11th And 12th > Physics
WAVE OPTICS MCQs
:
C
If one of the slits is closed then YDSE interference fringes are not formed on the screen, so you might be tempted to say 'A slit-shaped bright spot' will be all that you can see on the screen, but no, think about what you just learnt, a fringe pattern is observed due to diffraction from the single slit that is open.
:
A
n1λ1=n2λ2
60×5890=n25460
n2=65
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D
Sound wave and light waves both shows interference
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D
x=(2n–1)λ2
1029×10−7=(2n−1)6000×10−102
2×171.5=2n−1
344=2n
n=172
:
D
n=16,y=nλDd=n′λ′Dd
nλ=n′λ′
⇒ 16×600=n′×480
⇒ n′=20
A beam of light consisting of two wavelengths 6500 A and 5200 A is used to obtain interference fringes in Young’s double slit experiment. The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm. Find the distance of the 10th bright fringe from the central maximum corresponding to shorter wavelength
:
D
d=2mm.D=120cm,λ=5200˙A
ybright=nλDd
y10=10×5200×10−101.22×10−3
=52×6×10−5m=3.12mm
y10=10λDd
:
C
y=d2;y=nβ⇒d2=nλDd
⇒n=1.5 ∴ the total fringes=3
In a Young's double - slit experience, the slits are 2mm apart and are illuminated with a mixture of two wavelengths λ0=750nm and λ=900nm. The minimum distance from the common central bright fringe on a screen 2 m away from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other, is
:
C
y0=n[Dλd] and y′n=n′(Dλ′d)
Equating yn and y′n,we get
nn′=λ′λ=900750=65
Hence, the first position at which overlapping occurs is
y6=y6=6(2)(750×10−9)2×10−3=4.5mm
:
B
y=nβ
Here β=λDd=0.5cm
2 fringes.