ASSAM UNIVERSITY, SILCHAR
FYUG 3rd semester Physics DSC 201 Previous Year Question Papers Solutions
UNIT 4
2019
(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )
1.In Michelson's interferometer 100 fringes cross the field of view when the movable mirror is displaced through 0.02948 mm. Calculate the wavelength of monochromatic light used. (Mark:- 2)
Given:
For Michelson's interferometer,
Answer:
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2.Discuss the statement "A grating having higher dispersive power than another, does not necessarily has a higher resolving power". (Mark:- 2)
Dispersive power is the ability of a grating to separate different wavelengths.
Resolving power is the ability of a grating to distinguish two closely spaced spectral lines.
where N is the total number of illuminated lines.
A grating may have a large dispersive power because it has more lines per unit length. However, if only a small number of lines are illuminated, its resolving power may be low.
Similarly, a grating with lower dispersive power may have a larger illuminated width and hence a greater number of illuminated lines, giving higher resolving power.
Therefore, higher dispersive power does not necessarily imply higher resolving power because the two depend on different parameters.
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3. State the difference between the grating and the prism spectrum. (Mark:- 2)
| Grating Spectrum | Prism Spectrum |
|---|---|
| Produced by diffraction and interference. | Produced by refraction. |
| Red light deviates more than violet light. | Violet light deviates more than red light. |
| Spectrum is nearly uniformly distributed. | Spectrum is not uniformly distributed. |
| Several orders of spectra are obtained. | Only one spectrum is obtained. |
| Higher resolving power. | Lower resolving power. |
| Greater purity of spectrum. | Less pure spectrum. |
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4. Discuss the intensity distribution of Fabry-Perot interferometer fringes and the ratio of Imax to Imin. (Mark:- 4)
In a Fabry-Perot interferometer, multiple reflections occur between two partially reflecting parallel plates. The transmitted beams interfere and produce sharp circular fringes.
The intensity distribution is given by Airy's formula:
where
R is the reflectivity of the plates.
For maximum intensity,
For minimum intensity,
Hence,
5.Find the expression for the width of the central maximum in case of Fraunhofer diffraction pattern due to single slit. (Mark:- 4)
Fraunhofer diffraction is the diffraction phenomenon in which both the source of light and the screen are effectively at infinity. In practice, this condition is achieved by using two convex lenses. When monochromatic light passes through a narrow slit, diffraction takes place and a characteristic pattern consisting of a broad central bright band surrounded by alternate dark and bright bands is obtained on the screen.
Consider a slit AB of width a illuminated by monochromatic light of wavelength λ. A lens is used to focus the diffracted rays on a screen placed at its focal plane.
Let θ be the angle at which diffraction is observed. According to Huygens' principle, every point on the slit acts as a source of secondary wavelets. The resultant intensity at any point on the screen is obtained by superposition of these wavelets.
For a point corresponding to angle θ, the path difference between the extreme rays from the slit edges A and B is
The condition for minimum intensity (dark band) is obtained when the slit can be divided into equal parts whose contributions cancel each other.
Therefore, the condition for minima is
where
- n = 1, 2, 3, ...
- λ = Wavelength of light
- a = Width of the slit
For the first minimum,
Since θ is very small,
Hence,
The first minimum occurs at angle +θ on one side of the centre and at angle -θ on the other side.
The central maximum lies between these two first minima.
Therefore, the angular width of the central maximum is
Let D be the distance between the slit and the screen.
The linear distance of the first minimum from the central maximum is
For small θ,
Therefore,
Since the central maximum extends from the first minimum on one side to the first minimum on the other side, its linear width is
Thus, the width of the central maximum in Fraunhofer diffraction due to a single slit is
and the angular width of the central maximum is
From the expression, it is clear that the width of the central maximum is directly proportional to the wavelength of light and inversely proportional to the slit width. Hence, narrower slits produce broader diffraction patterns.
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6. What is a plane diffraction grating? In a plane transmission grating the angle of diffraction for 2nd order maxima for wavelength 5×10-5 cm is 30°. Calculate the number of lines in 1 cm of the grating surface.
A plane diffraction grating is an optical element consisting of a large number of parallel, equally spaced narrow slits ruled on a glass plate. It produces diffraction and interference of light and is used to obtain highly resolved spectra.
