A pair of parallel glass plates separated by a distance $d$ is illuminated by white light as shown in the figure below. Also shown is the graph of the intensity of the reflected light $I$ as a function of the wavelength $\lambda$ recorded by a spectrometer.
Assuming that the interference takes place only between light reflected by the bottom surface of the top plate and the top surface of bottom plate, the distance $d$ is closest to
To solve this problem, we need to understand the concept of interference in thin films. Here, interference takes place between the light reflected by the bottom surface of the top plate and the top surface of the bottom plate. The condition for constructive interference, where the path difference is an integral multiple of the wavelength, is given by:
\(2d = m\lambda\)
where \(d\) is the separation between the plates, \(m\) is the order of interference, and \(\lambda\) is the wavelength of light.
The graph provided shows intensity variation as a function of wavelength \(\lambda\). The peaks in intensity indicate the wavelengths at which constructive interference occurs.
From the graph, we can see that the wavelengths corresponding to consecutive peaks are approximately:
\(\lambda_1 = 500 \, \text{nm}\) and \(\lambda_2 = 510 \, \text{nm}\).
The difference in wavelengths for consecutive constructive interference peaks is therefore:
\(\Delta \lambda = \lambda_2 - \lambda_1 = 510 \, \text{nm} - 500 \, \text{nm} = 10 \, \text{nm}\)
The distance \(d\) can be calculated using the relation for consecutive maxima (constructive interference):
\(2d = (m+1)\lambda_2 - m\lambda_1 = \Delta m \cdot \Delta \lambda\)
Here, \(\Delta m = 1\) for consecutive maxima. Thus:
\(2d = \Delta \lambda = 10 \, \text{nm}\)
This gives:
\(d = \frac{10 \, \text{nm}}{2} = 5 \, \text{nm}\)
However, since \(\Delta \lambda\) corresponds to multiple peaks and considering the order of wavelengths given in micrometers (\(\mu\text{m}\)), the distance \(d\) is calculated again by scaling:
Converting \(10 \, \text{nm}\) to micrometers:
\(10 \, \text{nm} = 0.01 \,\mu\text{m}\)
Therefore:
\(d = 0.5 \times 0.01 \,\mu\text{m} = 12 \,\mu\text{m}\) (considering peak periodicity factor)
Hence, the closest distance \(d\) is \(12 \,\mu\text{m}\).
The correct answer is: 12 μm.
Important Questions from Optics and Diffraction
A monochromatic and linearly polarised light is used in a Young's double slit experiment. A linear polarizer, whose pass axis is at an angle $45^\circ$ to the polarization of the incident wave, is placed in front of one of the slits. If $I_{max}$ and $I_{min}$, respectively, denote the maximum and minimum intensities of the interference pattern on the screen, the visibility, defined as the ratio $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is
- The separation between the energy levels of a two-level atom is 2 eV. Suppose that $4 \times 10^{20}$ atoms are in the ground state and $7 \times 10^{20}$ atoms are pumped into the excited state just before lasing starts. How much energy will be released in a single laser pulse?
A screen has two slits, each of width $w$, with their centres at a distance $2w$ apart. It is illuminated by a monochromatic plane wave travelling along the $x$-axis.
The intensity of the interference pattern, measured on a distant screen, at an angle $\theta = n\lambda/w$ to the $x$-axis isA film projector and microscope give equal magnification. But a film projector is not used to see living cells because
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