n = tan i_p = tan 60^o = \(\sqrt{3}\)

2n t cos r = (2m + 1) \(\frac{\lambda}{2}\)
For maximum
m =  2 (For visible region), n - refractive index.
cos r = cos 0 = 1
t = 5 x 10^-5 x 10^-2 = 5 x 10^-7 m
\(n=\frac{(2m+1)\frac{\lambda}{2}}{2t}=\frac{5\lambda}{2 \times 2 \times t}\)
\(=\frac{5 \times5320\times10^{-10}}{4 \times 5 \times 10^{-7}}\)
\(=\frac{5 \times5320\times10^{-10}}{20}=1330 \times 10^3\)
n = 1.330

I = l₁ + l₂ + 2\(\sqrt{I_1I_2}\)cos θ
If cos θ = cos 0 = l, I is max
= I+ 4I + 2\(\sqrt{41^2}\) cos 0
= 5I + 4I = 91
If cos π = -1, I is min
I_min = I + 4I + 2\(\sqrt{41^2}\) cos π
= 5I + 4I(-1)
= 5I + 4I = I
(I_max, I_min)= (9I, I)

d' = 2d, β' = β, D' = ?
W.K.T, Fringe width
\(\beta = \frac{D\lambda}{d} \Rightarrow D' = \frac{Dd'}{d}\)
\(D' = \frac{D2d}{d}=2D\)

Refractive index for violet is more and wavelength for violet is very low comparing other colours. So, the letter which appears to be raised more is violet.

ß₁ = 7.2 mm, ß₂ = 8.1 mm, λ₁ = 600 nm, λ₂ =?
Dividing, we get  \(\beta = \frac{D\lambda}{d}(or)\beta∝\lambda\)
\(\frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2}\)
\(\therefore, { \lambda }_{ 2 }=\frac{\beta_2\times \lambda_1}{\beta_1}=\cfrac { 8.1\times10^{-3} \times 600 \times10^{-9}}{ 7.2 \times10^{-3}} \)
\(=\cfrac { 9 \times 600\times10^{-9} }{ 8 } =675\times 10^{-9 }\Rightarrow\lambda_2= 675nm\)

Heat waves are transverse or electromagnetic in nature whereas sound wave are not. Polarisation is possible only for transverse waves.

Yes, since magnification depends upon the focal length and focal length depends on the color and different colours have different wavelengths (i.e., different refractive indices.)
\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
(By Lens Makers Formula) Also, magnification of the compound microscope
\(M=\cfrac { -L }{ { f }_{ o } } \left( 1+\cfrac { D }{ { f }_{ e } } \right) \)

⋋₂ = 600 x 10^-9 m, ⇒ d = 1 x 10^-3 m, D =2m
n⋋ = dsinθ
For first dark fringe
n = 1 For minimum
d sinθ = ⋋, here
CO = Nc (approximately)

From Fig, sinθ \(=\frac{x/2}{D} =x/2D\)
\(sin \theta =\frac{ \lambda}{d} \)
\(\lambda/d=x/2D\)
\(\therefore x =\frac{2D\lambda}{d}=\frac{2 \times 2 \times 600\times 10^{-9}}{ 10^{-3}} \)
\(x=2.4 \times10^{-3}m =2.4\mathrm{~mm} \)

Hypermetropia:
A person suffering from farsightedness (or) hypermetropia (or) hyperopia cannot see closer object clearly.
Remedy:
Hypermetropia occurs when the eye lens has long focal length or shortening of the eyeball than usual.

Myopia:
A person suffering from nearsightedness or myopia cannot see distant objects clearly. This may due to short focal length of the eye lens or larger diameter of the eyeball than usual.
Remedy:
These people have difficulty in relaxing their eye more than what is needed to overcome this difficulty. They need correcting lens, which should be concave lens.

The phenomenon of addition or superposition of two light waves which produces increase in intensity at some points and decrease in intensity at some other points is called interference of light.

Quantum theory of light:
Quantum theory states that light waves consist of small packets of energy called photons. The energy associated with each photon is E = hv, Where 'h' is Planck's constant (h = 6.625 x 10^-34 J s) and v is frequency of electromagnetic radiation.

Electromagnetic wave theory of light:
Light is an electromagnetic wave which is transverse in nature carrying electromagnetic energy. No medium is necessary for the propagation of electromagnetic waves. All the phenomenon of light could be successfully explained by this theory.

