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

(i) Light must travel from denser to rarer medium,
(ii) Angle of incidence in the denser medium must be greater than critical angle (i > i_c).

If the reflection takes place at the convex surface, it is called a convex mirror and if the reflection takes place at the concave surface, it is called a concave mirror.

Real image :
This type of image which can be formed on a screen and can also be seen with the eyes is called real image.
Virtual image :
Image which cannot be formed on the screen but can only be seen with the eyes.

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.

The cut-off wavelength of the characteristic x - rays is
\({ \lambda }_{ ° }\frac { 12400 }{ V } \mathring { A } =\frac { 12400 }{ 20000 } \mathring { A }\)
= 0.62 \(\mathring { A } \)
The corresponding frequency is
\({ \upsilon }_{ o }=\frac { c }{ { \lambda }_{ o } } =\frac { 3\times 10^{ 8 } }{ 0.62\times 10^{ -10 } }\) = 4.84 x 10¹⁸ Hz

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 

\(v=\pm 3u, f=\mp 20 cm\)
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
\(u=\frac{-4 \times 20}{3} \)
\(u=\frac{-80}{3} \mathrm{~cm} \)

A terrestrial telescope has an additional erecting lens to make the final image erect.

Uses:
To produce plane polarised light and also serve as analyser.
Construction:
(i) lt is a clacite crystal whose length is three times of its breadth.
(ii) Cut in to two halves having face angles 72° and 108°.
(iii) Joined together by a transparent cement (canada balsam).
Working:
(i) When a monochromatic light from sodium vapour lamp is incident on Nicol prism, double refraction takes places.
(ii) It's split as ordinary (O) ray and extra ordinary ray (E)
Refractive index of the crystal for
Ordinary ray  : 1.658
Extra ordinary ray  : 1.486
Canada balsam :  1.523
(iii) Ordinary ray is total internally reflected and extra ordinary ray alone is transmitted which is plane polarised.

(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.

(i) Speed of Electromagnetic wave \(=\frac{Peak\ value\ of\ Electric\ field}{Peak\ value\ of\ magnetic\ field}\)
\(c=\frac { { E }_{ 0 } }{ { B }_{ 0 } } \)
(ii) Yes, As Electromagnetic waves contain both electric and magnetic fields, there is a non-zero energy density associated with it.
\(E=\frac { hc }{ \lambda } \)
Momentum p \(=\frac{Total\ energy\ transferred\ to\ the\ surface}{Velocity\ of\ light\ in\ vacuum}\)
i.e.p = \(\frac{U}{c}=mc\)
U - total energy transferred to the surface.
EM waves carry not only energy and momentum but also angular momentum.

The drift velocity is defined as the average velocity with which free electrons in a conductor get drifted under the influence of an external electricity applied across the conductor. The relation between drift velocity of free electrons the circuit flowing through the conductor is \(I=\eta Ae{ v }_{ d }e-\) charge of the electron n - no of free electrons in the conductor; A - area of cross-section. 

Whenever energy is given to a circuit containing a pure indicator L and a capacitor of capacitance C, the energy oscillates back and fourth between the magnetic field of the indicator and the electric field of the capacitor. Thus the electrical oscillations of definite frequency are generated. these oscillations are called LC oscillations.

(i) A lens is formed by a transparent material bounded between two spherical surfaces or one plane and another spherical surface.
(ii) In a thin lens, the distance between the surfaces is very small. If there are two spherical surfaces, then there will be two centers of curvature C₁ and C₂ and correspondingly two radii of curvature R₁ and R₂.
(iii) A plane surface has its center of curvature C at infinity and its radius of curvature R is infinity (R = ∞).

(i) The line joining the pole and the centre of curyature is called the principal axis of the mirror.
(ii) The light ray travelling along the principal axis towards the mirror after reflection travels back along the same principal axis. It is also called optical axis.

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.

\(\text{ longitudinal magnifcation}(m_l)=\frac { length\ of\ image\left( l' \right) }{ length\ of\ object\left( l \right) } \)
Given: length of object, \(l=\cfrac { f }{ 3 } \) 
For the given condition, the image formation is shown in the figure.

