We first consider the effect of the lens. For the
lens, we have
u = - 40 cm and f = + 20 cm
Using the lens formula, we get
\(\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ \left( -40 \right) } =\cfrac { 1 }{ 20 } \)   ஃv₁ = + 40 cm
Had there been the lens only the image would have been formed at Q₁. The plane mirror M is at a distance of 30 cm from lens 1. We can, therefore, think of Q₁, as a virtual object, located at a distance of 10cm, behind the plane mirror M. The plane mirror, therefore, forms a real image (of this virtual object Q₁) at Q, 10 cm in front of it.

P = P₁ + P₂ = (15 - 5) D = 10D
\(F=\cfrac { 1 }{ P } =\cfrac { 1 }{ 10 } m=10cm\) 
(b) \(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ F } or\cfrac { 1 }{ F } +\cfrac { 1 }{ u } \) 
or \(\cfrac { 1 }{ v } =\cfrac { 1 }{ 10 } +\cfrac { 1 }{ -30 } \) or v = 15cm
Again,\(\cfrac { I }{ O } =\cfrac { v }{ u } =\cfrac { 15 }{ -30 } =-\cfrac { 1 }{ 2 } \) 
or \(I=-\cfrac { 1 }{ 2 } \times 3cm=-1.5cm\) 
The negative sign indicates that the image is real and inverted.

For the convex lens 
Formula :\(\cfrac { 1 }{ { f }_{ 1 } } =\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ { u }_{ 1 } } \) 
\(\cfrac { 1 }{ +30 } =\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ { u }_{ 1 } } \) 
\(\cfrac { 1 }{ { v }_{ 1 } } =\cfrac { 1 }{ 30 } -\cfrac { 1 }{ 40 } =\cfrac { 1 }{ 120 } \) 
v_I = 120 cm a real image is formed.
On introducing a concave lens
f₂ = 50cm and u₂ = 120 - 20 = +100 cm from the concave lens
\(\cfrac { 1 }{ { f }_{ 2 } } =\cfrac { 1 }{ { v }_{ 2 } } -\cfrac { 1 }{ { u }_{ 2 } } \) 
\(\cfrac { 1 }{ -50 } =\cfrac { 1 }{ { v }_{ 2 } } -\cfrac { 1 }{ +100 } \) 
\(\therefore \cfrac { 1 }{ { v }_{ 2 } } =\cfrac { 1 }{ 580 } +\cfrac { 1 }{ 100 } =\cfrac { 1 }{ 100 } \)
v₂ = -100cm

Formula
We know \({ i }_{ c }=\cfrac { 1 }{ \mu } \)
 \(\mu =\cfrac { 1 }{ { sini }_{ c } } =\cfrac { 1 }{ { sin45 }^{ o } } =\sqrt { 2 } \) 
According to Brewster's law 
\({ i }_{ p }=\mu =\sqrt { 2 } \) 
\(\Rightarrow { i }_{ p }={ tan }^{ -1 }\sqrt { 2 } \) 
= tan^-1(1.414) ≅ 510 40^o
When light is incident at an angle i_p the corresponding angle of refraction 'r' is given by
i_p + r = 90^o
ஃ r = 90^o- (51^o40^') = (38^o 20')

