The P.D. between A and B is given by
\(V =\left[V_{\mathrm{A}}-V_{\mathrm{B}}\right]-V_{\mathrm{b}}(\mathrm{Si}) \)
\(=[3.3-(-7.4)]-0.7 \)
\(=10.7-0.7=10 \mathrm{~V} \)
The value of current flowing through AB can be obtained by using Ohm’s law
\(I=\frac { V }{ R } =\frac { 10}{ 1\times { 10 }^{ 3 } } ={ 10 }^{ -2 }A=10mA\)

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,\({ n }_{ 1 }\le 1.414\)

Infrared radiation waves are produced by hot bodies and molecules so it is referred as heat waves. (eg. Sun)
(i) Electromagnetic waves having shorter λ than Infrared radiation are visible, U - v, X-rays, and ૪ - rays.
(ii) Electromagnetic waves having longer λ than Infrared radiation are microwaves, radiowaves.

B = 4.1 x 10^–5 T; θ = 90^o – 47^o = 43° ;
A = 3m²
We know that   \(\Phi_{B}=B A \cos \theta\) 
\(\Phi_{\mathrm{B}}\) = 4.1 x 10^–5 x 3 x cos 43^o  
= 4.1 x 10^–5 x 3 x 0.7314
= 89.96 \(\mu \mathrm{Wb}\).

The geometrical length of the bar magnet is 12 cm
Magnetic length = \(\frac { 5 }{ 6 } \times \)(geometrical length)
= \(\frac { 5 }{ 6 } \times \)12 = 10 cm
In this figure, the dot implies the pole points.

β = 120
Base current, \({ I }_{ B }=\frac { 25V }{ 1M\Omega } =\frac { 25 }{ 1\times { 10 }^{ 6 } } =25\mu A\)
We know that
\(\beta =\frac { { I }_{ C } }{ { I }_{ B } } \) (or) 
I_C = β I_B = 120 x 25 μA
= 3000 μA = 3 mA
V_CE = V_CC - I_CR_C
= 25 - (3 mA x 5k) = 10 V 

Given, thickness of the slab, t = 0.25 m, refractive index, n = 1.5, angle of incidence, i = 60^o.
Using Snell’s law, 1 sin i = n sin r
\(sinr=\cfrac { sini }{ n } =\cfrac { sin60^o }{ 1.5 } =0.58\) 
\(r={ sin }^{ -1 }(0.58)=35.25^{ 0 }=35^o15'0''\) 
Lateral displacement is, \(L=t\left( \cfrac { sin\left( i-r \right) }{ cos\left( r \right) } \right) \) 
\(L=\left( 0.25 \right) \times \left( \cfrac { sin\left( 60-35.25 \right) }{ cos\left( 35.25 \right) } \right) =0.1281m\)
The  lateral displacement is, L = 12.81 cm

i) Consider a bar magnet of length 2l and pole strength q_m
ii) Force experienced by the magnet at each pole is q_m B (equal) in opposite direction. 
iii) So, magnet experiences a torque.

The force experienced by north pole,
\(\vec { { F }_{ N } } ={ q }_{ m }\vec { B } \)             
The force experienced by south pole,
\(\vec { { F }_{ S } } =-{ q }_{ m }\vec { B } \)
∴ The net force acting on the dipole is,
\(\vec { F } =\vec { { F }_{ N } } +\vec { F_{ S } } =\vec { 0 } \)
The moment of force or torque experienced by north and south pole about point O is,
\(\vec { \tau } =\vec { ON } \times \vec { { F }_{ N } } +\vec { OS } \times \vec { { F }_{ S } } \)
\(\vec { \tau } =\vec { ON } \times { q }_{ m }\vec { B } +\vec { OS } \times (-{ q }_{ m }\vec { B } )\)
By using right hand cork screw rule, we conclude that the total torque is pointing into the paper. Since the magnitudes \(|\vec { ON } |=|\vec { OS } |=l\)  and \(|{ q }_{ m }\vec { B } |=|-{ q }_{ m }\vec { B } |\).
The magnitudes of total torque about point O is
ፒ = l x q_mB sinθ + l x q_mB sinθ
ፒ = 2l x q_mB sinθ
ፒ = p_mB sinθ   (∴ q_m x 2l = p_m)
In vector notation, \(\vec { \tau } =\vec { { p }_{ m } } \times \vec { B } \). 

