Calculating Phase Difference in a Biological Specimen
This problem involves calculating the phase difference between two light beams that have passed through a biological specimen with different refractive indices for orthogonal polarizations.
Key Concepts
- The phase difference ($\Delta \phi$) arises from the difference in optical path lengths (OPL) travelled by the two polarized beams.
- Optical Path Length (OPL) is given by the product of the geometric path length (thickness, $d$) and the refractive index ($n$) of the medium: OPL = $n \times d$.
- The phase difference is related to the path difference ($\Delta L$) by the formula: $\Delta \phi = \frac{2 \pi}{\lambda_0} \Delta L$, where $\lambda_0$ is the wavelength in vacuum.
Step-by-Step Calculation
- Identify Given Values:
- Wavelength in vacuum, $\lambda_0 = 630$ nm $= 630 \times 10^{-9}$ m
- Specimen thickness, $d = 10$ $\mu$m $= 10 \times 10^{-6}$ m
- Refractive indices, $n_1 = 1.32$ and $n_2 = 1.333$
- Calculate Optical Path Lengths:
- OPL$_1 = n_1 d = 1.32 \times (10 \times 10^{-6} \text{ m}) = 13.2 \times 10^{-6}$ m
- OPL$_2 = n_2 d = 1.333 \times (10 \times 10^{-6} \text{ m}) = 13.33 \times 10^{-6}$ m
- Determine Path Difference:
The path difference ($\Delta L$) is the absolute difference between the OPLs:
$\Delta L = |OPL_2 - OPL_1| = |13.33 \times 10^{-6} \text{ m} - 13.2 \times 10^{-6} \text{ m}| = |0.13 \times 10^{-6}|$ m
Alternatively, $\Delta L = |n_2 - n_1| d = |1.333 - 1.32| \times (10 \times 10^{-6} \text{ m}) = 0.013 \times (10 \times 10^{-6} \text{ m}) = 0.13 \times 10^{-6}$ m.
- Calculate Phase Difference in Radians:
Using the formula $\Delta \phi = \frac{2 \pi}{\lambda_0} \Delta L$:
$\Delta \phi = \frac{2 \pi}{630 \times 10^{-9} \text{ m}} \times (0.13 \times 10^{-6} \text{ m})$
$\Delta \phi = \frac{2 \pi \times 0.13 \times 10^3}{630} = \frac{2 \pi \times 130}{630} = \frac{2 \pi \times 13}{63}$ radians
- Convert Phase Difference to Degrees:
To convert radians to degrees, multiply by $\frac{180^\circ}{\pi}$:
$\Delta \phi_{\text{deg}} = \left( \frac{2 \pi \times 13}{63} \right) \times \frac{180^\circ}{\pi}$
$\Delta \phi_{\text{deg}} = \frac{2 \times 13 \times 180}{63}$ degrees
$\Delta \phi_{\text{deg}} = \frac{2 \times 13 \times 20}{7}$ degrees (Simplifying by dividing 180 and 63 by 9)
$\Delta \phi_{\text{deg}} = \frac{520}{7}$ degrees
$\Delta \phi_{\text{deg}} \approx 74.2857^\circ$
Therefore, the two beams emerge out of phase by approximately 74.3°.
Important Questions from Interference
- A single slit of width $a$ is illuminated by a monochromatic light of wavelength $\lambda_1 = 6000 \text{ Å}$. The angular width of the central maximum observed in the Fraunhofer diffraction pattern is $\theta_1$. When the slit width is increased by $20\%$ and the light source is replaced with another monochromatic light of wavelength $\lambda_2$, the angular width of the central maximum becomes $\frac{3}{5}$ of its initial value, $\theta_1$. Determine the wavelength $\lambda_2$.
A system of three polarizers $P_1$, $P_2$, $P_3$ is set up such that the pass axis of $P_3$ is crossed with respect to that of $P_1$.
The pass axis of $P_2$ is inclined at $15^\circ$ to the pass axis of $P_1$.
When a beam of unpolarized light of intensity $I_0$ is incident on $P_1$, the intensity of light transmitted by the three polarizers is $I$. The ratio $(I_0/I)$ equals (nearly):The interference pattern is obtained with two coherent light sources. If the ratio of their amplitudes is $n$, then in the interference pattern, the ratio $\frac{{{I_{max}} - {I_{min}}}}{{{I_{max}} + {I_{min}}}}$ will be
- Two identical coherent waves are superimposed at a point. If the maximum possible resultant intensity from their interference is $I_{max}$, and the resultant intensity at this point is $I_{max}/4$, then find the phase difference between the two waves at this point.
Which of the following sources gives best monochromatic light?
