(i) The polaroid which polarises the light passing through it is called a polariser.
(ii) The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.

Polarisation by selective absorption
Selective absorption is the property of a material which transmits waves whose electric fields vibrate in a plane parallel to a certain direction of orientation and absorbs all other waves.

f_o = 125 cm; f_e = 2 cm; m = ? L = ?; θ_i = ?
(a) Equation for magnification of telescope,
\(m=\cfrac { { f }_{ o } }{ { f }_{ e } } \) 
Substituting, \(m=\cfrac { 125 }{ 2 } =62.5\) 
(b) Equation for approximate length of telescope, L = f_o+ f_e
Substituting, L = 125 + 2 = 127 cm = 1.27 m
(c) Equation for angular magnification,\(m=\cfrac { { \theta }_{ 1 } }{ { \theta }_{ 0 } } \) 
Rewriting, \({ \theta }_{ 1 }=m\times { \theta }_{ 0 }\) 
Substituting,
\({ \theta }_{ i }=62.5\times 1'=62.5'=\cfrac { 62.5 }{ 60 } =1.04^{ o }\) = 1^o2'30''

D = 25 cm;
The magnifying lens must be a convex lens of positive focal length
f = 5 cm;
For closest object distance u', the image distance, v is, –25 cm. (near point, v = –D)
For farthest object distance u', the corresponding image distance, v' is infinity.
(a) To find closest distance between lens and book, we can use lens equation,\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
Rewriting for closest object distance \(\cfrac { 1 }{ u } =\cfrac { 1 }{ v } -\cfrac { 1 }{ f } \) 
Substituting,
\(\cfrac { 1 }{ u } =\cfrac { 1 }{ -25 } -\cfrac { 1 }{ 5 } =\cfrac { 1 }{ 25 } =\cfrac { 1 }{ 5 } \left( \cfrac { -1-5 }{ 25 } \right) =-\cfrac { 6 }{ 25 } \)
\(u=-\cfrac { 25 }{ 6 } =4.167cm\)
The closest distance between the lens and the book is, u = –4.167 cm
To find farthest object distance, lens equation is, \(\cfrac { 1 }{ v' } -\cfrac { 1 }{ u' } =\cfrac { 1 }{ f' } \)
Rewriting for farthest object distance,\(\cfrac { 1 }{ u' } =\cfrac { 1 }{ v' } -\cfrac { 1 }{ f' } \)
Substituting,\(\cfrac { 1 }{ u' } =\cfrac { 1 }{ \infty } -\cfrac { 1 }{ 5 } ;u=-5cm\) 
The farthest distance at which the person can keep the book is, u' = -5 cm.
(b) To find magnification in near point focusing, \(m=1+\cfrac { D }{ f } =1+\cfrac { 25 }{ 5 } =6\)
To find magnification in normal focusing, \(m=\cfrac { D }{ f } =\cfrac { 25 }{ 5 } =5\)

λ = 5000 Å = 5000 x 10 ^-10 m; sin 30^o = 0.5; n = 1; a =?
Equation for diffraction minimum is,
asin θ = nλ
The central maximum is spread up to the first minimum. Hence, n = 1
Rewriting, \(a=\cfrac { \lambda }{ sin\theta } \) 
Substituting, \(a=\cfrac { 5000\times { 10 }^{ -10 } }{ 0.5 } \) 
a = 1 x 10^-6m = 0.001 x 10^-3 m = 0.001 mm.

λ = 500 nm = 500 x 10^-9 m;
a = 0.2 mm = 0.2 x 10^-3 m;
D = 60 cm = 60 x 10^-2 m
(i) Equation for diffraction minimum is, a sin θ = nλ
The central maximum is spread up to the first minimum. Hence, n = 1