A = 5 x 10^-2 m²; B = 0.2 T
(i) θ = 0°;
\({ \Phi }_{ B }=BAcos\theta =0.2\times 5\times { 10 }^{ -2 }\times { cos }0^{ o }\)
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }Wb\)
(ii) θ = 90° – 60° = 30°;
\(\Phi_B\) = BAcosθ  = 0.2 x 5 x 10^-2 x cos 30^o
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }\times \frac { \sqrt { 3 } }{ 2 } =8.66\times { 10 }^{ -3 }Wb\)
(iii) θ = 90°;
\({ \Phi }_{ B }\) = BA cos90^o = 0

\( \mathrm{i}=60^{\circ} \)
\( \mathrm{n}=\sqrt{3}\)
Applying Snell's law at A
\(\frac{\sin i}{\sin r_1}=\sqrt{3}\)
\(\sin \mathrm{r}_{\mathrm{t}}=\frac{\sin 60}{\sqrt{3}}=\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}}=\frac{1}{2}\)
r₁ = 30°
If r₁ = 30° & r₂ = 30°
Applying Snell's law at 'B'
\(\frac{\sin r_2}{\sin i_2}=\frac{1}{\sqrt{3}}\)
\(\sin i_2 =\sqrt{3} \times \sin 30^{\circ} \)
\(=\sqrt{3} \times \frac{1}{2} \)
\( \sin i_2 =\frac{\sqrt{3}}{2}=\sin 60^{\circ} \)
\( i_2 =60^{\circ} \)
At B,
\( \mathrm{r}_2=30^{\circ} \text { also } \theta =30^{\circ} \)
\(\therefore \theta+\alpha+\mathrm{i}_2 =180^{\circ}\)
\(\alpha =180^{\circ}-\left(\mathrm{i}_2+\theta\right) \)
\( =180^{\circ}-\left(60^{\circ}+30^{\circ}\right) \)
\( =180^{\circ}-90^{\circ} \)
\(\alpha =90^{\circ} \)

\(l=\frac{P}{V}=\frac{11000}{220}=50 \mathrm{~A}\)

\(Y_1=\overline{A+B}, y_2=\overline{A+B}=A+B, y=\overline{A+B}\)

Refractive index of water \(=\frac{Velocity \ of \ light \ in \ air(V_s)}{Velocity \ of \ light \ in \ water(V_w)}\)

\( { E } ={ E }_{ 0 }\hat { k } \) and \({ B } ={ B }_{ 0 }\hat { j } \)

Dispersion: It is splitting of white light into its constituent colours.
(i) Consider a beam of white light passes through a prism; it gets dispersed into its constituent colours as shown in Figure.
(ii) Let \(\delta_{v}, \delta_{R} \) are the angles of deviation for violet and red light. Let n_V and n_R are the refractive indices for the violet and red light respectively
(iii) The refractive index of the material of a prism is given by the equation
\(\mathrm{n}=\frac{\sin \left(\frac{\mathrm{A}+\mathrm{D}}{2}\right)}{\sin (\mathrm{A} / 2)}\)
(iv) Here A is the angle of the prism and D is the angle of minimum deviation. If the angle of prism is small of the order of 10^o, the prism is said to be a small angle prism.
(v) When rays of light pass through such prisms, the angle of deviation also becomes small. If A be the angle of a smitt angle prism and the angle of deviation then the prism formula becomes.

