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BTECH

(SEM I) THEORY EXAMINATION 2021-22

PHYSICS

**SECTION A**

Aktu Physics Previous Question paper solve here..

**Attempt all questions in brief. 2 x 10 = 20**

**a. What is inertial and non-inertial frame of references? **

In physics, a reference frame is a coordinate system used to describe the position and motion of objects in space. An inertial reference frame is one in which an object at rest remains at rest and an object in motion moves in a straight line with a constant speed unless acted upon by an external force. In other words, an inertial reference frame is a frame of reference that is not accelerating.

On the other hand, a non-inertial reference frame is a frame of reference that is accelerating or rotating. In a non-inertial reference frame, objects appear to experience fictitious forces, which are not real forces but are a consequence of the motion of the reference frame itself. For example, when you are in a car that is turning, you feel a force pushing you to the side, even though there is no real force acting on you.

**b. Show that the massless particle can exist only if they move with the speed oflight and their energy E and momentum p must have the relation E= pc.**

According to the theory of special relativity, the energy (E) and momentum (p) of a particle are related by the equation:

E^2 = (pc)^2 + (mc^2)^2

where c is the speed of light and m is the mass of the particle. This equation can be rearranged as:

E = pc * sqrt(1 + (m*c^2)/(pc)^2)

If the particle is massless, m = 0, and this reduces to:

E = pc * sqrt(1 + 0/(pc)^2) E = pc * sqrt(1) E = pc

Therefore, for a massless particle, its energy E and momentum p must have the relation E= pc.

Furthermore, if the particle is massless, then its rest mass is zero, which means that it always travels at the speed of light, c. This can be seen from the equation:

m = m0 / sqrt(1 – v^2/c^2)

where m0 is the rest mass of the particle, v is its velocity, and c is the speed of light. If the particle is massless, then m0 = 0, and the equation becomes:

m = 0 / sqrt(1 – v^2/c^2) m = 0.

This means that the particle has no rest mass and always travels at the speed of light, c. Therefore, we can conclude that massless particles can only exist if they move with the speed of light and their energy E and momentum p must have the relation E= pc.

**c. Write Maxwell’s equations in non-conducting medium.**

Maxwell’s equations describe the behavior of electric and magnetic fields in a vacuum or non-conducting medium. In their differential form, they can be written as:

- Gauss’s law for electric fields: ∇ · E = ρ/ε0
- where ∇ is the divergence operator, E is the electric field, ρ is the electric charge density, and ε0 is the electric constant (also known as the permittivity of free space).
- Gauss’s law for magnetic fields: ∇ · B = 0
- where B is the magnetic field.
- Faraday’s law of electromagnetic induction: ∇ × E = – ∂B/∂t
- where × represents the curl operator, t is time, and ∂/∂t denotes the partial derivative with respect to time.
- Ampere’s law (with Maxwell’s correction): ∇ × B = μ0J + μ0ε0 ∂E/∂t
- where μ0 is the magnetic constant (also known as the permeability of free space), J is the current density, and ∂E/∂t is the time derivative of the electric field.

**d. Define skin depth.**

Skin depth is a measure of how far an electromagnetic wave can penetrate into a material, and it is defined as the distance over which the amplitude of the wave is reduced to 1/e (about 37%) of its original value. Skin depth is a function of the frequency of the wave and the electrical properties of the material it is traveling through, such as its conductivity and permeability.

**e. Distinguish electromagnetic waves and matter waves? **

Electromagnetic waves and matter waves are two types of waves that have different properties and arise from different physical phenomena.

Electromagnetic waves are waves of oscillating electric and magnetic fields that propagate through space at the speed of light. They are produced by the acceleration of charged particles, such as in the oscillation of an antenna or in the movement of electrons in atoms. Electromagnetic waves have a wide range of frequencies, from radio waves with low frequencies to gamma rays with high frequencies. Electromagnetic waves do not require a medium to propagate through and can travel through a vacuum.

**f. What is de-Broglie hypothesis? **

According to the de Broglie hypothesis, every particle has a wavelength associated with it, known as the de Broglie wavelength, given by the equation λ = h/p, where λ is the wavelength, h is the Planck constant, and p is the momentum of the particle. This equation suggests that the wavelength of a particle is inversely proportional to its momentum, meaning that particles with higher momentum have shorter wavelengths.

**g. What are coherent sources? **

Coherent sources are two or more sources of waves that have a constant phase relationship between them. In other words, the waves from the different sources have the same frequency, wavelength, and amplitude, and their peaks and troughs are aligned in time and space.

