If I have seen a little further it is by standing on the shoulders
of Giants.

–Sir Isaac Newton

In Physical Optics the topics included are: (i) Interference and diffraction (ii) Dispersion of light and the electromagnetic spectrum. These topics carry 5% of the total points.

You must remember the following equations to make you strong in answering multiple choice questions involving

**:***interference and diffraction***(1)**

**Refractive index(**is given given by

*n*) of medium 2 with respect to medium 1

**where**

*n = n*_{2}/*n*_{1}=*λ*_{1}/*λ*_{2}=*v*_{1}/*v*_{2}*λ*

_{1 }and

*λ*

_{2}are the wave lengths and

*v*

_{1}and

*v*

_{2}are the velocities of light in medium 1 and medium 2 respectively].

**(2)**The resultant displacement produced (at a particular point in a medium) by a number of waves is the vector sum of the displacements produced by each of the waves

**.**For instance, if two waves arriving at a point P produce displacements

*y*

_{1 }and

*y*

_{2}, the resultant displacement at P is the vector sum of

*y*

_{1}and

*y*

_{2}. If the interfering waves have amplitudes

*a*and

_{1}*a*

_{2}_{ }and they arrive at the point P with a phase difference of

*φ*,

**the resultant amplitude at P will be**

**[**

**, in accordance with vector addition. If the interfering waves have intensities**

*a*2_{1}^{2}+ a_{2}^{2}+*a*cos_{1}a_{2 }*φ*]^{1/2}*I*and

_{1}*I*

_{2}_{ }the resultant intensity (

*I)*at P will be given by

**, since the intensity is directly proportional to the**

*I*=*I*_{1}+*I*+ [2√(_{2}*I*_{1}*I*)] cos_{2}*φ**square*

**of the amplitude.**

If

*φ*= 0, 2π, 4π, 6π,…..(an*even*multiple of π), the intensity is**given by***maximum (I*_{max})

*I*

_{max}**=**

*I*_{1}+*I*+ [2√(_{2}*I*_{1}*I*)]_{2}
If

*φ*= π, 3π, 5π, 7π,…..(an*odd*multiple of π),, the intensity is**given by***minimum (I*_{min})

*I*

_{min}**=**

*I*_{1}+*I*–_{2}**[2√(**

*I*_{1}*I*)]_{2}

**(3)**Two sources of light (or any wave) are said to be coherent if they produce waves of the same frequency with a constant phase difference between them. (The condition of same frequency is implied in the condition of constant phase difference; yet it is usually specified when we define coherent sources). If two coherent sources produce waves of equal amplitude (

*a*) and therefore equal intensity (

*I*

_{0}), then the resultant intensity (

*I*) when they interfere is given by

*I***=**

*I*_{0}+*I*_{0}+ [2√(*I*_{0}*I*_{0})]cos*φ*= 2*I*_{0}(1+cos*φ*)
If the phase difference between the interfering waves (of equal intensity) is zero or an

*even multiple*of π (in other words,*if the path difference between the waves is zero or an integral multiple of the wave length λ*), the intensity is**given by***maximum*(*I*_{max})

*I*_{max }=**2**…..(Path difference = nλ where n = 0,1,2,3,4,…..etc)

*I*_{0}(1+1) = 4*I*_{0 }
If the phase difference between the interfering waves (of equal intensity) is an

*odd multiple*of π (in otherwords,*if the path difference between the waves is an**odd multiple**of half wave length*), the intensity (*I*_{min}) is zero given by

*I*_{min }=**2**

*I*_{0}(1–**1) = 0**……(Path difference = (2n+1)λ/2 where n = 0,1,2,3,4,….etc)

**(4)**In the interference fringe pattern produced by

**, the distance of the**

*Young’s double slit**n*from the central (zero order) bright fringe is given by

^{th}bright fringe (x_{n})
where

*λ*is the wave length of light,*D*is the distance between the screen (on which the fringes are formed) and the double slit and ‘*d*’ is the separation between the two slits S_{1 }and S_{2}forming the double slit (fig.).
The distance of the

*n*from the central (zero order) bright fringe is given by^{th}dark fringe (x_{n})**, where**

*x*(2_{n}=*n*+ 1)*λD/*2*d***n = 0, ±1,**

**±2, ±3….etc. [We use ± sign for the integers to indicate the fringes on both sides of the central fringe]**

The

**is the distance between the centres of consecutive brigt fringes or dark fringes and is given by***fringe width β**β*=

*x*

_{n}_{+1 }–

*x*

_{n}
Therefore,

*β**=**λD/d***is independent of the distance of the screen and is given by**

*Angular fringe width***=**

*β*_{θ}

*β/D = λ/d***(5)**When light

**traveling through a rarer medium gets reflected from the surface of a denser medium, it suffers a phase change of**

*π*, which is equivalent to a path change (path difrference) of

*λ/2*

**(6)**Colours in thin transparent films can be observed in

*reflected*light as well as

*transmitted*light. The condition for

**is**

*brightness in the reflected light***2**

**=**

*μtcosr***(2**where

*n*+ 1)*λ/2**μ*is the refractive index of the film,

*t*is its thickness,

*r*is the angle of refraction in the film and n = 0,1,2,3,4,….etc.

[A film of negligible thickness cannot satisfy the above condition and hence it cannot appear bright in reflected light. (It will appear dark)].

The condition for

**is***brightness in transmitted light***2**

**=**

*μtcosr***where n = 0,1,2,3,4,….etc.**

*nλ*
[A film of negligible thickness can satisfy the above condition (for n = 0) and hence it will appear bright in transmitted light].

You

**should note that****in reflected light are produced by the interference between***Newton’s rings***the light waves reflected from the two surfaces of an air film contained in between a glass plate and a lens.****(7)**In the

*diffraction pattern*produced by a

*single slit*, the

**where**

*angular separation between the first minima is 2λ/a**a*is the

*width*of the slit. The angular width of the central maximum is therefore

**2**.

*λ/a*
The centre of the central maximum is at O (fig.). On either side of O, positions of

**are obtained when the angle***minimum intensity**θ =***±****where***nλ/a**n*= 1,2,3…etc.*secondary maxima*are obtained on either side of O corresponding to

*θ =*

*±*(2

*n*+ 1)

*λ/2a*where n = 1,2,3….etc.

Therefore, the secondary maxima are located at

*θ =*±3*λ/2a,*±5*λ/2a,*±7*λ/2a….*etc. The intensity of the secondary maxima goes on decreasing. [The first secondary maximum has intensity less than 5% of the intensity of the central maximum].**(8)**In a plane diffraction grating, the n

^{th}principal

*maximum*is obtained at an angle θ given by

**where**

*(a+b) sinθ = nλ**a*is the width of the slit and

*b*is the width of the opaque region between the slits. [

*a+b*is called the grating element]

In the next post questions on wave optics will be discussed. Here is a question which you must be able to work out in no time if you have understood the phenomenon of interference.

A student prepared a Young’s double slit by drawing two parallel lines (with a separation less than a millimeter) on a smoked glass plate. [For smoking, the plate was held above the flame of a kerosene lamp]. Accidentally he used two different pins to draw the lines so that the widths of the slits (regions from where smoke was removed by the pin) were in the ratio 4:1. What will be the ratio of the intensity at the interference maximum to that at the interference minimum?

(a) 4

(b) 2

(c) 3

(d) 6

(e) 9

Work out the above question. I’ll be back shortly with the solution and of course, more solved questions.

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