1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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.
7. 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.
8. 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.
9. 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.
10. What changes are observed in a 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
Explanation: As the whole apparatus is now immersed in water, the wavelength of the light will change.
λ’=\(\frac{\lambda}{\mu}\)
Therefore, as the refractive index of water is greater than the air, the wavelength of light will decrease.
Width of central maxima = \(\frac{2\lambda}{a}\)
Therefore, as the wavelength decreases, the width of the central maxima decreases.