Coherent sources are important in many areas of physics, including optics, acoustics, and quantum mechanics. In optics, for example, coherent sources are used to generate interference patterns, such as in the famous double-slit experiment, where two coherent sources of light are used to create an interference pattern on a screen. Coherent sources are also used in lasers, which produce coherent light that can be used for a wide range of applications, such as cutting, welding, and communication.

**SECTION B**

**a. What is length contraction? Derive the necessary expression for it. Show that x2+y2+z2-c2t2 is invariant. under Lorentz transformation**.

ength contraction is a phenomenon in special relativity where the length of an object appears to be shorter when it is moving at a relativistic speed relative to an observer. The closer the object’s speed is to the speed of light, the more pronounced the effect of length contraction becomes.

Let’s consider an object of length L in its rest frame. When it moves with a speed v relative to an observer, its length appears shorter along the direction of motion by a factor of γ, which is given by:

γ = 1/√(1 – v^2/c^2)

where c is the speed of light. Therefore, the length of the moving object, L’, as measured by the observer is given by:

L’ = L/γ

Now, let’s consider the quantity x^2 + y^2 + z^2 – c^2t^2. This quantity is known as the spacetime interval and is an invariant under Lorentz transformation. To see why, let’s consider two events, A and B, in spacetime. The coordinates of event A are (x1, y1, z1, t1) and the coordinates of event B are (x2, y2, z2, t2). The spacetime interval between these two events is given by:

Δs^2 = (x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2 – c^2(t2 – t1)^2

Now, let’s consider a Lorentz transformation that transforms the coordinates of event A to (x1′, y1′, z1′, t1′) and the coordinates of event B to (x2′, y2′, z2′, t2′). The transformed spacetime interval is given by:

Δs’^2 = (x2′ – x1′)^2 + (y2′ – y1′)^2 + (z2′ – z1′)^2 – c^2(t2′ – t1′)^2

We know that the spacetime interval is invariant under Lorentz transformation, which means that Δs^2 = Δs’^2. Substituting the Lorentz transformations for the coordinates, we get:

(x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2 – c^2(t2 – t1)^2 = (x2′ – x1′)^2 + (y2′ – y1′)^2 + (z2′ – z1′)^2 – c^2(t2′ – t1′)^2

Simplifying this equation using the Lorentz transformation equations, we get:

x2^2 + y2^2 + z2^2 – c^2t2^2 – (x1^2 + y1^2 + z1^2 – c^2t1^2) = x2’^2 + y2’^2 + z2’^2 – c^2t2’^2 – (x1’^2 + y1’^2 + z1’^2 – c^2t1’^2)

Since the left-hand side of the equation is equal to the right-hand side, and each term on the right-hand side is equal to its corresponding term on the left-hand side, we can conclude that x^2 + y^2 + z^2 – c^2t^2 is an invariant under Lorentz transformation.

**b. Show that the radiation pressure exerted by an electromagnetic wave is equal to the energy density. For a medium, conductivity = 58 106 seimen/m, r =1.Find out the conduction and displacement current densities if the magnitude ofelectric field intensity is given by E = 150 sin (1010 t) Volt/m.**

The radiation pressure exerted by an electromagnetic wave is given by:

P = (1/2)ε0cE^2

where ε0 is the permittivity of free space, c is the speed of light, and E is the magnitude of the electric field of the wave.

The energy density of the electromagnetic wave is given by:

u = (1/2)ε0E^2

To show that the radiation pressure is equal to the energy density, we need to express the speed of light in terms of ε0 and µ0 (the permeability of free space). Using the relation c = 1/√(ε0µ0), we can rewrite the radiation pressure as:

P = (1/2)ε0cE^2 = (1/2)E^2/µ0

The energy density can also be expressed in terms of µ0 as:

u = (1/2)ε0E^2 = (1/2)E^2/µ0

Comparing these two expressions, we see that the radiation pressure is equal to the energy density.

Now, let’s find the conduction and displacement current densities for an electric field given by E = 150 sin(1010t) V/m in a medium with conductivity σ = 58 × 10^6 S/m and relative permittivity εr = 1.

The conduction current density is given by:

J = σE

Substituting the given values, we get:

J = (58 × 10^6 S/m) × 150 sin(1010t) V/m = 8.7 × 10^9 sin(1010t) A/m^2

The displacement current density is given by:

Jd = ε0(∂E/∂t)

Taking the derivative of E = 150 sin(1010t) V/m with respect to time, we get:

∂E/∂t = 150 × 10^9 cos(1010t) V/s

Substituting the given values, we get:

Jd = (8.85 × 10^-12 F/m)(150 × 10^9 cos(1010t) V/s) = 1.33 cos(1010t) A/m^2

Therefore, the conduction current density is 8.7 × 10^9 sin(1010t) A/m^2 and the displacement current density is 1.33 cos(1010t) A/m^2.

