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We will continue the review starting with Faraday's Law

$\Phi_{B}=\int\vec{B}\cdot d\vec{A}$

or when the magnetic field is uniform

$\Phi_{B}=\vec{B}\cdot\vec{A}$

In the example above the magnetic flux is

$\Phi_{B}=\vec{B}\cdot\vec{A}=BA\cos\theta$

where $A=l^{2}$.

The unit of magnetic flux, is called a weber $\mathrm{Wb}$, where $1\mathrm{Wb}=1\mathrm{Tm^{2}}$

Faraday's law of induction states that the induced emf in a circuit is equal to rate of change of magnetic flux through the circuit.

$\mathcal{E}=-\frac{d\Phi_{B}}{dt}$

If the circuit is made of a number of loops $N$

$\mathcal{E}=-N\frac{d\Phi_{B}}{dt}$

The negative sign in Faraday's law represents Lenz's law which states that

“An induced emf is always in such a direction as to oppose the change in flux causing it”

A conductor moving in a magnetic field will experience a induced emf.

The emf produced is given by the change of flux

$\mathcal{E}=-\frac{d\Phi_{B}}{dt}=B\frac{dA}{dt}=-\frac{Blv\,dt}{dt}=-Blv$

But what about the case where the rails are not there?

In this case the electrons still feel the force and will collect at one end of rod, so there will be a potential difference across it.

It is important to remember that an emf is not a force, but rather a measure of the work done in a circuit, so we can write the emf in terms of an integral over a closed path of the electric field

$\mathcal{E}=\oint\vec{E}\cdot d\vec{l}$

and then

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$

Here we are taking an integral around the path that encloses the area in which magnetic flux is changing.

We should not that the implication of this is that in the presence of a time varying magnetic field the electric force is no longer a conservative force.

Consider the change in flux on a loop as it is rotated by some external torque:

$\mathcal{E}=\frac{d\Phi_{B}}{dt}=-\frac{d}{dt}\int\vec{B}\cdot d\vec{A}=-\frac{d}{dt}BA\cos\theta$

If the loop rotates with a constant angular velocity $\omega=\frac{d\theta}{dt}$ then $\theta=\theta_{0}=\omega t$ and we can say that

$\mathcal{E}=-BA\frac{d}{dt}(\cos\omega t)=BA\omega\sin\omega t$

of course if there are $N$ loops

$\mathcal{E}=-NBA\frac{d}{dt}(\cos\omega t)=NBA\omega\sin\omega t=\mathcal{E}_{0}\sin\omega t$

In a transformer two coils are coupled by an iron core so that the flux through the two coils is the same.

When an AC voltage is applied to the primary coil the magnetic flux passing through it is related to the applied field by

$V_{P}=N_{P}\frac{d\Phi_{B}}{dt}$

if we assume the coil has no resistance. The voltage induced in the secondary coil will have magnitude

$V_{S}=N_{S}\frac{d\Phi_{B}}{dt}$

We can thus see that

$\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}}$

If we assume there is no power loss (which is fairly accurate) then $I_{P}V_{P}=I_{S}V_{S}$ and

$\frac{I_{S}}{I_{P}}=\frac{N_{P}}{N_{S}}$

In general for two coils the relationship between the flux in one coil due to the current in another is described by a parameter called the mutual inductance.

$\Phi_{21}$ is the magnetic flux in each loop of coil 2 created by the current in coil 1. The total flux in the second coil is then $N_{2}\Phi_{21}$ and is related to the current in coil 1, $I_{1}$ by

$N_{2}\Phi_{21}=M_{21}I_{1}$

As, from Faraday's Law, the emf induced in coil 2 is $\mathcal{E}_{2}=-N_{2}\frac{d\Phi_{21}}{dt}$ so

$\mathcal{E}_{2}=-M_{21}\frac{dI_{1}}{dt}$

The mutual inductance of coil 2 with respect to coil 1, $M_{21}$ does not depend on $I_{1}$, but it does depend on factors such as the size, shape and number of turns in each coil, their position relative to each other and whether there is some ferromagnetic material in the vicinity.