(a) λ_e =\(\frac { h }{ { p }_{ e } } \) and λ_p=\(\frac { h }{ { p }_{ p } } \), λ_e = λ_p =1.00 nm.
So, \(\frac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\frac { { p }_{ p } }{ { p }_{ e } } =\frac { 1 }{ 1 } \Rightarrow \frac { { p }_{ p } }{ { p }_{ e } } =\frac { 1 }{ 1 } \)  = 1:1
(b) From relation K=\(\frac { 1 }{ 2 } mv^{ 2 }=\frac { { p }^{ 2 } }{ 2m } \)
K_e = \(\\ \frac { { p }_{ e }^{ 2 } }{ 2me } \) and K_p = \(\frac { { p }_{ e }^{ 2 } }{ 2m_{ p } } \)
\(\frac { { K }_{ p } }{ { K }_{ e } } =\frac { { p }_{ p }^{ 2 } }{ 2{ m }_{ p } } \times \frac { 2{ m }_{ e } }{ { p }_{ e }^{ 2 } } =\frac { { m }_{ e } }{ { m }_{ p } } \)
Since m_e <<< m_p, So K_p <<< K_e
\(\frac { { K }_{ p } }{ { K }_{ e } } =\frac { 9.1\times { 10 }^{ -31 } }{ 1.67\times 10^{ -27 } } \)
= 5.4 x 10^-4

\(\frac { 1 }{ \sqrt { V } } \)
\(\lambda =\frac { h }{ \sqrt { 2mqV } } =\frac { h }{ \sqrt { 2mq } } .\frac { 1 }{ \sqrt { V } } \)
This equation represents a straight line of slope \(\frac { h }{ \sqrt { 2mq } } \)
Clearly, the slope is inversely proportional to \(\sqrt{m}\). Since the slope of line A is less than the slope of line B therefore the line. A represents the particle of heavier mass.

Power of radiation, P =\(\frac { nhv }{ l } \) = Nhv, where N is a number of photons per sec.
or N =\(\frac { P }{ m } \)
= \(\frac { 2.0\times { 10 }^{ -3 } }{ 6.63\times 10^{ -34 }\times 6\times { 10 }^{ 14 } } \)
= 5 x 10¹⁵ photons per second.

Given λ = 589 nm = 5.89 x 10^-7 m
The de Broglie wavelength \(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2m{ E }_{ k } } } \Rightarrow { \lambda }^{ 2 }=\frac { { h }^{ 2 } }{ 2m{ E }_{ k } } \)
Kinetic energy \({ E }_{ k }=\frac { { h }^{ 2 } }{ 2m{ E }_{ k } } \)
(a) For electron \({ E }_{ k }=\frac { { h }^{ 2 } }{ 2m{ E }_{ k } } \)
\({ E }_{ k }=\frac { { (6.63\times 10 }^{ -34 })^{ 2 } }{ 2\times 9.1\times { 10 }^{ -31 }\times { ({ 5.89\times 10 }^{ -7 }) }^{ 2 } } \)
= 6.96 x 10^-25J
b) For neutron m = 1.67 x 10^-27 kg
\({ E }_{ k }=\frac { { (6.63\times 10 }^{ -34 })^{ 2 } }{ 2\times 1.67\times { 10 }^{ -31 }\times { ({ 5.89\times 10 }^{ -7 }) }^{ 2 } } \)
= 3.79 x1 0^-28J

i) X-rays are produced in x-ray tube which is essentially a discharge tube.
ii) A tungsten filament F is heated to incandescence by a battery. As a result, electrons are emitted from it by thermionic emission.

iii) The electrons are accelerated to high speeds by the voltage applied between the filament F and the anode.
iv) The target materials like tungsten, molybdenum are embedded in the face of the solid copper anode.
v) The face of the target is inclined at an angle with respect to the electron beam so that x-rays can leave the tube through its side.
vi) When high-speed electrons strike the target, they are decelerated suddenly and lose their kinetic energy.
vii) As a result, x-ray photons are produced. Since most of the kinetic energy of the bombarding electrons gets converted into heat, targets' made of high-melting-point metals and a cooling system are usually employed.

(i) The satellite communication is a mode of transmission of signal between transmitter and receiver via satellite.
(ii) The message signal from the Earth station is transmitted to the satellite on board via an uplink (frequency band 6 GHz), amplified by a transponder and then retransmitted to another earth station via a downlink (frequency band 4 GHz).
Applications:
Satellites are classified into different types based on their applications.
(i) Weather Satellites:
They are used to monitor the weather and climate of Earth. By measuring cloud mass, these satellites enable us to predict rain and dangerous storms like hurricanes, cyclones etc.
(ii) Communication satellites:
They are used to transmit television, radio, internet signals etc. Multiple satellites are used for long distance communication
(iii) Navigation satellites:
These are employed to determine the geographic location of ships, aircraft or any other object.