(i) If the scattering of light is by atoms and molecules which have size a very less than that of the wavelength λ of light a <.
(ii) The intensity of Rayleigh's scattering is inversely proportional to the fourth power of wavelength.
\(1\propto \cfrac { 1 }{ { \lambda }^{ 4 } } \)

(i) Similar to a rectangular slit, when a circular aperture or opening (like a lens or the iris of our eye) forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that become fainter while moving away from the center as shown in Figure. These are known as Airy's discs
\(asin\theta =1.22\lambda \) 

(ii) Here, the numerical value 1.22 comes for central "maximum formed by circular apertures.
For small angles, sin θ = θ
\(a\theta =1.22\lambda \) 
Rewriting further,
\(\theta =\cfrac { 1.22\lambda }{ a } \ and \cfrac { { r }_{ o } }{ f } =\cfrac { 1.22\lambda }{ a } \)
\({ r }_{ o }=\cfrac { 1.22\lambda f }{ a } \)

(i) An astronomic telescope used to get the magnification of distant astronomical objects like stars, planets, moon etc. 
(ii) The image formed by astronomicaltelescope willbe inverted. It has an objective of long focal length and a much larger aperture than the eyepiece as shown in Figure.
(iii) Light from a distant object enters the objective and a real image is formed in the tube at its second focal point.
(iv) The eyepiece magnifies this image producing a final inverted image.

Magnification of astronomical telescope:
The magnification m is the ratio of the angle β subtended by the image to the angle α which subtended by the object with the principal axis
\(m=\cfrac { \beta }{ \alpha } \)           ......(1)
From the diagram, ,\(\alpha=\frac{h}{f_e} \ and \ \beta=\frac{h}{f_e}\)                   ...................(2)
\(m=\cfrac { { f }_{ 0 } }{ { f }_{ e } } \)  .........................(3)
The length of the telescope is approximate, L = f₀ + f_e.

(i) Consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY as shown in Figure.

(ii) The incident wavefront is AB and the reflected wavefront is A'B' in the same medium. These wavefronts are perpendicular to the incident rays L, M and reflected rays L', M' respectively.
(iii) By the time point A of the incident wavefront touches the reflecting surface, point B is yet to travel a distance BB' to touch the reflecting surface a B'.
(iv) When point B falls on the reflecting surface at H', point A would have reached A.
(v) This is applicable to all the points on the wavefront. Thus, the reflected wavefront A'B' emanates as a plane wavefront. The two normals Nand N' are considered at the points where the rays Land Mfallon the reflecting surface.
(vi) As reflection happens in the same medium, the speed of light is the same before and after the reflection.
(vii) Hence, the time is taken for the ray to travel from B to B' is the same as the time taken for the ray to travel from A to A'.
(viii) Thus, the distance BB' is equal to the distance AA'; (A~A' = BB').
(a) The incident rays, the reflected rays, and the normal are in the same plane.
(b) Angle of incidence,\(\angle i=\angle NAL={ 90 }^{ o }-\angle NAB=\angle BAB'\) 
Angle of reflection,
∠r= ∠N' B' M' = 900 - ∠N' B' A' = A' B' A' 
(ix) For the two right-angle triangles, ΔABB' and ΔB' A' A', the right angles, ∠B and ∠A' are equal, (∠B and∠A = 900); the two sides, ∠A' and ∠B' are equal, (AA'= BB'); the side AB' is common.
(x) Thus, the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB' and A' B' A' must also be equal.
i = r
Hence, the laws of reflection are proved.

Reflection does not change wavelength.
Frequency remains unchanged both in reflection and refraction
ஃ λ =  5000 Å. Again \(v=\cfrac { c }{ \lambda } =\cfrac { 3\times { 10 }^{ 8 } }{ 5000\times { 10 }^{ -10 } } Hz\) 
= 6 x 1014 Hz
Again i = rand i + r = 90°.
ஃ i = 45°.

