From the mirror equation, Eq., we get \(v=\frac{f u}{u-f}\)
For convex mirror, since \(R=2 \mathrm{~m}, f=1 \mathrm{~m}\). Then for \(u=-39 \mathrm{~m}, v=\frac{(-39) \times 1}{-39-1}=\frac{39}{40} \mathrm{~m}\)
Since the jogger moves at a constant speed of \(5 \mathrm{~m} \mathrm{~s}^{-1}\), after 1 s the position of the image v (for \(u=-39+5=-34)\) is (34 / 35) m.
The shift in the position of image in 1 s is \(\frac{39}{40}-\frac{34}{35}=\frac{1365-1360}{1400}=\frac{5}{1400}=\frac{1}{280} \mathrm{~m}\)
Therefore, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is (1/280) m s^–1 Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the speed with which the image appears to move is
\(\frac{1}{150} \mathrm{~m} \mathrm{~s}^{-1}, \frac{1}{60} \mathrm{~ms}^{-1} \text { and } \frac{1}{10} \mathrm{~ms}^{-1} \text {, respectively. }\)
Although the jogger has been moving with a constant speed, the speed of his/her image appears to increase substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.

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.

A) Step 1: Find capacitance between the two plates. Formula used: \(C=\frac{\varepsilon_0 A}{d}\)
Given, Distance between the plates, d=5 cm=0.05m
Radius of each circular plate,
r = 12cm = 0.12m
Now, area of each plate, A=πr²
A = 3.14 (0.12)² = 0.045216 m²
Capacitance between two plates,
\(C=\frac{\varepsilon_0 A}{d}\)
Here, ϵ₀ = permittivity of free space =8.85 × 10⁻¹²C²/Nm²
C=8.85×10⁻¹²× 0.045216/0.05
C = 8.0032×10⁻¹²
Step 2: Find change of potential difference between the two plates. Formula used: q = CV
Given, charging current, I=0.15A
Charge on each plate, q=CV
Where, V = Potential difference across the plates
Differentiating both sides with respect to time (t), we get,
\(\begin{array}{rlr} \Rightarrow \quad \frac{d q}{d t} & =C \cdot \frac{d V}{d t} \end{array}\)
\(\begin{array}{rlr} \Rightarrow \quad I & =C \cdot \frac{d V}{d t} \end{array}\)      \(\left[\because \frac{d q}{d t}=I\right]\)      
\(\begin{array}{rlr} \Rightarrow \quad \frac{d V}{d t} & =\frac{I}{C}=\frac{0.15}{8.0032 \times 10^{-12}} \end{array}\)
Step 2 : Find change of potential difference between teh two plates.
Formula used: q = CV
dV/dt= 1.87 × 1010V/s
The rate of change of potential difference between the plates is 1.87×10¹⁰V/s
Final answer : C=8.0032×10⁻¹²,1.87×10¹⁰V/s
B) Formula used: id=ϵ0(dϕEdt)
Given, charging current or conduction current, I=0.15 A
The displacement current across the plates, id=ϵ₀(dϕE/dt)
Using Gauss's law, electric flux ϕE=qϵ₀
id=ϵ₀(1dq/ϵ₀dt)=dq/dt
id=dqdt = conduction current =0.15A
Hence, the displacement current, id is 0.15 A.
Final answer : 0.15 A.
C) Kirchhoff’s first rule (junction rule):
It states that at a junction in an electrical circuit, the sum of currents flowing into the junction is equal to the sum of currents flowing out of the junction.
Kirchhoff’s first rule is valid at each plate of the capacitor provided that we consider the current to be the sum of both conduction and displacement currents.

Here, Maximum magnetic field,
B₀ = 5 x 10^-9T;
charge on alpha particle, q = + 2e
= 2 x 1.6 x 10^-19 = 3.2 x 10^-19C, 
v = 3 x 10⁷ ms^-1 
Maximum electric field,
E₀ = cB₀ = (3 x 10⁸) x (5 x 10^-19) = 150 Vm^-1
Maximum force on alpha particle due to electric field = q x E₀ = (3.2 x 10^-19) x 150
= 4.80 x 10^-17 N
Force on alpha particle due to magnetic field = qvB₀ = (3.2 x 10^-19) x ( 3 x 10⁷) x (5 x 10^-9) 
= 4.80 x 10^-19 N

Here, v = 6 x 10¹⁴ Hz, E₀ = 2 V m^-1
(i) Average energy density of the electric field
\({ u }_{ E }=\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }=\frac { 1 }{ 4 } \times \left( 8.85\times { 10 }^{ -12 } \right) \times { 2 }^{ 2 }\)
\(=8.85\times { 10 }^{ -12 }J{ m }^{ -3 }\)
(ii) Average energy density of magnetic field
\({ u }_{ B }=\frac { { B }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { \left( { E }_{ 0 }/c \right) }^{ 2 } }{ { \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { E }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 }{ c }^{ 2 } } \)
\(=\frac { 1 }{ 4 } \times \frac { { 2 }^{ 2 } }{ \left( 4\pi \times { 10 }^{ -7 } \right) \times { \left( 3\times { 10 }^{ 8 } \right) ^{ 2 } } } \)
\(=8.85\times { 10 }^{ -12 }J{ m }^{ -3 }\)

