Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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The document Irodov Solutions: Diffraction of Light- 3 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q.138. Using a diffraction grating as an example, demonstrate that the frequency difference of two maxima resolved according to Rayleigh's criterion is equal to the reciprocal of the difference of propagation times of the extreme interfering oscillations, i.e. δv = = 1/δt.

Ans. For the just resolved waves the frequency difference

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

since N d sinθ is the path difference between waves emitted by the extremities o f the grating.

 

Q.139. Light composed of two spectral lines with wavelengths 600.000 and 600.050 nm falls normally on a diffraction grating 10.0 mm wide. At a certain diffraction angle θ these lines are close to being resolved (according to Rayleigh's criterion). Find θ.

Ans.

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.140. Light falls normally on a transparent diffraction grating of width l = 6.5 cm with 200 lines per millimetre. The spectrum under investigation includes a spectral line with λ = 670.8 nm consisting of two components differing by δλ = 0.015 nm. Find: (a) in what order of the spectrum these components will be resolved; (b) the least difference of wavelengths that can be resolved by this grating in a wavelength region λ ≈ 670 nm.

Ans. We see that

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.141. With light falling normally on a transparent diffraction grating 10 mm wide, it was found that the components of the yellow line of sodium (589.0 and 589.6 nm) are resolved beginning with the fifth order of the spectrum. Evaluate:
 (a) the period of this grating;
 (b) what must be the width of the grating with the same period for a doublet λ = 460.0 nm whose components differ by 0.13 nm to be resolved in the third order of the spectrum.

Ans. Here

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

so

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

(b) To resolve a doublet with  Irodov Solutions: Diffraction of Light- 3 Notes | EduRev in the third order we must have

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

This means that the grating is

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

wide = 6cm wide.

 

 

Q.142. A transparent diffraction grating of a quartz spectrograph is 25 mm wide and has 250 lines per millimetre. The focal length of an objective in whose focal plane a photographic plate is located is equal to 80 cm. Light falls on the grating at right angles. The spectrum under investigation includes a doublet with components of wavelengths 310.154 and 310.184 nm. Determine:
 (a) the distances on the photographic plate between the components of this doublet in the spectra of the first and the second order;
 (b) whether these components will be resolved in these orders of the spectrum.

Ans.

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

we get   Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.143. The ultimate resolving power λδλ of the spectrograph's trihedral prism is determined by diffraction of light at the prism edges (as in the case of a slit). When the prism is oriented to the least deviation angle in accordance with Rayleigh's criterion,

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

where b is the width of the prism's base (Fig. 5.28), and dnIdk is the dispersion of its material. Derive this formula.

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Ans. Suppose the incident light consists of two wavelengths λ and λ + δλ which are just resolved by the prism. Then by Rayleigh’s criterion, the maximum of the line of wavelength λ must coincide with the first minimum of the line of wavelength λ + δλ, Let us write both conditions in terms of the optical path differences for the extreme rays : For the light of wavelength λ

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.144. A spectrograph's trihedral prism is manufactured from glass whose refractive index varies with wavelength as n = A + B/λ2  where A and B are constants, with B being equal to 0.010 1μm2. Making use of the formula from the foregoing problem, find:
 (a) how the resolving power of the prism depends on k; calculate the value of λδλ in the vicinity of λ1 = 434 nm and λ2 = 656 nm if the width of the prism's base is b = 5.0 cm;
 (b) the width of the prism's base capable of resolving the yellow doublet of sodium (589.0 and 589.6 nm).

Ans.

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

(b) To resolve the D-lines we require

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.145. How wide is the base of a trihedral prism which has the same resolving power as a diffraction grating with 10 000 lines in the second order of the spectrum if Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Ans.

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

Q.146. There is a telescope whose objective has a diameter D = 5.0 cm. Find the resolving power of the objective and the minimum separation between two points at a distance l = 3.0 km from the telescope, which it can resolve (assume λ = 0.55 μm). 

Ans. Resolving power of the objective

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Let Irodov Solutions: Diffraction of Light- 3 Notes | EduRev be the minimum distance between two points at a distance of 3.0 km which the telescope can resolve. Then

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.147. Calculate the minimum separation between two points on the Moon which can be resolved by a reflecting telescope with mirror diameter 5 m. The wavelength of light is assumed to be equal to λ = 0.55 μm.

Ans. The limit of resolution of a reflecting telescope is determined by diffraction from the mirror and obeys a formula similar to that from a refracting telescope. The limit of resolution is

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.148. Determine the minimum multiplication of a telescope with diameter of objective D = 5.0 cm with which the resolving power of the objective is totally employed if the diameter of the eye's pupil is do  = 4.0 mm.

Ans. By definition, the magnification

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

where d0 = diameter of the pupil

Thus to avail of the resolution offered by the telescope we must have

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.149. There is a microscope whose objective's numerical aperture is sinα = 0.24, where α is the half-angle subtended by the objective's rim. Find the minimum separation resolved by this microscope when an object is illuminated by light with wavelength λ = 0.55 p,m.

