Lakshya Education MCQs

Question: 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



 
Options:
A.1.1 mm
B.2.6 mm
C.2.8 nm
D.3.12 mm
Answer: Option D
: D

d=2mm.D=120cm,λ=5200˙A
ybright=nλDd
y10=10×5200×10101.22×103
=52×6×105m=3.12mm
y10=10λDd

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More Questions on This Topic :

Question 1. In an interference pattern 16th order maximum corresponds to 600nm  .  What order will be visible if 480nm is used in the same position:
  1.    5
  2.    10
  3.    15
  4.    20
Answer: Option D
: D

n=16,y=nλDd=nλDd
nλ=nλ
16×600=n×480
n=20
Question 2. Two slits spaced 0.25 mm apart are placed at 0.75 m from a screen and illuminated by coherent light with a wavelength of 650 nm.  The intensity at the centre of the central maximum (θ=0) is I0.  The distance on the screen from the centre of central maximum to the point where the intensity has fallen to I02 is nearly
  1.    0.1 mm
  2.    25 mm
  3.    0.4 mm
  4.    0.5 mm
Answer: Option D
: D

I=I0cos2(πdsinθ)λ=I02
cos(πdsinθλ)=12
we get,
x=λD4d=650×109×0.754×0.25×103=487.5×106m
=0.4875mm=0.5mm
Question 3. In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is λ is K unit; λ being the wavelength of light used.  The intensity at a point where the path difference is λ4 will be:
  1.    K4
  2.    K2
  3.    K
  4.    zero
Answer: Option B
: B


K1=I+I+2Icosδ or K1=2I,K1=K2
For path difference λ4, phase difference, δ=2πλ.λ4=π2
Question 4. Monochromatic green light of wavelength 5×10-7  m illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed on a screen 2 m away is
  1.    0.25 mm
  2.    0.1 mm
  3.    1.0 mm
  4.    0.01 mm
Answer: Option C
: C

Question 5. In Young's double slit experiment, if one of the slit is closed fully, what will be visible on the screen?
  1.    A slit-shaped bright spot
  2.    A YDSE interference pattern
  3.    A diffraction pattern
  4.    Nothing will be visible
Answer: Option C
: C

If one of theslits 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 beall 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.
Question 6. 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
  1.    1.5mm
  2.    3 mm
  3.    4.5 mm
  4.    6 mm
Answer: Option C
: C

y0=n[Dλd] and yn=n(Dλd)
Equating yn and yn,we get
nn=λλ=900750=65
Hence, the first position at which overlapping occurs is
y6=y6=6(2)(750×109)2×103=4.5mm

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