Given data:
In first case, f_e.= 0.03 m
f₀= 0.04 m , L = 0.2 m
\(m=\cfrac { L }{ { f }_{ o } } \left( 1+\cfrac { 0.25 }{ { f }_{ o } } \right) \) 
For m to be the same in both cases,
= \(\cfrac { L }{ 0.03 } \left( 1+\cfrac { 0.25 }{ 0.04 } \right) \) 
= \(\cfrac { 0.2 }{ 0.04 } \left( 1+\cfrac { 0.25 }{ 0.04 } \right) \) 
= \(\cfrac { L }{ 0.03 } \times \cfrac { 29 }{ 4 } =\cfrac { 0.2 }{ 0.04 } \times \cfrac { 28 }{ 3 } \) 
\(L=\cfrac { 5.6 }{ 29 } =0.193m\)

\(\cfrac { 1 }{ -25 } -\cfrac { 1 }{ { u }_{ e } } =\cfrac { 1 }{ 5 } \) 
or \({ u }_{ e }=-\cfrac { 25 }{ 6 } cm\) 
\({ v }_{ o }=20-\cfrac { 25 }{ 6 } =\cfrac { 95 }{ 6 } cm\) 
\(\cfrac { 1 }{ { v }_{ 0 } } =\cfrac { 1 }{ { u }_{ 0 } } =\cfrac { 1 }{ { f }_{ 0 } } \) 
\(\cfrac { 6 }{ 95 } -\cfrac { 1 }{ { u }_{ 0 } } =\cfrac { 1 }{ 2 } ,{ u }_{ 0 }=2.29cm\) 
\(m=-\cfrac { 95/6 }{ 190/83 } \left( 1+\cfrac { 25 }{ 5 } \right) =-41.5\)

Formula:
From the condition of diffraction, 
a sin θ= nλ.. (for minima)
= \(\left( n+\cfrac { 1 }{ 2 } \right) \lambda \) 

Provided n = 1, 2, 3 ... and n = 0 for central
maximum
From the condition of minima
a sin θ = λ. (n = 1)
Since the value of λ. is nm, so
\(a.\theta =\lambda \Rightarrow a.\cfrac { y }{ D } =\lambda \left[ amgle=\cfrac { arc }{ radius } \right] \) 
= 2 x 10^-4 m.

Formula
We know \({ i }_{ c }=\cfrac { 1 }{ \mu } \)
 \(\mu =\cfrac { 1 }{ { sini }_{ c } } =\cfrac { 1 }{ { sin45 }^{ o } } =\sqrt { 2 } \) 
According to Brewster's law 
\({ i }_{ p }=\mu =\sqrt { 2 } \) 
\(\Rightarrow { i }_{ p }={ tan }^{ -1 }\sqrt { 2 } \) 
= tan^-1(1.414) ≅ 510 40^o
When light is incident at an angle i_p the corresponding angle of refraction 'r' is given by
i_p + r = 90^o
ஃ r = 90^o- (51^o40^') = (38^o 20')

i) The spectrometer is an optical instrument used to analyse the spectra of different sources of light, to measure the wavelength of different colours and to measure the refractive indices of materials of prisms.
ii) It basically consists of three parts namely. They are (i) collimator, (ii) prism table and (iii) Telescope
Adjustments of the spectrometer
(i) The following adjustments must be done in a spectrometer before doing the experiment.
(a) Adjustment of the eyepiece: 
The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.
(b) Adjustment of the telescope:
The telescope is adjusted to' receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.
(c) Adjustment of the collimator: 
The telescope is brought in line with the collimator. The distance between the illuminated slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire.
(d) Levelling the prism table:
The prism table is brought to the horizontal level by adjusting the levelling screws and it is ensured by using sprit level.

(i) A simple microscope is a single magnifying (converging) lens of small focal length. To get an erect, magnified and virtual image of the object.
(ii) For this the object is placed between the focal length Fand P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.
(a) Near point focusing:
The eye is least strained when image is formed at near point,i.e. 25 cm. The near point is also called as least distance of distinct vision. This is shown in Figure.

Magnification in near point focusing:
(i) Object distance u is less than f
(ii) The image distance is the near point D. The magnification m is given by the relation,
 \(m=\cfrac { v }{ u } \)                          ...............(1)
Substituting, V = - D and u= - u, as both the distances are measured to the left of the lens. Hence,
\(m=\cfrac { -D }{ -u }\)  
\(m=\cfrac { D }{ u } \)                          ...............(2)
Using lens equation, W.K.T, m = 1 - (v/f)
Substiuting v = -D gives, \(\\ m=1+\cfrac { D }{ f } \)                         ..................(3)
This is the magnification for near point focusing.
(b) Normal focusing :
(i) The eye is most relaxed when the image is formed at infinity. The focusing is called normal focusing when the image is formed at infinity. This is shown in Figure (b).

