Are you preparing for BEU Engineering Physics 2025? This post brings you a complete model paper, PYQ analysis, and the most expected questions based on the latest syllabus—all in one place.
Short Questions (2 Marks Each)
1. Define Huygens’ Principle. [Repeated in 2019, 2023]
Answer: Huygens’ Principle states that: Every point on a wavefront acts as a source of secondary wavelets. These wavelets spread out in all directions at the same speed as the original wave. The new wavefront is formed by joining the forward edges of these secondary wavelets.
2. Write the equation of damped harmonic oscillator. [Repeated in 2019, 2023]
Answer: The differential equation of a damped harmonic oscillator is:
$$ m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0 $$Where:
m = mass of the oscillatorb = damping constant (friction/resistance)
k = spring constant
x = displacement
t = time
3. What is population inversion in LASER? [2022, 2023]
Answer: Population inversion is a condition in which the number of atoms or molecules in a higher energy state becomes greater than the number in the lower energy state.
This condition is necessary to produce laser light through stimulated emission.
Energy Level Representation:
Before Inversion: N1 > N2
After Inversion: N2 > N1
(N1 = lower energy level population, N2 = upper energy level population)
4. Define Fermi level. [2019, 2022]
Answer: The Fermi level is the highest occupied energy level of electrons in a material at absolute zero temperature (0 Kelvin).This concept is important in understanding the electrical properties of conductors, semiconductors, and insulators.
5. What is Rayleigh’s criterion? [2019, 2022]
Answer: Rayleigh’s criterion defines the minimum angular separation at which two point sources can be distinctly resolved by an optical system.
According to the criterion, two sources are said to be just resolved when the principal maximum of one diffraction pattern coincides with the first minimum of the other.
Mathematical Expression:
$$
\theta = 1.22 \times \left( \frac{\lambda}{D} \right)
$$Where:
• θ = minimum angular resolution (in radians)
• λ = wavelength of light used
• D = diameter of the aperture (objective lens)re
6. Define wavefunction and its significance. [2021, 2022]
Answer: A wavefunction (denoted by $\psi$) is a mathematical function that describes the quantum state of a particle. It gives information about the position and behavior of a particle like an electron.
Significance:
- The square of the wavefunction, $|\psi|^2$, gives the probability of finding the particle at a particular point in space.
- It helps us understand where a particle is likely to be found at any given time.
- It is a key concept in quantum mechanics and used in Schrödinger’s equation.
7. What are intrinsic and extrinsic semiconductors? [Repeated in 2019, 2023]
Answer:
Intrinsic Semiconductor:
- It is a pure semiconductor without any added impurities.
- Electrical conductivity is low and depends only on temperature.
- The number of electrons = number of holes.
- Examples: Pure Silicon (Si) and Germanium (Ge).
Extrinsic Semiconductor:
- It is formed by adding impurities (called doping) to a pure semiconductor.
- Conductivity is higher due to increased charge carriers.
- There are two types:
- n-type: Doped with pentavalent elements (e.g., Phosphorus), adds extra electrons.
- p-type: Doped with trivalent elements (e.g., Boron), creates extra holes.
- Used in electronic devices like diodes and transistors.
8. Write time-dependent Schrödinger equation. [Repeated in 2021, 2022]
Answer: The time-dependent Schrödinger equation in one dimension is: $$
i\hbar \frac{\partial \psi(x, t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi(x, t)}{\partial x^2} + V(x)\psi(x, t)
$$Where:
- i = imaginary unit
- $\hbar$ = reduced Planck’s constant
- $\psi(x, t)$ = wavefunction
- m = mass of the particle
- $V(x)$ = potential energy function
- $x, t$ = position and time variables
9. What is the Quality Factor in oscillators? [New, High Chance]
Answer: The Quality Factor (Q) of an oscillator is a dimensionless number that measures the sharpness of resonance or energy loss in the system.
Formula:
$$ Q = \frac{f_0}{\Delta f} $$or$$ Q = 2\pi \times \frac{\text{Total Energy Stored}}{\text{Energy Lost per Cycle}} $$
10. Define gradient, divergence, and curl. [New, High Chance]
Answer:
1. Gradient (∇f):
The gradient of a scalar field represents the rate and direction of the fastest increase of the scalar quantity.
