**CBSE NOTES CLASS 12 PHYSICS **

**CHAPTER 10 WAVE OPTICS**

**Newton’s corpuscular theory**

- Light consists of very small invisible elastic particles which travel in vacuum with a speed of 3×10
^{8}m/s. - The theory
**could explain reflection and refraction**. - The size of corpuscles of different colours of light is different.
**Limitation:**It**could not**explain interference, diffraction, polarization, photoelectric effect and Compton effect. The theory failed as it could not explain why light travels faster in a rarer medium than in a denser medium.

**Wave theory - nature of electromagnetic waves**

According to wave theory of light, the light is a form of energy which travels through a medium or through vacuum in the form of **electromagnetic waves **whose nature is** transverse**. The speed of light in a medium depends upon the nature of medium.

**Wave front**

A **wave front** is defined as the **continuous locus** of all the particles of a medium, which are vibrating in the same phase. It can be defined as a surface of constant phases.

- Spherical wave front: It is wave front created by a point source. The locus of all the points which are vibrating in same phase lies on the surface of the sphere, of which at the centre, the point source is situated
- Cylindrical wave front: When the source of light is linear in shape then the locus of all the points which are equidistant from the linear source lie on a cylinder and the wave front formed is called cylindrical wave front.
- Plane wave front: It is small part of a big spherical or cylindrical wave front of which, the source of light is at a large distance from the wave front.

**Huygens principle**

- Every point on given wave front (called primary wave front) acts as a fresh source of new disturbance called secondary wavelets.
- The secondary wavelets travel in all the directions with the speed of light in the medium.
- A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new (secondary) wave front of that instant

If we wish to determine the shape of the wave front at **time = τ**, we draw spheres of radius **vτ** from each point on the spherical wave front where v represents the speed of the wave in the medium.

Common tangent to all these spheres gives the new position of the wave front at **t = τ**.

Huygens proposed that the amplitude of the secondary wavelets is **maximum in the forward** direction and **zero in the backward direction**. Huygens could explain the absence of the **back wave** using this assumption, but it is **not correct**. The easiest explanation is that the backward-propagating portions are **destructively interfered** with by the **wavelets from all of the other wave fronts**.

**Refraction of a plane wave from rarer to denser medium**

Let PP’ represent the surface separating medium 1 and medium 2. Let v_{1} and v_{2} represent the speed of light in medium 1 and medium 2, respectively. Let us take a plane wave front AB, propagating in the direction A’A, incident on the interface at an angle i. If τ be the time taken by the wave front to travel the distance BC and BC = v_{1}τ

By the time the point B on wave front travels to point C in medium 1, the point A would have travelled in medium 2, to E The new wave front will be CE and AE = v_{2}τ.

$$\mathrm{N}\mathrm{o}\mathrm{w}\mathrm{sin}\mathrm{i}=\frac{\mathrm{B}\mathrm{C}}{\mathrm{A}\mathrm{C}}={\mathrm{v}}_{1}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{sin}\mathrm{r}=\frac{\mathrm{A}\mathrm{E}}{\mathrm{A}\mathrm{C}}={\mathrm{v}}_{2}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}\frac{\mathrm{sin}\mathrm{i}\mathrm{}}{\mathrm{sin}\mathrm{r}}=\frac{{\mathrm{v}}_{1}}{{\mathrm{v}}_{2}}$$

This is **Snell’s law**.

If r < i then the ray bends toward the normal and the speed of the light wave in the second medium (v_{2}) will be less than the speed of the light wave in the first medium (v_{1}). **This is correct and proves the corpuscular theory wrong.**

The constants

$${\mathrm{n}}_{1}\mathrm{}=\frac{\mathrm{c}}{{\mathrm{v}}_{1}}\mathrm{}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}{\mathrm{n}}_{2}\mathrm{}=\frac{\mathrm{c}}{{\mathrm{v}}_{2}}\mathrm{}$$

are called refractive indices of medium 1 and 2 respectively.

Also, ${\mathrm{n}}_{1}\mathrm{sin}\mathrm{i}=\mathrm{}{\mathrm{n}}_{2}\mathrm{sin}\mathrm{r}$

$$\mathrm{a}\mathrm{n}\mathrm{d},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{v}}_{1}}{{\mathrm{v}}_{2}}=\frac{{\mathrm{\lambda}}_{1}}{{\mathrm{\lambda}}_{2}}$$

**Refraction of a plane wave from denser to rarer medium**

In this case the angle of refraction is greater than angle of incidence; but n_{1 }sin i = n_{2} sin r holds.

