Question

The Fraunhofer diffraction pattern of a circular aperture (of radius 0.5 mm) is observed on
the local plane of a convex lens of focal length 20cm.calculate the radii of the first and the
second dark rings. Assume λ = 5.5x105 cm.

Answer 1

Answer on Question #53235, Physics / Optics

The Fraunhofer diffraction pattern of a circular aperture (of radius $0.5\mathrm{mm}$ ) is observed on the focal plane of a convex lens of focal length $20\mathrm{cm}$ . Calculate the radii of the first and the second dark rings. Assume $\lambda = 5.5\times 10^{-5}\mathrm{cm}$ .

Solution:

The amplitude distribution for diffraction due to a circular aperture forms an intensity pattern with a bright central band surrounded by concentric circular bands of rapidly decreasing intensity (Airy pattern). The 1st maximum is roughly $1.75\%$ of the central intensity. $84\%$ of the light arrives within the central peak called the Airy disk.

Fig.. An image of an Airy disk.

Far away from the aperture, the angle at which the first minimum occurs, measured from the direction of incoming light, is given by the approximate formula:

$$\sin \theta \approx 1. 2 2 \frac {\lambda}{d}$$

or, for small angles, simply

$$\theta \approx 1. 2 2 \frac {\lambda}{d}$$

where $\theta$ is in radians, $\lambda$ is the wavelength of the light and $d$ is the diameter of the aperture.

We have $\lambda = 5.5 * 10^{-5} \, \text{cm}$ and $d = 2r = 2 * 0.5 = 1 \, \text{mm} = 0.1 \, \text{cm}$ .

Thus,

$$\theta \approx 1. 2 2 * \frac {5 . 5 * 1 0 ^ {- 5}}{0 . 1} = 6 7. 1 * 1 0 ^ {- 5} \mathrm {r a d}$$

Hence, the radius of the first dark ring

$$r _ {1} = f \theta = 2 0 * 6 7. 1 * 1 0 ^ {- 5} = 0. 0 1 3 4 2 \mathrm {c m} \approx 0. 1 3 \mathrm {m m}$$

Similarly for the second dark ring

$$\theta_ {2} \approx \sin \theta_ {2} = \frac {2 . 2 3 \lambda}{d} = \frac {2 . 2 3 * 5 . 5 * 1 0 ^ {- 5}}{0 . 1} = 1 2 2. 6 5 * 1 0 ^ {- 5} \mathrm {r a d}$$ $$r _ {2} = f \theta_ {2} = 2 0 * 1 2 2. 6 5 * 1 0 ^ {- 5} = 0. 0 2 4 5 3 \mathrm {c m} \approx 0. 2 5 \mathrm {m m}$$

Answer: $r_1 \approx 0.13 \, \mathrm{mm}$; $r_2 \approx 0.25 \, \mathrm{mm}$.

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