For total internal reflection,
sin i > sin c
\(n=\frac{1}{sin \ c}\)
\(sin \ c=\frac{1}{n}\)
\(sin \ i>\frac{1}{n}\)
\(n>\frac{1}{sin \ i}\)
n >\(\sqrt{2}\)
n >1.414 = 1.5

\(\frac{I}{f}=\left(\frac{\mu_{\mathrm{L}}}{\mu_L}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
When the biconvex lens of glass dipped in liquid, it acts as a plane sheet of glass.
\(\therefore \mathrm{f}=\infty, \frac{1}{\mathrm{f}}=0 \quad \frac{\mu_g}{\mu_{\mathrm{L}}}-1=0 ; \frac{\mu_{\mathrm{s}}}{\mu_{\mathrm{L}}}=1, \mu_{\mathrm{s}}=\mu_{\mathrm{L}}\)

v = nג
In an isotropic medium, there is no change in the frequency of the light. So, the speed of light depends on wavelength of light.

In the ray diagram, the incident ray is directed towards the first principal focus of the concave lens. After refraction, this ray becomes parallel to the principal axis.

(i) When lens is immersed in water, it behaves as a convex lens but its focal length will increase.
(ii) When convex lens is immersed in carbon - disulphide, it will behave as a concave lens.

The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.

The phenomenon in which a part of light from a source undergoing reflection and the other part of light from the same source undergoing refraction at the same surface is called simultaneous reflection or simultaneous refraction.

The plane through the focus and perpendicular to the principal axis is called the focal plane of the mirror.

When a wave undergoes reflection at a denser mediums, Its phase changes at 180^o.

Concave Lens:
1. It is a lens that diverges a light beam that falls on it.
2. It has at least one surface that is curved inside.

Air Bubble: 
1. These are circular spheres made up of thin water films and contain air.
2. The surface of an air bubble in water bulges outwards due to the air pressure inside the bubble.
How does an air bubble act like a concave lens:
1. The refractive index of water is greater than that of air, η_water>η_air. This implies that, water is a denser medium than air.
2. Let us consider that a light ray passes through water and enters an air bubble. So, the light ray enters from a denser to rarer medium. So, a ray diagram can be drawn as given below.
3. The ray of light entering the air bubble will diverge, since the ray of light entering from a denser to rarer medium bends away from the normal.
4. Hence, it behaves like a concave lens.

Hence, the air bubble inside water behaves like a concave lens.

Number of lines per cm = 4000 cm^-1; m = 2; θ = 30°; λ = ?
Number of lines per unit length
\(N=\cfrac { 4000 }{ 1\times { 10 }^{ -2 } } =4\times { 10 }^{ 5 }\) 
Equation for diffraction maximum in grating is, sinθ = Nmλ
After Rewriting, \(\lambda =\cfrac { sin\theta }{ Nm } \) 
Substituting,
\(\lambda =\cfrac { { \sin30 }^{ o } }{ 4\times { 10 }^{ 5 }\times 2 } =\cfrac { 0.5 }{ 4\times { 10 }^{ 5 }\times 2 } \) 
= \(\cfrac { 1 }{ 2\times 4\times { 10 }^{ 5 }\times 2 } =\cfrac { 1 }{ 16 \times 10^5} \)
λ = 6250 x 10^-10 m = 6205 Å

Given, focal length, f = 150 cm = 1.5 m
Equation for power of lens is, \(p=\cfrac { 1 }{ f } \) 
Substituting the values,
\(p=\cfrac { 1 }{ 1.5 } =0.67 D\) 
As the power is positive, it is a converging lens.

AB is the reflecting surface as shown in the Figure. Both the incident ray IO and the reflected ray OR₁ subtend angle i with the normal N as the angle of incidence is equal to angle of reflection. When the surface AB is tilted to A'B' by an angle θ, the normal N is also is tilted to N' by the same angle θ.Remember that the position of the incident ray IO remains unaltered. But the reflected ray now is OR₂.
Now, in the tilted system, the angle of incidence, ∠N'OI = i + θ and the angle of reflection, ∠N'OR₂ = i + θ are the same. The angle between ON' and OR₁ is, ∠N'OR₁ = i – θ. The angle tilted on the reflected light is the angle between OR₁ and OR₂ which is ∠R₁OR₂. From the geometry we can write,
\(\angle { R }_{ 1 }O{ R }_{ 2 }=\angle N'{ { OR }_{ 2 } }-\angle { NOR }_{ 1 }\) \(=\left( i+\theta \right) -\left( i-\theta \right) =2\) 
∠R₁ OR₂ = 2θ.