Critical angle i_{c} is defined by sin i_{c} = n_{2}/n_{1}

And sin r = 1

For **i > i**_{c} there will be total internal reflection.

**Reflection of a plane wave by a plane surface**

Consider a plane wave front AB incident at an angle i on a reflecting surface MN. If v represents the speed of the wave in the medium, τ represents the time taken by the wave front to advance from the point B to C then the distance BC = vτ

Let CE represent the new wave front. Now AE = BC = vτ

Triangles EAC and BAC are congruent and therefore, the angles **i** and **r** would be equal. This is the **law of reflection**.

**Behaviour of a plane wave front with different surfaces**

**Refraction through a prism:** Since the speed of light waves is less in glass, the lower portion of the incoming wave front (which travels through the greatest thickness of glass) will get delayed resulting in a tilt in the emerging wave front.

**Refraction through a lens: **In case of a** **plane wave incident on a thin convex lens; the central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most. The emerging wave front has a depression at the centre and therefore the wave front becomes spherical and converges to the point F which is known as the focus.

**Reflection at a mirror:** When a plane wave is incident on a concave mirror; on reflection we have a spherical wave converging to the focal point F.

In a similar manner, we can understand refraction and reflection by concave lenses and convex mirrors.

**The Doppler effect**

When the source moves away from the observer the frequency as measured by the observer will be smaller. Similarly when the source moves towards the observer, the frequency measured by the observer will be greater. This is known as the Doppler Effect. The increase in wavelength is called **red shift** and decrease in wavelength is called **blue shift.**

The fractional change in frequency or Doppler shift is given by

$$\frac{\mathrm{\Delta}\mathrm{\nu}}{\mathrm{\nu}}=\mathrm{}\u2013\frac{{\mathrm{v}}_{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{l}}}{\mathrm{c}}$$

Where, v_{radial} is the component of the **source velocity relative to the observer** along the line joining the observer to the source.

v_{radial} is considered **positive** when the **source moves away from the observer**.

**Superposition principle**

The resultant displacement produced by interaction of a number of waves is the vector sum of the displacements produced by each of the waves.

**Coherent sources**

If the phase difference between the displacements of waves produced by two sources does not change with time, the two sources are said to be coherent.

**Interference **

Superposition of two waves, travelling through the same medium, is called interference.

**Constructive interference** occurs when the two interfering waves have a displacement in the same direction and the amplitude of the resulting wave is the addition of amplitudes of interfering waves.

**Destructive interference** occurs when the two interfering waves have a displacement in different direction (out of phase) and the amplitude of the resulting wave is the difference of amplitudes of interfering waves.

Consider two coherent sources S_{1} and S_{2 }a point P for which S_{1}P = S_{2}P. Since the distances S_{1}P and S_{2}P are equal, waves from S_{1 }and S_{2} will take the same time to travel to the point P and waves that originate from S_{1} and S_{2} in phase will also arrive, at the point P, in phase.

If the displacement produced by the source S_{1} at the point P is given by

y_{1} = a cos ωt

then, the displacement produced by the source S_{2} (at the point P) will also be given by

y_{2} = a cos ωt

Thus, the resultant of displacement at P would be given by

y = y_{1} + y_{2} = 2 a cos ωt

Since the intensity is the proportional to the square of the amplitude, the resultant intensity will be given by,

I = 4I_{o}

where I_{o} represents the intensity produced by each one of the individual sources; I_{o} is proportional to a^{2}.

In fact for all points where the phase difference is 2nπ or path difference is nλ, the interference will be constructive.

If the phase difference is $(2\mathrm{n}+1)$π or path difference is $\left(\frac{2\mathrm{n}+1}{2}\right)$λ, the interference will be destructive.

For any other arbitrary point let the phase difference between the two displacements be φ.

y_{1} = a cos ωt ⇒ y_{2} = a cos (ωt + φ)

The resultant displacement will be given by

y = y_{1} + y_{2} = a [cos ωt + cos (ωt + φ]

$$=\mathrm{}2\mathrm{a}\mathrm{cos}\left(\frac{\phi}{2}\right)\mathrm{cos}\left(\mathrm{\omega}\mathrm{t}\mathrm{}+\frac{\phi}{2}\right)$$

Since φ is constant, the amplitude of the resultant displacement is 2a cos $\frac{\phi}{2}$.

