Answer to Question #122570 in Optics for Anelisa

Question

An object of height 7cm is placed a distance 25 cm in front of a thin converging lens
of focal length 35 cm. What is the height, location, and nature of the image? Suppose
that the object is moved to a new location a distance 90 cm in front of the lens. What
now is the height, location, and nature of the image?

Answer 1

Let, object height is $h_o=7cm$.

Given, $u=-25cm,f=35cm$

Thus, from lens maker formula

$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$

from, above data,

$$\frac{1}{35}=\frac{1}{v}-\frac{1}{-25}\implies v=-87.5cm$$

since, magnification

$$m=v/u=h_i/h_o$$

Thus, $h_i=24.5cm$ .

Image is magnified and virtual.

If $u=-90cm$ ,then again on applying the above formula we get,

$$\frac{1}{35}=\frac{1}{v}-\frac{1}{-90}\implies v=57.27$$

And,

$$v/u=h_i/h_o\implies h_i=-4.45cm$$

Image is real and diminished.