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Chapter 33: Q.65 (page 958)

Scientists use laser range-finding to measure the distance to the moon with great accuracy. A brief laser pulse is fired at the moon, then the time interval is measured until the "echo" is seen by a telescope. A laser beam spreads out as it travels because it diffracts through a circular exit as it leaves the laser. In order for the reflected light to be bright enough to detect, the laser spot on the moon must be no more than 1.0kmin diameter. Staying within this diameter is accomplished by using a special large diameter laser. If λ=532nm, what is the minimum diameter of the circular opening from which the laser beam emerges? The earth-moon distance is384,000km.

Short Answer

Expert verified

The minimum diameter of the circular opening from which the laser beam emerges0.50m

Step by step solution

01

Maximum Diameter

We should compute cartesian coordinates and for size of the light's "bullet" by using equations for the length of the main maximum:

w=2.44λLD⇒D=2.44λLw

localid="1650221494106" D=2.44×5.32×10−7m×3.84×108m1×103m=0.499m(Numerically)

02

Minimum Diameter

Lets remember and we want is to round right, still not down due, is because we're seeking for shortest length which might help to attain this. As a findings, we just round the result to D=0.50m.

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Most popular questions from this chapter

FIGURE shows two nearly overlapped intensity peaks of the sort you might produce with a diffraction grating . As a practical matter, two peaks can just barely be resolved if their spacing ∆yequals the width w of each peak, where wis measured at half of the peak’s height. Two peaks closer together than wwill merge into a single peak. We can use this idea to understand the resolution of a diffraction grating.

a. In the small-angle approximation, the position of the m=1peak of a diffraction grating falls at the same location as the m=1fringe of a double slit: y1=λL/d. Suppose two wavelengths differing by ∆lpass through a grating at the same time. Find an expression for localid="1649086237242" ∆y, the separation of their first-order peaks.

b. We noted that the widths of the bright fringes are proportional to localid="1649086301255" 1/N, where localid="1649086311478" Nis the number of slits in the grating. Let’s hypothesize that the fringe width is localid="1649086321711" w=y1/NShow that this is true for the double-slit pattern. We’ll then assume it to be true as localid="1649086339026" Nincreases.

c. Use your results from parts a and b together with the idea that localid="1649086329574" Δymin=wto find an expression for localid="1649086347645" Δλmin, the minimum wavelength separation (in first order) for which the diffraction fringes can barely be resolved.

d. Ordinary hydrogen atoms emit red light with a wavelength of localid="1649086355936" 656.45nm.In deuterium, which is a “heavy” isotope of hydrogen, the wavelength is localid="1649086363764" 656.27nm.What is the minimum number of slits in a diffraction grating that can barely resolve these two wavelengths in the first-order diffraction pattern?

A 0.50-mm-wide slit is illuminated by light of wavelength 500nm. What is the width (inmm)of the central maximum on a screen2.0mbehind the slit?

FIGURE Q33.1 shows light waves passing through two closely spaced, narrow slits. The graph shows the intensity of light on a screen behind the slits. Reproduce these graph axes, including the zero and the tick marks locating the double-slit fringes, then draw a graph to show how the light-intensity pattern will appear if the right slit is blocked, allowing light to go through only the left slit. Explain your reasoning.

In a single-slit experiment, the slit width is 200 times the wavelength of the light. What is the width (inmm)of the central maximum on a screen 2.0m behind the slit?

A diffraction grating produces a first-order maximum at an angle of 20.0°. What is the angle of the second-order maximum?
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