Rewriting, \(sin\theta =\cfrac { \lambda }{ a } \) (or) \(\theta =sin^{ -1 }\left( \cfrac { \lambda }{ a } \right) \) 
Substituting,
\(\theta ={ sin }^{ -1 }\left( \cfrac { 500\times { 10 }^{ -9 } }{ 0.2\times { 10 }^{ -3 } } \right) ={ sin }^{ -1 }\left( 2.5\times { 10 }^{ -3 } \right) \) 
\(\\ \theta =0.0025rad\) 
(ii) To find the value of y₁ for central maximum, which is spread up to first minimum with (n = 1) is, a sin θ = λ
As θ is very small, \(sin\theta \approx tan\theta =\cfrac { { y }_{ 1 } }{ D } \) 
\(a\cfrac { { y }_{ 1 } }{ D } =\lambda \) rewriting, \({ y }_{ 1 }=\cfrac { \lambda D }{ a } \) 
Substituting,
\({ y }_{ 1 }=\cfrac { 500\times { 10 }^{ -9 }\times { 10 }^{ -2 } }{ 0.2\times { 10 }^{ -3 } } =1.5\times { 10 }^{ -3 }=1.5mm\)
 To find the value of y₂ for second minimum with (n = 2) is, a sin θ = 2λ
\(a\cfrac { { y }_{ 2 } }{ D } =2\lambda \) rewriting ,\({ y }_{ 2 }=\cfrac { 2\lambda D }{ a } \) 
Substituting,
\({ y }_{ 2 }=\cfrac { 2\times 500\times { 10 }^{ -9 }\times 60\times { 10 }^{ -2 } }{ 0.2\times { 10 }^{ -3 } } =3 \times 10^{-3}=3mm\)
The distance between the central maximum and second minimum is, y₂ – y₁
y₂ – y₁ = 3 mm – 1.5 mm = 1.5 mm

Note: The above calculation shows that in the diffraction pattern caused by single slit, the width of each maximum is equal with central maximum as the double that of others. But the bright and dark fringes are not of equal width.

λ = 589 nm = 589 x 10⁻⁹ m
For the film to have strong reflection, the reflected waves should interfere constructively. The least optical path difference introduced by the film should be λ/2. The optical path difference between the waves reflected from the two surfaces of the film is 2μd. Thus, for strong reflection, 2μd = λ/2 [As given in equation 6.145. with n = 1]
Rewriting, \(d=\frac{\lambda}{4 \mu}\) 
Substituting, \(d=\frac{589 \times 10^{9}}{4 \times 1.25}=117.8 \times 10^{-9}\) 
d = 117.8 x 10^-9 = 117.8 nm
For the film to be anti-reflecting, the reflected rays should interfere destructively. The least optical path difference introduced by the film should be λ. The optical path difference between the waves reflected from the two surfaces of the film is 2μd. For strong reflection, 2μd = λ [As given in equation 6.146. with n = 1]
Rewriting, \(d=\cfrac { \lambda }{ 2\mu } \)
Substituting, \(d=\cfrac { 589\times { 10 }^{ 9 } }{ 2\times 1.25 } =235.6\times { 10 }^{ -9 }\) 
d = 235.6 x 10^-9 = 235.6 nm

λ₁ = 560 nm = 560 x 10^-9m;
λ₁ = 420 nm = 420 x 10^-9m;
D = 1 m; d = 3 mm = 3 x 10^-3m
Here, n and λ are inversely proportional for a given y.
Here, n^th order bright fringe of longer wavelength λ₁ coincides with (n+1)^th order bright fringe of shorter wavelength λ₂.
Equation for n^th bright fringe is, \({ y }_{ n }=n\cfrac { \lambda D }{ d } \) 
Here, \(n\cfrac { { \lambda }_{ 1 }D }{ d } =(n+1)\cfrac { { \lambda }_{ 2 }D }{ d } \) (as λ₁>λ₂)
\({ n\lambda }_{ 1 }=\left( n+1 \right) { \lambda }_{ 2 }\) (or) \(\cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\cfrac { (n+1) }{ n } ;1+\frac{1}{n}=\frac{\lambda_1}{\lambda_2}\) 
\(1+\cfrac { 1 }{ n } =\cfrac { 560\times { 10 }^{ -9 } }{ 420\times { 10 }^{ -9 } } \) (or) \(1+\cfrac { 1 }{ n } =\cfrac { 4 }{ 3 } \) 
\(\frac{1}{n}=\frac{1}{3}\) (or) n = 3
Thus, the 3^rd bright fringe of λ₁ and the 4^th bright fringe of λ₂ coincide at the least distance y.
The least distance from the central fringe where the bright fringes of the two wavelengths coincide is, \(y_{n}=n \frac{\lambda D}{d}\)  
\(y_{n}=3 \times \frac{560 \times 10^{-9} \times 1}{3 \times 10^{-3}}=560 \times 10^{-6} \mathrm{~m}\) 
\(y_{n}=0.560 \times 10^{-3} \mathrm{~m}=0.560 \mathrm{~mm}\) 

