Question

(a) What is the minimum angular spread of a \(633 - nm\) wavelength \({\rm{He - Ne}}\) laser beam that is originally \(1.00{\rm{ }}mm\) in diameter?

(b) If this laser is aimed at a mountain cliff \(15.0{\rm{ }}km\) away, how big will the illuminated spot be?

(c) How big a spot would be illuminated on the Moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round-trip time and, hence, distance.) Explicitly show how you follow the steps in Problem-Solving Strategies for Wave Optics.

Short answer

(a) The minimum angular spread of the laser beam is\(\theta = 7.723 \times {10^{ - 4}}{\rm{ }}rad\).

(b) The illuminated spot will be of length\(s = 11.585\;{\rm{ }}km\)on the mountain.

(c) The illuminated spot will be of length \(s = 29687\;{\rm{ }}km\)on the moon.

Step-by-step solution

Concept Introduction

The visible light is a part of the electromagnetic spectrum. The speed of the electromagnetic wave's speed is given by –

\(c = v\lambda \)

Also, the visible light has wavelengths between\(380{\rm{ }}nm\)and\(760{\rm{ }}nm\). The wavelength of the wavelength of electromagnetic wave in a medium is given by –

\({\lambda _n} = \dfrac{\lambda }{n}\)

Where,\(\lambda \)is the wavelength of the electromagnetic wave in vacuum and\(n\)is the redfraction index of the given medium. The thickness of any material is given by –

\(t = \dfrac{\lambda }{n}\)

Young's double slit experiment has unusual results. The light must interact with something small in dimension, such as the closely slits used by Young. To obtain constructive interference for a double slit, the path length difference must be an integral multiple of the wavelength –

\(d(\sin {\rm{ }}\theta ) = m\lambda \)

To obtain destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength –

\(d(\sin {\rm{ }}\theta ) = \left( {m + \dfrac{1}{2}} \right)\lambda \)

Where,\(d\)is the distance between the slits,\(\theta \)is the angle from the original direction of the beam and\({\rm{m}}\)is the order of the interference.

The Rayleigh criterion for the difdfraction limit to resolution states that two images are just resolvable whenthe centre of directly over the first minimum ofthe difdfraction pattern of the other. So, the first minimum is at an anglewhich is the angle that separates the two-point objects of –

\(\theta = 1.22\dfrac{\lambda }{D} = \dfrac{x}{d}\)

Where,\(D\)is the aperture's diameter of the lens and\(\theta \)is given by radians. The distance between two objects with distance\(r\)is given by –

\(s = r\theta \)

 Step 2: Information Provided

Diameter of flashlight beam is:

\(\begin{aligned}D &= 1.00{\rm{ }}mm\\ &= \dfrac{1}{{1000}}\\ &= 1 \times {10^{ - 3}}{\rm{ }}m\end{aligned}\).

Wavelength of the flashlight beam is:

\(\begin{aligned}\lambda &= 633{\rm{ }}nm\\ &= \dfrac{{633}}{{{{10}^9}}}\\ &= 633 \times {10^{ - 9}}{\rm{ }}m\end{aligned}\).

Distance of laser from mountain is:\(r = 15{\rm{ }}km\).

The distance between the Earth and the Moon is:\(r = 384.4 \times {10^3}{\rm{ }}km\).

Calculation for the angle

(a) Solve for the minimum angular spread.

The first minimum angle is represented as –

\(\begin{aligned}\theta &= 1.22\dfrac{\lambda }{D}\\ &= 1.22 \times \dfrac{{633 \times {{10}^{ - 9}}\;m}}{{1 \times {{10}^{ - 3}}\;m}}\\ &= 7.723 \times {10^{ - 4}}{\rm{ }}rad\end{aligned}\)

Therefore, the value for angle is obtained as\(\theta = 7.723 \times {10^{ - 4}}{\rm{ }}rad\).

Calculation for the length on mountain

(b) Solve for the length of the illuminated spot.

The distance between two objects with distance\(r\)is given by –

\(\begin{aligned}s &= r\theta \\ &= 15 \times {10^3}\;km \times 1.1 \times {10^{ - 13}}ly/km \times 7.723 \times {10^{ - 4}}rad\\ &= 1.274 \times {10^{ - 12}}{\rm{ }}ly\\ &= 11.585\;{\rm{ }}km\end{aligned}\)

Therefore, the value for length is obtained as\(s = 11.585\;{\rm{ }}km\).

Calculation for the length on moon

(c) Solve for the length of the illuminated spot.

The distance between two objects with distance\(r\)is given by –

\(\begin{aligned}s &= r\theta \\ &= 384.4 \times {10^3}{\rm{ }}km \times 7.723 \times {10^{ - 4}}{\rm{ }}rad\\ &= 29687\;{\rm{ }}km\end{aligned}\)

Therefore, the value for length is obtained as \(s = 29687\;{\rm{ }}km\).