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The biggest possible size of an object on the eye without the aid of an optical instrument is obtained by placing it at the near point of the eye $N$. If we bring an object closer than the near point then it will occupy a larger angle, but we won't be able to focus on it. This problem can be addressed by a magnifying glass. If we place an object closer than the focal length of the lens, it will produce a virtual image at a distance $d_{i}$. The maximum magnification is achieved by bringing the lens right up to your eye and then arranging the lens, object and your head so that the image is at the near point. To find the magnification we need to know $d_{o}$ which can be found from the lens equation, taking $d_{i}=-N$

$\frac{1}{d_{o}}=\frac{1}{f}-\frac{1}{d_{i}}=\frac{1}{f}+\frac{1}{N}$

In the small angle approximation

$\theta=\frac{h}{N}$ and $\theta'=\frac{h}{d_{o}}$

The angular magnification of the lens, also called the magnifying power is defined as $M=\frac{\theta'}{\theta}$ which we can see here is

$M=\frac{N}{d_{o}}=N(\frac{1}{f}+\frac{1}{N})=\frac{N}{f}+1$

It is not very convenient to use a magnifying glass with the eye focused at the near point, firstly as we are required to constantly maintain the correct positioning of lens, object and head, but also because our eye muscles are at maximum exertion, which is not very comfortable over long periods of time. An alternative way to use a magnifying glass is to place the object at the focal point of the lens producing an image at $\infty$.

In this case

$\theta'=\frac{h}{f}$

and the magnifying power is

$M=\frac{\theta'}{\theta}=\frac{N}{f}$

A magnifying glass is only useful for looking at objects nearby (the maximum object distance from the lens is the focal length). To view distant objects we need to use a telescope. A refracting telescope uses two lens, and objective and an eyepiece. The objective produces an image which is then magnified by the eyepiece.

The original apparent object size is

$\theta\approx\frac{h}{f_{o}}$

If we consider an object at infinity and an eyepiece which is adjusted so that the focus of the eyepiece is at the focus of the objective (this produces a final image at infinity, which is why I chose not to draw it this way on the diagram), then the apparent size of the final image is

$\theta'\approx\frac{h}{f_{e}}$

giving the magnification power of the telescope as

$M=\frac{\theta'}{\theta}=-\frac{f_{o}}{f_{e}}$

with the minus sign signifying that the image is inverted.

If we treat each of the slits as a point source of circular wavefronts. The condition for constructive interference (bright fringes) is

$d\sin\theta=m\lambda$ (m=0,1,2,..)

and for destructive interference (dark fringes)

$d\sin\theta=(m+\frac{1}{2})\lambda$ (m=0,1,2,..)

As in the case of a wave on a rope that is incident on a heavier rope and is reflected with a 180^{o} phase change when a light wave is reflected from a more optically dense media a 180^{o} phase change occurs. This effect is important when we want to consider interference effects in thin films.

As the two reflections both occur from more optically dense media they both experience a $\frac{\lambda}{2}$ phase change on reflection. To have the light be out of phase we need to light that goes through the coating to have advanced by $\frac{\lambda}{2}$ for destructive interference to occur. Critically, when destructive interference occurs the light is not lost, but is instead transmitted. As the wavelength of light in a medium is given by $\lambda=\frac{\lambda_{0}}{n}$ where $n$ is the refractive index of the medium and $\lambda_{0}$ is the wavelength of the light in free space, the thickness of the coating should be $\frac{\lambda}{4n_{2}}$.

In practice the light incident will not all be the same wavelength, so the thickness of the coating is typically chosen to work optimally in the center of the visible band (~550nm).

For a single wavelength dark rings will occur whenever

$2t=m\lambda$ (m=0,1,2,..)

and bright rings will occur whenever

$2t=(m+\frac{1}{2})\lambda$ (m=0,1,2,..)

For white light different colors will experience constructive interference at different thicknesses, leading to the colorful lines we see when an air gap is under normal light.

Rather than writing our equation in terms of k

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{Dk}{2}\sin\theta)$

it is convenient to write it in terms of the wavelength, using $k=\frac{2\pi}{\lambda}$

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)$

This function has a central maxima and then minima at

$D\sin\theta=m\lambda$ $m=\pm 1,\pm 2,\pm 3,..$

Now we have an expression for the intensity from a slit of width $D$

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)$

we can use consider slits as sources instead of the point sources we considered earlier which gave

$I_{\theta}=I_{0}\cos^{2}(\frac{d \pi}{\lambda}\sin \theta)$

The combination of these gives

$I_{theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)\cos^{2}(\frac{d\pi}{\lambda}\sin \theta)$

Irrespective of the number of slits $n$ the condition for maxima is the same as for the double slit

$d\sin\theta=m\lambda$ (m=0,1,2,..)

but that the larger the number of slits from which diffraction occurs the sharper the maxima will be.

The polarization of the light after it has passed through the polarizer is the direction defined by the polarizer, so the way to approach the calculation of the magnitude of the intensity that passes through the polarizer is by finding the component of the electric field which is in that direction

$E=E_{0}\cos\theta$

the intensity is then

$I=E^{2}=E_{0}^{2}\cos^{2}\theta=I_{0}\cos^{2}\theta$

This formula is known as Malus' Law.