Magnification in normal focusing (angular magnification):
(ii) The angular magnification is defined as the ratio of angle θ₁ subtended by the image with aided eye to the angle θ₀ subtended by the object with unaided eye.
\(m=\cfrac { { \theta }_{ 1 } }{ { \theta }_{ 0 } } \)                 .........(2)
For unaided eye shown in Figure (a),
\(tan\theta _{ 0 }\approx { \theta }_{ 1 }=\cfrac { h }{ D } \)                     ................(3)
For aided eye shown in Figure(b).
\(tan\theta _{ i }={ \theta }_{ i }=\cfrac { h }{ f } \)                        ...................(4)
The angular magnification is,
\(m=\cfrac { { \theta }_{ i } }{ { \theta }_{ o } } =\cfrac { h/f }{ h/D } \) 
\(m=\cfrac { D }{ f } \)              ..............(5)
This is the magnification for normal focusing.

(i) The effect of diffraction has an adverse effect in the sharpness of the image tormed.
(ii) There is always a spread of central maximum in the image for every point of the object, for every point of the object acts as a point source.
(iii) The condition for central maximum (or first minimum) produced by rectangular slit is given by the equation,
\(a \sin \theta=\lambda\)   ....(1)
(iv) But, a circular slit (aperture) produces diffraction pattern of concentric circles as shown in Figure.
(v) These are known as Airy's discs. Most of the optical instruments form images of objects only through the circular slits.

(vi) The condition for central maximum (or) first minimum for circular slit is,
\(\text { a } \sin \theta=1.22 \lambda\)  .....(2)
(vii) Here, the numerical value 1.22 appears in the expression for central maximum (or) first minimum formed by circular slits.
For small angles,\(sin\theta = \theta\), the above equation becomes,
\(a\theta = 1.22\lambda\)
Rewriting further,
\(\theta=\frac{1.22 \lambda}{a}\)   ......(3)
Form thegeometry, \(\theta=\frac{r_0}{f}\)
Substituting for in equation (3) and rearranging gives
\(r_0=\frac{1.22 \lambda f}{a}\)  ....(4)
(viii) For example, let two point-sources of light close to cach other form image on a screen. The diffraction pattern of one point-source may overlap with another and produce a blurred image (or) un-resolved image as shown in Figure (a). To obtain a quality image (or) well resolved image, the two point-sources must be kept apart in such a way that their diffraction patterns do not overlap as shown in Figure (c).
(ix) According to Rayleigh's criterion, the two points on an image are said to be just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other and vice-versa as shown in Figure (b).

(i) Gratting has multiple slits with equal widths of size comparable to the wavelength of diffracting light.
(ii) Grating is a plane sheet of transparent material on which opaque rulings are made with a fine diamond pointer.
(iii) The modern commercial grating contains about 6000 lines per centimeter. The rulings act as obstacles having a definite width b and the transparent space between the rulings act as slit of width a.
(iv) The combined width of a ruling and a slit is called Gratting element (e = a + b).
(v) points on slit separated by a distance equal to the grating element are called corresponding points.
(vi) A plane transmission grating is represented by AB in Figure. Let a plane wavefront of monochromatic light with wavelength λ be incident on the grating.
(vii) As the width of the slits is comparable to that of wavelength, the incident light undergoes diffraction.

(viii) A diffraction pattern is obtained on the screen when the diffracted waves are focused on a screen using a convex lens.
(ix) Let us consider a point P at an angle θ with the perpendicular drawn from the center of the grating to the screen.
(x) The path difference ઠ between the diffracted waves from one pair of corresponding points is,
\(\delta =(a+b)sin\theta \)               ........(1)
This path difference is the same for any pair of corresponding points. The point P on the screen will be maximum, when
ઠ= m λ where m = 0,1,2,3                ........(2)
Combining the above two equations, we get,
(a + b) sin θ = mλ                 ...............(3)
Here, m is called order of diffraction.
Condition for m^th order maximum :
(i) On the side of central maxima different higher orders of diffraction maxima are formed at different angular positions. If we take,
\(N=\cfrac { 1 }{ a+b } \)                     .................(4)
(ii) Then, N gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number N is specified on the grating itself. Now, the equation becomes,
\(\cfrac { 1 }{ N } sin\theta =m\lambda \) (or) \(sin\theta =Nm\lambda \)                        ...............(5)

Experimental setup:
(i) S is a source s₁ and s₂ the double slits which are at equidistances from 's'. Wavefronts from s₁ and s₂ spread out and overlap on other side of double slit.
(ii) When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.
(iii) Using an eyepiece the fringes can be seen directly. At the center point O on the screen, waves from s₁ and s₂ travel equal distances and arrive in-phase as shown in Figure.
(iv) These two waves constructively interfere and bright fringe is observed at O. This is called cental bright fringe.
(v) When one of the slits is closed, The fringes disappear and there in uniform illumination on the screen.
(vi) This shows clearly that the bands are due to interference.

