Chapter 23, Problem 25Q

To determine

(a) To determine:

If a thin converging lens is moved closer to a nearby object, the position of the real image formed, is changed.

Answer to Problem 25Q

Solution:

Yes, it is changed. As the thin converging lens is moved closer to a nearby object, the position of the real image formed, is changed.

Explanation of Solution

The given statement can be discussed using the thin lens equation.

The Thin lens equation is given by

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

Where,

d_i is the image distance

d_o is the object distance

f is the focal length.

As given in the statement, if the object distance is reduced by moving the lens closer to the object, $\frac{1}{d_o}$ will be large.

In order to comply with the above equation, $\frac{1}{d_i}$ decreases, this means d_i increases, the image distance increases. The image position is moved farther from the lens.

To determine

(b) To determine:

If a thin converging lens is moved closer to a nearby object, the size of the real image formed, is changed.

Answer to Problem 25Q

Solution:

Yes, it is changed. As the thin converging lens is moved closer to a nearby object, the size of the real image formed, is changed.

Explanation of Solution

The size of the image formed if the object position changes can be explained by the magnification equation.

$m=-\frac{d_i}{d_o}$

Where,

m is the magnification

d_i is the image distance

d_o is the object distance

As the lens is moved towards the object, the object distance is lowered and as discussed above the image distance increases.

If d_i is increased and d_o is lowered and from the above equation, the magnification becomes larger.

If the magnification is larger, the size of the image gets larger.