Unit 3 – Part B: Optics and Lasers

Q1: Explain the formation of interference fringes in an air-wedge shaped film. How is the thickness of the wire determined by this method?

1. Formation of Interference Fringes (Air Wedge)

An air wedge is a thin film of air with zero thickness at one end and progressively increasing thickness at the other. It is formed by placing a thin object (like a wire) between two optically flat glass plates.

[Place Diagram of interference in an air wedge film here]

2. Determination of Thickness of Wire

The setup is used to measure the thickness of a thin wire.

[Place Diagram of experimental setup here]

Calculation:
By measuring the average width of a specific number of fringes (e.g., 5 or 10), we determine the fringe width ($\beta$). The thickness of the wire ($t$) is related to fringe width ($\beta$), wavelength ($\lambda$), and the length of the wedge ($L$) by:

$$ \beta = \frac{\lambda L}{2t} $$

Rearranging for thickness:

$$ t = \frac{\lambda L}{2\beta} $$

Thus, knowing $\lambda$, $L$, and the measured $\beta$, the thickness $t$ is calculated.

Q2: Describe the construction of a Michelson interferometer and discuss the different type of interference fringes formed in it.

1. Principle and Construction

The Michelson Interferometer works on the principle of division of amplitude, splitting a light beam into two parts that travel different paths and recombine to interfere.

2. Types of Fringes

The pattern depends on the alignment of $M_1$ and $M_2$.

Q3: Explain the principle, construction, and working of an Nd-YAG laser.

The Nd-YAG is a four-level, solid-state laser.

1. Principle

The active medium is an Nd-YAG rod pumped by a krypton flash tube. Neodymium ions ($Nd^{3+}$) are raised to excited levels, achieving population inversion in a metastable state. The transition to a lower state emits a laser at $1.06 \mu m$.

2. Construction

3. Working

  1. Pumping: Flash tube light ($0.73 \mu m$, $0.80 \mu m$) excites $Nd^{3+}$ ions from ground state $E_0$ to pump bands $E_3$ and $E_4$.
  2. Metastable State: Ions decay non-radiatively to the metastable state $E_2$.
  3. Population Inversion: Ions accumulate at $E_2$, creating inversion between $E_2$ and $E_1$.
  4. Laser Output: Stimulated emission occurs from $E_2 \to E_1$, releasing photons at $1.06 \mu m$. These are amplified by the mirrors to form the laser beam.

Applications: Range finders, micro-machining (welding, drilling).

Q4: Explain the modes of vibrations of CO2 molecule. Explain the construction, functioning, and application of CO2 laser.

The $CO_2$ laser is a highly efficient four-level molecular gas laser.

1. Modes of Vibration

2. Construction

3. Working

  1. Excitation of Nitrogen: Electrons from the discharge collide with $N_2$ molecules: $$ N_2 + e^* \to N_2^* + e $$
  2. Energy Transfer: Excited $N_2^*$ transfers energy to $CO_2$ molecules (resonant transfer to level $E_5$): $$ N_2^* + CO_2 \to CO_2^* + N_2 $$
  3. Laser Transitions:
    • $E_5 \to E_4$ (Symmetric Stretch): $10.6 \mu m$ (Dominant).
    • $E_5 \to E_3$ (Bending): $9.6 \mu m$.
  4. Role of Helium: Conducts heat away to the tube walls to maintain efficiency.

Q5: With proper theory and diagrams, derive Einstein’s coefficients. Also discuss its inferences.

1. Interaction Processes

Consider two energy levels $E_1$ (Ground) and $E_2$ (Excited) with populations $N_1$ and $N_2$. Let $Q$ be the radiation energy density.

2. Derivation

At thermal equilibrium, Absorption Rate = Emission Rate:

$$ N_{ab} = N_{sp} + N_{st} $$ $$ B_{12} N_1 Q = A_{21} N_2 + B_{21} N_2 Q $$

Solving for energy density $Q$:

$$ Q (B_{12} N_1 - B_{21} N_2) = A_{21} N_2 $$ $$ Q = \frac{A_{21} N_2}{B_{12} N_1 - B_{21} N_2} $$

Divide numerator and denominator by $B_{21} N_2$:

$$ Q = \frac{\frac{A_{21}}{B_{21}}}{\frac{B_{12}}{B_{21}} \frac{N_1}{N_2} - 1} $$

Using Boltzmann’s law, $\frac{N_1}{N_2} = e^{h\nu / kT}$. Substituting this:

$$ Q = \frac{\frac{A_{21}}{B_{21}}}{\frac{B_{12}}{B_{21}} e^{h\nu / kT} - 1} \quad \text{--- (1)} $$

Compare this with Planck’s Radiation Law:

$$ Q = \frac{8\pi h \nu^3}{c^3} \left( \frac{1}{e^{h\nu / kT} - 1} \right) \quad \text{--- (2)} $$

3. Inferences

  1. $$ B_{12} = B_{21} $$ Probability of stimulated absorption equals stimulated emission.
  2. $$ \frac{A_{21}}{B_{21}} = \frac{8\pi h \nu^3}{c^3} $$ The ratio of spontaneous to stimulated emission depends on the cube of frequency.

Q6: Derive an expression for numerical aperture and acceptance angle of an optical fiber.

1. Principle

Light propagates via Total Internal Reflection (TIR). The core refractive index ($n_1$) must be greater than the cladding ($n_2$).

2. Derivation

Let light enter from a medium ($n_0$) at acceptance angle $\theta_0$, refracting into the core at $\theta_r$.

Step 1: At Fiber Entry (Snell’s Law)

$$ n_0 \sin \theta_0 = n_1 \sin \theta_r \quad \text{--- (1)} $$

Step 2: At Core-Cladding Interface
The ray strikes at angle $\phi$. From geometry, $\phi = 90^\circ - \theta_r$.
For the limiting case of TIR, $\phi$ equals the critical angle $\phi_c$, and the angle of refraction in cladding is $90^\circ$.

$$ n_1 \sin(90^\circ - \theta_r) = n_2 \sin 90^\circ $$ $$ n_1 \cos \theta_r = n_2 \implies \cos \theta_r = \frac{n_2}{n_1} \quad \text{--- (2)} $$

Step 3: Combining
Using trigonometric identity $\sin \theta_r = \sqrt{1 - \cos^2 \theta_r}$:

$$ \sin \theta_r = \sqrt{1 - \left(\frac{n_2}{n_1}\right)^2} = \frac{\sqrt{n_1^2 - n_2^2}}{n_1} $$

Substitute this back into Eq (1):

$$ n_0 \sin \theta_0 = n_1 \left( \frac{\sqrt{n_1^2 - n_2^2}}{n_1} \right) $$ $$ n_0 \sin \theta_0 = \sqrt{n_1^2 - n_2^2} $$

3. Final Expressions

Numerical Aperture (NA): ($NA = \sin \theta_0$)

$$ NA = \frac{\sqrt{n_1^2 - n_2^2}}{n_0} $$

If air is the launching medium ($n_0=1$):

$$ NA = \sqrt{n_1^2 - n_2^2} $$

Acceptance Angle ($\theta_0$):

$$ \theta_0 = \sin^{-1} \left( \frac{\sqrt{n_1^2 - n_2^2}}{n_0} \right) $$