Given, focal length of convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = -20 cm
Using the formula of combination of lenses,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\)
\(\Rightarrow\) f = -60 cm
Since, the focal length of combination is negative in nature. so, the combination behaves like a diverging lens, i.e. as a concave lens.

The electric field vector E_1A at A due to the positive charge q₁ points towards the right. Its magnitude,
E_1A = \(\frac { (9\times { 10 }^{ 9 }N-{ m }^{ 2 }c^{ -2 })({ 10 }^{ -7 }) }{ { (0.1\quad m) }^{ 2 } } =9\times { 10 }^{ 4 }\) NC^-1
E_A is directed towards right.
The electric field vector E_{2 A} due to q₂ points to the right and has the same magnitude. Hence, the magnitude of total electric field E_A at A,
E_A = E_{1 A} + E_2 A = 18 \(\times\) 10⁴ NC^-1
E_A is directed towards right.
The electric field E_{1 B} at B due to q₁ points towards the left and has a magnitude,
E_1B = \(\frac { (9\times { 10 }^{ 9 }N-{ m }^{ 2 }c^{ -2 })({ 10 }^{ -7 }C) }{ { (0.1\quad m) }^{ 2 } } =9\times { 10 }^{ 4 }\) NC^-1
The electric field E_2 B at B due to the negative charge q₂ points towards the right and has a magnitude
E_2B = \(\frac { (9\times { 10 }^{ 9 }N-{ m }^{ 2 }c^{ -2 })({ 10 }^{ -7 }C) }{ { (0.3\quad m) }^{ 2 } } =1\times { 10 }^{ 4 }\) NC^-1
The magnitude of the total electric field at B,
E_B = E_{1 B} - E_{2 B} = 8 \(\times\) 10⁴ NC^-1
E_B is directed towards the left.
The magnitude of each electric field vector at point C due to charge q₁ and q_2, 
E_{1 C} = E_{2 C} = \(\frac { (9\times { 10 }^{ 9 }N-{ m }^{ 2 }{ c }^{ -2 })({ 10 }^{ -7 }C) }{ { (0.2\quad m) }^{ 2 } } \)
= \(2.25\times10^4\) NC^-1
The resultant of these two vectors,
 E_C = E_{1 C} cos\(\frac { \pi }{ 3 }\) + E_2 C cos\(\frac { \pi }{ 3 }\)
 = \(2.25\times10^4\) NC^-1
E_C points towards the right.

(i)  Y-rays are used for the treatment of certain forms of cancer. Its frequency range is from 3 x 10¹⁹ Hz to 3 x 10²³ Hz.
(ii)  The thin ozone layer on top of stratosphere absorbs most of the harmful ultraviolet rays coming from the sun towards the earth. They include UVA, UVB and UVC radiations which can destroy the life system on the earth. Hence, this layer is crucial for human survival.
(iii) Momentum transferred = Energy/Speed of light
 \(=\frac { E }{ c } =\frac { hV }{ c } ={ 10 }^{ -22 }\)
Thus, the amount of the momentum transferred by the electromagnetic waves incident on the surface is very small.

Selectivity of radio receiver is the ability of a receiver to accept the wanted signal and to reject the unwanted signals. Sensitivity of a radio receiver is its ability to amplify the desired weak signal. 

7.07 A, 3.535 A
\(I_{ v }=5\ A,\ I_{ O }=\sqrt { 2 } I_{ v }=1.414\times 5=7.07\ A\)
From \(I=I_{ 0 }sin\quad \omega t\)
\(=7.07 \ sin \ \frac { \pi }{ 6 } =7.07\times \frac { 1 }{ 2 } =3.535\ A\)

Figure of merit, 
\(K=\frac { I }{ \theta } =\frac { V }{ R\theta } =\frac { 1 }{ (900+100)100 } \) 
= 10^-5 A/div.