d = 0.15 mm = 0.15 x 10^-3 m; D = 2 m;
λ = 450 nm = 450 x 10^-9 m; RI = 4/3
(a) Equation for n^th bright fringe,
\({ y }_{ n }=n\cfrac { \lambda D }{ d } \)
Distance of 2^nd bright fringe,
\({ y }_{ 2 }=2\times \cfrac { 450\times { 10 }^{ -9 }\times 2 }{ 0.15\times { 10 }^{ -3 } } \) 
y₂ = 12 x 10^-3m = 12 mm
Equation for n^th dark fringe,
\({ y }_{ n }=\cfrac { \left( 2n-1 \right) }{ 2 } \cfrac { \lambda D }{ d } \)
Distance of 3^rd dark fringe,
\({ y }_{ 2 }=\cfrac { 5 }{ 2 } \times \cfrac { 450\times { 10 }^{ -9 }\times 2 }{ 0.15\times { 10 }^{ -3 } } \) 
y₂ = 15 x 10^-3 m = 15 mm
(b) Equation for fringe width, \(\beta =\cfrac { \lambda D }{ d } \) 
Substituting, \(\beta =\cfrac { 450\times { 10 }^{ -9 }\times 2 }{ 0.15\times { 10 }^{ -3 } } \) 
\(\beta =6\times { 10 }^{ -3 }m=6mm\) 
(iii) The fringe width will increase as D is increased,
\(\beta =\cfrac { \lambda D }{ d } \) (or) \(\beta \propto D\)
(iv) The fringe width will decrease as the setup is immersed in water of refractive index 4/3
\(\beta =\cfrac { \lambda D }{ d } \) (or) \(\beta \propto \lambda\)
 We know that, \(\lambda '=\cfrac { \lambda }{ n } \) 
\(\cfrac { \beta ' }{ \beta } =\cfrac { \lambda ' }{ \lambda } =\cfrac { \lambda /RI}{ \lambda } =\cfrac { 1 }{ RI } \) (or) \(\beta '=\cfrac { \beta }{ RI } =\cfrac { 6\times { 10 }^{ -3 } }{ 4/3 } \) 
\(\beta '=4.5\times { 10 }^{ -3 }m=4.5mm\) 

Amplitudes, a₁ = 5, a₂ = 3
Resultant amplitude,
\(A=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+2{ a }_{ 1 }{ a }_{ 2 }cos\varphi } \) 
Resultant amplitude is maximum when,
\(\phi =0,cos0=1,{ A }_{ max }=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+{ 2a }_{ 1 }{ a }_{ 2 } } \) 
 \(\\ { A }_{ max }=\sqrt { \left( { a }_{ 1 }+{ a }_{ 2 } \right) ^{ 2 } } =\sqrt { \left( 5+3 \right) ^{ 2 } } =\sqrt { \left( 8 \right) ^{ 2 } } \) 
= 8 units
Resultant amplitude is minimum when
\(\phi =\pi,cos\pi=1,{ A }_{ max }=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+{ 2a }_{ 1 }{ a }_{ 2 } } \)
\({ A }_{ max }=\sqrt { \left( { a }_{ 1 }-{ a }_{ 2 } \right) ^{ 2 } } =\sqrt { \left( 5-3 \right) ^{ 2 } } =\sqrt { \left( 2 \right) ^{ 2 } } \)
= 2units
\(I\infty { A }^{ 2 }\) 
\(\cfrac { { I }_{ max } }{ { I }_{ min } } =\cfrac { \left( { A }_{ max } \right) ^{ 2 } }{ \left( { { A }_{ min } } \right) ^{ 2 } } \)
Substituting,
 \(\cfrac { { I }_{ max } }{ { I }_{ min } } =\cfrac { \left( 8 \right) ^{ 2 } }{ \left( 2 \right) ^{ 2 } } =\cfrac { 64 }{ 4 } 16\) (or)
\({ I }_{ max }:{ I }_{ min }=16:1\)

\(\mathrm{m}_{\alpha}=100 ; \mathrm{f}_{o}=0.5 \mathrm{~cm} ; \mathrm{f}_{\mathrm{e}}=? \)
\(\mathrm{~L}_{\alpha}=6.5 \mathrm{~cm}, \mathrm{D}=25 \mathrm{~cm} \)
When the image is formed at infinity,
\(\mathrm{m}_{\alpha}=\frac{\left(\mathrm{L}_{\alpha}-\mathrm{f}_{0}-\mathrm{f}_{\mathrm{c}}\right) \mathrm{D}}{\mathrm{f}_{0} \mathrm{f}_{\mathrm{e}}} \)
\(100 =\left(\frac{6.5-0.5-f_{e}}{0.5 \times f_{e}}\right) \times 25 \)
\(=\left(\frac{6-f_{e}}{0.5 f_{e}}\right) \times 25 \)
\(100 \times 0.5 f_{e} =150-25 f_{e} \)
\(50 f_{e} =150-25 f_{e} \)
\(75 f_{e} =150 \)
\(f_{e} =150 / 75=2 \mathrm{~cm} \)

Advantages:
Only one surface it to be polished and maintained. Support can be given from the entire back of the miror rather than only at the rim for lens. Mirrors weigh much less compared to lens.
Disadvantages:
The objective mirror would focus the light inside the telescope tube.

