(i) The incident ray, refracted ray and normal to the refracting surface are all coplanar (ie. lie in the same plane).
(ii) The ratio of sine of angle of incident i in the first medium to the sine of angle of refraction r in the second medium is equal to the ratio of refractive index of the second medium n₂ to that of the refractive index of the first medium n₁.
\(\frac{sin \ i}{sin \ r}=\frac{n_2}{n_1}(Snell's \ law)\)
n₁ sin i = n₂ sin r 

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

The centre of the shadow will be bright.

Yes, there is a difference between coloured light obtained from prism and soap bubble Dispenion takes place in prism. Interference takes place in soap bubbles.

No, Diffraction does not take place at the young's double slit. Interference takes place in young's double slit.

Two independent monochromatic sources cannot act as coherent sources because they emit wanes of same frequency, same amplitude but not with same phase.

Sky appears blue :
when white light from the sun enters the earth's atmosphere scattering takes place.
According to Ray light's scattering
\(\text { Law, } \operatorname{S\alpha } \frac{1}{\lambda_{4}}\)

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.

(i) Dish antennas are curved because it receives parallel signal rays coming from the same direction.
(ii) These rays are reflected by parabolic dish and gathered at maintained antenna part which increases directivity of antenna and gives sufficient amplitude signal.

Given, n_v = 1.632; n_R = 1.613; n_G = 1.620
Equation for dispersive power is,
\(\omega =\cfrac { \left( { n }_{ v }-{ n }_{ g } \right) }{ \left( { n }_{ G }-1 \right) } \) 
Substituting the values,
\(\omega =\cfrac { 1.632-1.613 }{ 1.620-1 } =\cfrac { 0.019 }{ 0.620 } =0.0306\) 
The dispersive power of the prism is,\(\\ \omega =0.0306\)

Given, A = 60°; D = 37°
Equation for refractive index is,
\(n=\cfrac { \sin\left( \frac { A+D }{ 2 } \right) }{ \sin\left( \frac { A }{ 2 } \right) } \) 
Substituting the values,
\(n=\cfrac { \sin\left( \frac { 60^{ o }+37^{ o } }{ 2 } \right) }{ \sin\left( \cfrac { { 60 }^{ o } }{ 2 } \right) } =\cfrac { \sin\left( 48.5^{ o } \right) }{ \sin\left( { 30 }^{ o } \right) } =1.5\)
The refractive index of the material of the prism is, n = 1.5 

Since, the prism is equilateral, A = 60^o;
Given, i₁ = 30^o;i₂ = 75^o
Equation for angle of deviation, d = i₁ + i₂ – A
Substituting the values, d = 30°+ 75°– 60°= 45°
The angle of deviation produced d = 45^o

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.

Given, n₁ = 1.68, n₂ = 1.44, n₃ = 1
Acceptance angle, \(\\ { i }_{ a }={ sin }^{ -1 }\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) 
\({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { \left( 1.68 \right)^2 -\left( 1.44 \right) ^{ 2 } } \right) ={ sin }^{ -1 }\left( 0.865 \right) \) 
\({ i }_{ a }\simeq { 60 }^{ o }\) 
If there is no cladding then, n₂ = 1
Acceptance angle, \({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { { n }_{ 1 }^{ 2 }-1 } \right) \)
\({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { \left( 1.68 \right) ^{ 2 }-1 } \right) ={ sin }^{ -1 }\left( 1.35 \right) \) 
sin⁻¹(more than 1) is not possible. But, this includes the range 0^o to 90^o. Hence, all the rays entering the core from flat surface will undergo total internal reflection.
Note: If there is no cladding then there is a condition on the refractive index (n₁) of the core
\({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { { n }_{ 1 }^{ 2 }-1 } \right) \) 
Here, as per mathematical rule,
\(\left( { n }_{ 1 }^{ 2 }-1 \right) \le 1\) or \(\left( { n }_{ 1 }^{ 2 } \right) \le 2\) or \({ n }_{ 1 }\le \sqrt { 2 } \)
Hence, in air (no cladding) the refractive index n₁ of the core should be,\({ n }_{ 1 }\le 1.414\)

When seen from (air medium) on top, the coin will still appear to be at the bottom with each medium appearing to have shrunk with respect to the air medium outside. This situation is illustrated below.

