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A coherent beam of light strikes a single slit and produces a spread-out diffraction pattern beyond. The number of photons detected per unit time a detector in the very center of the pattern isX. Now two more slits are opened nearby, the same width as the original. equally spaced on either side of it, and equally well. illuminated by the beam. How many photons will be detected per unit time at the center detector now: Why?

Short Answer

Expert verified

The number of photons per unit time detected at the center detector is9X.

Step by step solution

01

Concept used

A low-intensity beam of light is sent toward a narrow single slit. On the far side, individual flashes are seen sporadically at detectors over a broad area that has an order of magnitude wider than the slit width. Two more slits equidistant from the given slit are opened on either side of the given slit. All the slits are equally illuminated (same magnitude of the electric field).

02

Diffraction Pattern

As the number of slits of equal width increases, the diffraction pattern becomes sharper and more intense. As the number of slits increases, the constructive and destructive interferences also increase with the maximum constructive interference occurring at the peak positions. The intensity of light in these peak positions is proportional to the number of photons that hit per unit of time. Also, the number of photons hitting per unit of time is proportional to the square of the sum of electric fields.

03

Number of Photons

For a single slit with an electric field E , the number of photons X is given as:

XE2X=kE2

Where, K is the constant of proportionality.

Now, when there are three slits of equal illumination, the sum of electric fields is 3E.

So, the number of photons hitting per unit of time X1 is given as:

X1=k(3E)2X1=9kE2X1=9X

04

Conclusion 

Therefore, the number of photons per unit time detected at the center detector is 9X when a total of three slits are used for observing the diffraction pattern.

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Most popular questions from this chapter

A bedrock topic in quantum mechanics is the uncertainty principle. It is discussed mostly for massive objects in Chapter 4, but the idea also applies to light: Increasing certainty in knowledge of photon position implies increasing uncertainty in knowledge of its momentum, and vice versa. A single-slit pattern that is developed (like the double-slit pattern of Section 3.6) one photon at a time provides a good example. Depicted in the accompanying figure, the pattern shows that pho tons emerging from a narrow slit are spreadall-over; a photon's x-component of momentum can be any value over a broad range and is thus uncertain. On the other hand, the x -coordinate of position of an emerging photon covers a fairly small range, for w is small. Using the single-slit diffractionformula =wsinθ , show that the range of likely values of px, which is roughly psinθ , is inversely proportional to the range w of likely position values. Thus, an inherent wave nature implies that the precisions with which the particle properties of position and momentum can be known are inversely proportional.

A photon scatters off of a free electron. (a) What is the maximum possible change in wavelength? (b) Suppose a photon scatters off of a free proton. What is the maximum possible change in wavelength now? (e) Which more clearly demonstrates the particle nature of electromagnetic radiation--collision with an electron or collision with a proton?

Photons from space are bombarding your laboratory and smashing massive objects to pieces! Your detectors indicate that two fragments each of mass m0 depart such a collision moving at 0.6c at 60o to the photon’s original direction of motion. In terms of m0 what are the energy of the cosmic ray photon and the massMof the particle being struck (assumed stationary initially).

An isolated atom can emit a photon and the atom's internal energy drops. In fact, the process has a name:spontaneous emission. Can an isolated electron emit a photon? Why or why not?

According to Wien's Law, the wavelengthλmax of maximum thermal emission of electromagnetic energy from a body of temperature Tobeys

localid="1660036169367" λmaxT=2.898×10-3m·K

Show that this law follows from the spectral energy density obtained in Exercise 13. Obtain an expression that, when solved, would yield the wavelength at which this function is maximum. The transcendental equation cannot be solved exactly, so it is enough to show thatlocalid="1660036173306" λ=2.898×10-3m·KTsolves it to a reasonable degree of precision.

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