The focal length f = -15/2 cm = -7.5 cm
(i) The object distance u = -10 cm. Then Eq gives
\(\frac { 1 }{ v } +\frac { 1 }{ 10 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 10\times 7.5 }{ -2.5 } =-30\) cm
The image is 30 cm from the mirror on the same side as the object
Also, magnification m = \(\frac { v }{ u } =-\frac { (-30) }{ (-10) } =-3\)
The image is magnified, real and inverted.
(ii) The object distance u = -5 cm. Then from Eq
\(\frac { 1 }{ v } +\frac { 1 }{ -5 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 5\times 7.5 }{ (7.5-5) } =15\) cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = 15 \(-\frac { v }{ u } =-\frac { 15 }{ (-5) } =3\)
The image is magnified, virtual and erect.

The required energy band diagram are as shown:


(a) When the semiconductor is doped with an acceptor impurity there is an additional energy level a little above the top of the valence band.
(b) The donor impurity results in an additional energy level a little below the bottom of the conduction band.
(Also accept diagrammatic representations) In the first case, electrons, from the valence band, easily jump over to the acceptor level, leaving 'holes' behind. Hence, 'holes' become the majority charge carriers.
In the second case, electrons from the donor level, easily 'jump over' to the conduction band. Hence, electrons become the majority charge carriers.
The two processes, involved in the formation of the p-n junction are.
(i) Diffusion
(ii) Drift

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

Here q = \(1.6\times 10^{-19}C\)
\(2a=4\times10^{-10}m, E=3\times 10^5NC^{-1}, \theta=30^o\)
\(p=q\times2a=1.6\times10^{-19}\times4\times10^{-10}\)
= \(6.4\times10^{-29}Cm\)
\(\tau=pEsin\theta=6.4\times10^{-29}\times3\times10^5\times sin 30^o\)
\(=9.6\times 10^{-24}Nm\)

Rest mass of electron,
\({ m }_{ 0 }=9.1\times { 10 }^{ -31 }kg\)
\(v=0.6c=0.6,\times 3\times { 10 }^{ 8 }=1.8\times { 10 }^{ 8 }{ ms }^{ -1 }\)
As v is comparable to c, hence mass of the electron in motion will be a relativistic mass. So,
\(m=\frac { { m }_{ 0 } }{ \sqrt { 1-{ v }^{ 2 }/{ c }^{ 2 } } } =\frac { { m }_{ 0 } }{ \sqrt { 1-\frac { { \left( 0.6 \right) }^{ 2 } }{ { c }^{ 2 } } } } =\frac { { m }_{ 0 } }{ 0.8 } \)
De-Broglie wavelength,
\(\lambda =\frac { h }{ mv } =\frac { h }{ \left( { m }_{ 0 }/0.8 \right) \times v } =\frac { h\times 0.8 }{ { m }_{ 0 }v } \)
\(=\frac { \left( 6.6\times { 10 }^{ -34 } \right) \times 0.8 }{ \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 1.8\times { 10 }^{ 8 } \right) }\)
\( =0.322\times { 10 }^{ -11 }m=0.0322\times { 10 }^{ -10 }m\)
\(=0.032\overset { \circ }{ A } \)

A = 60⁰ ,  \(\delta \)_m = 30⁰
i = e = ¾ A = 450
as A + \(\delta \) = i + e
60 + \(\delta \) = 45 +45
or \(\delta \) = 30⁰
Refractive index,
\(\mu \) = sin a + \(\delta \)m /2/sin A/2 = sin 60⁰+30⁰/2/sin 60⁰/2
= sin 45⁰/sin30⁰ = 1\(\surd 2\) 1/2 = \(\surd 2\) = 1.414

R

\(22*102 Ω ± 10%\)

(a) Modulation index =10 / 20 = 0.5
(b) The side bands are at (1000+10 kHz)=1010 kHz and (1000 –10 kHz) = 990 kHz.

Y₁ Microwaves
Microwave oven, Aircraft Navigator or any other
Y₂ Ultraviolet waves
Sterilize surgical instruments, food preservation or any other