Definition
Geometric optics is a branch of optics that deals with the behavior of light as it propagates in straight lines and interacts with various optical elements such as lenses and mirrors. It simplifies the study of light by neglecting the wave nature of light and considering only the paths of light rays.
Formulas
Snell's Law
It describes the relationship between the angles of incidence and refraction when light passes through the interface between two different media.
$$n_{1} sinθ_{1} = n_{2} sinθ_{2}$$
Where:
$n_{1}$ and $n_{2}$ are the refractive indices of the first and second media, respectively.
$θ_{1}$ and $θ_{2}$ are the angles of incidence and refraction, respectively, with respect to the normal of the interface.
Lens-Maker's formula
It relates the focal length of a lens to the refractive indices of the lens material and the medium surrounding it, as well as the radii of curvature of the lens surfaces.
$$\frac{1}{f} = (n - 1) \cdot (\frac{1}{R_{1}} - \frac{1}{R_{2}}$$
Where:
f is the focal length of the lens.
n is the refractive index of the lens material.
$R_{1}$ and $R_{2}$ are the radii of curvature of the first and second surfaces of the lens.
Mirror Equation
It relates the object distance $d_{o}$, the image distance $d_{i}$, and the focal length $f$ of a mirror.
$$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$$
Where:
$d_{o}$ is the object distance (distance from the object to the mirror).
$d_{i}$ is the image distance (distance from the image to the mirror).
$f$ is the focal length of the mirror.
Magnification (m)
It describes the ratio of the size of the image to the size of the object.
$$m = - \frac{d_{i}}{d_{o}}$$
The negative sign indicates that the image is inverted if m is negative.
Now, let's solve an example using these formulas:
Problem 1
An object is placed 20 cm in front of a converging lens with a focal length of 10 cm. Calculate the position and magnification of the image formed by the lens.
Solution
Given:
$d_{o} = - 20 \ cm $(since the object is in front of the lens)
$f = 10 \ cm$
Using the lens formula, we can find the image distance $d_{i}$:
$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$
Substituting the given values:
$\frac{1}{- 20} + \frac{1}{d_{i}} = \frac{1}{10}$
Simplifying:
$- \frac{1}{20} + \frac{1}{d_{i}} = \frac{1}{10}$
$\frac{1}{d_{i}} = \frac{1}{10} + \frac{1}{20}$
$\frac{1}{d_{i}} = \frac{3}{20}$
$d_{i} = \frac{20}{3} \ cm ≈ 6.67 \ cm$
The image distance is approximately 6.67 cm.
To calculate the magnification, we can use the magnification formula:
$m = - \frac{d_{i}}{d_{o}}$
Substituting the values:
$m = - \frac{6.67 \ cm}{- 20 \ cm}$
$m ≈ 0.333$
The magnification is approximately 0.333.
Therefore, the image is formed at a distance of approximately 6.67 cm from the lens, and it is one-third the size of the object.