In the reverse situation where a current flows in coil 2

$\mathcal{E}_{1}=-M_{12}\frac{dI_{2}}{dt}$

but in fact $M_{12}=M_{21}=M$

The mutual inductance is measured in Henrys ($\mathrm{H}$), $1\mathrm{H}=1\mathrm{\frac{Vs}{A}}=1\mathrm{\Omega s}$

The magnetic flux $\Phi_{B}$ passing through the coil is proportional to the current, and as we did for mutual inductance we can define a constant of proportionality between the current and the flux, the self-inductance $L$

$N\Phi_{B}=LI$

The emf $\mathcal{E}=-N\frac{d\Phi_{B}}{dt}=-L\frac{dI}{dt}$

The self-inductance is also measured in henrys.

A component in a circuit that has significant inductance is shown by the symbol.

We can can calculate the self-inductance of a solenoid from it's field

$B=\mu_{0}\frac{NI}{l}$

The flux in the solenoid is

$\Phi_{B}=BA=\mu_{0}\frac{N_{1}IA}{l}$

so

$L=\frac{N\Phi_{B}}{I}=\frac{\mu_{0}N^{2}A}{l}$

When we take a resistor an inductor in series and connect it to a battery then Kirchoff's loop rule tells us that

$V_{0}-IR-L\frac{dI}{dt}=0$

which we can rearrange and integrate

$\int_{I=0}^{I}\frac{dI}{V_{0}-IR}=\int_{0}^{t}\frac{dt}{L}$

$-\frac{1}{R}\ln(\frac{V_{0}-IR}{V_{0}})=\frac{t}{L}$

$I=\frac{V_{0}}{R}(1-e^{-t/\tau})=I_{0}(1-e^{-t/\tau})$ where $\tau=\frac{L}{R}$

If we then switch back to the closed loop that does not include the battery then Kirchoff's loop rule gives us

$L\frac{dI}{dt}+RI=0$

$\int_{I_{0}}^{I}\frac{dI}{I}=-\int_{0}^{t}\frac{R}{L}dt$

$\ln\frac{I}{I_{0}}=-\frac{R}{L}t$

$I=I_{0}e^{-t/\tau}$

The current changes exponentially according to $e^{-t/\tau}$. For an RC circuit the time constant is $\tau=RC$, for an LR circuit the time constant is $\tau=\frac{L}{R}$.

For a charged capacitor in series with an inductor, according to Kirchoff's loop rule:

$-L\frac{dI}{dt}+\frac{Q}{C}=0$

The current comes from the capacitor so $I=-\frac{dQ}{dt}$ so

$\frac{d^{2}Q}{dt^{2}}+\frac{Q}{LC}=0$

This is like a simple harmonic oscillator, but with charge in place of displacement! A solution to this differential equation is

$Q=Q_{0}\cos(\omega t+\phi)$

To find $\omega$ we substitute the solution in to the equation

$-\omega^{2}Q_{0}\cos(\omega t + \phi)+\frac{Q_{0}}{LC}\cos(\omega t+\phi)=0$

$(-\omega^{2}+\frac{1}{LC})\cos(\omega t+ \phi)=0$

Which requires $\omega=\sqrt{\frac{1}{LC}}$

As we saw previously an actual inductor has some resistance, and thus a real LC circuit is better represented by an LRC circuit in which we represent the resistance of the inductor, or other resistances in the circuit by an in series resistor $R$

Now Kirchoff's loop rule when the circuit is closed is

$-L\frac{dI}{dt}-IR+\frac{Q}{C}=0$

Which can be written in terms of the charge as before

$L\frac{d^{2}Q}{dt^{2}}+R\frac{dQ}{dt}+\frac{Q}{C}=0$

and we can compare this to the differential equation for a damped harmonic oscillator.