It gives a vector that points in the direction of the greatest change.
$$ \vec{\nabla}f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) $$
2. Divergence (∇·𝐀):
The divergence of a vector field represents the net rate of flow (outflow) from a point.
It gives a scalar value indicating how much the field is spreading out at a point.
$$ \vec{\nabla} \cdot \vec{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} $$
3. Curl (∇×𝐀):
The curl of a vector field measures its tendency to rotate or circulate around a point.
It gives a vector that represents the axis of rotation and magnitude of circulation.
$$ \vec{\nabla} \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \\ \end{vmatrix} $$
11. What is magnetic vector potential? [New, Medium Chance]
Answer: The magnetic vector potential (𝐩) is a vector field used in electromagnetism to describe the magnetic field. It is defined such that:
$$ \vec{B} = \vec{\nabla} \times \vec{A} $$
Where:
- $ \vec{A} $ = magnetic vector potential
- $ \vec{B} $ = magnetic field
- $ \vec{\nabla} \ $ = curl operator
11. State Lenz’s Law. [New, Medium Chance]
Answer: “The direction of an induced current is such that it always opposes the change in magnetic flux that caused it.”
This means the induced magnetic field created by the current will oppose the motion or change in the original magnetic field.
Mathematically:
$$ \mathcal{E} = -\frac{d\Phi_B}{dt} $$
Where:
- 𝓔 = induced electromotive force (emf)
- ΦB = magnetic flux
The negative sign indicates that the induced emf opposes the change in flux, according to the law of conservation of energy.
BEU Engineering Physics 2025 | New Syllabus Model Paper, PYQ Analysis & Most important Questions
Long Questions (7 or 14 Marks each):
Q1. Derive the expression for dark ring diameter in Newton’s Ring with diagram. [Repeated in 2019, 2021, 2022, 2023]
Answer: Newton’s Ring Experiment is based on the interference of light waves reflected from the two surfaces of a thin air film formed between a plano-convex lens and a flat glass plate.
# Why Rings Are Formed?
- When monochromatic light falls on the setup, light reflects from both the curved surface of the lens and the flat glass plate.
- These two reflected rays interfere with each other.
- At some points, destructive interference (dark rings) occurs and at other points, constructive interference (bright rings) occurs.
Click to See Diagram
BEU Engineering Physics 2025 | New Syllabus Model Paper, PYQ Analysis & Most important Questions
Derivation:
# Path Difference Condition for Dark Rings: For dark rings (destructive interference), the condition is: $$
2t = n\lambda
$$
where:
- t = thickness of air film
- λ = wavelength of light
- n = ring number
# Geometry of the Air Film: From geometry, the thickness of the air film at a distance r from the center is: $$
t = \frac{r^2}{2R}
$$
where R : the radius of curvature
# Substituting into the interference condition$$
2 \cdot \frac{r^2}{2R} = n\lambda \Rightarrow \frac{r^2}{R} = n\lambda
$$Multiplying both sides by 4: $$
4r^2 = 4n\lambda R
$$So, the diameter of the ring becomes: $$
D_n = 2r = \sqrt{4n\lambda R}
$$
diameter of the n-th dark ring in Newton’s Rings: $$
\boxed{D_n^2 = 4n\lambda R}
$$
Q2. Derive intensity expression for Fraunhofer single slit diffraction. [Repeated in 2019, 2021, 2022]
Answer:
Click to see complete solution
BEU Engineering Physics 2025 | New Syllabus Model Paper, PYQ Analysis & Most important Questions
Q3. Derive and solve Schrödinger Equation for a particle in 1D box. [2021, 2022, 2023]
Answer: We consider a particle of mass m confined inside a 1D box of length L. The potential inside the box is:
$$ V(x) = \begin{cases} 0, & \text{if } 0 < x < L \\ \infty, & \text{otherwise} \end{cases} $$
Inside the box, where ( V(x) = 0 ), Schrödinger’s equation becomes:
$$ -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} = E\psi(x) $$
Or,
$$ \frac{d^2\psi(x)}{dx^2} + \frac{2mE}{\hbar^2} \psi(x) = 0 $$
Let:
$$ k^2 = \frac{2mE}{\hbar^2} $$
Then the equation becomes:
$$ \frac{d^2\psi(x)}{dx^2} + k^2 \psi(x) = 0 \tag{1} $$
The general solution of equation (1) is:
$$ \psi(x) = A \sin(kx) + B \cos(kx) $$
Since the potential is infinite at x = 0 and x = L, the wavefunction must be zero at both ends.