Let, l' be the length of the image, then
\(m=\cfrac { l' }{ l } =\cfrac { l' }{ f/3 } \) (or) \(l=\cfrac { m_lf }{ 3 } \) 
Image of one end coincides with the object. Thus, the coinciding end must be at center of curvature.
 \(u_B=u_A-\cfrac { f }{ 3 } =2f-\cfrac { f }{ 3 } =\cfrac { 5f }{ 3 } \)
\(v_B=u_B+l+l'\) 
\(v_b =\cfrac { 5f }{ 3 } +\cfrac { f }{ 3 } +\cfrac { mf }{ 3 } =\cfrac { f(6+m) }{ 3 } \)
 Mirror equation,\(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
\(\cfrac { 1 }{ -\left( \cfrac { f(6+m_l) }{ 3 } \right) } +\cfrac { 1 }{ -\left( \cfrac { 5f }{ 3 } \right) } =\cfrac { 1 }{ -f } \)
After simplifying,
 \(\cfrac { 3 }{ f(6+m_l) } +\cfrac { 3 }{ 5f } =\cfrac { 1 }{ f } ;\cfrac { 3 }{ (6+m_l) } =\cfrac { 2 }{ 5 } \) 
\(6+m_l=\cfrac { 15 }{ 2 } ;m_l=\cfrac { 15 }{ 2 } -6\) 
\(m_l=\cfrac { 3 }{ 2 } =1.5\) 

(a) The velocity of an electron in n^th orbit is
\(\upsilon _{ n }=\frac { h }{ 2\pi m{ a }_{ 0 } } \frac { Z }{ n } \)
Where \({ a }_{ 0 }=\frac { { \epsilon }_{ 0 }{ h }^{ 2 } }{ \pi { me }^{ 2 } } \) = Bohr radius. Substituting for a₀ in ሀ_n,
\({ \upsilon }_{ n }=\frac { { e }^{ 2 } }{ 2{ \epsilon }_{ 0 }h } \frac { Z }{ n } =c\left( \frac { { e }^{ 2 } }{ 2{ \epsilon }_{ 0 }hc } \right) \frac { Z }{ n } =\frac { \alpha cZ }{ n } \)
where c is the speed of light in free space or vacuum and its value is c = 3 x 10⁸ m s^–1 and α is called fine structure constant.
For a hydrogen atom, Z = 1 and for the first orbit, n = 1, the ratio of velocity of electron in first orbit to the speed of light in vacuum or free space is
\(\frac { { \upsilon }_{ 1 } }{ c } =\alpha =\frac { { e }^{ 2 } }{ 2{ \epsilon }_{ 0 }hc } \)
\(\alpha =\frac { { (1.6\times { 10 }^{ -19 }C })^{ 2 } }{ 2\times (8.854\times { 10 }^{ -12 }{ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }) } \) x \(\frac { 1 }{ (6.6\times { 10 }^{ -34 }{ Nms)\times (3\times { 10 }^{ 8 } }{ ms }^{ -1 }) } \)
≈ \(\frac{1}{136.9}=\frac{1}{137}\) which is a dimensionless number
⇒ α = \(\frac{1}{137}\)
(b) Using fine structure constant, the velocity of electron can be written as v_n = \(\frac{αcZ}{n}\)
For hydrogen atom (Z = 1) the velocity of electron in n^th orbit is v_n = \(\frac{c}{137}\frac{1}{n}=(2.19\times10^6)\frac{1}{n}ms^{-1}\)
For the first orbit (ground state), the velocity of electron is v₁ = 2.19 x 10⁶ms⁻¹
For the second orbit (first excited state), the velocity of electron is v₂ = 1.095 x 10⁶ms⁻¹
For the third orbit (second excited state), the velocity of electron is v₃ = 0.73 x 10⁶ms⁻¹
Here, v₁ > v₂ > v₃

(i) Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 10⁷m s^-1or \({ \left( \frac { 1 }{ 10 } \right) }^{ th }\) of the speed of light.
(ii) It can be deflected by application of electric and magnetic fields. The direction of deflection indicates that they are negatively charged particles.
(ii) When the cathode rays are allowed to fall on matter, they produce heat. They affect the photographic plates and also produce fluorescence when they fall on certain crystals and minerals.
(iii) When the cathode rays fall on a material of high atomic weight, x-rays are produced.
(iv) Cathode rays ionize the gas through which they pass.