For a plane transmission grating, the grating equation is
where,
- n = Order of spectrum
- λ = Wavelength of light
- θ = Angle of diffraction
- (a + b) = Grating element
Given:
Using the grating equation,
Number of lines per cm,
Answer: The number of lines in 1 cm of the grating surface is
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7. Derive an expression for the resolving power of a plane transmission grating.
The resolving power of a diffraction grating is its ability to distinguish two spectral lines whose wavelengths differ by a very small amount.
If two wavelengths λ and λ + dλ are just resolved, then according to Rayleigh's criterion, the principal maximum of one wavelength coincides with the first minimum of the other.
The resolving power (R) of a grating is defined as
where dλ is the minimum difference between two wavelengths that can be resolved.
Consider a plane transmission grating having N illuminated lines and let the spectrum be observed in the nth order.
The grating equation is
Differentiating with respect to λ,
For principal maxima, the path difference between light from adjacent slits is
The angular separation between two nearby wavelengths λ and λ + dλ is obtained from the above relation.
According to Rayleigh's criterion, two wavelengths are just resolved when the principal maximum of one coincides with the first minimum of the other.
For a grating having N illuminated slits, the condition for the first minimum near the principal maximum is
The angular half-width of the principal maximum is therefore inversely proportional to N.
Using Rayleigh's criterion, the condition for just resolution becomes
Therefore,
Hence, the resolving power of a plane transmission grating is
where
- n = order of the spectrum
- N = total number of illuminated grating lines
Thus, the resolving power of a grating increases with:
- Increase in the order of spectrum (n).
- Increase in the number of illuminated lines (N).
Therefore, a grating with a large number of ruled lines and operated in higher orders possesses a high resolving power and can distinguish very closely spaced spectral lines.
Final Result:
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2022
(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )
8. What is Fraunhofer and Fresnel class of diffraction?
Fresnel Diffraction:
The diffraction that occurs when the source of light and the screen are at finite distances from the diffracting aperture or obstacle is called Fresnel diffraction.
- The incident wavefront is spherical.
- No lenses are required.
- The diffraction pattern changes with the distance between the aperture and the screen.
- Example: Diffraction at a straight edge.
Fraunhofer Diffraction:
The diffraction that occurs when the source and the screen are effectively at infinite distances from the diffracting aperture is called Fraunhofer diffraction.
- The incident wavefront is plane.
- Convex lenses are used to obtain parallel rays.
- The diffraction pattern is independent of the distance between the aperture and the screen.
- Example: Single slit Fraunhofer diffraction pattern.
Thus, Fresnel diffraction is observed with source and screen at finite distances, whereas Fraunhofer diffraction is observed with source and screen effectively at infinite distances using lenses.
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9. Mention the differences between interference and diffraction pattern.
| Interference Pattern | Diffraction Pattern |
|---|---|
| Produced by superposition of light from two or more coherent sources. | Produced by superposition of wavelets from different parts of the same wavefront. |
| Fringes are equally spaced. | Fringes are unequally spaced. |
| Bright fringes have nearly equal intensity. | Central maximum is brightest and widest. |
| All fringes have nearly equal width. | Central maximum is about twice the width of secondary maxima. |
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10.What is Rayleigh's criterion for resolution?
According to Rayleigh's criterion, two closely spaced objects are said to be just resolved when the principal maximum of one diffraction pattern coincides with the first minimum of the other.
This criterion gives the limit of resolution of an optical instrument.
13. What are Fresnel integrals?
Here, C(u) and S(u) are called the Fresnel cosine integral and Fresnel sine integral respectively.
These integrals are extensively used in determining the intensity distribution in Fresnel diffraction phenomena.
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11. Explain how Michelson interferometer can be used to measure (a) the wavelength of monochromatic light and (b) the difference in wavelengths between the D-lines of sodium light.
(a) Measurement of Wavelength of Monochromatic Light
Michelson interferometer is based on the principle of interference produced by dividing a beam of light into two parts and then recombining them after travelling different optical paths.
A monochromatic light source is allowed to fall on a semi-silvered glass plate G. The incident beam is divided into two parts.
- One beam is reflected towards mirror M1.
- The other beam is transmitted towards mirror M2.
