The ray diagram for the formation of the image of the phone is shown in Fig. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B'C = BC. You can yourself realise why the image is distorted.

(i) Pulse Amplitude Modulation : Amplitude of the pulse varies in accordance with the modulating signal.

(ii) (ii) Pulse Position Modulation. : Pulse position (ie) time of rise or fall of the pulse ) changes with the modulating signal.

AND Gate using NAND GATE:

The fringe width in the two cases will be \(\beta\) = D\(\lambda \)/d
And \(\beta\)' = D' \(\lambda \)/d
\(\beta\) - \(\beta\) '  = (D - D' )\(\lambda \)/d
or wavelength \(\lambda \) = (\(\beta\) -\(\beta\)' )d /(D - D')
But D -D'  = 5 x 10^-2 m
And  \(\beta\) - \(\beta\)'  = 3 x 10^-5 m, d = 10^-3 m
\(\lambda \) = 3 x 10^-5 x 10^-3 / 5 x 10^-2 = 6 x 10^-7 m = 6000A

(i) As the lens forms a real iamge, it must be a convex lens.
(ii) From the graph, when u = 20 cm , we have v = 20 cm.
For the convex lens forming a real iamge, u is negative and v and f are positive.
U = -20 cm v = +20cm
Using this lens formula,
1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm

r = mv/qB .........(i)
Also P = mv = \(\sqrt { 2mE } \)..........(ii)
By equ (i) and equ (ii)
As the radius is \(r\frac { \sqrt { 2mE } }{ qB } \) proportional to square root of kinetic energy, so if the kinetic energy is halved the radius become \(\sqrt { 1 } /2\) times of its initial value 

(i) Dipole has two equal and opposite charges. In the uniform electric field they will experience equal and opposite force. Net force is zero, So there can't be any translatory motion
(ii) Torque \(\tau \)F = 21 sin \(\theta\) qE = p x E

High frequency carrier waves are used to increase operating range, to reduce antenna length and convert the wide band signal into narrow band signal. Then the signal can be easily recovered and distinguished from other signals at the receiving station.

(i) Attenuation The loss of strength of signal during its propagation through the transmission medium is called attenuation. Repeater is used to compensate the attenuation.
(ii) Demodulation The process of recovering the original information signal from the modulated wave at the receiver end is called demodulation. It is the reverse process of modulation.

During landing or take off, the tyres of aircraft's get charged due to the friction between tyres and ground. In case, the tyres are slightly conducting, the charge developed on the tyres will not stay on them and it finds its way to the earth.

(a) As Q = CV, if V doubles, charge also gets doubled.
(b) As \(U={1\over 2}CV^2\), if V is doubled, U is quadrupled.

Range of frequency used for TV transmission is 54 MHz to 890 MHz (VHF and UHF).These waves and light waves are electromagnetic waves.The ionosphere is unable to reflect back these waves to earth.

Let R₀ be the resistance of the conductor at 0^oC. Then resistance of conductor B at 0^oC = n R₀.
Resistance of conductor A at \(\theta^oC\),
R₁ = R₀(1 + \(\alpha_1 \theta\))
Resistance of conductor A at  \(\theta^oC\) ,
R₂ = n R₀(1 + \(\alpha_2 \theta\))
Total resistance,
R_s = R₁ + R₂ 
= R₀(1 + \(\alpha_1 \theta\)) +  n R₀(1 + \(\alpha_2 \theta\))
= \((1+n)R_0[1+\frac{\alpha_1+n \alpha_2}{1+n}\theta]\)
Comparing this relation with
R_s = R_s0 [ 1 + \(\alpha_s \theta\))
We have, resistance of series combination at \(\theta^oC\) 
R_s0 = (1 + n)R₀
Temperature coefficient of resistance of the series combination is
\(\alpha_s=\frac{\alpha_1+n\alpha_2}{1+n}\) 

(i) Like charges repel and unlike charges attract each other.
(ii) Charge is quantized
(iii) Charge is conserved 
(iv) Charge on a body is not affected by its motion.