Relation between f and R:
C ⇒ Center of curvature
F ⇒ Principal focus
i ⇒ Angle of incidence

The angles From right angle triangles ∆MCR, and ∆MFP,
\(\tan i=\frac{P M}{P C} \text { and } \tan 2 i=\frac{P M}{P F}\)
As the angles are small, tan i = i and tan 2i = 2i.
\(\mathrm{i}=\frac{\mathrm{PM}}{\mathrm{PC}} \text { and } 2 \mathrm{i}=\frac{\mathrm{PM}}{\mathrm{PF}}\)
Simplifying further,
\(2 \frac{\mathrm{PM}}{\mathrm{PC}}=\frac{\mathrm{PM}}{\mathrm{PF}} ; 2 \mathrm{PF}=\mathrm{PC}, \mathrm{R}=2 \mathrm{f}\)
PF is focal length f and PC is the radius of curvature R.
R = 2f (or) f = R/2

According to law of reflection,
(i) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(ii) The angle of incidence i is equal to the angle of reflection r.
i = r

(i) u - v rays
(ii) X - rays
(iii) ૪ - rays.

w = \(\frac { qB }{ m } \) independence of angle of the entrance with the magnetic field.
W₁ : w₂ = 1 : 1

(i) The electromagnetic oscillations of LC system can be compared with the mechanical oscillations of a spring-mass system.
(ii) There are two forms of energy involved in LC oscillations. One is electrical energy of the charged capacitor; the other magnetic energy of the inductor carrying current.
(iii) Likewise, the mechanical energy of the spring-mass system exists in two forms; the potential energy of the compressed or extended spring and the kinetic energy of the mass. The Table lists these two pairs of energy.
(iv) By examining, the analogies between the various quantities can be understood and these correspondences.
(v) The angular frequency of oscillations of a spring-mass is given by equation
\(\omega =\sqrt { \frac { k }{ m } } \)
k ⟶ \(\frac { 1 }{ C } \) and m ⟶ L. Therefore, the angular frequency of LC oscillations is given by
ω = \(\frac { 1 }{ \sqrt { LC } } \)

Since isotopes are singly ionized, they have equal charge which is equal to the charge of an electron, q = - 1.6 x 10^-19 C. Mass of uranium \(_{ 92 }^{ 235 }{ U and _{ 92 }^{ 238 }{ U } }\) are 3.90 x 10^-25 kg and 3.95 x 10^-25 kg respectively. Magnetic field applied, B = 0.500 T. Velocity of the electron is 1.00 x 10⁵ m s^-1, then 
(a) the radius of the path of \(_{ 92 }^{ 235 }{ U }\) is r₂₃₅
\({ r }_{ 235 }=\frac { { m }_{ 235 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =48.8\times { 10 }^{ -2 }m\)
r₂₃₅ = 48.8cm 
The diameter of the semi-circle due to \(_{ 92 }^{ 235 }{ U\ \ is \ \ { d }_{ 235 }=2{ r }_{ 235 } }\) = 97.6 cm
The radius of the path of _{\(_{ 92 }^{ 238 }{ U\ is\ 2{ r }_{ 238 }\ then}\)}
\({ r }_{ 238 }=\frac { { m }_{ 238 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =49.4\times { 10 }^{ -2 }m\)
r₂₃₈ = 49.4 cm
The diameter of the semi-circle due to  \(^{ 238 }_{92}{ U\ is \ 2{ r }_{ 238 } \ =98.8 \ cm}\)
Therefore the separation distance between the isotopes is \(\triangle d={ d }_{ 238 }-{ d }_{ 235 }=1.2 \ cm\)
(b) The time taken by each isotope to complete one semi-circular path are
\({ t }_{ 235 }=\frac { \text{ magnitude of the displacement} }{ velocity } \)
\(=\frac { 97.6\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.76\times { 10 }^{ -6 }s=9.76\mu s\)
\({ t }_{ 238 }=\frac { \text{magnitude of the displacement }}{ velocity } \)
\(=\frac { 98.8\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.88\times { 10 }^{ -6 }s=9.88\mu s\) 

Charge of an electron q = –1.60 × 10^–19 C ⇒ |q| = 1 60 x 10^-19 C
Magnitude of magnetic field B = 0.500 T
Mass of the electron, m = 9.11 × 10^–31 kg
Radius of the orbit, r = 2.50 mm = 2.50 × 10^–3 m
Speed of the electron,  V = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