At ( x = 0 ):
$$ \psi(0) = A\sin(0) + B\cos(0) = B = 0 $$
At ( x = L ):
$$ \psi(L) = A \sin(kL) = 0 \Rightarrow \sin(kL) = 0 \Rightarrow kL = n\pi, \quad n = 1, 2, 3, \dots $$
So,
$$ k = \frac{n\pi}{L} $$
Substitute back into $( \psi(x) ):$
$$ \psi_n(x) = A \sin\left(\frac{n\pi x}{L}\right) $$
Normalize it:
$$ \int_0^L |\psi_n(x)|^2 dx = 1 \Rightarrow A = \sqrt{\frac{2}{L}} $$
So, the normalized wavefunction is:
$$ \boxed{\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)} $$
Using $( k = \frac{n\pi}{L} )$,
$$ E = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2}{2m} \cdot \left( \frac{n\pi}{L} \right)^2 $$
So, the quantized energy levels are:
$$ \boxed{E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}}, \quad n = 1, 2, 3, \dots $$
Q4. State and derive Heisenberg’s Uncertainty Principle. [2019, 2023]
Answer: It is impossible to determine precisely both the position and momentum of small moving particles simultaneously.
→ The more precisely the position is known, the more uncertain the momentum, and vice versa.
→This states that there is inherent uncertainty in the act of measuring a variable of particle.
Derivation:
$
\text{Step 1: From Fourier theory, we have:}
$
$$
\Delta x \cdot \Delta k \geq \frac{1}{2}
$$
$
\text{Step 2: Use relation between momentum and wave number: } $ $$p = \hbar k \Rightarrow \Delta p = \hbar \Delta k
$$
$
\text{Step 3: Substitute }$ $$ \Delta k = \frac{\Delta p}{\hbar} \text{ into Step 1:}
$$
$$
\Delta x \cdot \frac{\Delta p}{\hbar} \geq \frac{1}{2}
$$
$$
\Rightarrow \Delta x \cdot \Delta p \geq \frac{\hbar}{2}
$$
$
\text{Step 4: Since } $ $$\hbar = \frac{h}{2\pi}, \text{ we get:}
$$
$$
\Delta x \cdot \Delta p \geq \frac{h}{4\pi}
$$
$$
\boxed{\Delta x \cdot \Delta p \geq \frac{h}{4\pi}}
$$
Where:
- $\Delta x$ = uncertainty in position
- $\Delta p$= uncertainty in momentum
- $h$ = Planck’s constant
Q5. Explain Einstein A & B coefficients. Describe Ruby LASER. [2021, 2022, 2023]
Answer:
Einstein A and B Coefficients: Einstein explained how atoms interact with light using three processes:
- Spontaneous Emission:
When an atom is in a higher energy level, it can come down to a lower level on its own and emit light.
The probability of this happening is given by the A coefficient (A₍₂₁₎). - Stimulated Emission:
If an atom is already in an excited state, and a photon hits it, the atom gets “stimulated” to emit another photon that is exactly the same (same energy, direction, and phase).
This is called stimulated emission and is explained by B coefficient (B₍₂₁₎). - Absorption:
When a photon hits an atom in a lower energy level, it can absorb the energy and jump to a higher level.
This is called absorption, explained by B₍₁₂₎.
Einstein also gave a formula that relates spontaneous and stimulated emissions: $$\frac{A_{21}}{B_{21}} = \frac{8\pi h \nu^3}{c^3}$$
Ruby LASER:
Ruby LASER is a type of solid-state LASER. It uses a ruby crystal as the active material. Ruby is made of aluminium oxide (Al₂O₃) doped with chromium ions (Cr³⁺).
- A flash lamp is used to provide energy (this is called optical pumping).
- The chromium ions go to a high energy level, then quickly fall to a middle (metastable) level.
