Advertisements
Advertisements
Question
The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
Advertisements
Solution
Given:-
Separation between consecutive dark fringes = fringe width (β) = 1 mm = 10−3 m
Distance between screen and slit (D) = 2.5 m
The separation between slits (d) = 1 mm = 10−3 m
Let the wavelength of the light used in experiment be λ.
We know that
\[\beta = \frac{\lambda D}{d}\]
\[{10}^{- 3} m = \frac{2 . 5 \times \lambda}{{10}^{- 3}}\]
\[ \Rightarrow \lambda = \frac{1}{2 . 5} {10}^{- 6} m\]
\[= 4 \times {10}^{- 7} m = 400 \text{ nm}\]
Hence, the wavelength of light used for the experiment is 400 nm.
APPEARS IN
RELATED QUESTIONS
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters same?
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Write two characteristics features distinguish the diffractions pattern from the interference fringes obtained in Young’s double slit experiment.
Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing λ = 400 nm). Describe the nature of the fringe pattern observed.
In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 × 10−7 m and the distance of the screen from the slits is 1.0m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimetre width on the screen?
A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
In a Young's double slit experiment, using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron (1 micron = 10−6 m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.
A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?
A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits 4/9 of the light energy falling on it. (a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern. (b) How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.
In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the centre of the central maximum is 0.20 W m−2, what will be the intensity at a point 0.5 cm away from this centre along the width of the fringes?
The line-width of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the line-width of a bright fringe in a Young's double slit experiment in terms of \[\lambda,\] d and D where the symbols have their usual meanings.
Consider the arrangement shown in the figure. By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When \[z = \frac{D\lambda}{2d},\] the intensity measured at P is I. Find the intensity when z is equal to
(a) \[\frac{D\lambda}{d}\]
(b) \[\frac{3D\lambda}{2d}\] and
(c) \[\frac{2D\lambda}{d}\]
In Young’s double-slit experiment, show that:
`beta = (lambda "D")/"d"` where the terms have their usual meaning.
In Young’s double slit experiment, what should be the phase difference between the two overlapping waves to obtain 5th dark band/fringe on the screen?
Draw the intensity distribution as function of phase angle when diffraction of light takes place through coherently illuminated single slit.
Young's double slit experiment is made in a liquid. The 10th bright fringe lies in liquid where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately
Two slits, 4mm apart, are illuminated by light of wavelength 6000 A° what will be the fringe width on a screen placed 2 m from the slits?
Using Young’s double slit experiment, a monochromatic light of wavelength 5000Å produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000Å is used and the separation between the slits is doubled, then the new fringe width will be ______.
In Young's double slit experiment, show that:
`β = (λ"D")/"d"`
Where the terms have their usual meaning.