Or the intensity I = 4I_{o} cos^{2 }$\frac{\phi}{2}$

If the two sources are not coherent the average intensity will be given by

<I> = 4I_{o} <cos^{2} $\frac{\phi}{2}$>

where angular brackets represent time averaging.

The function cos^{2 }$\frac{\phi}{2}$ will randomly vary between 0 and 1 and the average value will be $\frac{1}{2}$.

The resultant intensity will be given by,

I = 2I_{o}

**Young’s double slit experiment**

Thomas Young made two pinholes S_{1} and S_{2} (very close to each other) on an opaque screen. These were illuminated by another pinhole that was in turn, lit by a bright source. Light waves spread out from S and fall on both S_{1} and S_{2}. S_{1} and S_{2} then behave like two coherent sources because light waves coming out from S_{1} and S_{2} are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S_{1} and S_{2}.

Thus, the two sources S_{1} and S_{2} will be locked in phase; i.e., they will be coherent like the two vibrating needles.

_{}

Spherical waves originating from S_{1} and S_{2} will produce interference fringes on the screen GG’.

For an arbitrary point P on the line GG’ to have a maximum,

S_{2}P – S_{1}P = nλ; n = 0, 1, 2 ...

From the geometry of the figure,

$$\mathrm{}{\left({\mathrm{S}}_{2}\mathrm{P}\right)}^{2}\mathrm{}\u2013\mathrm{}{\left({\mathrm{S}}_{1}\mathrm{P}\right)}^{2}=\left[{\mathrm{D}}^{2}\mathrm{}+\mathrm{}{\left(\mathrm{x}+\frac{\mathrm{d}}{2}\right)}^{2}\mathrm{}\u2013\mathrm{}\left\{{\mathrm{D}}^{2}+{\left(\mathrm{x}-\frac{\mathrm{d}}{2}\right)}^{2}\right\}\right]=2\mathrm{x}\mathrm{d}$$

which gives

$$\left({\mathrm{S}}_{2}\mathrm{P}-{\mathrm{S}}_{2}\mathrm{P}\right)\left({\mathrm{S}}_{2}\mathrm{P}+{\mathrm{S}}_{2}\mathrm{P}\right)=2\mathrm{x}\mathrm{d}$$

$$\Rightarrow {\mathrm{S}}_{2}\mathrm{P}-{\mathrm{S}}_{2}\mathrm{P}=\frac{2\mathrm{x}\mathrm{d}}{{\mathrm{S}}_{2}\mathrm{P}+{\mathrm{S}}_{2}\mathrm{P}}$$

If x << D and d << D, then,

$${\mathrm{S}}_{2}\mathrm{P}+{\mathrm{S}}_{2}\mathrm{P}\mathrm{}\mathrm{}2\mathrm{D}$$

Hence,

$${\mathrm{S}}_{2}\mathrm{P}-{\mathrm{S}}_{2}\mathrm{P}=\frac{\mathrm{x}\mathrm{d}}{\mathrm{D}}$$

For constructive interference or bright fringes,

$${\mathrm{x}}_{\mathrm{n}}=\frac{\mathrm{n}\mathrm{\lambda}\mathrm{D}}{\mathrm{d}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}[\mathrm{n}\mathrm{}=\mathrm{}0,\mathrm{}1,\mathrm{}2\dots .\mathrm{}\mathrm{}]$$

For destructive interference or dark fringes,

$${\mathrm{x}}_{\mathrm{n}}=\left(\mathrm{n}+\frac{1}{2}\right)\frac{\mathrm{\lambda}\mathrm{D}}{\mathrm{d}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}[\mathrm{n}\mathrm{}=\mathrm{}0,\mathrm{}1,\mathrm{}2\dots .\mathrm{}\mathrm{}]$$

**Fringe width**

Dark and bright fringes are equally spaced and the distance between two consecutive bright fringes or two consecutive dark fringes is given by

$$\mathrm{f}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{w}\mathrm{i}\mathrm{d}\mathrm{t}\mathrm{h},\mathrm{}\mathrm{\beta}=\mathrm{}{\mathrm{x}}_{\mathrm{n}+1}-{\mathrm{x}}_{\mathrm{n}}=\mathrm{}\frac{\mathrm{\lambda}\mathrm{D}}{\mathrm{d}}$$

The central point O will be bright because S_{1}O = S_{2}O and it will correspond to n = 0. If we consider the line perpendicular to the plane of the paper and passing through O [i.e., along the y-axis] then all points on this line will be equidistant from S_{1} and S_{2} and we will have a bright central fringe which is a straight line.