- From here, they stay for a while and then come down to the ground state by releasing energy as laser light.
- The setup has mirrors on both ends — one fully reflective, one partially reflective — to build up the light.
The light produced is:
- Ruby LASER gives red light (wavelength = 694.3 nm).
- It works in pulsed mode.
- It is a three-level laser system.
Q6. State Bloch’s Theorem and derive Kronig-Penney Model. [2019, 2021, 2022, 2023]
Answer: Bloch’s theorem states that The wavefunction of an electron moving in a periodic potential (such as in a crystal lattice) is the product of a plane wave and a periodic function.
Mathematically, $$\psi(x) = u(x) \cdot e^{ikx}$$
Where:
- u(x) is a periodic function with the same period as the crystal, i.e., $u(x + a) = u(x)$
- k is the wave number
- a is the lattice constant (distance between atoms)
Kronig-Penney Model (Simplified Derivation)
The Kronig-Penney model is used to understand how energy bands and band gaps are formed in solids.
Assumption:
We consider a 1D periodic potential which consists of alternating regions:
- Free region where $V(x)=0$ (width = a)
- Barrier region where $V(x)=V_0$ (width = b)
Schrödinger Equation in Free Region (V = 0) $$\frac{d^2 \psi}{dx^2} + \frac{2mE}{\hbar^2} \psi = 0 \Rightarrow \psi_1(x) = A e^{i\alpha x} + B e^{-i\alpha x}$$
where $\alpha = \sqrt{\frac{2mE}{\hbar^2}}$
Schrödinger Equation in Barrier Region (V = V₀) $$\frac{d^2 \psi}{dx^2} + \frac{2m(E – V_0)}{\hbar^2} \psi = 0 \Rightarrow \psi_2(x) = C e^{\beta x} + D e^{-\beta x}$$
where $\beta = \sqrt{\frac{2m(V_0 – E)}{\hbar^2}} \quad \text{(assuming } E < V_0 \text{)}$
Apply Boundary Conditions and Bloch’s Theorem
We apply:
- Continuity of wavefunction ψ and its derivative at region boundaries
- Bloch’s condition:
$$\psi(x + a + b) = \psi(x) \cdot e^{ik(a + b)}$$
After applying these, we arrive at the Kronig-Penney equation: $$\cos(k(a + b)) = \cos(\alpha a)\cosh(\beta b) – \frac{\alpha^2 + \beta^2}{2\alpha\beta} \cdot \sin(\alpha a)\sinh(\beta b)$$
Q7. State and prove Gauss’s Divergence Theorem. [New, High Chance]
Answer: Gauss’s Divergence Theorem states:
The total flux of a vector field outward through a closed surface is equal to the volume integral of the divergence of the vector field over the region enclosed by the surface.
Mathematical Form:
$$ \iint_{S} \vec{A} \cdot \hat{n} \, dS = \iiint_{V} (\nabla \cdot \vec{A}) \, dV $$
Where:
- $ \vec{A} $ = vector field
- $\hat{n} $ = outward unit normal vector to the surface
- $S $ = closed surface
- $V$ = volume enclosed by surface \( S \)
- $ \nabla \cdot \vec{A} $ = divergence of the vector field
Proof (In Cartesian Coordinates):
Let $( \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} )$ be a vector field.
Consider a small cube inside the volume with sides $ dx, dy, dz $.