(i) An oscillator basically converts de energy into ac energy of high frequency ranging from a few Hz to several MHz.
(ii) There are two types of oscillators
a) Sinusoidal
b) Non-sinusoidal.
(iii) Sinusoidal oscillators generate oscillations in the form of sine wave at constant amplitude and frequency.
(iv) Non-sinusoidal oscillators generate complex non-sinusoidal waveforms like square wave, Triangular - wave or Sawtooth -wave

(v) Sinusoidal oscillations can be of two types:
a) Damped
b) undamped.
(vi) If the amplitude of the electrical oscillations decreases with time due to energy loss, is called damped oscillations
(vii) The amplitude of the electrical oscillations remains constant with time in undamped oscillations.

Decay constant \(\lambda =\frac { 0.6931 }{ T } =\frac { 0.6931 }{ 24.1 } \) day^-1
Also, \(\frac { N }{ { N }_{ o } } =\frac { 10 }{ 100 } =\frac { 1 }{ 10 } \)
Now, \(\frac { N }{ { N }_{ o } } ={ e }^{ -\lambda t };\frac { 1 }{ 10 } ={ e }^{ -\lambda t };{ e }^{ -\lambda t }=10\)
Taking logs, we get λt = log, 10 = 2.3
t = \(t=\frac { 23 }{ \lambda } =\frac { 2.3\times 24.1 }{ 06931 } \) days = 19.97 days

(i) The smoke detector uses around 0.2 mg of a man-made weak radioactive isotope called americium \((_{ 95 }^{ 241 }{ Am })\)

(ii) This radioactive source is placed between two oppositely charged metal plates and α radiations from \(_{ 95 }^{ 241 }{ Am }\) continuously ionize the nitrogen, oxygen molecules in the air space between the plates
(iii) As a result, there will be a continuous flow of small steady currents in the circuit.
(iv) If smoke enters, the radiation is being absorbed by the smoke particles rather than air molecules.
(v) As a result, the ionization and along with it the current is reduced. This drop-in current is detected by the circuit and the alarm starts.
(vi) The radiation dosage emitted by americium is very much less than the safe level, so it can be considered harmless.

Given: The amplitude of magnetic field of an Electromagnetic wave B = 1.6 x 10^-7 T
To find:
The amplitude of electric field of an Electromagnetic wave E = ?
frequency ૪ = 30 Mhz = 30 x 10⁶ Hz.
To find: Angle velocity ω =?
Wavelength of Electromagnetic wave λ = ?
(i) Ampere of electric field E = ?
\(\frac { E }{ B } =C\Rightarrow E=C.B\Rightarrow 3\times { 10 }^{ 8 }\times 1.6\times { 10 }^{ -7 }\)
E = 48Vm^-1.
(ii) Angle velocity, ω = 2π૪
ω = 2 x 3.14 x 30 x 10⁶
ω = 1.885 x 10⁸ rad /s.
(iii) Wavelength of Electromagnetic wave, λ = \(\frac{C}{\gamma}\)
\(\gamma=\frac{3\times 10^8}{30\times 10^6}\) = 10m
λ = 10m

I₁ = 4A, I₂= 6A, r = 3cm = 0.03A
\(\frac { F }{ l } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2{ I }_{ 1 }{ I }_{ 2 } }{ 0.03 } \)
\(\frac { F }{ 1 } =\frac { 10^{ -7 }\times 2\times 4\times 6 }{ 0.03 } \)
F = 1.6 x 10^-4 N/m.

