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Chapter 35: Q102P (page 1080)

In a phasor diagram for any point on the viewing screen for the two slit experiment in Fig 35-10, the resultant wave phasor rotates60.0掳in 2.50脳10-16鈥�s. What is the wavelength?

Short Answer

Expert verified

Thus, the wavelength of light is 450nm.

Step by step solution

01

According to the question.

It is given that

螖蠒=60掳=蟺3rad

The figure according to the question is:

The phasors rotate with constant angular velocity

蝇=螖蠒螖迟=蟺3脳2.5脳10-16鈥�s=4.19脳1015鈥�rad/s

02

The wavelength of light in medium. 

Since, light waves travelling in a medium (air) where the wave speed is approximately c, then kc=蝇,wherek=2蟺位, which leads to

位=2蟺c蝇=2蟺脳3脳108m/s4.19脳1015rad/s=4.498脳10-7m=450nm

Hence, the wavelength of light is, 450nm.

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Most popular questions from this chapter

In a double-slit experiment, the fourth-order maximum for a wavelength of 450 nm occurs at an angle of 胃=90掳. (a) What range of wavelengths in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change is needed?

In the double-slit experiment of Fig. 35-10, the viewing screen is at distance D=4.00m, point P lies at distance role="math" localid="1663143982922" y=20.5cmfrom the center of the pattern, the slit separation d is 4.50mm, and the wavelength 位is 580 nm. (a) Determine where point P is in the interference pattern by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies. (b) What is the ratio of the intensitylPat point P to the intensitylcen at the centerof the pattern?

Find the slit separation of a double-slit arrangement that will produce interference fringes0.018鈥�radapart on a distant screen when the light has wavelength位=589鈥�nm.

A double-slit arrangement produces interference fringes for sodium light(位=589nm)that are 0.200Capart. What is the angular separation if the arrangement is immersed in water (n=1.33)?

Transmission through thin layers. In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of and interfere, r3and r4here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1,n2and n3, the type of interference, the thin-layer thickness Lin nanometers, and the wavelength in nanometers of the light as measured in air. Where 位is missing, give the wavelength that is in the visible range. Where Lis missing, give the second least thickness or the third least thickness as indicated.
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