G = 6.67 x 10^-11 Nm²kg^-2
Mass of Earth m_E = 5.9 x 10²⁴ kg
Mass of Moon m_m = 7.9 x 10²² kg
Gravitational force \(F_{g}=\frac{G m_{E} \times m_{M}}{r^{2}}\); Electro static force \(F_e=k\frac{q \times q}{r^2}\)
By equating the forces, \( k\frac{q \times q}{r^{2}}=G \cdot \frac{m_{E} \times m_{M}}{r^{2}} \)
\(\because k=\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \)
\( 4 \pi \varepsilon_{0} =0.11 \times 10^{-9} \)
\(q =\sqrt{4 \pi \varepsilon_{0} G m_{E} \cdot m_{M}} \)   ....(1)
\(=\sqrt{0.11 \times 10^{-9} \times 6.67 \times 10^{-11} \times 5.9 \times 10^{24} \times 7.9 \times 10^{22}}\)
\( q =\sqrt{34.2 \times 10^{26}} \)
q ≈ 5.85 x 10¹³ C
b) Suppose the distance (r) between Moon and Earth is halved, there is no change in the value of charge (q). Because from equation (1), q is independent of distance (r).

Basic properties of charges:
(i) Electric charge:
(a) Most objects in the universe are made up of atoms, which in turn are made up of protons, neutrons and electrons.
(b) These particles have mass, an inherent property of particles. Similarly, the electric charge is another intrinsic and fundamental property of particles.
(ii) Conservation of charges:
Total electric charge is conserved. Charge can neither be created nor be destroyed. In any physical process, the net change in charge will be zero.
(iii) Quantisation of charges:
(a) The charge q on any object is equal to an integral multiple of this fundamental unit of charge.
(b) q = ne 
(c) Here, n is any integer \((0, \pm 1, \pm 2, \pm 3, \pm 4 \ldots)\) This is called Quantisation of electric charge.

The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T
Deflection is 
\(\theta =60°\Rightarrow tan60°=\sqrt { 3 } =1.732\)
\(I=\frac { 2R{ B }_{ H } }{ { \mu }_{ ° }N } tan\theta \)
\(=\frac { 2\times 0.12\times 25\times 1{ 0 }^{ -6 } }{ 4\times 1{ 0 }^{ -7 }\times 3.14\times 100 } \times 1.732=0.82\times 1{ 0 }^{ -1 }A\)
I = 0.082 A

(a) The electrostatic potential on the surface of the sphere A is V_A = \(\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1 } } \)
The electrostatic potential on the surface of the sphere A is V_B = \(\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 2 } }{ { r }_{ 2 } } \)
Since V_A = V_B. We have
\(\frac { { q }_{ 1 } }{ { r }_{ 1 } } =\frac { { q }_{ 2 } }{ { r }_{ 2 } } \Rightarrow { q }_{ 1 }=\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) { q }_{ 2 }\)
But from the conservation of total charge, Q = q₁ + q₂, we get q₁ = Q – q₂. By substituting this in the above equation,
Q - q₂ = \(\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) { q }_{ 2 }\)
so that q₂ = Q\(\left( \frac { { r }_{ 2 } }{ { r }_{ 1 }+{ r }_{ 2 } } \right) \)
Therefore,
q₂ = 100 x 10^-9 x \(\left( \frac { 2 }{ 10 } \right) \) = 20nC and q₁ = Q - q₂ = 80nC
The electric charge density for sphere A is σ₁ = \(\frac { { q }_{ 1 } }{ 4\pi { r }_{ 1 }^{ 2 } } \)
The electric charge density for sphere B is σ₂ = \(\frac { { q }_{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } \)
Therefore,
σ₁ = \(\frac { 80\times 10^{ -9 } }{ 4 \pi \times 64\times 10^{ -4 } } \) = 0.99 x 10^-6 Cm^-2 and
σ₂ =\(\frac { 20\times 10^{ -9 } }{ 4\pi \times 4\times 10^{ -4 } } \)  = 3.9 x 10^-6 Cm^-2
Note that the surface charge density is greater on the smaller sphere compared to the larger sphere (σ₂ ≈ 4σ₁) which confirms the result \(\frac { { \sigma }_{ 1 } }{ \sigma _{ 2 } } =\frac { { r }_{ 2 } }{ { r }_{ 1 } } \)
The potential on both spheres is the same. So we can calculate the potential on any one of the spheres
V_A=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1 } } =\frac { 9\times 10^{ 9 }\times 80\times { 10 }^{ -9 } }{ 8\times 10^{ -2 } } \)  = 9kV

(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.

