Interference fringes are observed with a biprism of refracting angle 2° and refractive index 1.5 on a screen 100 cm away from it. If the distance between the source and the biprism is 20 cm and the fringe width is 0.10 mm, the wavelength of light used is:
5820 Å
Biprism Interference Fringes Explained
Interference fringes are a fascinating phenomenon in optics that occurs when two coherent light waves combine. A biprism is a clever device used to produce two virtual coherent sources from a single real light source, which then create an interference pattern on a screen. This problem requires us to determine the wavelength of the light used, given various parameters of the biprism setup and the observed fringe width.
Biprism Setup and Key Parameters
To begin, let's list all the given information and convert them into consistent units, typically meters (SI units), to ensure accurate calculations.
- Refracting angle of the biprism (\(\alpha\)): \(2^\circ\)
- Refractive index of the biprism (\(\mu\)): \(1.5\)
- Distance between the screen and the biprism (\(\text{D'}\)): \(100 \text{ cm} = 1.0 \text{ m}\)
- Distance between the source and the biprism (\(\text{d}_1\)): \(20 \text{ cm} = 0.2 \text{ m}\)
- Fringe width (\(\beta\)): \(0.10 \text{ mm} = 0.10 \times 10^{-3} \text{ m}\)
A crucial step for calculations involving small angles, like the refracting angle of a biprism, is to convert the angle from degrees to radians. This is important because the formulas used in biprism theory rely on the small angle approximation where \(\sin \theta \approx \theta\) (in radians).
Conversion of refracting angle to radians:
\[ \alpha_{\text{radians}} = \alpha_{\text{degrees}} \times \frac{\pi}{180^\circ} = 2^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{90} \text{ radians} \]
Calculating Virtual Source Separation
The biprism splits the light from a single source into two beams, making them appear to originate from two virtual coherent sources. The distance between these two virtual sources, denoted as 'd', is a key component in the fringe width formula. This separation 'd' depends on the distance of the real source from the biprism, the refractive index of the biprism material, and its refracting angle.
The formula for the separation 'd' between the virtual sources in a biprism experiment is:
\[ d = 2 d_1 (\mu - 1) \alpha_{\text{radians}} \]
Now, let's substitute the values we have into this formula:
\[ d = 2 \times (0.2 \text{ m}) \times (1.5 - 1) \times \frac{\pi}{90} \]
\[ d = 0.4 \text{ m} \times 0.5 \times \frac{\pi}{90} \]
\[ d = 0.2 \times \frac{\pi}{90} \text{ m} \]
Using the approximate value of \( \pi \approx 3.14159 \):
\[ d = 0.2 \times \frac{3.14159}{90} \approx \frac{0.628318}{90} \approx 0.0069813 \text{ m} \]
Determining Total Optical Path Distance
For interference calculations, the total distance 'D' refers to the distance from the coherent sources (in this case, the virtual sources created by the biprism) to the screen where the interference fringes are observed. In a biprism arrangement, this total distance is simply the sum of the distance from the real source to the biprism and the distance from the biprism to the screen.
\[ D = d_1 + D' \]
Substituting the given values:
\[ D = 0.2 \text{ m} + 1.0 \text{ m} = 1.2 \text{ m} \]
Wavelength Calculation using Fringe Width
The fringe width (\(\beta\)) is a directly measurable quantity in an interference pattern. It is directly related to the wavelength of the light (\(\lambda\)), the total distance to the screen (D), and inversely related to the separation between the coherent sources (d). This relationship is a fundamental equation in wave optics:
\[ \beta = \frac{\lambda D}{d} \]
Our goal is to find the wavelength (\(\lambda\)), so we need to rearrange this formula to solve for \(\lambda\):
\[ \lambda = \frac{\beta d}{D} \]
Now, we can substitute the known values for fringe width, virtual source separation, and total optical path distance into this rearranged formula:
\[ \lambda = \frac{(0.10 \times 10^{-3} \text{ m}) \times (0.0069813 \text{ m})}{1.2 \text{ m}} \]
\[ \lambda = \frac{0.00069813 \times 10^{-3}}{1.2} \text{ m} \]
\[ \lambda = 0.000581775 \times 10^{-3} \text{ m} \]
\[ \lambda = 5.81775 \times 10^{-7} \text{ m} \]
Final Wavelength Result
The wavelength of light is often expressed in Ångströms (Å). The conversion factor is \(1 \text{ Å} = 10^{-10} \text{ m}\). To convert our calculated wavelength from meters to Ångströms, we multiply by \(10^{10}\).
\[ \lambda = 5.81775 \times 10^{-7} \times 10^{10} \text{ Å} \]
\[ \lambda = 5817.75 \text{ Å} \]
When we round this calculated value to the nearest option provided, the wavelength of light used in this biprism experiment is approximately 5820 Å.
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?
