A point source of light illuminates an aperture $2.0\mathrm{m}$away. A $12.0-c\mathrm{m}-$wide bright patch of light appears on a screen $1.0\mathrm{m}$behind the aperture. How wide is the aperture?

The Ray Structure of Light: Light passes at $v=c/n$via neat lines called light ray. The light ray will persist indefinitely except if it interacts with substance, in which case it will reflects, refract, disperse, or be consumed. Light rays are emitted by things. Rays are sent out in every dimensions from every spot on the item. When divergent rays are gathered either by pupil and centered just on retina, the person perceives an item (or a picture). If lens, reflectors, and apertures are bigger over $\approx 1\mathrm{mm}$, ray optics is acceptable.

$2.0\mathrm{m}$is the spacing in between the aperture and point.

$12.0c\mathrm{m}$is the width of the brilliant spot of light upon that screen.

$1.0\mathrm{m}$is the distance seen between aperture and the display.

Lets $x$have been the aperture's width. Its luminous area and aperture width are connected by, as seen by the identical triangles as in picture under.

$\frac{12.0\mathrm{cm}}{1.0\mathrm{m}+2.0\mathrm{m}}=\frac{x}{2.0\mathrm{m}}$

Calculate for$x$

$x=\frac{\left(12.0\mathrm{cm}\right)\left(2.0\mathrm{m}\right)}{1.0\mathrm{m}+2.0\mathrm{m}}$

$\mathrm{x}=8.0\mathrm{cm}$