**c. Define wave function with its physical significance. Derive Schrodinger’s time independent wave equation.**

a) Wave function and its physical significance:

In quantum mechanics, the wave function, denoted by the symbol Ψ, is a mathematical function that describes the state of a particle. The square of the absolute value of the wave function, |Ψ|^2, gives the probability density of finding the particle at a certain position in space.

In other words, the wave function provides a complete description of the quantum state of a particle, including its position, momentum, and energy. The wave function can be used to calculate various physical properties of the particle, such as its average position, energy, and momentum.

b) Derivation of Schrodinger’s time-independent wave equation:

Schrodinger’s time-independent wave equation describes the behavior of a non-relativistic quantum system in terms of its wave function. The equation is given by:

ĤΨ = EΨ

where Ĥ is the Hamiltonian operator, Ψ is the wave function, and E is the total energy of the system.

The Hamiltonian operator is defined as:

Ĥ = – (h^2/2m) ∇^2 + V

where h is the Planck constant, m is the mass of the particle, ∇^2 is the Laplacian operator, and V is the potential energy of the system.

To derive Schrodinger’s time-independent wave equation, we start with the time-dependent Schrodinger equation:

ĤΨ = iħ (∂Ψ/∂t)

where ħ is the reduced Planck constant.

Assuming that the wave function can be written as a product of a time-dependent factor and a time-independent factor, Ψ(x, t) = ψ(x)φ(t), we can separate the time-dependent and time-independent parts of the equation:

Ĥψ(x)φ(t) = iħ (1/φ(t)) (∂φ(t)/∂t)ψ(x)

Dividing both sides by Ψ(x, t) = ψ(x)φ(t), we get:

(1/φ)Ĥψ = (iħ/φ) (∂φ/∂t)

The left-hand side of the equation depends only on x, while the right-hand side depends only on t. Therefore, both sides must be equal to a constant, which we denote by E:

(1/φ)Ĥψ = Eψ

and

(iħ/φ) (∂φ/∂t) = E

Solving the second equation for φ(t), we get:

φ(t) = exp(-iEt/ħ)

Substituting this into the first equation and rearranging terms, we obtain Schrodinger’s time-independent wave equation:

Ĥψ = Eψ

This equation relates the Hamiltonian operator and the wave function to the total energy of the system. By solving this equation, we can determine the wave function and calculate various physical properties of the particle, such as its energy and probability density.

**d. Prove that reflection and transmission are complimentary in thin filminterference.**

In thin film interference, a beam of light is incident on a thin film, and part of the beam is reflected and part is transmitted through the film. The reflected and transmitted beams interfere with each other, leading to interference fringes.

The reflection and transmission coefficients for the thin film can be calculated using the Fresnel equations. Let r be the amplitude reflection coefficient and t be the amplitude transmission coefficient. Then, we have:

r = (n1 cosθ1 – n2 cosθ2) / (n1 cosθ1 + n2 cosθ2)

t = 2n1 cosθ1 / (n1 cosθ1 + n2 cosθ2)

where n1 and n2 are the refractive indices of the two media, θ1 is the angle of incidence, and θ2 is the angle of refraction.

The total amplitude of the reflected beam is given by R = r^2, and the total amplitude of the transmitted beam is given by T = t^2. The intensity of the reflected and transmitted beams is proportional to the square of their amplitude.

Now, let’s consider the interference between the reflected and transmitted beams. The phase difference between the two beams is given by:

δ = (2π/λ) * 2t * d * cosθ1

where λ is the wavelength of the incident light, d is the thickness of the thin film, and θ1 is the angle of incidence.

At certain angles of incidence, the phase difference δ is an integer multiple of 2π, leading to constructive interference and bright fringes. At other angles of incidence, the phase difference is an odd multiple of π, leading to destructive interference and dark fringes.

If we sum up the intensities of all the reflected and transmitted beams for all angles of incidence, we get:

R + T = 1

This is known as the law of energy conservation, which states that the total energy of the incident beam is either reflected or transmitted.

Now, if we consider the interference fringes, we can show that the total intensity of the reflected and transmitted beams at any point along the fringes is also constant:

R + T = constant

This means that if the intensity of the reflected beam increases at a certain point, the intensity of the transmitted beam must decrease by the same amount, and vice versa. Therefore, reflection and transmission are complementary in thin film interference.

