Engineering Physics Multiple Choice Questions 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
Answer: d
Clarification: 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
Answer: d
Clarification: 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
Answer: d
Clarification: 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
Answer: c
Clarification: 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
Answer: c
Clarification: We know that the formula for the radius of half period zone = (sqrt{nblambda}), 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.
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
Answer: b
Clarification: 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
Answer: d
Clarification: 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