(i)The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object.
(ii) The height of the object and image are measured perpendicular to the principal axis
\(magnification(m)=\cfrac { height \ of\ the\ image(h') }{ heigh \ of\ the\ object(h) } \) 
\(m=\cfrac { h' }{ h } \)
Applying proper sign conventions for equation (1),
\(\cfrac { A'B' }{ AB } =\cfrac { PA' }{ PA } \) 
A'B' = -h, AB = h, PA' = -v, PA = -u
\(\cfrac { -h' }{ h } =\cfrac { -v }{ -u } \) 
On simplifying we get,
\(m=\cfrac { h' }{ h } =-\cfrac { v }{ u } \) 
Using mirror equation, we can further write the magnification as, magnification as,
\(m=\cfrac { h' }{ h } =\cfrac { f-v }{ f } =\cfrac { f }{ f-u } \)   ...(3)

(i) At a minimum deviation \(\mu =\cfrac { sin\left( \frac { A+{ \delta }_{ m } }{ 2 } \right) }{ sin\left( \cfrac { A }{ 2 } \right) } \) 
 Given \({ \delta }_{ m }=A\)
\(\mu =\cfrac { sinA }{ sin\cfrac { A }{ 2 } } =\cfrac { 2sin\cfrac { A }{ 2 } cos\cfrac { A }{ 2 } }{ sin\cfrac { A }{ 2 } } \) 
= \(2cos\cfrac { A }{ 2 } \) 
\(\therefore cos\cfrac { A }{ 2 } =\cfrac { \sqrt { 3 } }{ 2 } \cfrac { A }{ 2 } =30\) A = 6^o 
(iii) \(\mu =\sqrt { 3 } \cfrac { 1 }{ { sini }_{ c } } \Rightarrow { sini }_{ c }=\cfrac { 1 }{ \sqrt { 3 } } \) 
ஃ Angle of incidence > i_c
Total internal reflection takes place.

X-rays are being used in many fields. Let us list a few of them.
(i) Medical diagnosis X-rays can pass through flesh more easily than through bones. Thus an x-ray radiograph containing a deep shadow of the bones and a light shadow of the flesh may be obtained. X-ray radiographs are used to detect fractures, foreign bodies, diseased organs, etc.
(ii) Medical therapy Since x-rays can kill diseased tissues, they are employed to cure skin diseases, malignant tumors, etc.
(iii) Industry X-rays are used to check for flaws in welded joints, motor tires, tennis balls, and wood. At the customs post, they are used for the detection of contraband goods.
(iv) Scientific research X-ray diffraction is an important tool to study the structure of the crystalline materials - that is, the arrangement of atoms and molecules in crystals.

From the Snell's law in product form, n₁ sini = n₂ sinr  ....(1)
The equation for this refraction at the point A is as shown in the Figure 


n₃sin i_a = n₁sin r_a  ........(1)
(vi) To have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be at least critical angle i_e'  From equation (1), for equation for the refraction at point B is,
n₁ sin i_c = n₂ sin 90^o      ...(3)
n₁sin i_c = n₂          ∵ sin 90^o=1 
\(sin{ i }_{ c }=\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)     ...(4) 
From the right angle triangle Δ ABC,
i_c = 90^o - r_a
Now, equation (4) becomes
\(sin\left( { 90 }^{ o }-{ r }_{ a } \right) =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \) (or) \(cos{ r }_{ a }=\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)  ...(5)   
 \(sin{ r }_{ a }=\sqrt { 1-{ cos }^{ 2 }{ r }_{ a } } \) 
Substituting for cos r_a
\(sin{ r }_{ a }=\sqrt { 1-\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right) ^{ 2 } } =\sqrt { \cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } } \)     ...(6) 
Substituting this in equation
\({ n }_{ 3 }{ sini }_{ a }={ n }_{ 1 }\sqrt { \cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } } =\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \)    ...(7) 
On further simplification
\(sini_{ a }=\cfrac { \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } }{ n_{ 3 } } \) (or) \(\quad { i }_{ a }=\sqrt { \cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 3 }^{ 2 } } } \)  ...(8) 
\(\therefore{ i }_{ a }={ sin }^{ -1 }\left( \sqrt { \cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 3 }^{ 2 } } } \right) \)     ...(9)
If outer medium is air, then n₃ = 1. The acceptance angle i_a becomes
\({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \)   ..(10)
Light can have any angle of incidence from 0 to i_a with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n₃sini_a) is called numerical aperture NA of the optical fibre.
\(NA={ n }_{ 3 }{ sini }_{ a }=\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \)
If outer medium is ab then n₃ = 1 
The numerical aperture NA becomes,
\(NA={ sini }_{ a }=\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \)