**e. Develop the expressions for acceptance angle and numerical aperture of anoptical fiber. A step index fiber has core refractive index 1.466, claddingrefractive index 1.46. If the operating wavelength of the rays is 0.85 µm,calculate the cut – off parameter and the number of modes, which the fibre willsupport. The diameter of the core = 50µm**.

An optical fiber is a thin, flexible strand of glass or plastic that can transmit light signals over long distances with minimal loss of signal quality. The acceptance angle and numerical aperture are two important parameters that determine the performance of an optical fiber.

The acceptance angle of an optical fiber is the maximum angle of incidence at which light can enter the fiber and still be transmitted down its length. It depends on the refractive indices of the core and cladding materials, and is given by:

sin θmax = NA = √(n1^2 – n2^2)

where NA is the numerical aperture, n1 is the refractive index of the core, and n2 is the refractive index of the cladding.

The numerical aperture is a measure of the light-gathering ability of the fiber and is given by the same equation as the acceptance angle. A larger numerical aperture indicates that the fiber can gather more light and transmit it over a longer distance.

For a step-index fiber with a core refractive index n1 and cladding refractive index n2, the cut-off parameter is given by:

V = (2π/λ) * a * √(n1^2 – n2^2)

where λ is the wavelength of the light, a is the radius of the core, and V is the V-number. The V-number is a dimensionless parameter that determines the number of modes that the fiber can support.

The number of modes that the fiber can support is given by:

M = (V^2 / 2)

where M is the number of modes.

Now, let’s apply these equations to the given example. We have a step-index fiber with a core refractive index of 1.466 and a cladding refractive index of 1.46. The operating wavelength of the rays is 0.85 µm, and the diameter of the core is 50 µm.

The cut-off parameter is:

V = (2π/0.85 µm) * 25 µm * √(1.466^2 – 1.46^2) = 14.26

The number of modes is:

M = (14.26^2 / 2) = 101

Therefore, the fiber will support 101 modes.

To calculate the acceptance angle and numerical aperture, we first need to calculate the critical angle of the fiber. The critical angle is the angle of incidence at which light is totally internally reflected at the core-cladding interface. It is given by:

sin θc = n2 / n1 = 1.46 / 1.466 = 0.9976

The acceptance angle and numerical aperture are then:

sin θmax = NA = √(1.466^2 – 1.46^2) = 0.239

θmax = 13.8°

Therefore, the acceptance angle is 13.8° and the numerical aperture is 0.239.

**SECTION C**

**Attempt any one part of the following:**

**a. By using Lorentz transformation equations, derive time dilation. Show thattime dilation is a real effect.**

Time dilation is a fundamental concept in special relativity that refers to the phenomenon of time appearing to pass more slowly for an observer who is in motion relative to another observer who is at rest. Time dilation arises due to the fact that the speed of light is always constant in all inertial reference frames, regardless of the relative motion between the observer and the light source.

To derive time dilation, we start with the Lorentz transformation equations, which describe how the coordinates of events are transformed from one inertial reference frame to another. The Lorentz transformation equations are given by:

x’ = γ(x – vt) y’ = y z’ = z t’ = γ(t – vx/c^2)

where x, y, z, t are the coordinates of an event in the rest frame, x’, y’, z’, t’ are the coordinates of the same event in a moving frame, v is the velocity of the moving frame relative to the rest frame, c is the speed of light, and γ is the Lorentz factor given by:

γ = 1 / √(1 – v^2/c^2)

Let’s consider two observers, Alice and Bob, with Alice at rest in the rest frame and Bob moving with velocity v relative to Alice. Alice observes a clock that is stationary in her frame, while Bob observes the same clock as moving with velocity v in his frame. Let t0 be the time interval between two ticks of the clock as observed by Alice in her frame, and t be the time interval between two ticks of the same clock as observed by Bob in his frame. We want to derive an expression for t in terms of t0.

From the Lorentz transformation equations, we can see that the time interval t’ between two events as observed in the moving frame is given by:

t’ = γ(t – vx/c^2)

For the clock, the events are two successive ticks, and we can assume that they occur at the same position in both frames (x = x’). Therefore, we can set x = x’ and solve for t:

t = (t’ + vx’/c^2) / γ

The position x’ of the clock in the moving frame can be obtained from the first equation of the Lorentz transformation:

x’ = γ(x – vt)

Since the clock is at rest in Alice’s frame, we have x = 0. Substituting x’ and simplifying, we get:

t = γt’ / (1 – v^2/c^2)

This is the time dilation formula, which shows that the time interval t as observed in the moving frame is longer than the time interval t0 as observed in the rest frame by a factor of γ. This means that moving clocks appear to run more slowly than stationary clocks.

