In a Newton’s rings experiment, light of wavelength λ and a lens of radius of curvature R are used. The difference in diameters of 25th and 16th dark rings is:
Newton's Rings Experiment Overview
The Newton's rings experiment is a classic demonstration of optical interference, specifically observed when a plano-convex lens with a large radius of curvature is placed on a flat glass plate. A thin air film is formed between the lens and the plate, which varies in thickness radially from the point of contact. When monochromatic light is shone on this setup, concentric bright and dark rings are observed due to constructive and destructive interference of light waves reflecting from the top and bottom surfaces of the air film. These concentric patterns are known as Newton's rings.
In this Newton's rings experiment, we are given the wavelength of light ($\lambda$) and the radius of curvature of the lens (R). Our goal is to determine the difference in diameters of the 25th and 16th dark rings.
Understanding Dark Rings Diameter
For dark rings in a Newton's rings setup, the condition for destructive interference is met. The radius of the $n^{\text{th}}$ dark ring is given by the formula:
$r_n = \sqrt{n \lambda R}$
Where:
- $r_n$ is the radius of the $n^{\text{th}}$ dark ring.
- $n$ is the order of the dark ring (e.g., 1 for the first dark ring, 2 for the second, and so on).
- $\lambda$ is the wavelength of the light used.
- $R$ is the radius of curvature of the plano-convex lens.
Since the diameter ($D_n$) is twice the radius ($r_n$), the formula for the diameter of the $n^{\text{th}}$ dark ring is:
$D_n = 2r_n = 2\sqrt{n \lambda R}$
We will use this formula to calculate the diameters of the 25th and 16th dark rings.
Calculating 25th Dark Ring Diameter
To find the diameter of the 25th dark ring, we set $n = 25$ in the formula for $D_n$:
$D_{25} = 2\sqrt{25 \lambda R}$
We know that $\sqrt{25} = 5$. Substituting this value:
$D_{25} = 2 \times 5 \sqrt{\lambda R}$
$D_{25} = 10\sqrt{\lambda R}$
Calculating 16th Dark Ring Diameter
Similarly, to find the diameter of the 16th dark ring, we set $n = 16$ in the formula for $D_n$:
$D_{16} = 2\sqrt{16 \lambda R}$
We know that $\sqrt{16} = 4$. Substituting this value:
$D_{16} = 2 \times 4 \sqrt{\lambda R}$
$D_{16} = 8\sqrt{\lambda R}$
Difference in Diameters of Dark Rings
Now, we need to find the difference between the diameters of the 25th and 16th dark rings. This is calculated as $D_{25} - D_{16}$.
Difference $= D_{25} - D_{16}$
Difference $= 10\sqrt{\lambda R} - 8\sqrt{\lambda R}$
Difference $= (10 - 8)\sqrt{\lambda R}$
Difference $= 2\sqrt{\lambda R}$
Therefore, the difference in diameters of the 25th and 16th dark rings in the Newton's rings experiment is $2\sqrt{\lambda R}$.
| Parameter | Value/Formula |
|---|---|
| Radius of $n^{\text{th}}$ dark ring ($r_n$) | $\sqrt{n \lambda R}$ |
| Diameter of $n^{\text{th}}$ dark ring ($D_n$) | $2\sqrt{n \lambda R}$ |
| Diameter of 25th dark ring ($D_{25}$) | $10\sqrt{\lambda R}$ |
| Diameter of 16th dark ring ($D_{16}$) | $8\sqrt{\lambda R}$ |
| Difference in diameters ($D_{25} - D_{16}$) | $2\sqrt{\lambda R}$ |
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?
