it becomes the object for the diverging lens.

As the virtual image of the converging lens becomes the object for the diverging lens, the object of the diverging lens is the virtual object.

The virtual object for the diverging lens, whose focal length is negative, can produce a real image formed behind the diverging lens.

This can be obtained from the lens equation for the diverging lens.

$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$

Where,

d₀ is the object distance for the diverging lens, and is negative

d_i is the image distance, which is to be found.

f is the focal length of the diverging lens, which is negative.

$d_o=-d_o$ (The object is the virtual object, an image formed by the converging lens)

$f=-f$ (The focal length is negative for the diverging lens).

Then, from the equation

$\frac{1}{-d_o}+\frac{1}{d_i}=-\frac{1}{f}$

$\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}$

The image distance is positive, which is formed behind the diverging lens. The image produced is the real image.

Thus, the diverging lens, if it can obtain a virtual object by another lens or lens system, can create a real image.