FIGURE Q33.5 shows the light intensity on a viewing screen behind a single slit of width a The light's wavelength isλ. Isλ<a,λ=a,λ>a, or is it not possible to tell? Explain

Short Answer

Expert verified

a) The light wavelength is remains the same

b)The lambda less than a is remains the same

c) The lambda equal to a is decrease

d)The lambda greater than a is increase

Step by step solution

01

The light wavelength(part a)

We will use some of the expressions from Young's Double-Slit Experiment chapter in the book in this problem.

Slits are replaced with slits, and the spacing d between adjacent slits remains constant.

Expression from the angles of bright fringes:

θm=m×λd

For constant angular positions of the bright fringes in the interference pattern and spacing d, m represents the number of fringes that will remain constant.

02

Find fringe spacing(part b)

Fringe spacing expression:

Fringe spacing is independent of the number of slits, and for constant spacing d between slits, fringe spacing remains constant.

Δy=λ×Ld

03

 step3: Find λ=a (part c) 

We don't need a physics expression to assume that as the number of slits increases from to, the width of each fringe decreases while the spacing d remains constant.

04

Findλ>a(part d)

d) Intensity from expression:

Imax=N2×I1

Because brightness intensity is proportional to the square of the amplitude N but not to the number of slits, the brightness of each fringe increases for constant spacing d.

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Most popular questions from this chapter

FIGURE shows two nearly overlapped intensity peaks of the sort you might produce with a diffraction grating . As a practical matter, two peaks can just barely be resolved if their spacing yequals the width w of each peak, where wis measured at half of the peak’s height. Two peaks closer together than wwill merge into a single peak. We can use this idea to understand the resolution of a diffraction grating.

a. In the small-angle approximation, the position of the m=1peak of a diffraction grating falls at the same location as the m=1fringe of a double slit: y1=λL/d. Suppose two wavelengths differing by lpass through a grating at the same time. Find an expression for localid="1649086237242" y, the separation of their first-order peaks.

b. We noted that the widths of the bright fringes are proportional to localid="1649086301255" 1/N, where localid="1649086311478" Nis the number of slits in the grating. Let’s hypothesize that the fringe width is localid="1649086321711" w=y1/NShow that this is true for the double-slit pattern. We’ll then assume it to be true as localid="1649086339026" Nincreases.

c. Use your results from parts a and b together with the idea that localid="1649086329574" Δymin=wto find an expression for localid="1649086347645" Δλmin, the minimum wavelength separation (in first order) for which the diffraction fringes can barely be resolved.

d. Ordinary hydrogen atoms emit red light with a wavelength of localid="1649086355936" 656.45nm.In deuterium, which is a “heavy” isotope of hydrogen, the wavelength is localid="1649086363764" 656.27nm.What is the minimum number of slits in a diffraction grating that can barely resolve these two wavelengths in the first-order diffraction pattern?

If sunlight shines straight onto a peacock feather, the feather appears bright blue when viewed from15on either side of the incident beam of light. The blue color is due to diffraction from parallel rods of melanin in the feather barbules, as was shown in the photograph on page 940. Other wavelengths in the incident light are diffracted at different angles, leaving only the blue light to be seen. The average wavelength of blue light is 470nm. Assuming this to be the first-order diffraction, what is the spacing of the melanin rods in the feather?

Light of630nmwavelength illuminates a single slit of width 0.15mm. FIGURE EX33.17shows the intensity pattern seen on a screen behind the slit. What is the distance to the screen?

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?

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?

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