Flux in x-direction:
- From face at x : $ A_x(x) \cdot dy\,dz $
- From face at ( x + dx ): $A_x(x + dx) \cdot dy\,dz $
Net flux in x-direction:
$$ [A_x(x + dx) – A_x(x)] dy\,dz = \frac{\partial A_x}{\partial x} dx\,dy\,dz $$
Similarly:
- Net flux in y-direction: $\frac{\partial A_y}{\partial y} dx\,dy\,dz $
- Net flux in z-direction: $\frac{\partial A_z}{\partial z} dx\,dy\,dz $
Total net flux from the cube:
$$ \left( \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} \right) dx\,dy\,dz = (\nabla \cdot \vec{A}) \cdot dV $$
Add the flux from all small cubes inside the volume:
$$ \text{Total flux out of surface} = \iiint_{V} (\nabla \cdot \vec{A}) \, dV $$
And that is equal to:
$$ \iint_{S} \vec{A} \cdot \hat{n} \, dS $$
Hence Proved:
$$ \boxed{ \iint_{S} \vec{A} \cdot \hat{n} \, dS = \iiint_{V} (\nabla \cdot \vec{A}) \, dV } $$
Q8. Derive Laplace’s and Poisson’s equations. [New, High Chance]
Answer: Starting with Gauss’s Law in Differential Form: We know from electrostatics:
$$ \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0} $$
Where:
- $\vec{E} $ is the electric field
- $\rho $ is the charge density
- $\varepsilon_0 $ is the permittivity of free space
Electric field is the negative gradient of electric potential V :
$$ \vec{E} = -\nabla V $$
Now substitute this into Gauss’s law:
$$ \nabla \cdot (-\nabla V) = \frac{\rho}{\varepsilon_0} \Rightarrow -\nabla^2 V = \frac{\rho}{\varepsilon_0} $$
So,
$$ \nabla^2 V = -\frac{\rho}{\varepsilon_0} $$
This is known as the Poisson’s Equation.
Poisson’s Equation:
$$ \boxed{\nabla^2 V = -\frac{\rho}{\varepsilon_0}} $$
Poisson’s equation applies in regions with charge
When there is no charge in the region (i.e., $ \rho = 0 $):
$$ \nabla^2 V = 0 $$
This is called the Laplace’s Equation.
Laplace’s Equation:
$$ \boxed{\nabla^2 V = 0} $$
It applies in free space or regions where there are no charges.
Q9. Discuss boundary conditions at dielectric interface. [New, High Chance]
Answer: When two different dielectric materials (like plastic, glass, etc.) are in contact, the behavior of the electric field and electric flux changes at their boundary or interface.
To understand how they change, we use some rules called boundary conditions.
Let the two dielectrics have permittivities $ \varepsilon_1 $ and $ \varepsilon_2 $.
1. Tangential part of the electric field that is parallel to the surface stays the same on both sides.
$$ E_{1t} = E_{2t} $$
This means the direction of the field does not suddenly change along the surface.
2. Normal part of the electric flux (D) that is perpendicular to the surface remains the same if there is no free charge at the surface:
$$ D_{1n} = D_{2n} $$
Since $D = \varepsilon E $, this becomes:
$$ \varepsilon_1 E_{1n} = \varepsilon_2 E_{2n} $$
So, the electric field (E) changes if permittivities are different.
3. If there is any free surface charge (like a charged layer) at the boundary, then:
$$ D_{2n} – D_{1n} = \sigma_f $$
Where $\sigma_f $ is the surface charge density.
4. Direction of Field Bends Because $E $ changes due to different $\varepsilon $, the electric field lines bend at the boundary.
The bending is given by:
$$ \frac{\tan \theta_1}{\tan \theta_2} = \frac{\varepsilon_1}{\varepsilon_2} $$
This is similar to how light bends (refraction) when passing between materials.
Q10. Derive Maxwell’s equations and state physical meaning. [New, Medium Chance]
Answer: Maxwell’s equations are four fundamental laws that describe how electric and magnetic fields behave and interact with charges and currents.
1. Gauss’s Law for Electricity : This law says that electric field lines begin from positive charges and end at negative charges. The total electric flux coming out of a closed surface depends on the total charge enclosed within that surface.
$$ \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0} $$
2. Gauss’s Law for Magnetism : There are no isolated magnetic poles (no magnetic monopoles). Magnetic field lines always form closed loops, so the net magnetic flux through a closed surface is zero.
$$ \nabla \cdot \vec{B} = 0 $$
3. Faraday’s Law of Electromagnetic Induction : A changing magnetic field with time produces an electric field. This is the basic working principle behind generators and electromagnetic induction.
$$ \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} $$
4. Ampère–Maxwell Law : A current or a time-varying electric field produces a magnetic field. The second term (displacement current) is important when the electric field is changing, especially in capacitors and electromagnetic waves.
$$ \nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t} $$
Q11. Prove Poynting Theorem and explain energy flow. [New, High Chance]
Answer: Poynting’s theorem gives the law of conservation of electromagnetic energy. It explains how energy is transferred and stored in an electric and magnetic field.
