Engineering Physics Questions and Answers – Fresnel and Fraunhofer Diffraction

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

1. How many lenses are used in Fraunhofer Diffraction?
a) Two Convex lenses
b) Two Concave lenses
c) One Convex lens
d) No lens used

Explanation: In Fraunhofer Diffraction, two convex lenses are used. One convex lens renders the incident rays parallel and the other focuses the diffracted ray on the screen.

2. 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

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.

3. In Fresnel diffraction, the relative phase difference between the curved wavefront is ___________
a) Constant
b) Zero
c) Linearly increasing
d) Non-constant

Explanation: Since the radii of each half period zone are different, the distance traveled by each wavefront is different. Thus, the relative phase difference turns out to be non-constant.

4. In Fresnel Diffraction, the incident wavefront is _________
a) Hyperbolic
b) Linear
c) Spherical
d) Elliptical

Explanation: In Fresnel Diffraction, the interference takes place between the light waves reaching a point from different parts of the same wavefront. Thus, the incident wavefront is spherical or cylindrical.

5. The radius of the half period zone is proportional to __________
a) The wavelength of light
b) The square root of the frequency of light
c) The square root of the wavelength light
d) The frequency of light

Explanation: We know that the formula for the radius of half period zone = $$\sqrt{nb\lambda}$$, where n is a natural number. Thus, it is proportional to the square root of wavelength light and inversely proportional to the square root of the frequency of light.
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6. 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

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.

7. Light of 5000 Å is incident on a circular hole of radius 1 cm. How many half period zones are contained in the circle if the screen is placed at a distance of 1 m?
a) 20
b) 200
c) 2000
d) 20000

Explanation: In this case, λ = 5000 Å = 5 X 10-5 cm, b = 1 m = 100 cm
Therefore, Number of half period zones = $$\frac{1}{λ}$$
= 1/5 X 10-5
= 20000.

8. Light of 6000 Å is incident on a circular hole and is received on a screen 50 cm away. What is the radius of the hole, if the intensity of light on the screen is 4 times the intensity without the hole?
a) 0.025 cm
b) 0.047 cm
c) 0.054 cm
d) 0.089 cm

Explanation: The intensity will be 4 times than in its absence if the radius of the hole is equal to that of the first half period zone.
Therefore, radius, r = $$\sqrt{b\lambda}$$
Here, b = 50 cm and λ = 6000 Å = 6 X 10-5 cm
r = 0.0548cm.

9. The zone plate behaves like a ___________
a) Concave Lens with multiple foci
b) Convex Lens with multiple foci
c) Convex Lens with single foci
d) Concave Lens with single foci

Explanation: In a zone plate, a much brighter image of an object is obtained at the screen, which shows the converging action of a zone plate. Also, it’s equation resembles that of a lens. Thus, the zone plate behaves like a convex lens with multiple foci.

10. 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, …

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

11. Which graph is shown in the figure?

a) Amplitude variation with the number of exposed zones
b) Intensity variation with the number of exposed zones
c) Frequency variation with the number of exposed zones
d) Phase variation with the number of exposed zones

Explanation: The figure shows the variation of intensity with the number of exposed zones. We know, the resultant intensity at O is $$\frac{m_1^2}{4}$$. Only the first half period zone is effective in producing the illumination at O.

12. 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

Explanation: Given: f = 2 m, x = 5 X 10-3 m, λ = 4 X 10-7 m, n=1
sin θ = $$\frac{n\lambda}{a}$$, we have
a = $$\frac{n\lambda}{sin⁡ \theta}$$
= 1.6 X 10-4 m
= 0.16 mm.
Location for first maximum, x1 = $$\sqrt{\frac{3b(a+b)}{a}}$$
= 1.01 cm.

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