Given: I = 1m, r = \(\frac22\) 1 cm = 10-2 m
Nl = 2000, N² = 1000, A = ㅠr²
A = n(10^-2)²m²
To find:
M = ?
Mutual inductance between the coils is given by
M = \(\frac { { \mu }_{ o }{ N }_{ 1 }{ N }_{ 2 }{ A }_{ 2 } }{ l } \)
= \(\frac { 4\pi \times { 10 }^{ -7 }\times 2000\times 1000\times { 10 }^{ -4 } }{ 1 } \)
M =78.89 x 10^-5 H.

(i) Consider a charged particle of charge 'q' having mass m enters into a region of a uniform magnetic field \(\vec { B } \) with velocity \(\vec { v } \).
(ii) Such that velocity is perpendicular to the magnetic field and velocity \(\vec{v}\).
(iii) The charged particle moves in a circular orbit.
(iv) Lorentz force.

\(\vec { F } =q(\vec { v } \times \vec { B } )\)
In magnitude F = qVB
(v) This Lorentz force acts as centripetal force for the particle to execute circular motion. Therefore,
qvB = m\(\frac { { v }^{ 2 } }{ r } \)
The radius of the circular path is
r = \(\frac { mv }{ qB } =\frac { p }{ qB } \)               ..........(1)
(vi) where p = mv is the magnitude of the linear momentum of the particle. Let T be the time taken by the particle to finish one complete circular motion, then
T = \(\frac { 2\pi r }{ v } \)                  .............(2)
Hence substituting (1) in (2), we get,
T = \(\frac { 2\pi m }{ qB } \)        .............(3)
(vii) Equation (3) is called the cyclotron period. The reciprocal of time period is the frequency f, which is
f = \(\frac { 1 }{ T } \)
f = \(\frac { qB }{ 2\pi m } \)                  ...........(4)
In terms of angular frequency ω,
ω = 2πf = \(\frac { q }{ m } \)B          ...........(5)
(viii) Equations (4) and equation (5) are called cyclotron frequency or gyrofrequency.
(ix) Time period and frequency depend only on charge-to-mass ratio (specific charge) and independent of velocity or radius.

Given: Radius of the to roid r = 15cm
= 15 x 10^-2m

Initral magnetic flux Φ_i = NBA
= 500 x 3 x 10^-5 x π (0.1)²
= 15π x 10^-5 wb.
an turning by 180⁰
Final flux Φ_f= -NBA = -15π x 10^-15wb.
Formula:
Magnitude of induced emf e = - \(\frac{dΦ}{dt}\)
Solution:
e = \(\frac { 2\times 15\pi \times { 10 }^{ -5 } }{ 0.25 } \) = 120π x 10^-5V
= 376.8 x 10^-5= 0.038V
Induced current = \(\frac{ε}{e}\) = \(\frac{0.038}{200}\)
= 19 x 10^-5A

Diameter d 0.24 cm = 0.24 x 10^-2 m
radius \(r=\cfrac { d }{ 2 } =0.12\times { 1 }^{ -2 }m\)
Current, I = 10A
Current density \(J=\cfrac { 1 }{ A } =\cfrac { 1 }{ { \pi r }^{ 2 } } \)
= \(\cfrac { 10 }{ 3.14\times \left( 0.12\times { 10 }^{ -2 } \right) ^{ 2 } } \)
= 2.2 x 106 Am^-2

At the point x = 0, the electric field due to all the charges are in the same negative X-direction and hence get added up, i.e.,
\(E=\frac { q }{ 4\pi { \varepsilon }_{ 0 } } \left[ \frac { q }{ { (1) }^{ 2 } } +\frac { q }{ { (2) }^{ 2 } } +\frac { q }{ { (4) }^{ 2 } } +\frac { q }{ { (8) }^{ 2 } } +.... \right] \)
\(=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left[ \frac { 1 }{ 1 } +\frac { 1 }{ 4 } +\frac { 1 }{ 16 } +\frac { 1 }{ 64 } +..... \right] \)
\(=\frac { q }{ 4\pi { \varepsilon }_{ 0 } } \left[ \frac { 1 }{ 1-\frac { 1 }{ 4 } } \right] =\frac { q }{ 3\pi { \varepsilon }_{ 0 } } \) (along negative X-axis)