The rate of energy flow per unit area in an electromagnetic field is equal to the Poynting vector, which is given by:
$$ \vec{S} = \vec{E} \times \vec{H} $$
Where:
- $ \vec{S} $ = Poynting vector (energy flow per unit area per unit time)
- $ \vec{E} $ = Electric field
- $ \vec{H} $ = Magnetic field
Proof: Start with Maxwell’s equations (differential form):
- $ \nabla \cdot \vec{D} = \rho $
- $ \nabla \cdot \vec{B} = 0 $
- $ \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} $
- $ \nabla \times \vec{H} = \vec{J} + \frac{\partial \vec{D}}{\partial t} $
Take dot product of equation (4) with $ \vec{E} $:
$$ \vec{E} \cdot (\nabla \times \vec{H}) = \vec{E} \cdot \vec{J} + \vec{E} \cdot \frac{\partial \vec{D}}{\partial t} $$
Take dot product of equation (3) with $ \vec{H} $:
$$ \vec{H} \cdot (\nabla \times \vec{E}) = -\vec{H} \cdot \frac{\partial \vec{B}}{\partial t} $$
Now subtract both equations:
$$ \vec{E} \cdot (\nabla \times \vec{H}) – \vec{H} \cdot (\nabla \times \vec{E}) = \vec{E} \cdot \vec{J} + \vec{E} \cdot \frac{\partial \vec{D}}{\partial t} + \vec{H} \cdot \frac{\partial \vec{B}}{\partial t} $$
Using vector identity:
$$ \vec{E} \cdot (\nabla \times \vec{H}) – \vec{H} \cdot (\nabla \times \vec{E}) = \nabla \cdot (\vec{E} \times \vec{H}) = \nabla \cdot \vec{S} $$
Also:
- $ \vec{D} = \varepsilon \vec{E} \Rightarrow \vec{E} \cdot \frac{\partial \vec{D}}{\partial t} = \frac{\partial}{\partial t} \left( \frac{1}{2} \varepsilon E^2 \right) $
- $ \vec{B} = \mu \vec{H} \Rightarrow \vec{H} \cdot \frac{\partial \vec{B}}{\partial t} = \frac{\partial}{\partial t} \left( \frac{1}{2} \mu H^2 \right) $
Substituting all into the equation:
$$ \nabla \cdot \vec{S} = -\vec{E} \cdot \vec{J} – \frac{\partial}{\partial t} \left( \frac{1}{2} \varepsilon E^2 + \frac{1}{2} \mu H^2 \right) $$
$$ \boxed{ \nabla \cdot \vec{S} + \frac{\partial u}{\partial t} + \vec{E} \cdot \vec{J} = 0 } $$
Where:
- $ u = \frac{1}{2} \varepsilon E^2 + \frac{1}{2} \mu H^2 $ is the electromagnetic energy density
- $ \vec{E} \cdot \vec{J} $ = energy lost as heat or mechanical work
- $ \nabla \cdot \vec{S} $ = energy flowing out of a region
Q12. Derive group velocity and phase velocity from wave packet. [New, High Chance]
Answer: When two waves of slightly different frequencies and wavelengths combine, they form a wave packet. A wave packet helps us understand how particles behave like waves in quantum mechanics.Let’s take two waves with similar wave numbers and frequencies:
First wave:
$$ \psi_1 = A \cos(k_1 x – \omega_1 t) $$
Second wave:
$$ \psi_2 = A \cos(k_2 x – \omega_2 t) $$
Total wave = sum of both:
$$ \psi = \psi_1 + \psi_2 $$
Using the identity:
$$ \cos a + \cos b = 2 \cos\left(\frac{a + b}{2}\right) \cdot \cos\left(\frac{a – b}{2}\right) $$
We get:
$$ \psi = 2A \cos\left[\frac{(k_1 + k_2)x – (\omega_1 + \omega_2)t}{2}\right] \cdot \cos\left[\frac{(k_1 – k_2)x – (\omega_1 – \omega_2)t}{2}\right] $$
$$ k = \frac{k_1 + k_2}{2}, \quad \omega = \frac{\omega_1 + \omega_2}{2} $$
$$ \Delta k = k_1 – k_2, \quad \Delta \omega = \omega_1 – \omega_2 $$
The phase velocity is the speed of the carrier wave (the fast wave):
$$ v_p = \frac{\omega}{k} $$
The group velocity is the speed at which the envelope or the wave packet moves:
$$ v_g = \frac{d\omega}{dk} $$
This is usually the speed at which energy or particles move.