When $R^{2}<\frac{4L}{C}$ the system is underdamped and the solution to this equation is

$Q=Q_{0}e^{-\frac{R}{2L}t}\cos(\omega' t+\phi)$

where $\omega'=\sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}$

When $R^{2}>\frac{4L}{C}$ the system is overdamped and the charge will decay slowly.

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We can begin our analysis of this circuit by applying Kirchoff's loop rule to the find potential at any given time.

$V=V_{R}+V_{L}+V_{C}$

However an important consequence of the voltages not being at the same phase in the different components is that the peak voltage is not experienced at the same time in each component, and so the peak source voltage $V_{0}$

$V_{0}\neq V_{R0}+V_{L0}+V_{C0}$

The condition of continuity of current demands that the current throughout the circuit should always be in phase, so at any point in this circuit the current will

$I=I_{0}\cos\omega t$

To understand the Phasor approach to AC circuits we can follow a very nice set of animations from Physclips.

A phasor is a way of representing the voltage across a component taking into account the phase difference between the voltage and the current.

As we saw in our last lecture a resistor has a voltage in phase with the current flowing through it. So if we now represent the current as vector moving in a plane we can also represent the voltage across the resistor as a vector of magnitude $V_{R}=I{R}$ which points in the same direction as the current flowing through it.

We saw in our last lecture that in a capacitor the current leads the voltage by 90^{o}.

Also we saw that the reactance $X_{C}=\frac{1}{\omega C}=\frac{1}{2\pi f C}$ depends on frequency and so the size of the voltage phasor $V_{0}=I_{0}X_{C}$ also should.

We saw in our last lecture that in an inductor the current lags the voltage by 90^{o}.

Also we saw that the reactance $X_{L}=\omega L=2\pi f L$ depends on frequency and so the size of the voltage phasor $V_{0}=I_{0}X_{L}$ also should.

To find the total voltage in the circuit at time $t$ we add the phasors for the different components together as we would vectors. We can see that for a current $I=I_{0}\cos\omega t$ flowing through through the circuit the voltage will be offset by a phase $\phi$

$V=V_{0}\cos(\omega t + \phi)$

The peak voltage $V_{0}$ is linked to the peak current $I_{0}$ through the impedance $Z$

$V_{0}=I_{0}Z$ and we can also say $V_{rms}=I_{rms}Z$

The value of $Z$ is found by considering the vector sum of the voltages

$V_{0}=\sqrt{V_{R0}^{2}+(V_{L0}-V_{C0})^{2}}=I_{0}\sqrt{R^{2}+(X_{L}-X_{C})^2}$

so

$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}$

The phase difference between the current and voltage $\phi$ is obtained from

$\tan \phi=\frac{V_{L0}-V_{C0}}{V_{R0}}=\frac{I_{0}(X_{L}-X_{C})}{I_{0}R}=\frac{X_{L}-X_{C}}{R}$

To find the total voltage in the circuit at time $t$ we add the phasors for the different components together as we would vectors. We can see that for a current $I=I_{0}\cos\omega t$ flowing through through the circuit the voltage will be offset by a phase $\phi$

$V=V_{0}\cos(\omega t + \phi)$

The peak voltage $V_{0}$ is linked to the peak current $I_{0}$ through the impedance $Z$

$V_{0}=I_{0}Z$ and we can also say $V_{rms}=I_{rms}Z$

The value of $Z$ is found by considering the vector sum of the voltages

$V_{0}=\sqrt{V_{R0}^{2}+(V_{L0}-V_{C0})^{2}}=I_{0}\sqrt{R^{2}+(X_{L}-X_{C})^2}$

so

$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}$

The phase difference between the current and voltage $\phi$ is obtained from

$\tan \phi=\frac{V_{L0}-V_{C0}}{V_{R0}}=\frac{I_{0}(X_{L}-X_{C})}{I_{0}R}=\frac{X_{L}-X_{C}}{R}$

Power in an LRC circuit is only dissipated in the resistor, and so the average power dissipated is given by

$\bar{P}=I_{RMS}^{2}R$

but we may want to express this in terms of the impedance of the circuit or the $V_{RMS}$ which is applied.