All other fringes will be hyperbolas. If D >> β then they will be almost straight lines.

**Diffraction**

Diffraction is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle. It is due to interference of waves according to the Huygens–Fresnel principle. These characteristic behaviors are exhibited when a wave encounters an obstacle or a slit that is comparable in size to its wavelength.

**Single slit experiment**

When **single narrow slit** is illuminated by a **monochromatic source**, a broad pattern with a **central bright region** is seen. On both sides, there are alternate dark and bright regions, the intensity becoming weaker away from the centre.

Let us consider a parallel beam of light falling normally on a single slit LN of width a. The diffracted light meets a screen. The midpoint of the slit is M. MC is perpendicular to the slit.

Consider the intensity at a point P on the screen.

The slit is further divided into smaller parts, whose midpoints are M_{1}, M_{2} etc. Straight lines joining P to the different points L, M, N, etc., can be treated as parallel, making an angle θ with the normal MC.

Different parts of the wave front can be treated as secondary sources.

Since the incoming wave front is parallel to the plane of the slit, these sources are in phase.

The path difference NP – LP between the two edges of the slit will be

NP – LP = NQ = a sin θ ≈ aθ

Similarly, if two points M_{1} and M_{2} in the slit plane are separated by y, the path difference will be

M_{2}P – M_{1}P = yθ.

Equal, coherent contributions from a large number of sources, each with a different phase need to be summed up.

**Explanation for position of minima in diffraction pattern**

The minima (zero intensity) occurs at,

$$\theta =\frac{n\lambda}{a}$$

Where, n = ±1, ±2, ±3, ....

Consider the angle θ where the path difference *a*θ is λ. Then,

$$\theta \approx \frac{\lambda}{a}$$

Now, divide the slit into two equal halves LM and MN each of size $\frac{a}{2}$. For every point M_{1} in LM, there is a point M_{2} in MN such that M_{1}M_{2} = $\frac{a}{2}$. The path difference between M_{1} and M_{2} at P,

$${\mathrm{M}}_{2}\mathrm{P}\mathrm{}\u2013\mathrm{}{\mathrm{M}}_{1}\mathrm{P}\mathrm{}=\mathrm{}\frac{\mathrm{a}\mathrm{\theta}}{2}=\frac{\mathrm{\lambda}}{2}$$

That is, the contributions from M_{1} and M_{2} are 180º out of phase and cancel each other, when,

$$\theta =\frac{\lambda}{a}$$

Therefore, contributions from the two halves of the slit LM and MN, cancel each other. The intensity is also zero for,

$$\theta =\frac{n\lambda}{a}$$

where, *n *is any integer (≠ 0).

- The angular size of the central maximum increases when the slit width
*a*decreases.

**Explanation for position of maxima in diffraction pattern**

At the central point C on the screen, the angle θ is zero. All path differences are zero and hence all the parts of the slit contribute in phase. This gives maximum intensity at C. Other secondary maxima are shown at,

$$\theta =\left(n+\frac{1}{2}\right)\frac{\lambda}{a}$$

and they go on becoming weaker and weaker with increasing *n*.

Consider an angle

$$\theta =\frac{3\lambda}{2a}$$

which is midway between two of the dark fringes.

Divide the slit into three equal parts. If we take the first two thirds of the slit, the path difference between the two ends would be

$$\frac{2}{3}a=\frac{2a}{3}\times \frac{3\lambda}{2a}=\lambda $$

The first two-thirds of the slit can therefore be divided into two halves which have a path difference of $\frac{\lambda}{2}$. The contributions of these two halves cancel as described earlier. Only the remaining one-third of the slit contributes to the intensity at a point between the two minima. This will be much weaker than the central maximum, where the entire slit contributes in phase.

We can similarly show that there are maxima at

$$\theta =\left(n+\frac{1}{2}\right)\frac{\lambda}{a}$$

* *Where, *n *= 2, 3, etc.

These become weaker with increasing *n*, since only one-fifth, one-seventh, etc., of the slit contributes in these cases.

**Double slit vs single slit patterns**

In the double-slit experiment, we must note that the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole, and the double-slit interference pattern. It shows a broader diffraction peak in which there appear several fringes of smaller width due to double-slit interference.

The number of interference fringes occurring in the broad diffraction peak depends on the ratio $\frac{d}{a}$, that is the ratio of the path difference to the width of a slit. In the limit of ‘a’ becoming very small, the diffraction pattern will become very flat and we will observe the two-slit interference pattern.

