The Lens Camera

The aperture stop, which is the limiting diameter of the aperture thought which light enters the camera (given as fraction of focal length F of lens) is also called F Number.

This F Number; is the fraction of focal length of the lens given as focal length divide by lens diameter.

F number = Focal length, F/Lens diameter, d

$FN = \frac{F}{d}$

Where d = is lens diameter.

1. The Number Indicates the Number of times the focal length F of times the focal length F of the lens diameter ( or stop)
2. The smaller the F - Number for a given focal length the larger the lens diameter
3. The lens with a larger diameter has a greater light- gathering power or speed
4. This for such a lens the shutter allows light in the camera for a short interval of time.

Magnification of a lens camera is obtained as the ratio of the Image distance and the object distance.

But from the lens formula:

Thus $M = v/U$

$\frac{1}{U} + \frac{1}{V} = \frac{1}{F}$

$\frac{1}{V} = \frac{1}{F} - \frac{1}{U}$

$\left(\frac{1}{V}\right) - 1 = \left(\frac{U - F}{FU}\right) - 1$

$V = \frac{FU}{U - F}$

A lens camera is to be used to take a picture of a man 2m tall if the lens of the camera Has a focal length of 10cm, calculate the minimum size of the film frame required, given that the man is 20.1m from the camera. Solution:

Magnification is given by:

$M = \frac{f}{u-f}$

Where

F= 10cm U = 201/m / 2010cm

$M = ( 10/2010 - 10)$

$M = 1/20$ ....................................i

Let the size of the frame be h when the height of man is 2m.

Then

$M = 1/200$

But $h_1/h = 1/200$

$h_1 = (1/200) \cdot 2$

$h_1 = (1/200) \cdot 2$

$h_1 (2/200)$

$h_1 = (1/100)$ m

$h_1 = 1$ cm or 10mm

The film frame should be at least 10mm square.