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Two light rays, initially in phase and with a wavelength of 500 nm, go through different paths by reflecting from the various mirrors shown in Fig. 35-49. (Such a reflection does not itself produce a phase shift.) (a) What least value of distance will put the rays exactly out of phase when they emerge from the region? (Ignore the slight tilt of the path for ray 2.) (b) Repeat the question assuming that the entire apparatus is immersed in a protein solution with an index of refraction of 1.38.

Short Answer

Expert verified

(a) The least distance of d is 50nm.

(b) The least value of the distance if the apparatus is immersed in protein solution is 36.2nm

Step by step solution

01

Given data

Wavelength of light λ=500 nm

02

Concept used

A light ray is a line (straight or curved) that is perpendicular to the light's wave fronts; its tangent is collinear with the wave vector. Light rays in homogeneous media are straight. They bend at the interface between two dissimilar media and may be curved in a medium in which the refractive index changes

03

(a) Determine the least value of distance 

Distance is the total path travelled from initial point to final point.

Calculate the least value of d to put the rays exactly out of the phase when they emerge for the region.

Ray 1 travelled a total distance of 7d

Ray 2 travelled a total distance of 2d

The path difference between ray 1 and ray 2 is

7d-2d=5d

For the rays to be exactly out of phase this path different must be equal λ2, that is,

5d=λ2

Here, d is the least distance and λ is the wavelength of the light.

Rearrange the above equation for d.

d=λ10

Substitute 500 nm for λin above equation.

d=500nm10

Therefore, the least distance of d is 500 nm.

04

(b) Determine the least value of distance

Calculate least value of the, d to put the rays exactly out of the phase when they emerge from the region if the apparatus is immersed in protein solution.

When the apparatus is immersed in protein solution with a reactive index of n,

Then the path difference becomes n5d

For the rays to be exactly out of phase his path difference must be equal to λ2, that is,

n5d=λ2

Here, n is the reactive index of the protein solution d is the least distance and λis the wavelength of the light.

Rearrange the above equation for d.

d=λn10

Substitute 1.38 for n and 500 nm for λ in above equation.

d=500nm101.38=36.2nm

Therefore, the least value of the distance if the apparatus is immersed in protein solution is 36.2nm.

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Most popular questions from this chapter

Figure 35-22 shows two light rays that are initially exactly in phase and that reflect from several glass surfaces. Neglect the slight slant in the path of the light inthe second arrangement.

(a) What is the path length difference of the rays?

In wavelengthsλ,

(b) what should that path length difference equal if the rays are to be exactly out of phase when they emerge, and

(c) what is the smallest value of that will allow that final phase difference?

Figure 35-29 shows the transmission of light through a thin film in the air by a perpendicular beam (tilted in the figure for clarity). (a) Did rayr3undergo a phase shift due to reflection? (b) In wavelengths, what is the reflection phase shift for rayr4? (c) If the film thickness is L, what is the path length difference between raysr3andr4?

Ocean waves moving at a speed of 4.0 m/s are approaching a beach at angle θ1=30°to the normal, as shown from above in Fig. 35-55. Suppose the water depth changes abruptly at a certain distance from the beach and the wave speed there drops to 3.0 m/s. (a) Close to the beach, what is the angle θ2between the direction of wave motion and the normal? (Assume the same law of refraction as for light.) (b) Explain why most waves come in normal to a shore even though at large distances they approach at a variety of angles.

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°in 2.50×10-16 s. What is the wavelength?

In Fig. 35-45, two microscope slides touch at one end and are separated at the other end. When light of wavelength 500 nm shines vertically down on the slides, an overhead observer sees an interference pattern on the slides with the dark fringes separated by 1.2 mm. What is the angle between the slides?

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