Here, \({ B }_{ 0 }=300\mu T=3\times { 10 }^{ -4 }T\)
(i) Maximum value of electric field, \({ E }_{ 0 }=c{ B }_{ 0 }\)
\(\therefore { E }_{ 0 }=\left( 3\times { 10 }^{ 8 } \right) \times \left( 3\times { 10 }^{ -4 } \right) =9\times { 10 }^{ 4 }V{ m }^{ -1 }\)
(ii) Average energy density corresponding to electric field is
\({ u }_{ E }=\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }=\frac { 1 }{ 4 } \times \left( 8.85\times { 10 }^{ -12 } \right) \times \left( 9\times { 10 }^{ 4 } \right) ^{ 2 }\)
\(=19.91\times { 10 }^{ -4 }=1.99\times { 10 }^{ -3 }J{ m }^{ -3 }\)

Uses of electromagnetic waves
Following are some important uses of electromagnetic waves:
1. Radio waves. Radio waves are electromagnetic waves and are used in radio and television communication systems.
2. Microwaves. Microwaves are used in radar and other communication systems.
3. Infrared radiations. Infrared rays are used in:
(a) solar water heater and solar cooker.
(b) weather forecasting
(c) taking photographs during fog, smoke etc.
(d) dehydrating fruits.
(e) the treatment of muscular strain.
(f) greenhouse to keep the plants warm.
4. Ultraviolet rays.  ultraviolet rays are used:
(a) for checking mineral samples by making use of its property of causing fluorescence and also used for the study of molecular structure.
(b) for sterilizing the surgical instruments because U.V.-rays destroy bacteria.
(c) in food preservations
(d) in detection of invisible writing 
5. X-rays. X-rays are used:
(a) in surgery
(b) in radiotherapy.
(c) in medical diagnosis to detect the fracture in bones etc.
(d) in detective departments to detect gold, diamond etc concealed in bags etc without opening them.
(e) in scientific research to study the crystal structure etc.
6. \(\gamma \)-rays. Gamma rays are used to get information of structure of atomic nucleus.

(i) The given equation signifies that the electromagnetic wave is moving along Y-axis and also in negative direction, so it moves in - \(\hat{j}\) direction.
(ii) The electric part of electromagnetic wave in vacuum.
E = [3.1 cos{1.8 y + (5.4 \(\times\)10⁶ t)}] \(\hat{i}\)
Comparing with standard equation,
E = E₀ cos (ky + \(\omega\)t), we get
Angular frequency, \(\omega\) = 5.4 \(\times\)10⁶ rad/s
Wave number, k = 1.8 rad/m
The amplitude of the electric field part of the wave,
E₀ = 3.1 N/C
\(\begin{array}{rlrl} \lambda & =\frac{2 \pi}{k}=\frac{2 \pi}{1.8}=3.491 \mathrm{~m} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \lambda =3.5 \mathrm{~m} \end{array}\)
(iii) Angular frequency, \(\omega\) = 2\(\pi\)v
\(v=\frac{\omega}{2 \pi}=\frac{5.4 \times 10^6 \times 7}{2 \times 22}\)
= 0.86 \(\times\)10⁶ Hz
(iv) As, \(c=\frac{E_0}{B_0}\)
Amplitude of magnetic field,
\(\begin{aligned} B_0 & =\frac{E_0}{c}=\frac{3.1}{3 \times 10^8} \end{aligned}\)
\(\begin{aligned} =1.03 \times 10^{-8} \mathrm{~T} \end{aligned}\)
(v) Expression for the magnetic field part of wave,
B = B₀ cos (ky + \(\omega\)t) \(\hat{k}\)
B = 1.03 \(\times\)10^-8 cos (1.8 y + 5.4 \(\times\)10⁶ t) \(\hat{k}\)

Given, amplitude of an electromagnetic wave,
E₀ = 120 N/C
Frequency of wave, v = 50 MHz = 50 \(\times\) 10⁶ Hz
(i) Speed of light in vacuum, \(c=\frac{E_0}{B_0}\)
\(\begin{aligned} B_0=\frac{E_0}{c}= & \frac{120}{3 \times 10^8}=40 \times 10^{-8} \end{aligned}\)
= 400 \(\times\) 10^-9 T = 400 nT
Angular frequency of electromagnetic wave,
\(\omega=2 \pi \nu=2 \times 3.14 \times 50 \times 10^6\)
\(\omega=3.14 \times 10^8 \mathrm{rad} / \mathrm{s}\)
Wave number of electromagnetic wave,
\(k=\frac{\omega}{c}=\frac{3.14 \times 10^8}{3 \times 10^8}=1.05 \mathrm{rad} / \mathrm{m}\)
Wavelength of electromagnetic wave,
\(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{50 \times 10^6}=6.00 \mathrm{~m}\)
(ii) Expression of electric field, E = E₀ sin (kx - \(\omega\)t)
E = 120 sin (1.05x - 3.14 \(\times\)10⁸ t)
Expression of magnetic field B,
B = B₀ sin (kx - \(\omega\)t)
E = 120 sin (kx - \(\omega\)t)
B = 4 \(\times\)10^-7 sin (1.05x - 3.14 \(\times\) 10⁸ t)