Answer to Question #110002 in Optics for Nya

Question

A 0.31 m tall object is placed 0.45 m from a converging lens with a 0.04 m focal length. How tall is the image?

Answer 1

Optics

We need to find height of the image.

Solution:

Given,

Height of the Object = $h =0.31 m =0.31 \times 100 = 31 \space cm$

Focal length =$f = -20.0 \space cm$

Distance of Object from a converging lens = $p =0.45 \space m = 0.45 \times 100 = 45 cm$

The form of lens equation is

$$\frac {1}{f} =\frac {1}{p} + \frac {1}{q}$$

$$\frac { -1}{20} =\frac {1}{45} + \frac {1}{q}$$

$$\frac {1}{q} = \frac {-1}{20} - \frac {-1}{45} = \frac {-9+4}{180} =\frac {-5}{180} = \frac {-1}{36}$$

$$q = -36$$

Now, the magnification is

$$m = \frac {-q}{p} =\frac {-(-36)}{45} = 0.8$$

Let the height of the image is H, then the formula with m, h and H is

$$m = \frac {-H}{h}$$

This can write as, $H = - m \times h = - 0.8 \times 31 = 24.8 cm = 0.248 m$

Answer: Height of the image is 0.248 m