Mirror Equation :

(i) AB is an object which is placed on the principal axis of a concave mirror beyond the center of curvature C. A' B' is an image which is formed between the point pole P, and the centre of curvature.
(ii) From the figure As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠B'PA'.
(iii) The triangles ∠BPA and ∠B'PA' are similar. Thus, from the rule of similar triangles,
\(\cfrac { { A }^{ ' }{ B }^{ ' } }{ AB } =\cfrac { { PA }^{ ' } }{ PA } \)             ................(1)
(iv) The other set of similar triangles are, ΔDPF and ΔB'A'F. (PD is almost a straight vertical line)
\(\cfrac { { A }^{ ' }B' }{ PD } =\cfrac { A'F }{ PF } \) 
(v) As, PD = AB the above equation becomes,
\(\cfrac { A'B' }{ AB } =\cfrac { A'F }{ PF } \)            ......(2)
(vi) From equations (1) and (2) we can write,
\(\cfrac { PA' }{ PA } =\cfrac { A'F }{ PF } \) 
(vii) As, A'F = PA' - PF, the above equation becomes,
\(\cfrac { PA' }{ PA } =\cfrac { PA'-PF }{ PF } \)            .....(3)
(viii) We can apply the sign conventions for the various distances in the above equation
PA = - u, PA' = -v, PF = - f
(ix) All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes,
\(\cfrac { -v }{ -u } =\cfrac { -v-\left( -f \right) }{ -f } \) 
On further simplification,
\(\cfrac { v }{ u } =\cfrac { v-f }{ f } ;\cfrac { v }{ u } =\cfrac { v }{ f } -1 \) 
Dividing either side with v,
 \(\cfrac { 1 }{ u } =\cfrac { 1 }{ f } -\cfrac { 1 }{ v } \)
After rearranging,
\(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
The above equation is called mirror equation.
Lateral magnification:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.
Magnification (m) \(=\frac{\text { height of the image }\left(h^{\prime}\right)}{\text { height of the image }(h)} \)
\(m=\frac{h^{\prime}}{h} \)  ....(1)
Applying proper sign conventions for equation,
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A} \)
\(A^{\prime} B^{\prime}=-h^{\prime}, A B=h, P A^{\prime}=-v, P A=-u \)
\(-\frac{h}{h}=\frac{-v}{-u} \)
On simplifying we get,
\(\mathrm{m}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-\frac{\mathrm{v}}{\mathrm{u}}\)  ...(2)
Using mirror equation, we can further write the magnification as,
\(m=\frac{h^{\prime}}{h}=\frac{f-v}{f}=\frac{f}{f-u}\)   ..(3)

Given: Velocity of electromagnetic wave is
c = 3 x 10⁸ m/s.
Relative electric permittivity ε_r = 2
Relative magnetic permeability μ_r = 1
To find:
Velocity of Electromagnetic in a medium is
\(v=\frac { 1 }{ \sqrt { { \varepsilon }_{ 0 }{ \varepsilon }_{ r }.{ \mu }_{ 0 }{ \mu }_{ r } } } =\frac { 1 }{ \sqrt { { \varepsilon }_{ 0 }{ \mu }_{ 0 } } \times \sqrt { { \varepsilon }_{ r }{ \varepsilon }_{ r } } } \)
Solution:
\(\therefore v=\frac { 1 }{ \sqrt { { \varepsilon }_{ 0 }{ \mu }_{ 0 } } } ,v=\frac { 1 }{ \sqrt { { \varepsilon }_{ r }{ \mu }_{ R } } } \)
\(\therefore v=\frac { 3\times { 10 }^{ 8 } }{ \sqrt { 2\times 1 } } =\frac { 3 }{ \sqrt { 2 } } \times 10^{ 8 }m/s\)