\(n=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin (A / 2)}\)
For small angles of  \(A\ and \ \delta \)
\(\sin \frac{A+\delta}{A} \approx \frac{A+\delta}{A} \)
\(\sin \frac{A}{2} \approx \frac{A}{2} \)
\(n=\frac{(A+\delta / 2)}{(A / 2)}=\frac{A+\delta}{A}=1+\frac{\delta}{A} \)
Further simplifying, 
\(\frac{\delta}{A} =n-1 \)
\(\delta =(n-1) A \)    .....(1)
(vi) When white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours
For Violet colour, \( \delta_{\mathrm{v}}=\left(\mathrm{n}_{\mathrm{v}}-1\right) \mathrm{A} \)   ...(2)
For Red colour, \(\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{R}}-1\right) \mathrm{A} \)   ....(3)
(vii) As, angle of deviation for violet colour \(\delta_{v}\) is greater the angle of deviation for red colour \(\delta_{\mathrm{R}}\) the refractive index for violet colour n_v is greater than the refractive index for red colour n_R Subtracting \(\delta_{v}\)  from \(\delta_{\mathrm{R}}\) we get
\(\delta_{\mathrm{v}}-\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{R}}\right) \mathrm{A}\) ....(4)
(viii) The term \(\left(\delta_{v}-\delta_{R}\right)\)  is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. If we take \(\delta\) is the angle of deviation for any middly ray (green or yellow) and the corresponding refractive index. Then,
\(\delta=(n-1) A\)    .....(5)
Dispersive power (ω):
It is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour.
Dispersive power
\(\omega=\frac{\text { Angular dispersion }}{\text { Mean deviation }}=\frac{\delta_{v}-\delta_{R}}{\delta}\)  ....(6)
Substituting \(\left(\delta_{\mathrm{v}}-\delta_{\mathrm{R}}\right) \text { and }(\delta) \)
\(\omega=\frac{\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{R}}}{(\mathrm{n}-1)} \)  .......(7)
(ix) Dispersive power is a dimensionless quality It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.

(i) Galvanometer to an Ammeter:

(i) Ammeter is an instrument used to measure current flowing in the electrical circuit.
(ii) The Ammeter must offer low resistance such that it will not change the current passing through it. So, ammeter is connected in series to measure the circuit current.
(iii) A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
(iv) Let I be the current passing through the circuit. When current I reaches the junction A, it divides into two components.
a) I_g → Current passing through the galvanometer
b) I - I_g → Current passing through the shunt resistance.
(v) The potential difference across the galvanometer is same as the potential difference across the shunt resistance.
\(\mathrm{V}_{\text {galvanometer }} =\mathrm{V}_{\text {shunt }} \)
\(\Rightarrow \mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}} =\left(\mathrm{I}-\mathrm{I}_{g}\right) \mathrm{S} \)
\(\mathrm{S} =\frac{I_{g}}{\left(I-I_{g}\right)} R_{g} \)  (or)
\(\mathrm{I}_{\mathrm{g}}=\frac{S}{S+R_{g}} I \Rightarrow I_{g} \propto I\)
Since, the deflection in the galvanometer is proportional to the current passing through it.
\(\theta=\frac{1}{G} I_{g} \Rightarrow \theta \propto I_{g} \Rightarrow \theta \propto I\)
Where, R_g → Galvanometer resistance, S → Shunt resistance.
Since shunt resistance is connected in parallel to galvanometer,
Effective resistance,\(\frac{1}{R_{e f f}}=\frac{1}{R_{g}}+\frac{1}{S} \Rightarrow R_{e f f}=\frac{R_{g} S}{R_{g}+S}=R_{a}\)
R_a ⇒ low resistance. An ideal ammeter has zero resistance.
The percentage error in measuring a current through an ammeter is,
\(\frac{\Delta I}{I} \times 100 \%=\frac{I_{i d e a l}-I_{a c t u a l}}{I_{a c t u a l}} \times 100 \%\)
(ii) Galvanometer to a voltmeter:
i) A voltmeter is an instrument used to measure potential difference across any two points in the electrical circuits. 
ii) Voltmeter must have high resistance and when it is connected in parallel, it will rot draw appreciable current so that it will indicate the true potential difference.
iii) A galvanometer is converted into a voltmeter by connecting high resistance  R_h in series with galvanometer.
iv) Let R_g be the resistance of galvanometer and I_g be the current with which the galvanometer produces full scale deflection.
v) Since the galvanometer is connected in series with high resistance, the current in the electrical circuit is same as the current passing through the galvanometer.

\(\mathrm{I}=\mathrm{I}_{\mathrm{g}} \)
\(\mathrm{I}=I_{g} \Rightarrow I_{g}=\frac{\text { potential difference }}{\text { total resistance }} \)
Since the galvanometer and high resistance are connected in series, the voltmeter resistance is,
\(R_{v} =R_{g}+R_{h} \)
Therefore,
\(I_{g} =\frac{V}{R_{g}+R_{h}} \)
\(\Rightarrow R_{h} =\frac{V}{I_{g}}-R_{g} \)
Note that \(I_{g} \propto V\)
R_h is very large. An ideal voltmeter has infinite resistance