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A diffraction grating has resolving power R=λavgΔλ=Nm. (a) Show that the corresponding frequency range f that can just be resolved is given by ∆f=c±·³¾Î». (b) From Fig. 36-22, show that the times required for light to travel along the ray at the bottom of the figure and the ray at the top differ by ∆t=(NdC)sinθ. (c) Show that (Δ´Ú)(Δ³Ù), this relation being independent of the various grating parameters. Assume N≫1.

Short Answer

Expert verified

(a) it is proved that corresponding frequency range f that can just be resolved is given byΔf=cNmλ

(b) The lens does not affect the travel time, so it is proved the times required for light to travel along the ray at the bottom of the figure and the ray at the top differ by Δt=Ndcsinθ.

(c) It is proved that the multiplication of frequency and time change is constant so, it is independent of diffraction grating parameters

Step by step solution

01

Identification of given data

The given data can be listed below,

  • The diffracting grating is,R=λavgΔλ=Nm
  • The frequency range is,Δf=cNmλ
02

Concept/Significance of diffraction grating.

A diffraction grating is an optical element with a prism-like effect that divides white light into its individual hues. To divide light into portions based on wavelength, a diffraction grating functions similarly to a prism. The incident light angle is distorted by its tiny, typically periodic characteristics.

03

The corresponding frequency range

(a)

The wavelength range that can only be resolved in order m is given by,

Δλ=λNm …â¶Ä¦â¶Ä¦(¾±)

Here, N is the grating's number of rulings, andλ stands for the average wavelength.

The relationship between frequency and wavelength is given by,

f=cλ ……..… (ii)

Here, c is the speed of light,

Consequently,

fΔλ+λΔf=0 …â¶Ä¦â¶Ä¦.(¾±¾±¾±)

The change in the wavelength is given by,

Δλ=-λfΔf=-λ2cΔf  .......... iv

Equate equation (i) and (iv) change in frequency is given by,

λNm=-λ2cΔfΔf=cλNm

Thus, it is proved that corresponding frequency range f that can just be resolved is given byΔf=cNmλ.

04

Time required for light to travel along the ray

(b)

When waves are going along the two extreme rays, the difference in travel time is given by,

Δt=ΔLc

Here, L is the route length difference.

The path difference is given by

ΔL=N-1dsinθ,

Since the waves begin at slits that are separated by (N - 1)d.

Substitute all the values in the above and N is large, the time difference is given by

N-1dsinθ=cΔtΔt=N-1dsinθc

If the Nis large the time difference is given by,

Δt=Ndcsinθ

Thus, the lens does not affect the travel time, so it is proved the times required for light to travel along the ray at the bottom of the figure and the ray at the top differ by

Δt=Ndcsinθ

05

The relation (Δf)(Δt) being independent of the various grating parameters

(c)

Brag’s law of diffraction is given by,

dsinθ=mλ

Here, dis the intermolecular distance,λ is the wavelength.

The change in frequency and time is given by,

Δf=cλNm

The time difference is given by,

Δt=Ndcsinθ

The product of the frequency and time change is given by,

ΔfΔt=cλNmNdcsinθ=dsinθmλ=mλmλ=1

Thus, it is proved that the multiplication of frequency and time change is constant so, it is independent of diffraction grating parameters.

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Most popular questions from this chapter

(a) Figure 36-34a shows the lines produced by diffraction gratingsA and B using light of the same wavelength; the lines are of the same order and appear at the same angles θ. Which grating has the greater number of rulings? (b) Figure 36-34b shows lines of two orders produced by a single diffraction grating using light of two wavelengths, both in the red region of the spectrum. Which lines, the left pair or right pair, are in order with greater m? Is the center of the diffraction pattern located to the left or to the right in(c) Fig. 36-34a andd) Fig. 36-34b?

A diffraction grating 3.00 cm wide produces the second order at 33.0° with light of wavelength 600 nm. What is the total number of lines on the grating?

A diffraction grating having is illuminated with a light signal containing only two wavelengths and . The signal in incident perpendicularly on the grating. (a) What is the angular separation between the second order maxima of these two wavelengths? (b) What is the smallest angle at which two of the resulting maxima are superimposed? (c) What is the highest order for which maxima of both wavelengths are present in the diffraction pattern?

(a) For a given diffraction grating, does the smallest difference Δλin two wavelengths that can be resolved increase, decrease, or remain the same as the wavelength increases? (b) For a given wavelength region (say, around 500 nm), is Δλ greater in the first order or in the third order?

If first-order reflection occurs in a crystal at Bragg angle3.4°, at what Bragg angle does second-order reflection occur from the same family of reflecting planes?

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