To do this we write

$\cos \phi =\frac{V_{R0}}{V_{0}}=\frac{I_{0}R}{I_{0}Z}=\frac{R}{Z}$

which means that $R=Z\cos\phi$ and

$\bar{P}=I_{RMS}^{2}Z\cos\phi$

or, as $V_{RMS}=I_{RMS}Z$

$\bar{P}=I_{RMS}V_{RMS}\cos\phi$

The RMS current in the circuit we are considering is given by

$I_{RMS}=\frac{V_{RMS}}{Z}=\frac{V_{RMS}}{\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}}$

We can see that the current should be frequency dependent and have a maximum when

$(\omega L - \frac{1}{\omega C})=0$

which gives the resonant frequency

$\omega_{0}=\sqrt{\frac{1}{LC}}$

One can get a feel for the circuit response here and here.

Here is a demo that shows how the resonant frequency is varied by the inductor and capacitor values.

Maxwell's equations, named after James Clerk Maxwell who first expressed them together are a set of four equations from which all electromagnetic theory can be derived.

The integral form of Maxwell's equations in free space (ie., in the absence of dielectric or magnetic materials) are

$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$ (Gauss's Law)

$\oint \vec{B}\cdot d\vec{A}=0$ (Magnetic equivalent of Gauss's Law)

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$ (Faraday's Law)

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ (Modified form of Ampere's Law).

From Maxwell's equations

$\frac{\partial^{2} E_{y}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} E_{y}}{\partial x^{2}}$ and $\frac{\partial^{2} B_{z}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} B_{z}}{\partial x^{2}}$

tell that electromagnetic waves can be described by an electric and magnetic field in phase with one another, but at right angles to one another

$E_{y}(x,t)=E_{0}\sin\frac{2\pi}{\lambda}(x-\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$

$B_{z}(x,t)=B_{0}\sin\frac{2\pi}{\lambda}(x-\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$

with velocity

$v=\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}=3.00\times10^{8}\,\mathrm{m/s}=c$

The energy stored in an electric field is

$u_{E}=\frac{1}{2}\varepsilon_{0}E^{2}$

and in a magnetic field

$u_{B}=\frac{1}{2}\frac{B^{2}}{\mu_{0}}$

The total energy of an EM wave is

$u=u_{E}+u_{B}=\frac{1}{2}\varepsilon_{0}E^{2}+\frac{1}{2}\frac{B^{2}}{\mu_{0}}$

and using

$\frac{E}{B}=c=\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}$

$u=\varepsilon_{0}E^{2}=\frac{B^{2}}{\mu_{0}}=\sqrt{\frac{\varepsilon_{0}}{\mu_{0}}}EB$

In an electromagnetic wave the fields are moving with velocity $c$ the amount of energy passing through a unit area at any given time is

$S=\varepsilon_{0}cE^{2}=\frac{cB^{2}}{\mu_{0}}=\sqrt{\frac{EB}{\mu_{0}}}$

More generally, the Poynting vector is a vector $\vec{S}$ which represents the flux of energy in an electromagnetic field

$\vec{S}=\frac{1}{\mu_{0}}\vec{E}\times\vec{B}$

We define an angle of incidence $\theta_{i}$ relative to the surface normal and find that the angle of reflection $\theta_{r}$, also defined relative to the surface normal is given by