To show that time dilation is a real effect, we can consider the example of muons, which are subatomic particles that are created in the upper atmosphere by cosmic rays. Muons have a very short half-life of only 2.2 microseconds, which means that they should decay before reaching the surface of the Earth. However, due to time dilation, muons that are created at high altitudes and travel at close to the speed of light are observed to reach the Earth’s surface without decaying.

**b. Derive Einstein’s mass-energy relation Calculate the amount of work to bedone to increase the speed of an electron from 0.6C to 0.8C. Given that the restmass energy of electron= 0.5 MeV.**

Einstein’s mass-energy relation, also known as the famous equation E = mc^2, is a fundamental result of special relativity that relates an object’s mass to its energy. The equation states that the energy E of an object with mass m is equal to its mass multiplied by the square of the speed of light c:

E = mc^2

To derive this equation, we start with the relativistic energy-momentum relation:

E^2 = (pc)^2 + (mc^2)^2

where p is the momentum of the object. In the case of a particle with zero momentum (such as an object at rest), this equation reduces to:

E = mc^2

which is Einstein’s famous equation.

To calculate the amount of work required to increase the speed of an electron from 0.6C to 0.8C, we can use the kinetic energy formula:

K = (1/2)mv^2

where m is the mass of the electron, v is its velocity, and K is the kinetic energy. We can calculate the initial kinetic energy of the electron as:

K1 = (1/2)(m)(0.6C)^2

where we have used the fact that the speed of light c is the maximum velocity that any object can attain, so we have expressed the electron’s velocity as a fraction of the speed of light.

Similarly, we can calculate the final kinetic energy of the electron as:

K2 = (1/2)(m)(0.8C)^2

The work required to increase the speed of the electron from 0.6C to 0.8C is equal to the difference in kinetic energy between the two states:

W = K2 – K1

Substituting the rest mass energy of the electron, which is given as 0.5 MeV, and using the fact that 1 MeV = 1.6 x 10^-13 Joules, we can calculate the work required as:

W = (1/2)m[(0.8C)^2 – (0.6C)^2]c^2 = (1/2)(9.1 x 10^-31 kg)[(0.8 x 3 x 10^8 m/s)^2 – (0.6 x 3 x 10^8 m/s)^2](3 x 10^8 m/s)^2 = 3.8 x 10^-13 J

Therefore, the amount of work required to increase the speed of an electron from 0.6C to 0.8C is 3.8 x 10^-13 J.

**Attempt any one part of the following:**

**a. Derive the Poynting or work-energy theorem for the flow of energy in anelectromagnetic field. Also give the physical interpretation.**

The Poynting theorem describes the flow of energy in an electromagnetic field. It is based on the principle of conservation of energy, and it relates the rate of change of energy density in a volume of space to the energy flux density, which is the amount of energy flowing through a unit area per unit time.

To derive the Poynting theorem, we start with the equations for the electric field E and the magnetic field B:

∇ x E = -∂B/∂t ∇ x B = μ0(J + ε0∂E/∂t)

where J is the current density, ε0 is the permittivity of free space, and μ0 is the permeability of free space.

Taking the dot product of the first equation with B and the second equation with E, and using vector calculus identities, we obtain:

B · (∇ x E) = -B · (∂B/∂t) E · (∇ x B) = E · μ0(J + ε0∂E/∂t)

The left-hand sides of these equations can be rewritten using the vector identity:

A · (∇ x B) = ∇ · (A x B) – B · (∇ x A)

where A and B are vectors. Applying this identity, we obtain:

∇ · (E x B) = -B · (∂B/∂t) – ε0E · (∂E/∂t) – J · E ∇ · (E x B) = E · μ0(J + ε0∂E/∂t) – ε0B · (∂B/∂t)

Taking the difference of these two equations and rearranging, we get:

-∇ · (E x B) – ε0∂(E^2 + B^2)/∂t = J · E – B · (∂B/∂t) – E · (∂E/∂t)

The left-hand side of this equation represents the rate of change of energy density, while the right-hand side represents the work done on charges by the electromagnetic fields. Therefore, the Poynting theorem can be written as:

∂(u + 1/ε0 E^2 + 1/2μ0 B^2)/∂t + ∇ · S = -J · E

where u is the energy density of the electromagnetic field, S is the Poynting vector representing the energy flux density, and the last term on the right-hand side represents the rate of work done by the electric field on the charges.