Two narrow slits 80μmapart are illuminated with light of wavelength localid="1649170764860" 620nm. What is the angle of thelocalid="1649170756737" m=3bright fringe in radianslocalid="1649170769758" ?In degreeslocalid="1649170772775" ?

Short Answer

Expert verified

The Angle of m=3bright fringe in radian is0.0225rad,in degrees 1.29°.

Step by step solution

01

Formula for angle

For double-slit interference, the m-th order angle is given by

sinθm=d

The distance between the slits is d.

And the wavelength is λ.

02

Calculation for angle

Angle is,

θ=sin-1d

=sin-13·6·10-78·10-5

=sin(0.0225)

=0.0225rad

In degrees,

θdeg=2π360·θrad

=1.29°

As can be seen, the small angle approximation holds to a high degree of accuracy at this angle.

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Most popular questions from this chapter

Optical computers require microscopic optical switches to turn signals on and off. One device for doing so, which can be implemented in an integrated circuit, is the Mach-Zender interferometer seen in FIGURE. Light from an on-chip infrared laser (λ=1.000μm)is split into two waves that travel equal distances around the arms of the interferometer. One arm passes through an electro-optic crystal, a transparent material that can change its index of refraction in response to an applied voltage. Suppose both arms are exactly the same length and the crystal’s index of refraction with no applied voltage is1.522.

a. With no voltage applied, is the output bright (switch closed, optical signal passing through) or dark (switch open, no signal)? Explain.

b. What is the first index of refraction of the electro-optic crystal larger than 1.522that changes the optical switch to the state opposite the state you found in part a?

A double-slit experiment is set up using a helium-neon laser (λ=633nm). Then a very thin piece of glass (n=1.50) is placed over one of the slits. Afterward, the central point on the screen is occupied by what had been the m=10 dark fringe. How thick is the glass?

A radar for tracking aircraft broadcasts a 12GHzmicrowave beam from a 2.0m-diameter circular radar antenna. From a wave perspective, the antenna is a circular aperture through which the microwaves diffract.

a. What is the diameter of the radar beam at a distance of 30km?

b. If the antenna emits 100kWof power, what is the average microwave intensity at 30km?

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0cm. You have a helium neon laser (λ=633nm)and a0.12-mm-diameter pinhole. How far behind the pinhole should you place the screen that’s to be photographed?

The pinhole camera of FIGURE images distant objects by allowing only a narrow bundle of light rays to pass through the hole and strike the film. If light consisted of particles, you could make the image sharper and sharper (at the expense of getting dimmer and dimmer) by making the aperture smaller and smaller. In practice, diffraction of light by the circular aperture limits the maximum sharpness that can be obtained. Consider two distant points of light, such as two distant streetlights. Each will produce a circular diffraction pattern on the film. The two images can just barely be resolved if the central maximum of one image falls on the first dark fringe of the other image. (This is called Rayleigh’s criterion, and we will explore its implication for optical instruments in Chapter 35.)

a. Optimum sharpness of one image occurs when the diameter of the central maximum equals the diameter of the pinhole. What is the optimum hole size for a pinhole camera in which the film is 20cmbehind the hole? Assume localid="1649089848422" λ=550nman average value for visible light.

b. For this hole size, what is the angle a (in degrees) between two distant sources that can barely be resolved?

c. What is the distance between two street lights localid="1649089839579" 1kmaway that can barely be resolved?

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