A thin flake of mica (n = 1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe (m = 7). If λ=550nm , what is the thickness of the mica?

Short Answer

Expert verified

The thickness of mica is 6.64μm.

Step by step solution

01

Identification of given data

The index of refraction for mica is n=1.58

The order of seventh fringe is m=7

The wavelength of light is λ=550nm

The thickness of the mica is found by using the formula for thin film interference for bright fringes.

02

Determination of thickness of the mica

The thickness of the mica is given as:

t=mλn-1

Substitute all the values in equation.

t=7550nm1.58-1t=6.64×10-6mt=6.64×10-6m1μm10-6mt=6.64μm

Therefore, the thickness of mica is 6.64μm.

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Most popular questions from this chapter

Figure 35-57 shows an optical fiber in which a central platic core of index of refractionn1=1.58-is surrounded by a plastic sheath of index of refractionn2=1.53. Light can travel along different paths within the central core, leading to different travel times through the fiber, resulting in information loss. Consider light that travels directly along the central axis of the fiber and light that is repeatedly reflected at the critical angle along the core-sheath interface, reflecting from side to side as it travels down the central core. If the fiber length is 300 m, what is the difference in the travel times along these two routes?

In a double-slit experiment, the fourth-order maximum for a wavelength of 450 nm occurs at an angle of θ=90°. (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?

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, n2andn3, 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 Lis missing, give the second least thickness or the third least thickness as indicated.

A thin film suspended in air is 0.410 μmthick and is illuminated with white light incident perpendicularly on its surface. The index of refraction of the film is 1.50. At what wavelength will visible light that is reflected from the two surfaces of the film undergo fully constructive interference?

White light is sent downward onto a horizontal thin film that is sandwiched between two materials. The indexes of refraction are 1.80for the top material, 1.70for the thin film, and 1.50for the bottom material. The film thickness is5×10-7m . Of the visible wavelengths (400 to 700nm ) that result in fully constructive interference at an observer above the film, which is the (a) longer and (b) shorter wavelength? The materials and film are then heated so that the film thickness increases. (c) Does the light resulting in fully constructive interference shift toward longer or shorter wavelengths?

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