By the concept of Drift velocity \(I=nAeu_{ d }\)
\({ \mu }_{ d }=\cfrac { eE }{ m } \tau \)
\(\rho =\cfrac { m }{ { ne }^{ 2 }\tau } \)
\(\therefore\) Current density
\(J=\cfrac { I }{ A } ={ \cfrac { nA{ ev }_{ d } }{ A } =ne.\cfrac { em\tau }{ m } }=\left( \cfrac { { ne }^{ 2 }\tau }{ A } \right) E\)
\(J=\cfrac { 1 }{ \rho } .E\)

(i) The resistivity of a material is dependent on temperature. The resistivity of a conductor increases with increase in temperature according to the expression 
\({ \rho }_{ r }={ \rho }_{ 0 }[I+\alpha (T-{ T }_{ 0 })\)
(ii) Where PT is the resistivity of a conductor at T^oC, is the resistivity of the conductor at some reference temperature T_o (usually at 20°C), and a is the temperature coefficient of resistivity.
(iii) It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T_o
From equation (1), we can write
\({ \rho }_{ r }-{ \rho }_{ 0 }=\alpha { \rho }_{ 0 }(T-{ T }_{ 0 })\)
\(\therefore \alpha =\cfrac { { \rho }_{ r }-{ \rho }_{ 0 } }{ { \rho }_{ 0 }(T-{ T }_{ 0 }) } =\cfrac { \Delta \rho }{ { \rho }_{ 0 }\Delta T } \)
where \(\Delta \rho ={ \rho }_{ r }-{ \rho }_{ 0 }\)  is change in resistivity for a change in temperature \(\Delta T=T-{ T }_{ 0 }\) Its unit is per ^oC \(\alpha \) of conductor:
(iv) For conductors a is positive. If the temperature of a conductor increases, the average kinetic energy of electrons in the conductor increases. This results in more frequent collisions and hence the resistivity increases.
(v) The graph of the Even though, the resistivity of conductors like metals varies linearly for wide range of temperatures, there also exists a nonlinear region at very low temperatures.
(vi) The resistivity approaches some finite values the temperature approaches absolute zero
(vii) As the resistance is directly proportional to the resistivity of the material, we can also write the resistance of a conductor at temperature T °C as
\({ R }_{ T }={ R }_{ 0 }\left[ 1+\left( T-{ T }_{ 0 } \right) \right] \)
\(\alpha =\cfrac { { R }_{ T }-{ R }_{ 0 } }{ { R }_{ 0 }\left( T-{ T }_{ 0 } \right) } =\cfrac { I }{ { R }_{ 0 } } \cfrac { \Delta R }{ \Delta T } \)
\(\alpha =\cfrac { I }{ { R }_{ 0 } } \cfrac { \Delta R }{ \Delta T } \)
where \(\Delta R={ R }_{ r }-{ R }_{ 0 }\)  is the change in resistance during the change in temperature \(\Delta T=T-{ T }_{ 0 }\)
(viii) An of semiconductors For semiconductors, the resistivity decreases with increase in temperature. As the temperature increases, more electrons will be liberated from their atoms. Hence the current increases and therefore the resistivity decreases. A semiconductor with a negative temperature coefficient of resistance is called a thermistor.

\({ V }_{ S }-{ V }_{ R }=\frac { W }{ q } \)
Work W = 50μJ = 50 x 10^-6 J
Charge q = 2μC = 2 x 10^-6 C
V = V_S - V_R \(=\frac { W }{ q } =\frac { 50\times { 10 }^{ -6 } }{ 2\times { 10 }^{ -6 } } =25V\)
V = 25V