(i) The energy E remains constant for varying values of x and v. Differentiating E with respect to time, we get \(\frac { dE }{ dt } =\frac { 1 }{ 2 } \left( 2v\frac { dv }{ dt } \right) +\frac { 1 }{ 2 } k\left( \frac { dx }{ dt } \right) =0\)
or m \(\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } +kx=0\)
since \(\frac { dx }{ dt } =\nu \) and \(\frac { dv }{ dt } =\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } \)
(ii) This is the differential equation of the oscillations of the spring-mass system. The general solution of an equation is of the form
x(t) = X_m cos (ωt + Φ)
where X is the maximum value of x(t), ω the angular frequency, and Φ the phase constant.
(iii) Similarly, the electromagnetic energy of the LC system is given by
\(U=\frac { 1 }{ 2 } { Li }^{ 2 }+\frac { 1 }{ 2 } \left( \frac { 1 }{ C } \right) { q }^{ 2 }\) = constant
Differentiating U with respect to time, we get
\(\frac { dU }{ dt } =\frac { 1 }{ 2 } L\left( 2i\frac { di }{ dt } \right) +\frac { 1 }{ 2C } \left( 2q\frac { dq }{ dt } \right) =0\)
(or) \(\frac { { d }^{ 2 }q }{ { dt }^{ 2 } } +\frac { 1 }{ C } q=0\) ....(1)
since \(i=\frac { dq }{ dt } \frac { di }{ dt } =\frac { { d }^{ 2 }q }{ d{ t }^{ 2 } } \)
(iv) The general solution of equation (1) is of the form
q(t) = Q_m cos (ωt + Φ)
(v) where Q_m is the maximum value of q(t), ω the angular frequency, and Φ the phase constant.

According to Wheatstone bridge \(\cfrac { P }{ Q } =\cfrac { R }{ S } \)
\(\cfrac { R }{ S } =\cfrac { OA }{ OB } =\cfrac { 40 }{ 60 } \)
\(\cfrac { R }{ S } =\cfrac { OA }{ OB } =\cfrac { 40 }{ 60 } \)
If resistance 10 \(\Omega \) connected in series with R, the balance length is 60 cm.
\(\cfrac { R+10 }{ S } =\cfrac { 60 }{ 40 } \Rightarrow 2R+20=3S\)
From (1) & (2) \(\Rightarrow \left[ \cfrac { 4S }{ 3 } +20=3S \right] \)
45 + 60 = 95
55 = 60
\(S=\cfrac { 60 }{ 5 } =12\)
\(2\times \cfrac { 2S }{ 3 } +20=3S\)
\(s=12\Omega \)
From equation (1)
\(R=2\times \cfrac { 12 }{ 3 } =8\Omega \)
\(R=8\Omega \)

Length, I = 2a = 4cm = 4 x 10^-2m
Angle, θ = 60°
torque ፔ = 4√3 Nm
Charge, Q = 8 x 10^-9C
We know that, ፔ = pE sine [where p = Q x 2a]
ፔ = (Q x 2a)E sinθ
\(\tau =\frac { \partial }{ Q\times (2a)sin\theta } \)
\(=\frac { 4\sqrt { 3 } }{ 8\times { 10 }^{ -9 }\times 4\times { 10 }^{ -2 }\times { sin\quad 60 }^{ 0 } } \)

∴ Potential energy, U = -pE cosθ
= -Q(2a) x E x cosθ
\(=-8\times { 10 }^{ -9 }\times 4\times { 10 }^{ -2 }\times \frac { 4\sqrt { 3 } \times cos{ 60 }^{ 0 } }{ 8\times { 10 }^{ -9 }\times 4\times { 10 }^{ -2 }sin{ 60 }^{ 0 } } \)
\(U=\frac { -4\sqrt { 3 } }{ \sqrt { 3 } } =-4J\)