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Chapter 20: Q3P (page 525)

Would you use a tungsten or a deuterium lamp as a source of 300-nm radiation? What kind of lamp provides radiation at 4-mm wavelength?

Short Answer

Expert verified

We would use a deuterium lamp as a source of 300-nm radiation.

Silicon carbide globar lamp provides radiation at wavelength.

Step by step solution

01

Step: 1 Explanation of tungsten lamp, deuterium lamp and radiation:

Radiation is the release of energy, whether it is in the form of waves or particles.

A deuterium lamp emits light in the near ultraviolet and ultraviolet regions from 150 nm to 400 nm.

Tungsten lamp, a lamp in which light is produced by a tungsten filament heated to incandescence by an electric current.

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Most popular questions from this chapter

The exitance (power per unit area per unit wavelength) from a blackbody (Box 20-1) is given by the Planck distribution: Mλ=2πhc2λ5(1ehc/Akt-1)where λis wavelength, Tis temperature K), his Planck’s constant, Cis the speed of light, and kis Boltzmann’s constant. The area under each curve between two wavelengths in the blackbody graph in Box 20-1is equal to the power per unit area(W/m2)emitted between those two wavelengths. We find the area by integrating the Planck function between wavelengthsλ1andλ2

role="math" localid="1664862982839" Poweremitted=∫λ1λ2Mλdλ

For a narrow wavelength range, ∆λ1the value of Mλis nearly constant and the power emitted is simply the product Mλ∆λ2.

(a) Evaluate Mhatλ=2.00μmand atλ=10.00μmatT=1000K

(b) Calculate the power emitted per square meter atin the intervalby evaluating the product

(c) Repeat part (b) for the interval9.99to10.01μm

(d) The quantityM2μm/M10μmis the relative exitance at the two wavelengths. Compare the relative exitance at these two wave-lengths atwith the relative exitance at 100K. What does your answer mean?

Calculate the power per unit area (the exitance, W/m2) radiating from a blackbody at 77 K (liquid nitrogen temperature) and at 298 K (room temperature).

Explain why the transmission spectrum is calculated from the quotient (sample transform) M(background transform) instead of the difference (sample transform) - (background transform).

Find the minimum angleθifor total reflection in the optical fiber in Figure 20-21 if the index of refraction of the cladding is 1.400 and the index of refraction of the core is

(a) 1.600 or

(b) 1.800.

Consider a reflection grating operating with an incident angle of 408 in Figure 20-7. (a) How many lines per centimeter should be etched in the grating if the first-order diffraction angle for 600 nm (visible) light is to be 2308? (b) Answer the same question for 1 000 cm21 (infrared) light.
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