Equation for path difference :
(i) The Let d be the distance between the double slits s₁ and s₂ which act as coherent sources of wavelength λ.
(ii) A screen is placed parallel to the double slit at a distance D from it.
(iii) P is any point at a distance y from O.
(iv) The waves from S₁ and S₂ meet at P either in-phase or out-of-phase depending upon the path difference between the two waves.

The path difference \(\delta\) between the light waves from s₁ and s₂ to the point p is,
\(\delta=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}\)
\(\delta=\mathrm{S}_{2} \mathrm{P}-\mathrm{MP}=\mathrm{S}_{2} \mathrm{M}\)                        .........(1)
\(\angle \mathrm{OCP}=\angle \mathrm{S}_{2} \mathrm{~S}_{1} \mathrm{M}=\theta\)
In right angle triangle \(\Delta \mathrm{S}_{1} \mathrm{S}_{2} \mathrm{M}\), the path difference S₂M = d sin \(\theta\)
\(\delta=d \sin \theta\)                 ...........(2)
If the angle \(\theta\) is small, \(\sin \theta \approx \tan \theta \approx \theta\)
From the right angle triangle \(\Delta \mathrm{OCP}, \tan \theta=\frac{\mathrm{y}}{\mathrm{D}}\)
The path differences \(\delta=\frac{d y}{D}\)     ...........(3)
Based on the condition of the path difference, the point P may have a bright (or) dark fringe

(i) Let us  Consider a parallel beam of light is incident on a refracting plane surface XY such as a glass surface as shown in Figure.
(ii) The incident wavefront AB is in rarer medium (1) and the refracted wavefront A'B' is in denser medium (2).
(iii) These wavefronts are perpendicular to the incident rays L, M and refracted rays L', M' respectively.
\(t=\cfrac { BB' }{ { v }_{ 1 } } =\cfrac { AA' }{ { v }_{ 2 } } \) or \(\cfrac { BB' }{ AA' } =\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } \) 
(i) The incident rays, the refracted rays and the normal are in the same plane.
(ii) Angle of incidence
i = ∠ NAL = 90^o - ∠NAB = ㄥ BAB'
Angle of refraction,
r = ∠ N'B'M = 90^o-ㄥN'B'A' =∠ A'B'A
For the two right angle triangles ∆ABB' and ∆AA'B',
\(\cfrac { sini }{ sinr } =\cfrac { \frac { BB' }{ AB' } }{ \frac { AA' }{ AB' } } =\cfrac { BB' }{ AA' } =\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\cfrac { \frac { c }{ { v }_{ 2 } } }{ \frac { c }{ { v }_{ 1 } } } \)
(iv) Here, C is speed of light in vacuum. The ratio \(\cfrac { c }{ v } \) is the constant, called refractive index of the medium. The refractive index of medium (1) is,\(\cfrac { c }{ { v }_{ 1 } } ={ n }_{ 1 }\) and that of medium (2) is,\(\cfrac { c }{ { v }_{ 1 } } ={ n }_{ 2 }\) In ratio form,
\(\cfrac { sini }{ sinr } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)                .............(1)
In product form,
n₁ sin i = n₂ sin r               ..........(2)
Hence, the laws of refraction are proved.

(i) Let us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY.
(ii) The incident wavefront is AB and the reflected waterfront is A'B'.
(iii) These wavefronts are perpendicular to the incident rays L, M and reflected rays L', M', respectively.

(i) The incident rays, the reflected rays and the normal are in the same plane.
(ii) Angle of incidence, ∠i = ∠NAL = 90°- ∠NAB = ∠BAB'
Angle of reflection ∠r = ∠N'B'M = 90°- ∠N'B'A'= ∠A'B'A
(a) For the two right angle triangles, ∆ABB' and ∆B'A'A, the two right angles, ∠B and ∠A' are equal, (∠B and ∠A' = 90°); the two sides., AA' and BB' are equal, (AA' = BB'); the side AB' is common
(b) Thus the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB' and ∠A'B'A must also be equal.
i = r
Hence, the laws of reflection are proved.