If the scattering of light is by atoms and molecules which have size a yery less than that of the wavelength \({\lambda}\) of light a << \({\lambda}\) the scattering is called Rayleigh's scattering.
The intensity of Rayleigh's scattering is inversely proportional to fourth power of wavelength
\(\mathrm{I} ∝ \frac{1}{\lambda^{4}}\)

(i) Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁. Let R₁ and R₂ be the radii of curvature of two spherical surfaces (i) and (ii) respectively and P be the pole.
(ii) Consider a point object 'O' on the principal axis. A paraxial ray from 'O' which falls very close to P, after refraction at the surface (i) forms image at I'.
(iii) Before it does so, it is again refracted by the surface (ii). Therefore the final image is formed at I.
(iv) The general equation for the refraction at a single spherical surface is given by the equation is,
\(\cfrac { { n }_{ 2 } }{ v } =\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { \quad n }_{ 2 }-{ { n }_{ 1 } } \right) }{ R } \)   ....(i)
(v) For the refracting surface (1), the light goes from n₁ to n₂
 \(\cfrac { { { n }_{ 2 } } }{ v' } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( n_{ 2 }-{ n }_{ 1 } \right) }{ { R }_{ 1} } \)   ......(ii)

(vi) For the refracting surface (ii), the light goes from medium n₂ to n₁
 \(\cfrac { { { n }_{ 1 } } }{ v' } -\cfrac { { n }_{ 2 } }{ v' } =\cfrac { \left( n_{ 1 }-{ n }_{ 2 } \right) }{ { R }_{ 2 } } \)   ....(iii)
(vii) Adding the above two equations (ii) and (iii)
\(\cfrac { { { n }_{ 1 } } }{ v } =\cfrac { { n }_{ 1 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)
On further simplifying and rearranging
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { n }_{ 1 } } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)    ....(iv) 
If the refractive of the lens is n₂ and is placed in air, then n₂= n and n₁= 1. So the equation (iv) becomes,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{u } =(n-1) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ...(v)
According to lens makers formula
\(\cfrac { 1 }{ f } =(n-1) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)   ...(vi) 
Comparing the two equations (v) and (vi), we find
 \(\cfrac { 1 }{ v } -\cfrac { 1 }{ u }=\cfrac { 1 }{ f }\)          .....(vii)
This is called lens equation

Critical angle:
The angle of incidence in the denser medium for which the angle reflection is 90^o or the reflected ray graces the boundary between the two media is called critical angle.
Total Internal reflection:
For any angle of incidence greater than the critical angle, the center light is reflected back into the denser medium itself. This phenomenon is called Total internal reflection.

We first consider the effect of the lens. For the
lens, we have
u = - 40 cm and f = + 20 cm
Using the lens formula, we get
\(\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ \left( -40 \right) } =\cfrac { 1 }{ 20 } \)   ஃv₁ = + 40 cm
Had there been the lens only the image would have been formed at Q₁. The plane mirror M is at a distance of 30 cm from lens 1. We can, therefore, think of Q₁, as a virtual object, located at a distance of 10cm, behind the plane mirror M. The plane mirror, therefore, forms a real image (of this virtual object Q₁) at Q, 10 cm in front of it.

For minimum deviation position
\({ \mu }_{ red }=\cfrac { sin\left( \frac { A+{ \delta }_{ red } }{ 2 } \right) }{ sin\frac { A }{ 2 } } \) 
or \(sin\left( \cfrac { A+{ { \delta }_{ red } } }{ 2 } \right) ={ n }_{ red }\) 
\(sin\cfrac { 2 }{ A } =1.622\times 0.5=0.811\) 
\(\therefore \cfrac { 60+{ \lambda }_{ red } }{ 2 } ={ 54 }^{ o }12'\) 
\({ \delta }_{ red }={ 108 }^{ o }24'-{ 60 }^{ o }={ 48 }^{ o }24'\)
Similarly, \(sin\left( \cfrac { A+{ \delta }_{ viloet } }{ 2 } \right) =1.663\times 0.5'\) 
= 0.8315 or \(\cfrac { { 60 }^{ o }+{ \delta }_{ viloet } }{ 2 } =56^{ o }15'\) 
\({ \delta }_{ viloet }=112^{ o }30'-{ 60 }^{ ' }={ 52 }^{ o }30'\) 
\(\therefore { \delta }_{ violet }-{ \delta }_{ red }=\left( 52-30' \right) -\left( 48^{ o }24' \right) ={ 4 }^{ o }6'\) 
It is not advisable to use the formula,
\({ \delta }_{ v }-{ \delta }_{ r }=\left( { \mu }_{ v }-{ \mu }_{ r } \right) \) in the above solution.

Since the angle of incidence is equal to the angle of emergence, therefore, the ray of light passes symmetrically through the prism. So, the prism is in a minimum deviation position
Now,\(\\ \\ A+{ \delta }_{ m }=i+e\) 
or \({ \delta }_{ m }=e+e-A=2e-A=-2\left( \cfrac { 3 }{ 2 } A \right) -A\) 
= \(\cfrac { 3 }{ 2 } A-A=\cfrac { A }{ 2 } =\cfrac { { 60 }^{ o } }{ 2 } ={ 30 }^{ o }\) 
\({ n }_{ 21 }=\cfrac { sin\left( \frac { A+{ \delta }_{ m } }{ 2 } \right) }{ sin\left( \frac { A }{ 2 } \right) } =\cfrac { sin\left( \cfrac { { 60 }^{ o }+30^{ o } }{ 2 } \right) }{ sin\cfrac { { 60 }^{ o } }{ 2 } } \) 
= \(\cfrac { sin{ 45 }^{ o } }{ sin{ 30 }^{ o } } -\cfrac { 1 }{ \sqrt { 2 } } \times \cfrac { 2 }{ 1 } =\sqrt { 2 } =1.414\) 

Formula:
Refractive index,\(\mu =\cfrac { sini }{ sinr } \) 
\(\sqrt { 3 } =\cfrac { sin{ 60 }^{ o } }{ sinr } \) 
\(sinr=\cfrac { \sqrt { 3 } }{ 2 } \times \cfrac { 1 }{ \sqrt { 3 } } =\cfrac { 1 }{ 2 } \) 
sin r = sin30^o
⇒ Angle of refraction = 30^o

Angle of deviation Produced by Prism:
(i) Let light ray PQ is incident on one of the refracting faces of the prism.
(ii) The angles of incidence and refraction at the first face AB are i₁ and r_l. The path of the light inside the prism is QR.
(iii) The angle of incidence and refraction at the second face AC is r₂ and i₂ respectively.
(iv) RS is the ray emerging from the second face. Angle i₂ is also caned angle of emergence.
(v) The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
(vi) The two normals drawn at the point of incidence Q and emergence R meet at point N. They meet at point N.
(vii) The extended incident ray and the emergent ray meet at a point M.

The angle of deviation d₁ at the surface AB is,
ㄥRQM = d = i₁ - r₁ ...(1)       
The angle of deviation d₂ at the surface AC is
ㄥQRM = d₂ = i₂ - r₂          .......(2)
Total angle of deviation d produced is, 
d = d₁ + d₂             .....(3)
Substituting for d₁ and d₂ in equation (3)
d = (i₁ - r₁) + (i₂ - r₂)
After rearranging,
d = (i₁ - r₁) + (i₂ - r₂)                    ........(4)
 In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
\(\angle A+\angle QNR={ 180 }^{ 0 }\)         .........(5)
From the triangle ΔQNR 
\({ r }_{ 1 }+{ r }_{ 2 }+\angle QNR={ 180 }^{ o }\)                ......(6)
Comparing these two equations (5) and (6) we get,
r₁ + r₂ = A              .......(7)
Substituting this in equation (4) for angle of deviation,
d = i₁+ i₂ - A                .............(8)
(viii) Thus, the angle of deviation depends on the angle of incidence i₁, angle of emergence i₂ and the angle for the prism A. 
(ix) For a given angle of incidence the angle of emergence is decided by the refractive index of the material of the prism. Hence the angle of deviation depends on these following factors.
(i) the angle of incidence
(ii) the angle of the prism.
(iii) the refractive index of the material of the prism (which decides the angle of emergence).
Refractive index of the material of the prism:

At minimum deviation, i₁ = i₂ = i and r₁ = r₂ = r
Now, the equation (8) becomes,
D - i₁ + i₂ - A = 2_i - A (or) \(i=\cfrac { \left( A+D \right) }{ 2 } \) 
The equation (7) becomes
r₁ + r₂ = A ⇒ 2r = A (or) \(r=\cfrac { A }{ 2 } \)
Substituting i and r in Snell's law
\(n=\cfrac { sini }{ sinr } \) 
\(n=\cfrac{\cfrac{sin(A+D)}{2}}{sin(A/2)}\)

(i) The angle of view for water animals is restricted to twice the critical angle 2i_c. The critical angle for water is 48.6°. Thus the angle of view is 97.2°.
(ii) The radius R of the circular area depends on the depth d from which it is seen and also the refractive indices of the media.
(iii) The radius R of Snell's window can be deduced with the illustration as shown in Figure.

(iv) Light is seen from a point A at a depth 'd'.
(v) From the Snell's law in product form, n₁ sini = n₂ sinr
(vi) The equation for the refraction happening at the point B on the boundary between the two media is,
n₁ sin i_c = n₂ sin90^o    ..(1)
n₁sini_c = n₂ (∵ sin90^o = 1)
\(sin{ i }_{ c }=\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)     ...(2)
From the right angle triangle ΔABC,
\({ sini }_{ c }=\cfrac { CB }{ AB } =\cfrac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \)      ....(3)
Equating the above two equation 
\(\cfrac { R }{ \sqrt { { d }^{ 2 }+{ R }_{ 2 } } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \) 
Squaring on both sides
\(\cfrac { { R }^{ 2 } }{ { R }^{ 2 }+d^{ 2 } } \left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right) ^{ 2 }\) 
Taking reciprocal,
\(\cfrac { { R }^{ 2 }+{ d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 }\) 
On further simplifying
\(1+\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 };\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 }-1;\) 
\(\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\cfrac { { n }_{ 1 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } -1=\cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 2 }^{ 2 } } \) 
Again taking reciprocal and rearranging
\(\cfrac { { R }^{ 2 } }{ { d }^{ 2 } } =\cfrac { { { n }_{ 2 }^{ 2 } } }{ { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } { R }^{ 2 }={ d }^{ 2 }\left( \cfrac { { n }_{ 2 }^{ 2 } }{ { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) 
∴ The radius of illumination is,
\(R=d\sqrt { \cfrac { { n }_{ 2 }^{ 2 } }{ \left( n_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } \right) } } \)   ...(4)
If the rarer medium outside is air, then, n₂ = 1, and we can take n₁ = n
\(R=d\left( \cfrac { 1 }{ \sqrt { { n }^{ 2 }-1 } } \right) \) or \(R=\cfrac { d }{ \sqrt { { n }^{ 2 }-1 } } \)   ....(5)

(i) When light rays incident on a spherical mirror, after reflection, the rays converge at a point on the principal axis for concave mirror or appear to diverse from a point for convex Mirror.
(ii) This point is called the focus or focal point (F) of the mirror.

The given situation is shown in the figure
Given, A = 60^o ; i₁ = 0^o; i₂ = 90^o
(a) Equation for angle of deviation,
d = i₁ + i₂ - A
Substituting the values
d = 0^o + 90^o- 60^o = 30^o
The angle of deviation produced is, d = 30°
(b) The light inside the prism must be falling on the second face at critical angle as it graces the boundary. i_c = 90° – 30° = 60°
Equation for critical angle is, \(\sin i_{c}=\frac{1}{n}\) 
\(n=\frac{1}{\sin i_{c}} ; \quad n=\frac{1}{\sin 60^{\circ}}=\frac{1}{\sqrt{3} / 2}=\frac{2}{\sqrt{3}}=1.15\) 
The refractive index of the material of the prism is, n = 1.15

This lens is called convexo-concave lens
Given, n = 1.52, R₁ = 10 cm and R₂ = 20 cm
Both R₁ and R₂ are positive
Lens makers formula,
\(\cfrac { 1 }{ f } =\left( n-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
Substituting the values,
\(\cfrac { 1 }{ f } =\left( 1.52-1 \right) \left( \cfrac { 1 }{ 10 } -\cfrac { 1 }{ 20 } \right) \) 
\(\cfrac { 1 }{ f } =\left( 0.52 \right) \left( \cfrac { 2-1 }{ 20 } \right) =\left( 0.52 \right) (\frac{1}{20})=\cfrac { 0.52 }{ 20 } \) 
\(f=\cfrac { 20 }{ 0.52 } =38.46cm\)
As the focal length is positive, the lens is a converging lens.

Case i) \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
applying sign convention, u = –10 cm, R = 15 cm
\(\cfrac { 2 }{ v } -\cfrac { 1 }{ -10 } =\cfrac { \left( 2-1 \right) }{ 15 } ;\cfrac { 2 }{ v } +\cfrac { 1 }{ 15 } =\cfrac { \left( 1 \right) }{ 10 } \) 
∴ =− 60 cm
[a virtual image is formed 60 cm, to the left of the surface]

Case ii) \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
applying sign convention, u = –30 cm, R = 15 cm
\(\cfrac { 2 }{ v } -\cfrac { 1 }{ -30 } =\cfrac { \left( 2-1 \right) }{ 15 } ;\cfrac { 2 }{ v } +\cfrac { 1 }{ 30 } =\cfrac { \left( 1 \right) }{ 15 } \)
∴ = 60 cm [a real image is formed 60 cm, to the right of the surface]

Given, thickness of glass slab, d = 50 cm = 0.5 m, refractive index, n = 1.5
refractive index, \(n=\cfrac { c }{ v } \) 
(a) speed of light in the glass slab is,
\(v=\cfrac { c }{ n } =\cfrac { 3\times { 10 }^{ 8 } }{ 1.5 } =2\times { 10 }^{ 8 }{ ms }^{ -1 }\) 
(b) time taken by light to travel through the glass slab is,
\(t=\cfrac { d }{ v } =\cfrac { 0.5 }{ 2\times { 10 }^{ 8 } } =2.5\times { 10 }^{ -9 }{ s }\)
(c) optical path,
d' = nd = 1.5 x  0.5 = 0.75 m = 75 cm
Light would have traveled an additional 25 cm (75 cm – 50 cm) in vacuum at the same time had there been no glass slab in its path.

\(P_{a}=\frac{1}{f_{a}}=\left(\frac{\mu_{g}}{\mu_{a}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)  ..(1)
\(P_{w}=\frac{1}{f_{w}}=\left(\frac{\mu_{g}}{\mu_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)  ...(2)
\(p_{a}=5 D, f_{w}=-100 \text { (Diverging lens) } \)
\(\mu_{\mathrm{g}}=1.5, \mu_{\mathrm{a}}=1 \)
\(5=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)   ...(3)
\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\frac{5}{0.5}=\frac{50}{5}=10 \)
\(\frac{1}{f_{w}}=\left(\frac{1.5}{n_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)    ...(4)
\(\frac{-1}{100 \times 10^{-2}}=\left(\frac{1.5}{n_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)
\(-1=\left(\frac{1.5}{n_{w}}-1\right)(10) \Rightarrow \frac{-1}{10}=\frac{1.5}{n_{w}}-1 \)
\(\frac{1.5}{n_{w}}=\frac{-1}{10}+1 \)
\(\frac{1.5}{n_{w}}=\frac{9}{10} \)
\(n_{w}=\frac{1.5 \times 10}{9}=\frac{15}{9}=\frac{5}{3} \)
\(n_{w}=\frac{5}{3} \)