The equations for apparent depth for each medium is,
\({ d' }_{ 1 }=\cfrac { { d }_{ 1 } }{ { n }_{ 1 } } ;{ d }_{ 2 }^{ ' }=\cfrac { { d }_{ 2 } }{ { n }_{ 2 } } ;{ d }_{ 3 }^{ ' }=\cfrac { { d }_{ 3 } }{ { n }_{ 3 } } \) 
\({ d }^{ ' }={ d }_{ 1 }^{ ' }+{ d }_{ 2 }^{ ' }+{ d }_{ 3 }^{ ' }=\cfrac { { d }_{ 1 } }{ n_{ 1 } } +\cfrac { { d }_{ 2 } }{ { n }_{ 2 } } +\cfrac { { d }_{ 3 } }{ n_{ 3 } } \) 
\(d'=\cfrac { 30 }{ 1.3 } +\cfrac { 1.6 }{ 1.4 } +\cfrac { 30 }{ 1.5 } =23.1+11.4+13.3\) 
d' = 47.8 cm

Given, n_go = 1.25 and n_g = 1.5
Refractive index of glass with respect to oil,
\({ n }_{ go }=\cfrac { { n }_{ g } }{ { n }_{ 0 } } \) 
Rewriting for refractive index of oil,
\({ n }_{ p }=\cfrac { { n }_{ g } }{ { n }_{ go } } =\cfrac { 1.5 }{ 1.25 } =1.2\) 
The refractive index of oil is, n_o = 1.2

\(n=\cfrac { c }{ v } ;\quad v=\cfrac { c }{ n } \) 
\(v=\cfrac { 3\times { 10 }^{ 8 } }{ 1.33 } =2.26\times { 10 }^{ 8 }{ ms }^{ -1 }\)
Light travels with a speed of 2.26 x 10⁸ m s^-1 through pure water.

Let us assume a person of height h is standing in front of a vertical plane mirror. The person could see his/her head when light from the head falls on the mirror and gets reflected to the eyes. Same way, light from the feet falls on the mirror and gets reflected to the eyes.

If the distance between his head H and eye E is h₁ and distance between his feet F and eye E is h₂. The person’s total height is, h Here it is, h = h₁ + h₂ 
By the law of reflection, the angle of incidence and angle of reflection are the same in the two extreme reflections. The normals are now the bisectors of angles between incident and reflected rays in the two points. By geometry, the height of the mirror needed is only half of the height of the person. \( \frac { { h }_{ 1 }+{ h }_{ 2 } }{ 2 } =\frac { h }{ 2 } \).
Does the height depend on the distance between the person and the mirror?

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.

\(v=\pm 3u, f=\mp 20 cm\)
Case i :
V = -3u,
\(\frac{1}{f} =\frac{1}{u}+\frac{1}{v} =\frac{1}{u}-\frac{1}{3 u} =\frac{2}{3 u} \)
\(\)3u = 2f
\(u=\frac{2 f}{3}=\frac{2 \times(-20)}{3} \)
\(u=\frac{-40}{3} \mathrm{~cm} \)
Case ii :
v = 3u
\(\frac{1}{f} =\frac{1}{u}+\frac{1}{3 u} \)
\(=\frac{4}{3 u} \)
3u = 4f = -4 x 20
\(u=\frac{-4 \times 20}{3} \)
\(u=\frac{-80}{3} \mathrm{~cm} \)

\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(m=\frac{-v}{u}=-4, f=20 \mathrm{~cm} \text { (Given) } \)
V = 4u
\(\frac{1}{f} =\frac{1}{4 u}-\frac{1}{u} \)
\(=\frac{1-4}{4 u}=\frac{-3}{4 u} \)
\(\frac{1}{f} =\frac{-3}{4 u} \)
4u = -3 x f
\(u=\frac{-3}{4} \times 20=-15 \mathrm{~cm}\)

(i) A simple microscope is a single magnifying (converging) lens of a small focal length.
(ii) To get an erect, magnified, and virtual image of the object, the object is placed between F and P on one side of the Jens and viewed from another side of the lens.
(iii) There are two magnifications for two kinds of focusing.
(iv) The image is formed at infinity.
(v) The angular magnification is defined as the ratio of angle e, subtended by the image with an aided eye to the angle 80 subtended by the object with the unaided eye.
(vi) This is the magnification for normal focusing.
(vii) The magnification for normal focusing is one less than that for near point focusing.

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) If light is scattered by large particles like dust and water droplets present in the atmosphere which have size a greater than the wavelength \(\lambda\) of light, a > > \(\lambda\), the intensity of scattering is equal for all the wavelengths.
(ii) It is happening in clouds which contains large amount of dust and water droplets. Thus, in clouds all the colours get equally scattered irrespective of wavelength. so, the clouds appears white.