$\theta_{r}=\theta_{i}$

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

The magnification of a mirror $m$

$m=\frac{h_{i}}{h_{o}}=-\frac{d_{i}}{d_{o}}$

**Sign Conventions:**

- The image height $h_i$ is positive if the image is upright, and negative if inverted.
- $d_i$ or $d_o$ is positive if image/object is in front of the mirror (real image).
- $d_i$ is negative if image is behind mirror (virtual image)
- $r$ and $f$ are also negative when behind the mirror
- Magnification is positive for an upright image and negative for an inverted image

A useful way to describe a material in terms of it's refractive index. The velocity of light in a medium is related to it's velocity on free space by the refractive index $n$ through the equation

$v=\frac{c}{n}$

Refractive indices of materials are typically somewhere between 1 (vacuum) and ~2.5 (diamond, strontium titanate). Glass will typically have a refractive index of about 1.5 though the exact value depends on the type of glass.

We can recall that the velocity of a wave is given $v=f\lambda$, when light is traveling in a medium, the frequency $f$ does not change, the wavelength $\lambda$ changes according to

$\lambda=\frac{\lambda_{0}}{n}$

$n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}$

The refractive index of the medium the light is traveling out of is $n_{1}$, the refractive index of the material the light is traveling in to is $n_{2}$ and both the angle of incidence $\theta_{1}$ and refraction $\theta_{2}$ are defined relative to the normal to the surface.

For light leaving a more optically dense medium and entering a less optically dense one there is a maximum incident angle, the critical angle $\theta_{C}$ above which light is completely reflected, which we refer to as total internal reflection.

The critical angle can be found from the condition that the refracted angle is $90^{o}$

$\sin\theta_{C}=\frac{n_{2}}{n_{1}}\sin 90^{o}=\frac{n_{2}}{n_{1}}$

A lens which is thicker in the center than at the edges is a converging lens, an incoming parallel beam of light will be focused to a point $F$ at $x=f$ from the center of the lens.

To obtain a parallel beam of light light should be radially propagating outwards from the point $F'$ a distance $x=-f$ from the center of the lens.

A lens which is thinner in the center than at the edges is a diverging lens, an incoming parallel beam of light will be focused to a point $F$ at $x=-f$ from the center of the lens, which means that the rays appear to diverge outwards from that point.

To obtain a parallel beam of light light the incoming rays should have a path such that they would go to the point $F'$ a distance $x=f$ from the center of the lens if the lens were not there..

To find the image position for a lens we can use a technique called raytracing. We only need to use 3 rays to find an image for a given object (provided we know the focal length of the lens).

- A ray that leaves the object parallel to the axis and then goes through $F$
- A ray that passes through the center of the lens and is not bent
- A ray that passes through $F'$ and exits parallel to the axis

To find the image position for a lens we can use a technique called raytracing. We only need to use 3 rays to find an image for a given object (provided we know the focal length of the lens).

- A ray that leaves the object parallel to the axis and then goes through $F$
- A ray that passes through the center of the lens and is not bent
- A ray that passes through $F'$ and exits parallel to the axis

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

This equation also works for diverging lens (with $d_{i}$ and $f$ negative).

As with mirrors, the magnification $m$ is

$m=\frac{h_{i}}{h_{0}}=-\frac{d_{i}}{d_{o}}$

Combinations of lens can be treated sequentially, by first finding the image produced by the first lens and then then using it as the object for the next lens.

Applying the lens equation to the first lens

$\frac{1}{d_{iA}}=\frac{1}{f_{A}}-\frac{1}{d_{oA}}$

and to the second lens

$\frac{1}{d_{iB}}=\frac{1}{f_{B}}-\frac{1}{d_{oB}}$

and then using $d_{oB}=l-d_{iA}$ allows the determination of the final image position.

The first lens produces an image that has height $-\frac{d_{iA}}{d_{oA}}h_{o}$, which will then be used as the object height in the next lens, so the final image has height

$\frac{d_{iA}}{d_{oA}}\frac{d_{iB}}{d_{oB}}h_{o}$

When considering the multiplying power of lens combinations we can simply multiply the effects of the individual lens.

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.