Engineering Physics Questions and Answers – Fraunhofer Diffraction

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This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Fraunhofer Diffraction”.

1. What is the effective distance between the source of light and the screen in Fraunhofer Diffraction?
a) Focal length of the convex lens
b) Less than Focal Length of the convex lens
c) Greater than the focal length of the convex lens and less than infinite
d) Infinite
View Answer

Answer: d
Explanation: In Fraunhofer Diffraction, the source of light and the screen are effectively placed at infinite distance. Two convex lenses are used for achieving such a condition. Thus, the incident waveform is plane and the secondary wavelets are in the same phase at every point in the plane of the aperture.
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2. What happens with the Fraunhofer single slit diffraction pattern if the whole apparatus is immersed in water?
a) The Wavelength of light increases
b) Width of central maximum increases
c) Width of central maximum decreases
d) Frequency of light decreases
View Answer

Answer: c
Explanation: As the whole apparatus is now immersed in water, the wavelength of the light will change.
\(\lambda^{‘}=\frac{\lambda}{\mu}\)
Therefore, as the refractive index of water is greater than the air, the wavelength of light will decrease.
Width of central maxima = \(\frac{2λ}{a}\)
Therefore, as the wavelength decreases, the width of the central maxima decreases.

3. How shall the pattern change when white light is used in Fraunhofer Diffraction instead of monochromatic light?
a) The pattern will no longer be visible
b) The shape of the pattern will change from hyperbolic to circular
c) The colored pattern will be observed with a white bright fringe at the center
d) The bright and dark fringes will change position
View Answer

Answer: c
Explanation: When white light is used instead of monochromatic light in Fraunhofer Diffraction, then the central maximum remains white as all seven wavelengths meet there in the same phase. The first minimum and second maximum will be formed by violet color due to its shortest wavelength while the last is due to the red color as it has the longest wavelength. Thus, a colored pattern is observed.
However, after the first few colored bands, the clarity of the band is lost, due to overlapping.

4. If the separation between the two slits in Double Slit Fraunhofer Diffraction is changed, what change will be observed in the diffraction pattern?
a) The fringe length will increase
b) The fringe length will decrease
c) Fringes will be colored
d) No change
View Answer

Answer: d
Explanation: The separation between the two slits only affects the interference pattern in Double Slit Fraunhofer Diffraction. The diffraction pattern does not change.
For Diffraction, e sin θ = ±mλ
Where e = Width of slits
m = Any integer
For interference, (e + d) sin θ = ±nλ
Where e = Width of slits
d = Separation between the two slits
m = Any integer
Hence, there is no change in the Diffraction Pattern.

5. In Double Slit Fraunhofer Diffraction, some orders of interference pattern are missing. It is called _____
a) Missing Spectra
b) Absent Spectra
c) End Spectra
d) Emission Spectra
View Answer

Answer: b
Explanation: In Double Slit Fraunhofer Diffraction, there are certain angles where the interference maxima and Diffraction minima overlap. These orders of interference pattern are missing in the pattern. It is known as Absent Spectra.
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6. A screen is placed 2m away from the lens to obtain the diffraction pattern in the focal plane of the lens in a single slit diffraction experiment. What will be the slit width if the first minimum lies 5 mm on either side of the central maximum when plane light waves of wavelength 4000 Å are incident on the slit?
a) 0.16 mm
b) 0.26 mm
c) 0.36 mm
d) 0.46 mm
View Answer

Answer: a
Explanation: Given: f = 2 m, x = 5 X 10-3m, λ = 4 X 10-7m, n=1
sin θ = \(\frac{nλ}{a}\), we have
a = \(\frac{n\lambda}{sin⁡\theta} \)
= 1.6 X 10-4 m
= 0.16 mm.

7. Find the missing order for a double-slit Fraunhofer Diffraction pattern if the slit widths are 0.2 mm separated by 0.6 mm.
a) 1st, 5th, 9th, ….
b) 2nd, 6th, 10th, …
c) 3rd, 7th, 11th, ….
d) 4th, 8th, 12th, …
View Answer

Answer: d
Explanation: We know, for interference maxima (e + d) sin θ = ±nλ and diffraction minima e sin θ = ±mλ.
Therefore, \(\frac{a+b}{a}=\frac{n}{m} \)
\(\frac{0.2+0.6}{0.2}=\frac{n}{m}\)
n = 4 m
As m = 1, 2, 3, …. Therefore, 4th, 8th, 12th, …. order interference maximum will be missing.

8. The following pattern is observed for which experiment?
engineering-physics-questions-answers-fraunhofer-diffraction-q8
a) Fabry-Perot Interferometer
b) Double Slit Fraunhofer Diffraction
c) Single Slit Fraunhofer Diffraction
d) Fresnel Diffraction
View Answer

Answer: b
Explanation: In Fraunhofer double slit diffraction, interference and diffraction operate simultaneously. They give their own respective pattern on the screen as shown in the figure.
In the figure, the outer pattern is the pattern obtained by diffraction while the pattern enclosed within it (the one with higher frequency) is obtained by the interference of light.

9. What is the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of a slit of width 12 X 10-7m when the slit is illuminated by monochromatic light of wavelength 6000 Å.
a) 20°
b) 30°
c) 40°
d) 50°
View Answer

Answer: b
Explanation: We know sin θ =\(\frac{\lambda}{a}\), where θ is the half angular width of the central maximum.
Given: a = 12 X 10-7 m, λ = 6000 Å = 6.0 X 10-7 m
sinθ = \(\frac{\lambda}{a}\) = 0.5 = 30°
Thus, the half angular width of the central bright maximum is 30°.

10. The angular separation between the central maximum and first order minimum of the diffractions pattern due to a single slit of width 0.25 mm, when light of wavelength 5890 Å is incident normally on the slit, is ________
a) 7.1 minute
b) 8.1 minute
c) 9.1 minute
d) 10.1 minute
View Answer

Answer: b
Explanation: a = 0.25 mm = 0.25 X 10-3m, λ = 5890 Å = 5.89 X 10-7 m.
Angular separation between the central maximum and the first order minimum is given by:
sin θ = \(\frac{\lambda}{a}\) = 0.02356
sin θ ≈ θ
θ = 0.02356 radian
θ = 8.1 minute.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn