Lens
#Physics
Topics
| $s>0$ | $s'>0$ | $R>0$ | |
|---|---|---|---|
| If the X is on the same side as Y light rays | object; incoming | image; outgoing | center of curvature; outgoing |
$m = \frac{y'}{y} = -\frac{s'}{s}=-\frac{n_as'}{n_bs}$
The lateral magnification of an image
$m<0$ corresponds to an inverted image
$y'$ is the image height
$y$ is the object height
$s$ is the distance from the object to the mirror surface
$s'$ is the distance from the object to the mirror surface
$s > 0$ when the object’s on the same side as the incident ray
$s' > 0$ when the image’s on the same side as the reflected or refracted ray
- $m_\text{total}=m_1m_2\ldots m_n$
The total magnification of a series of lens is the product of each lens’ magnifications - $\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}=\frac{2}{R}$
Image position for a spherical mirror of radius $R$
$f$ is the focal length of the spherical mirror
Same sign conventions as in above equation
$R>0$ when the center of the spherical mirror is on the same side as the reflected light
Essentially assumes $R \ll s$ - $\frac{n_a}{s}+\frac{n_b}{s'}=\frac{n_b-n_a}{R}$
Spherical lens equation for refraction between two media with different refractive indexes
$n_a$ is the refractive index of the medium that the incoming light first passes through
$n_b$ is the refractive index of the medium that the refracted light passes through
Same sign rules as above equation
Essentially assumes $R \ll s$ - $f\text{-number}=\frac{f}{D}$
The $f\text{-number}$ of a camera depends on the focal length $f$ and the diameter of the aperature $D$
$f\text{-numbers}$ are generally given in the format of $\frac{f}{f\text{-number}}$. For example, $\frac{f}{1.4}$ would correspond to an $f\text{-number}$ of $1.4$
$I\propto\frac{1}{(f\text{-number})^2}$ - $P=\frac{1}{f}$
$P$ is the optical power of the lens
$f$ is the focal length of the lens and $=\frac{R}{2}$, where $R$ is the radius of the lens
The optical power of a lens determines how much bends light toward a point
$P>0$ means $f>0$ and that the lenses are converging for hyperopia (far-sighted, short eyeballs)
$P<0$ means $f<0$ and that the lenses are diverging for myopia (near-sighted, long eyeballs)
Increasing $|P|$ means lowering $f$ and $R$, which means more converging or divering of light - $M=\frac{\theta'}{\theta}=\frac{s_\text{nearpoint}}{f}$
$M$ is the angular magnification of
$\theta'$ is the angle subtended by the image height at the point of observation
$\theta$ is the angle subtended by the object height at the point of observation
$s_\text{nearpoint}$ is often taken as 25 cm for the human eye
$f$ is the focal length of the lens
Assumes that the object angle is small to allow for $\theta'\approx\theta\approx \sin\theta=\frac{y}{f}$ - $M_\text{microscopes}=m_1M_2=\frac{s_\text{nearpoint}s_1'}{f_1 f_2}$
Microscopes use two converging lens: the objective lens and the eyepiece
$s_\text{nearpoint}$ is often taken as 25 cm for the human eye
$s_1'$ is the length between the objective lens and the focal point of the eyepiece
$f_1$ and $f_2$ are the focal length of the lenses - $M_\text{telescopes}=-\frac{f_1}{f_2}$
Telescopes use two converging lens like microscopes