Question

A pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image without a lens. A certain pinhole camera is a box that is 20.0 cmsquare and 22.0 cm deep, with the hole in the middle of one of the 20.0 cm × 20.0 cm faces. If this camera is used to photograph a fierce chicken that is 18.0 cm high and 2.00 m in front of the camera, how large is the image of this bird on the film?What is the magnification of this camera?

Answer 1

Answer:

Size of image: 1.8 cm inverted

Magnification of camera: -0.1

Explanation:

Given:

• $h_o$ = size of bird = 18 cm

• $u$ = distance of the bird from the camera = 2 m = 200 cm

• $v$ = distance of the film from the camera = 20 cm

Assume:

• $m$ = magnification of the camera

• $h_i$ = size of image of the bird

Using sign convention, we have

$u = -200\ cm\\v = 20\ cm\\h_o = 18\ cm$

The hole in the pinhole camera behaves as a convex lens. The other face opposite to the hole face behaves like a film where the image is formed to be seen. So, using the formula of magnification, we have

$m = \dfrac{h_i}{h_o}=\dfrac{v}{u}\\\Rightarrow \dfrac{h_i}{h_o}=\dfrac{v}{u}\\\Rightarrow h_i=\dfrac{v}{u}h_o\\\Rightarrow h_i=\dfrac{20}{-200}\times 18\\\Rightarrow h_i=-1.8 cm$

This means the image of the bird measures 1.8 cm in length where negative sign in the calculation represents that the image formed is inverted.

Hence, the image of the bird on the film is 1.8 cm large.

Now, again using the formula of magnification, we have

$m = \dfrac{h_i}{h_o}\\\Rightarrow m = \dfrac{-1.8}{18}\\\Rightarrow m =-0.1$

Hence, the magnification of the camera is -0.1.