After reflection from the mirrors, the two beams return and recombine, producing interference fringes.
Suppose the movable mirror M1 is displaced through a distance d.
Since the light travels to the mirror and back, the optical path difference changes by
If N fringes cross the field of view during this displacement, then the change in path difference is equal to N wavelengths.
Therefore,
Thus, by measuring the displacement d of the movable mirror and counting the number of fringes N crossing the field of view, the wavelength of monochromatic light can be determined.
(b) Measurement of Difference in Wavelengths Between the Sodium D-Lines
Sodium light consists mainly of two closely spaced wavelengths:
Each wavelength produces its own interference pattern. Since the wavelengths are slightly different, the two fringe systems do not exactly coincide.
As the movable mirror is displaced, the visibility of the fringes changes periodically.
- At certain positions, the fringes are very distinct (maximum visibility).
- At other positions, the fringes disappear (minimum visibility).
Let d be the displacement of the mirror between two successive positions of maximum visibility.
For successive maxima of visibility,
Since the two wavelengths are very close,
Hence,
Thus, the difference between the wavelengths of the sodium D-lines is obtained by measuring the distance through which the mirror moves between two successive positions of maximum fringe visibility.
Results
For wavelength measurement:
For difference between sodium D-lines:
Hence Michelson interferometer provides a highly accurate method for measuring both the wavelength of monochromatic light and the small wavelength difference between the sodium D-lines.
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12. What is meant by resolving power and limit of resolution of an optical instrument? Find the expression for resolving power of a diffraction grating.
Resolving Power and Limit of Resolution of an Optical Instrument
Resolving Power: The resolving power of an optical instrument is its ability to distinguish two closely spaced objects or spectral lines as separate.
Resolving Power = 1 / (Limit of Resolution)
Limit of Resolution: The limit of resolution is the minimum separation between two objects (or wavelengths) that can be just distinguished by the instrument.
A smaller limit of resolution means a greater resolving power.
Resolving Power of a Diffraction Grating
Let,
- m = Order of diffraction
- N = Total number of illuminated grating lines
- λ = Wavelength of light
- dλ = Smallest difference in wavelength that can be resolved
According to Rayleigh's criterion, two wavelengths λ and (λ + dλ) are just resolved when the principal maximum of one coincides with the first minimum of the other.
Grating equation:
(a + b) sinθ = mλ
Differentiating,
(a + b) cosθ dθ = m dλ
For a grating, the angular half-width of the principal maximum is
dθ = λ / [N(a + b) cosθ]
Substituting this value in the previous equation,
λ / [N(a + b) cosθ] = m dλ / [(a + b) cosθ]
Therefore,
λ = mN dλ
Hence, the resolving power of a diffraction grating is
R = λ/dλ = mN
Result
Resolving Power of a Diffraction Grating:
R = λ/dλ = mN
where,
R = Resolving Power
m = Order of Spectrum
N = Total Number of Illuminated Grating Lines
Thus, the resolving power increases with both the order of the spectrum and the number of illuminated lines on the grating.
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2023
(FYUG PHY-DSC201 Same as CBCS PHҮНСС–202T )
14. What is meant by the visibility of fringes in Michelson Interferometer?
The visibility of fringes is a measure of the contrast between the bright and dark fringes in an interference pattern. It is defined as
V = (Imax − Imin)/(Imax + Imin)
where Imax is the intensity of the bright fringe and Imin is the intensity of the dark fringe. The value of visibility ranges from 0 to 1. A value of 1 indicates perfectly distinct fringes.
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15. Name the devices or optical elements used to produce interference and diffraction.
Interference:
- Young's Double Slit
- Fresnel Biprism
- Lloyd's Mirror
- Michelson Interferometer
- Fabry-Perot Interferometer
Diffraction:
- Single Slit
- Double Slit
- Diffraction Grating
- Zone Plate
- Circular Aperture
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16. How is the resolving power of an optical instrument related to the aperture of the lens?
The resolving power of an optical instrument is directly proportional to the aperture of its lens or objective.
A larger aperture produces a smaller diffraction pattern and enables the instrument to distinguish finer details.
Therefore,
Resolving Power ∝ Aperture
Hence, increasing the aperture of the lens increases the resolving power of the optical instrument.
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17. Give expression for the angular position of the first minimum in terms of width of the slit and wavelength of light.