**Interference vs diffraction due to single slit**

(i) The interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima. The intensity falls as we go to successive maxima away from the centre, on either side.

(ii) We calculate the interference pattern by superposing two waves originating from the two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.

(iii) For a single slit of width a, the first null of the interference pattern occurs at an angle of $\frac{\lambda}{a}$. At the same angle of $\frac{\lambda}{a}$, we get a maximum (not a null) for two narrow slits separated by a distance ‘a’.

**Constraints for diffraction due to single slit**

- a and d have to be small
- D has to be large
- Source must be monochromatic

**When can we consider the light beam to be parallel beam in single slit experiment? **

- We can consider the two or more paths from the slits to the screen as parallel only if the screen on which the fringes are formed is at a large distance, i.e.
*a*<< D. - Alternatively, this situation can be obtained by placing a convex lens after the slit and putting the
**screen at its focus**. - The lens does not introduce any extra path differences in a parallel beam.
- This arrangement gives more intensity than placing the screen far away.
- The separation of the central maximum from the first null of the diffraction pattern is by an angle $\frac{\lambda}{a}$. This size on the screen will be $\frac{f\mathrm{\lambda}}{a}$.

**Viewing the diffraction pattern**

The diffraction pattern can be viewed by putting two razor blades touching each other in front of a filament. If filter for red or blue light are used (for monochromatic source), the fringes can be seen clearly.

- We should not use direct sunlight – it can damage the eye and will not give fringes anyway as the Sun subtends an angle of ${\frac{1}{2}}^{\mathrm{o}}$ on the eye.

**Energy is conserved during interference and diffraction**

In **interference and diffraction**, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the **principle of conservation of energy**.

**Resolving power of an objective lens**

The resolving power of an objective lens is measured by its ability to differentiate two lines or points in an object. The greater the resolving power, the smaller the minimum distance between two lines or points that can still be distinguished.

A parallel beam of light falling on a convex lens, because of diffraction, instead of getting focused to a point gets focused to a **spot of finite area**.

The pattern on the focal plane consists of a central bright region surrounded by concentric dark and bright rings.

The radius of the central bright region is approximately

$${\mathrm{r}}_{\mathrm{o}}=\frac{1.22\mathrm{\lambda}f}{2\mathrm{a}}=\frac{0.61\mathrm{\lambda}f}{\mathrm{a}}$$

Although the size of the spot is very small, it affects the limit of resolution of optical instruments like a telescope or a microscope.

For the two stars to be just resolved,

$$f\mathrm{\Delta}\mathrm{}\mathrm{}{\mathrm{r}}_{\mathrm{o}}=\frac{0.61\mathrm{\lambda}f}{\mathrm{a}}\mathrm{}\mathrm{}\Rightarrow \mathrm{}\mathrm{\Delta}\mathrm{}=\mathrm{}\frac{0.61\mathrm{\lambda}}{\mathrm{a}}$$

Thus Δθ will be small if the diameter of the objective is large. It is for this reason that for better resolution, a telescope must have a **large diameter objective**.

**For a microscope** the object is placed slightly beyond *f*, so that a real image is formed at a distance v. The magnification – ratio of image size to object size – is given by,

$$\mathrm{m}\mathrm{=}\frac{\mathrm{v}}{f}$$

Also, $\frac{D}{f}$ ≈ 2tan β, where 2β is the angle subtended by the diameter of the objective lens at the focus of the microscope.

When the separation between two points in a microscopic specimen is comparable to the wavelength λ of the light, the diffraction effects become important. The image of a point object will again be a diffraction pattern whose size in the image plane will be

$$\mathrm{v\theta}\mathrm{}\mathrm{=\; v}\frac{1.22\mathrm{\lambda}}{\mathrm{D}}\mathrm{}\mathrm{}$$

Two objects whose images are closer than this distance will not be resolved and will be seen as one. The corresponding minimum separation, **d**_{min}, in the object plane is given by

$${\mathrm{d}}_{\mathrm{m}\mathrm{i}\mathrm{n}}=\mathrm{}\frac{1.22\mathrm{\lambda}}{\mathrm{D}}\mathrm{}\times \frac{\mathrm{v}}{\mathrm{m}}$$

$$=\frac{1.22f\mathrm{\lambda}}{\mathrm{D}}\mathrm{}$$

$$=\frac{1.22\mathrm{\lambda}}{2\mathrm{tan}\beta}\mathrm{}\mathrm{}\mathrm{}\frac{1.22\mathrm{\lambda}}{2\mathrm{sin\; \beta}}$$