Ans. 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Let A and B be two points in the field of a microscope which is represented by the lens C D. Let A' ,2T be their image points which are at equal distances from the axis of the lens CD. Then all paths from A to A' are equal and the extreme difference of paths from A to B' is equal to

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

From the theory o f diffraction by circular apertures this distance must be equal to 1.22 λ

when B’ coincides with the minimum o f the diffraction due to A and A' with the minimum o f the diffraction due to B . Thus

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Here 2α is the angle subtended by the objective of the microscope at the object.
Substituting the values

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

Q.150. Find the minimum magnification of a microscope, whose objective's numerical aperture is sinα = 0.24, at which the resolving power of the objective is totally employed if the diameter of the eye's pupil is do  = 4.0 mm.

Ans. Suppose dmn - minimum separation resolved by the microscope ψ = angle subtended at the eye by this object when the object is at the least distance of distinct v ision l0 ( = 2 5 cm ) .

ψ' minimum angular separation resolved by the eye = Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Now

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Thus Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.151. A beam of X-rays with wavelength λ falls at a glancing angle 60.0° on a linear chain of scattering centres with period a. Find the angles of incidence corres- ponding to all diffraction maxima if λ = 2a/5.

Ans.   Path difference 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

and we get

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Other values o f k are not allowed as they lead toIrodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

Q.152. A beam of X-rays with wavelength λ = 40 μm falls normally on a plane rectangular array of scattering centres and produces a system of diffraction maxima (Fig. 5.29) on a plane screen re- moved from the array by a distance l = 10 cm. Find the array periods a and b along the x and y axes if the distances between symmetrically located maxima of second order are equal to Δx = 60 mm (along the x axis) and Δy = 40 mm (along the y axis).

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Ans. We give here a simple derivation of the condition for diffraction maxima, known as Laue equations. It is easy to see form the above figure that the path difference between waves scattered by nearby scattering centres Pand P2 is 

 

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

maxima this path difference must be an integer multiple of λ, for any two neighbouring atoms. In the present case of two dimensional lattice with X - rays incident normallyIrodov Solutions: Diffraction of Light- 3 Notes | EduRev Taking successively nearest neighbours in the x - & y - directions

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Here cosα and cosβ are the direction cosines o f the ray with respect to the x & y axes o f the two dimensional crystal.

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.153. A beam of X-rays impinges on a three-dimensional rectangular array whose periods are a, b, and c. The direction of the incident beam coincides with the direction along which the array period is equal to a. Find the directions to the diffraction maxima and the wavelengths at which these maxima will be observed. 

 

Ans. Suppose α, β, and γ are the angles between the direction to the diffraction maximum and the directions of the array along the periods a, b and c respectively ( call them x, y, & z axes). Then the value of these angles can be found from the following familiar conditions

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

(These formulas are, in effect, Laue equations, see any text book on modem physics). Squaring and adding we g et on using Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Thus Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

Q.154. A narrow beam of X-rays impinges on the natural facet of a NaCl single crystal, whose density is p = 2.16 g/cm3  at a glancing angle θ = 60.0°. The mirror reflection from this facet produces a maximum of second order. Find the wavelength of radiation.

Ans. The unit cell of NaCl is shown below. In an infinite crystal, there are four Irodov Solutions: Diffraction of Light- 3 Notes | EduRev and four Cl- ions per unit cell. (Each ion on the middle of the edge is shared by four unit cells; each ion on the face centre by two unit cells, the ion in the middle of the cell by one cell only and finally each ion on the comer by eight unit cells.) Thus

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.155. A beam of X-rays with wavelength λ = 174 pm falls on the surface of a single crystal rotating about its axis which is parallel to its surface and perpendicular to the direction of the incident beam. In this case the directions to the maxima of second and third order from the system of planes parallel to the surface of the single crystal form an angle θ = 60° between them. Find the corresponding interplanar distance.

Ans. When the crystal is rotated, the incident monochromatic beam is diffracted from a given crystal plane of interplanar spacing d whenever in the coursVof rotation the value of θ satisfies the Bragg equation.

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

 

 

Q.156. On transmitting a beam of X-rays with wavelength λ = 17.8 pm through a polycrystalline specimen a system of diffraction rings is produced on a screen located at a distance l = 15 cm from the specimen. Determine the radius of the bright ring corresponding to second order of reflection from the system of planes with interplanar distance d = 155 pm.

Ans. In a polycrystalline specimen, microcrystals are oriented at various angles with respect to one another. The microcrystals which are oriented at certain special angles with respect to the incident beam produce diffraction maxima that appear as rings.
The radial of these rings are given by

Irodov Solutions: Diffraction of Light- 3 Notes | EduRev

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