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In Fig, monochromatic light of wavelength diffracts through narrow slit S in an otherwise opaque screen. On the other side, a plane mirror is perpendicular to the screen and a distance h from the slit. A viewing screen A is a distance much greater than h. (Because it sits in a plane through the focal point of the lens, screen A is effectively very distant. The lens plays no other role in the experiment and can otherwise be neglected.) Light travels from the slit directly to A interferes with light from the slit that reflects from the mirror to A. The reflection causes a half-wavelength phase shift. (a) Is the fringe that corresponds to a zero path length difference bright or dark? Find expressions (like Eqs. 35-14 and 35-16) that locate (b) the bright fringes and (c) the dark fringes in the interference pattern. (Hint: Consider the image of S produced by the minor as seen from a point on the viewing screen, and then consider Young’s two-slit interference.)

Short Answer

Expert verified
  1. A zero path difference would correspond to a dark fringe.
  2. For bright fringes,2hsinθ=m+12λ
  3. For dark fringes,2hsinθ=mλ

Step by step solution

01

The given data

Given that monochromatic light of wavelength diffracts through narrow slit S in an otherwise opaque screen.

On the other side, a plane mirror is perpendicular to the screen and a distance h from the slit. A viewing screen A is a distance much greater than h.

Light travels from the slit directly to A interferes with light from the slit that reflects from the mirror to A. The reflection causes a half-wavelength phase shift.

02

Concept and Formula used

For bright fringes, the distance is either zero or an integer number of wavelengths dsinθ=mλfor m=0,1,2,...

And for dark fringes, the distance is odd multiple of half of wavelength,dsinθ=m+12λform=0,1,2,...

Here, d is distance, and λis wavelength.

Also, reflected ray has a phase change of πthat causes the location of dark and bright fringes to interchange.

03

Interpret whether the fringe is dark or bright

(a)

The combination of the direct ray and reflected ray from the mirror will produce an interference pattern on the screen, like the double-slit experiment. However, here, the reflected ray has a phase change of π, causing the locations of the dark and bright fringes to be interchanged. Thus, a zero path difference would correspond to a dark fringe.

04

Write expression for bright fringe

(b)

Here, the distance is 2h and has a phase change ofπ occurs, So

Condition to locate bright fringe in interference pattern is 2hsinθ=m+12λ

05

Write expression for dark fringe

(c)

Here, the distance is 2h and has a phase change of πoccurs, So

Condition to locate dark fringe in interference pattern is 2hsinθ=mλ

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Most popular questions from this chapter

Reflection by thin layers. In Fig. 35-42, 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.) The waves of rays r1and r2interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35- 2 refers to the indexes of refraction n1, n2and n3, the type of interference, the thin-layer thickness Lin nanometres, and the wavelength λin nanometres of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where localid="1663142040666" Lis missing, give the second least thickness or the third least thickness as indicated


Does the spacing between fringes in a two-slit interference pattern increase, decrease, or stay the same if

(a) the slit separation is increased,

(b) the color of the light is switched from red to blue, and

(c) the whole apparatus is submerged in cooking sherry?

(d) If the slits are illuminated with white light, then at any side maximum, does the blue component or the red component peak closer to the central maximum?

In Fig. 35-45, a broad beam of light of wavelength 620 nm is sent directly downward through the top plate of a pair of glass plates touching at the left end. The air between the plates acts as a thin film, and an interference pattern can be seen from above the plates. Initially, a dark fringe lies at the left end, a bright fringe lies at the right end, and nine dark fringes lie between those two end fringes. The plates are then very gradually squeezed together at a constant rate to decrease the angle between them. As a result, the fringe at the right side changes between being bright to being dark every 15.0 s.

(a) At what rate is the spacing between the plates at the right end being changed?

(b) By how much has the spacing there changed when both left and right ends have a dark fringe and there are five dark fringes between them?

Reflection by thin layers. In Fig. 35-42, 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.) The waves of rays r1and r2interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35- 2 refers to the indexes of refraction n1, n2, and n3, the type of interference, the thin-layer thickness in nanometres, and the wavelength λ in nanometres of the light as measured in air. Where λis 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.

57 through 68 64, 65 59 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 r3 and r4interfere, and here 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,n2andn3, the type.

Of interference, the thin-layer thickness L in nanometres, and the wavelength λ in nanometres of the light as measured in air.

Where λ is missing, give the wavelength that is in the visible range.

Where L is missing, give the second least thickness or the third least thickness as indicated?

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