(i) Polaroids are used in goggles and cameras to avoid glare of light.
(ii) Polaroids are useful in 3D pictures i.e., in holography.
(iii) Polaroids are used to improve contrast in old oil paintings.
(iv) Polaroids are used in optical stress analysis.
(v) Polaroids are used as window glasses to control the intensity of incoming light.
(vi) Polarised laser beam acts as needle to read/ write in compact discs (CDs).
(vii) Polarised lights is used in liquid crystal display (LCD).

When a beam of plane polarised light of intensity (I_o) is incident on an analyser, the intensity of light (I) transmitted from the analyser varies directly as the square of the cosine of angle between the transmission axes of polariser and analyser.
\(I={ I }_{ o }cos^{ 2 }\theta \)
Consider the plane of polariser and analyser are inclined to each other at an angle ፀ. Let I_o be the intensity and 'a' be the amplitude of the electric vector transmitted by the polariser. The amplitude 'a' of the incident light has two rectangular components, (acosθ) and (asinθ) which are the parallel and perpendicular components to the axis of transmission of the analyser. Only the component (acosθ) will be transmitted by the analyzer.
According to Malus's law
\(I\propto \left( acos\theta \right) ^{ 2 }\) 
\(I=k\left( acos\theta \right) ^{ 2 }\) 
Where k is constant of proportionality,
I = ka² cos² θ
I = I_o = cos² θ
Where I_o = ka²  is the maximum intensity of light transmitted from the analyser.

Plane Polarised Light:
In plane polarised light the intensity varies from maximum to zero for every rotation of 90° of the analyser.
Partially Polarised Light:
If the intensity of light varies between maximum and minimum for every rotation of 90° of the analyser, the light is said to be partially polarised light.
Unpolarise Light:
The light is coming out from the source before entering a polarised is called unpolarized light.

Fresnel's distance is the distance upto which the ray optics is valid in terms of rectilinear propagation of light.
(or)
Fresnel's distance is the distance upto which ray optics is obeyed and beyond which ray optics is not obeyed but, wave optics becomes significant,

The diffraction equation for first minimum is, sinθ \(=\frac{ \lambda}{2};\)
When θ is small, θ \(=\frac{ \lambda}{2}\)
From the definition of Fresnel's distance, 2θ\(=\frac{a}{z}\) (or) θ \(=\frac{a}{2z}\)
Equating the above two equation for θ gives, \(\frac{\lambda}{a}=\frac{a}{2z}\)
After rearranging, we get Fresnel's distance z as,
\(z=\cfrac { { a }^{ 2 } }{ 2\lambda } \)

Condition for bright fringe (or) maxima :
The condition for the point P to have a constructive interference (or) be a bright fringe Is,
Path diference, δ = nλ    Where, n = 0, 1, 2,....
\(\therefore\frac{dy}{D}=n\lambda\)
\(y=n\frac{\lambda D}{d}(or)y_n=n\frac{\lambda D}{d}\)   .....(4)
This is the condition for the point P to have a bright fringe. The distance y_n is the distance or the n^th bright fringe from the point O.
Condition for dark fringe (or) minima:
The condition for the point P to have a destructive interference (or) be a dark fringe is,
Path difference, δ = \((2n-1)\frac{\lambda}{2}\)    Where, n = 1, 2, 3....
\(\therefore\frac{dy}{D}=(2n-1)\frac{\lambda}{2}\)
\(y=\left(\frac{(2n-1)}{2} \frac{\lambda D}{d}\right)(or)\left(\frac{(2 n-1)}{2} \frac{\lambda D}{d}\right) \)    .....(5)
This is the condition for the point P to have a dark fringe. The distance y_n is the distance of the n^th dark fringe from the point O
Bandwidth:
The bandwidth \((\beta)\) is defined as the distance between any two consecutive bright or dark fringes. 
\(\beta=y_{(n+1)}-y_{n}=\left((n+1) \frac{\lambda D}{d}\right)-\left(n \frac{\lambda D}{d}\right) \)
\(\beta=\frac{\lambda D}{d} \)    .....(6)
Bright and Dark tinges are of same width equally spaced on either side of the central bright fringe.