If the medium between the object and the objective lens has a refractive index n

$${\mathrm{d}}_{\mathrm{m}\mathrm{i}\mathrm{n}}=\mathrm{}\frac{1.22\mathrm{\lambda}}{2\mathrm{n}\mathrm{sin\; \beta}}$$

**Fresnel distance**

The angle of diffraction due to an aperture (i.e., slit or hole) of size *a* illuminated by a parallel beam is $\frac{\lambda}{a}$. This is the angular size of the bright central maximum. In travelling a distance *z*, the diffracted beam acquires a width $\frac{z\lambda}{a}$ due to diffraction.

The distance beyond which the divergence of beam of width ‘*a*’ becomes significant is called Fresnel distance, at which point, is denoted by z_{F}

$$\frac{{\mathrm{z}}_{\mathrm{F}}\mathrm{\lambda}}{\mathrm{a}}=\mathrm{a}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}{\mathrm{z}}_{\mathrm{F}}\mathrm{\lambda}=\mathrm{}{\mathrm{a}}^{2}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}{\mathrm{z}}_{\mathrm{F}}=\mathrm{}\frac{{\mathrm{a}}^{2}}{\mathrm{\lambda}}\mathrm{}$$

For distances much smaller than z_{F}, the spreading due to diffraction is smaller compared to the size of the beam. It becomes comparable when the distance is approximately z_{F}. For distances much greater than z_{F}, the spreading due to diffraction dominates over that due to ray optics (i.e., the size ‘*a*’ of the aperture). The ray optics is valid in the limit of wavelength tending to zero.

**Polarisation**

A light wave is an electromagnetic wave that travels through the vacuum of outer space. Light waves are produced by vibrating electric charges.

Ordinary light has electric vectors in all possible directions in a plane perpendicular to the direction of propagation of light. Hence it is called **unpolarised light.**

It can be considered a combination of vertical and horizontal components. Average of half its vibrations are in a horizontal plane and half of its vibrations are in a vertical plane.

The phenomenon of restructuring of electric vectors of light into a single direction is called **polarisation.**

When ordinary light is passed through a tourmaline, calcite or quartz crystal the transmitted light has electric vectors in a particular direction parallel to the axis of crystal. This light is called plane **polarised light**.

A plane containing the vibrations of polarised light is called **plane of vibration**.

A plane perpendicular to the plane of vibration is called **plane of polarisation**.

**Polarisation** can take place only in **transverse waves**.

**Polarisation by transmission**

When unpolarized light is transmitted through a **Polaroid filter**, it emerges with one-half the intensity and with vibrations in a single plane; it emerges as polarized light.

**Polarisation by scattering **

When we look at the light scattered by the molecules at 90^{o }to direction of incident light, we see only the perpendicular component of the light. The radiation scattered by the molecules is polarized perpendicular to the plane of the figure.

**Polarisation by reflection - Brewster’s law**

When unpolarised light is incident at an angle of **polarisation ( i**

_{B}

**)**on the interface separating air from a medium of refractive index

*n*, then reflected light becomes fully polarised, provided

*n = *tan* i*_{B}

Refractive index

*n* = tan *i*_{B} = $\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{C}}$,

where, **C** = critical angle.

If angle of refraction is r then,

*i*_{B }+ r = 90°

**Law of Malus**

When a beam of completely plane polarised light is incident on an analyser, the intensity of transmitted light from analyser is directly proportional to the square of the cosine of the angle between plane of transmission of analyser and polariser, i.e.,

I ∝ cos^{2 }θ

When ordinary light is incident on a polariser the intensity of transmitted light is half of the intensity of incident light.

When a polariser and analyser are perpendicular to each other, then intensity of transmitted light from analyser becomes 0.

**Polaroid**

It is a polarising film mounted between two glass plates.

**Uses of polaroid**

- It is used to produce polarised light.
- A polaroid is used to avoid glare of light in spectacles. Polaroids are used in sun glasses. They protect the eyes from glare.
- The polaroids are used in window panes of a trains and an aeroplane. They help to control the light entering through the window.
- The pictures taken by a stereoscopic camera, when seen with the help of polarized spectacles, create three dimensional effect.
- The windshield of an automobile is made of polaroid. Such a wind shield protects the eyes of the driver of the automobile from the dazzling light of the approaching vehicles.