# Engineering Physics Questions and Answers – Young’s Experiment

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Young’s Experiment”.

1. In Young’s Double Slit Experiment, if instead of monochromatic light white light is used, what would be the observation?
a) The pattern will not be visible
b) The shape of the pattern will change from hyperbolic to circular
c) Colored fringes will be observed with a white bright fringe at the center
d) The bright and dark fringes will change position

Explanation: When white light is used instead of monochromatic light in Young’s Double Slit experiment, all the seven constituent colors produce their interference pattern. At the center of the screen, all the wavelengths meet in phase and, therefore, a white bright fringe is formed. Then the next fringe will be formed due to violet color as the wavelength is shortest for violet color. This will be followed by indigo, blue till red color.

2. Young’s Double Slit Experiment was conducted in ______________
a) 1802
b) 1889
c) 1925
d) 1930

Explanation: Interference of light was first discovered by Thomas Young, in 1802. He established the wave nature of light.

3. What kind of sources are required for Young’s Double Slit experiment?
a) Coherent
b) Incoherent
c) Intense
d) Bright

Explanation: For the interference pattern to be visible, coherent sources are required. If the sources are coherent, only then would there be constructive interference.

4. If the distance between the two slits is doubled, the fringe width _________
a) Doubles
b) Halves
c) Four-times
d) Remains same

Explanation: The fringe width is inversely proportional to the distance between the two slits. As the distance between the slits doubles, the fringe width becomes half of its original value.

5. What change is observed when the whole pattern is immersed in water?
a) Fringe width decreases
b) Fringe width increases
c) Colored fringes
d) The bright and dark fringes will change position

Explanation: When the whole apparatus of Young’s Double slit experiment is immersed in water, the only change is the pattern shrinks a bit. The width of the fringes decreases, as an external path difference is present because of the water.
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6. In YDS, the width of the fringes obtained from a light of wavelength 500 nm is 3.6 mm. What is the fringe width id the apparatus is immersed in a liquid of refractive index 1.2?
a) 2 mm
b) 2.6 mm
c) 3 mm
d) 3.2 mm

Explanation: AS we know, Fringe width, β = λD/d, where D is the distance between the slits and the screen and d is the distance between the slits.
Now, β’ = λ’D/d
β’/ β = λ’/λ
β’ = β/μ
= 3.6 mm/1.2
= 3 mm.

7. There is no effect on the interference pattern when the width of the slit is increased.
a) True
b) False

Explanation: As the width of the slit increases, the pattern becomes less sharp. Until a point, where the width of the slit is large enough that the interference pattern disappears.

8. In a Young’s double slit experiment, the distance between the two slits is 0.5 mm and the distance between the screen and the slits is 1 m. When a light of wavelength 500 nm is incident on the slits, what would be distance between the two second bright fringes?
a) 1 mm
b) 2 mm
c) 3 mm
d) 4 mm

Explanation: As we know, β = λD/d. In this question, we need to find 2 X β.
Here, d = 0.5 mm and D = 1 m
Therefore, β = 500 nm / 0.5 mm
= 1 mm
Now, Separation between the second bright fringe on both sides of the central maxima = 2 X 1 mm
= 2 mm.

9. A thin sheet of refractive index 1.25 and thickness 0.5 cm is placed in the path of light from one source in the Young’s double slit experiment. What is the path difference observed?
a) 1.25 mm
b) 2.5 mm
c) 2.78 mm
d) 3.25 mm

Explanation: As we know, the path difference introduced by the sheet = (μ – 1) t, where t is the thickness of the sheet.
Here, μ = 1.5 and t = 0.01 m
Therefore, Δx = 0.25 X 0.005 m
= 0.0012 m
= 1.25 mm.

10. A Young’s double slit apparatus is immersed in a liquid of refractive index 1.25. What is the ratio of the fringe width in air and liquid?
a) 1: 2
b) 4: 5
c) 5: 4
d) 2: 1

Explanation: In air, β = λaD/d
In liquid, β’ = λlD/d
Now, β: β’ = λa: λl
= μ : 1
= 5 : 4.

11. The following intensity variation is observed in a Young’s Double Slit Experiment.

a) True
b) False

Explanation: The intensity of the pattern remains same for every fringe in a Young’s Double Slit experiment. The given intensity variation with phase graph is true.

12. Two narrow and parallel slits 0.08 cm apart are illuminated by a light of frequency 6 X 1011 kHz. At what distance from the slits should the screen be placed to obtain fringe width of 0.6 mm?
a) 0.98 m
b) 1.06 m
c) 1.28 m
d) 1.74 m

Explanation: Here, d = 0.08 cm = 8 X 10-4 m, β = 0.6mm = 6 X 10-4 m
λ = c/v
= 3 X 108/8 X 1014
= 3.75 X 10-7 m
D = βd/λ
= 1.28 m.

13. When a plate of thickness 0.05 mm is placed in the path of a Michaelson Interferometer, a shift of 100 fringes is observed for a light of wavelength 5000 Å. What is the refractive index of the plate?
a) 1
b) 1.5
c) 2
d) 2.5

Explanation: We know, μ = nλ/2t + 1
Here, λ = 5000 Å = 5 X 10-7 m, n = 100, t = 0.05 mm = 5 X 10-5 m
Therefore, μ = 100 X 5 X 10-7/2 X 5 X 10-5 + 1
μ = 1.5.

14. The visibility of fringes is given by the expression ___________
a) lmax/lmin
b) lmin/lmax
c) lmin + lmax / lmin – lmax
d) lmin – lmax / lmin + lmax

Explanation: The visibility of fringes observed in Young’s double slit experiment is given by the expression: lmin– lmax / lmin + lmax. Maximum intensity for the bright fringes while minimum intensity is for the dark fringes.

15. In Young’s Double Slit experiment, the angular width of the fringe is 0.1o. If the light used has a wavelength of 600 nm, what is the spacing between the slits?
a) 0.17 mm
b) 0.23 mm
c) 0.34 mm
d) 0.49 mm

Explanation: Now, as we know, β = λD/d
βθ = θn+1 – θn = β/D
βθ = λ/d
Here, λ = 6 X 10-7 m and βθ = 0.1 X π/180 radians
Therefore, d = 6 X 10-7 m X 180 = 0.1 X π
= 0.344 mm.

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