(i) During sunrise and sunset, the light from sun travels a greater distance through the atmosphere.
(ii) Hence, the blue light which has shorter wavelength is scattered away and the less scattered red light of longer wavelength manages to reach our eye.
(iii) This is the reason for the reddish appearance of sky during sunrise and sunset.

\(\mathrm{I} ∝ \frac{1}{\lambda^{4}}\)
According to Rayleigh's scattering equation, violet colour which has the shortest wavelength gets much scattered during day time. The next scattered colour is blue. As our eyes are more sensitive to blue colour than violet colour the sky appears blue during day time.

Rainbows is an example of dispersion of sunlight through droplets of water during rainy days. Rainbow is formed, where sunlight falls on the water drop of rainfall suspended in air. It splits into its constituent seven colours. Primary rainbow is formed when light entering the drop undergoes arc total internal reflection.

Dispersion is splitting of white light into its constituent colours. This band of Colours of light is called its spectrum.

Angle of minimum deviation:
(i) The angle between the direction of incident ray and the emergent of a prism called the 'Angle of deviation' 'd'.
(ii) The minimum value of angle of deviation is called 'Angle of minimum deviation' 'D'.
At 'D' i₁ = i₂ , r₁ = r₂

The power of a lens P is defined as the reciprocal of its focal length (in metre).
\(P=\frac{1}{f}\)

Sign conventions for lens
(i) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right.
(ii) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.
(iii) The other sign conventions for object. distance, image distance, radius of curvature, object height and image height remain the same for thin lenses as that of spherical mirrors.

Primary focus:

(i) The primary focus F_I is defined as a point where an object should be placed to give parallel emergent rays to the principal axis after passing through lens.
Secondary focus:

(ii) The secondary focus F₂ is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis after passing through lens.

An endoscope is an instrument used by doctors which has a bundle of optical fibres that are used to see inside a patient's body. Endoscopes work on the phenomenon of total internal reflection. The optical fibres are inserted in to the body through mouth, nose or a special hole made in the body.

Cartesian sign conventions for a spherical mirror

(i) The incident light is taken as if it is travelling from left to right (i.e. object on the left of mirror).
(ii) All the distances are measured from the pole of the mirror (pole is taken as origin).
(iii) The distances measured to the right of pole along the principal axis are taken as positive.
(iv) The distances measured to the left of pole along the principal axis are taken as negative.
(v) Heights measured upwards perpendicular to the principal axis are taken as positive.
(vi) Heights measured downwards perpendicular to the principal axis are taken as negative.

(i) Optical fibers consist of the inner part called core and the outer part called cladding (or) sleeving. The refractive index of core is higher than that of cladding.
(ii) Signal in the form of light is made to incident inside the core-cladding boundary at an angle greater than the critical angle. So, that total internal reflection happens without undergoing any refraction. The light travels without appreciable loss of intensity.

When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle i_c. The restricted illuminated circular area is called Snell's window

Prisms can be designed to reflect light by 90^o or by 180^o by making use of total internal reflection from the Figures (a) and (b). In the first two cases, the critical angle i_c for the material of the prism must be less than 45^o. Prisms are also used to invert images without changing their size as shown in Figure(c).

Mirage:
Mirage is an optical illusion caused by atmospheric conditions especially the appearance of sheet of water in a desert caused by total internal reflection (or) refraction of light from the sky by heated air.
Looming:
Looming is an optical illusion caused by bending of light which appear an object floating high above its actual position specially in polar region.

Diamond appears glittering because the total internal reflection of light.
Inside the diamond the refractive index of diamond is about 2.417 which is greater than the refractive index of glass. (μ_g =1.5). The critical angle of diamond is 24.4^d which is much less than that of glass (g_crown = 40.5°, g_flint 31.9°).
So, when the light enters the diamond, the total internal reflection of light happens inside the diamond before getting out. This gives a sparkling effect for diamond.

Stars appear twinkling because of the movement of the atmospheric layer with varying refractive indices due to refraction.

\(\cfrac { sini }{ sinr } =\cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \) (Snell's law)
The term \(\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \) is called relative refractive index of second medium with respect to the first medium \(n_{21}=\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \)
(i) Inverse rule: \(n_{21}=\frac{1}{n_{21}}(or)\cfrac { { n }_{ 1 } }{ { n }_{ 2 } }=\frac{1}{(n_2/n_1)}\)
(ii) Chain rule: \(n_{32}=n_{31} \times n_{21}(or)\frac{n_3}{n_2}=\frac{n_3}{n_1}\times\frac{n_1}{n_2}\)

The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

The angle between the incident and deviated light is called Angle of deviation due to refraction. When light travels from
(i) rarer to denser medium, d = i - r
(ii) denser to rarer medium, d = r - i