Q13. Differentiate metals, insulators, and semiconductors with band theory. [New, High Chance]
Answer: This is the difference of metals, semiconductors, and insulators using band theory:
| Property | Metals | Semiconductors | Insulators |
|---|---|---|---|
| Band Structure | Valence and conduction bands overlap | Small gap between valence and conduction band | Large gap between valence and conduction band |
| Band Gap | Zero or negligible | ~1 eV | > 5 eV |
| Valence Band | Partially filled | Completely filled | Completely filled |
| Conduction Band | Partially filled or overlaps | Empty at 0 K, partly filled at room temp | Completely empty |
| Electron Movement | Free electrons move easily | Few electrons move at room temp | Electrons can’t jump to conduction band |
| Electrical Conductivity | Very high | Moderate (increases with temp) | Very low |
| Temperature Effect | Slight decrease in conductivity | Conductivity increases | No major effect |
| Examples | Copper, Silver, Iron | Silicon, Germanium | Glass, Rubber, Wood |
Q14. Explain Free Electron Theory and significance of Fermi energy. [Repeated in 2022, 2023]
Answer: Free Electron Theory is used to explain how electrons move inside metals, and how they help in conducting electricity.
This theory assumes that:
- Metals have free electrons (not tightly bound to atoms).
- These electrons move randomly, like gas molecules.
- When an electric field is applied, they start moving in one direction, which creates electric current.
Types of Free Electron Theory
1. Classical Free Electron Theory (Drude Model)
- Electrons behave like gas particles.
- It explains electrical conductivity and Ohm’s law.
Problems:
- Cannot explain why some materials are good conductors and some are not.
- Cannot explain heat behavior properly.
2. Quantum Free Electron Theory (Sommerfeld Model)
- It uses quantum rules instead of classical ones.
- Electrons are still free, but they now follow the Pauli exclusion principle and Fermi-Dirac statistics.
Fermi energy is the maximum energy that an electron can have at absolute zero temperature (0 K).
- All the energy levels below it are completely filled with electrons.
- All the energy levels above it are empty.
It is usually written as $E_F$.
Significance
- It tells us how electrons are distributed in a material.
- Only the electrons close to the Fermi level can move and take part in conduction.
- Helps in understanding:
- How metals conduct electricity.
- Why some materials behave as semiconductors or insulators.
- Heat capacity and thermal conductivity.
Q15. Show that the diameter $D_n$ of the $n^{\text{th}}$ dark ring in Newton’s Rings experiment is given by the relation: $D_n = 2\sqrt{n\lambda R}$ [PYQ]
where
- n=1,2,3,…
- $\lambda$ = wavelength of the incident light
- R = radius of curvature of the plano-convex lens
Answer: Refer to Q.1 (the answer is similar to the Q.1(Long))
BEU Engineering Physics 2025 | New Syllabus Model Paper, PYQ Analysis & Most important Questions
Q16. In a Newton’s Rings experiment, the diameter of the 15th dark ring was found to be 0.590 cm and that of the 5th ring was 0.336 cm.
If the radius of curvature of the plano-convex lens is 100 cm, calculate the wavelength of the light used.
Answer:
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BEU Engineering Physics 2025 | New Syllabus Model Paper, PYQ Analysis & Most important Questions
📘 Conclusion
This post covers the most important questions for BEU Engineering Physics 2025 | New Syllabus Model Paper, PYQ Analysis & Most important Questions, aligned with the new syllabus and based on PYQ trends. Regular practice and understanding of these key concepts will help students perform confidently in the semester exam.
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The questions and solutions presented here are derived from previous year question patterns, model papers, and subject analysis. While we strive to provide accurate and helpful content, we do not guarantee that these questions will appear in the actual BEU examination.
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