Angular Position of the First Minimum in Single-Slit Fraunhofer Diffraction
For a single slit of width a, the condition for minima in the Fraunhofer diffraction pattern is
a sinθ = nλ
where,
- a = width of the slit
- θ = angular position of the minimum
- λ = wavelength of light
- n = 1, 2, 3, ...
For the first minimum, n = 1.
Therefore,
a sinθ₁ = λ
Hence, the angular position of the first minimum is
sinθ₁ = λ/a
For small diffraction angles,
θ₁ ≈ λ/a (in radians)
Thus, the angular position of the first minimum is directly proportional to the wavelength of light and inversely proportional to the slit width.
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18. Discuss the 'construction and theory of spectrum formed by a diffraction grating
Construction and Theory of Spectrum Formed by a Diffraction Grating
Introduction
A diffraction grating is an optical device used to disperse light into its constituent wavelengths. It consists of a very large number of equally spaced parallel slits ruled on a glass plate or a reflecting surface. Due to the combined effects of diffraction and interference, a grating produces sharp and well-defined spectra.
Construction of a Diffraction Grating
A plane glass plate is ruled with a large number of fine, parallel and equidistant lines by a ruling engine. The ruled portions are opaque, while the spaces between them are transparent and act as slits.
The distance between the centers of two successive slits is called the grating element and is denoted by
(a + b)
where,
- a = width of the transparent slit
- b = width of the opaque spacing
If N lines are ruled per unit length, then
(a + b) = 1/N
Theory of Formation of Spectrum
Consider monochromatic light of wavelength λ incident normally on the grating. Each slit acts as a source of secondary wavelets according to Huygens' principle.
The diffracted light from all the slits interferes with one another. Constructive interference occurs only in certain directions where the path difference between waves from adjacent slits is an integral multiple of the wavelength.
If θ is the angle of diffraction, then the path difference between waves from adjacent slits is
(a + b) sinθ
For principal maxima,
(a + b) sinθ = nλ
where,
- n = 0, 1, 2, 3, ... (order of spectrum)
- λ = wavelength of light
This is known as the grating equation.
Formation of Spectra with White Light
When white light is incident on the grating, each wavelength satisfies the grating equation at a different angle.
For n = 0,
θ = 0
All wavelengths overlap and form the central white image called the zero-order spectrum.
For n = 1, 2, 3, ...
Different wavelengths are diffracted at different angles and separate from one another, producing first-order, second-order and higher-order spectra on both sides of the central image.
Since the angle of diffraction increases with wavelength, red light is deviated more than violet light.
Thus, the order of colours in the spectrum is:
Violet → Indigo → Blue → Green → Yellow → Orange → Red
Characteristics of Grating Spectrum
- The central image is white and undeviated.
- Spectra are formed symmetrically on both sides of the central image.
- Higher-order spectra are more dispersed.
- The spectral lines are sharp and well resolved.
- Red light is deviated more than violet light.
- The grating possesses high resolving and dispersive power.
Advantages of Diffraction Grating
- Produces pure and sharp spectra.
- Provides greater resolving power than a prism.
- Gives accurate wavelength measurements.
- Produces spectra of higher orders.
Conclusion
A diffraction grating consists of a large number of equally spaced parallel slits. The spectrum is formed due to the interference of diffracted light from these slits. The condition for principal maxima is
(a + b) sinθ = nλ
and the separation of different wavelengths gives rise to highly resolved spectra, making diffraction gratings important instruments in spectroscopy.
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2024
(FYUG PHY-DSC201 )
20. Explain visibility of fringes ?
21. Differentiate between Fresnel diffraction's and fraunhofer's diffraction.
22. What do you mean by resolving power of an optical instrument? Write the formula for resolving power of a diffraction grating.
23. What is the principle of michelson interferometer for formation of fringes? Explain how Michelson interferometer be used to determine the wavelength of source . If the movable mirror in Michelson interferometer is
shifted through 0.0575 mm parallel to its self; 180 number of fringes are found to shift pass a reference mark , find the wavelength of light used.
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24. Discuss the theory of a diffraction grating to explain how you world use it
to determine the wavelenght of light. Using revelant formula, explain how the resolving
power of a grating is different from resolving power of a telescope.
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