Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
Given:-
The thickness of the strips = \[t_1 = t_2 = t = 0 . 5 mm = 0 . 5 \times {10}^{- 3} m\]
Separation between the two slits,
\[d = 0 . 12 cm = 12 \times {10}^{- 4} m\]
The refractive index of mica, μm = 1.58 and of polystyrene, μp = 1.58
Wavelength of the light,
\[\lambda = 590 nm = 590 \times {10}^{- 9} m,\]
Distance between screen and slit, D = 1 m
(a)
We know that fringe width is given by
\[\beta = \frac{\lambda D}{d}\]
\[\Rightarrow \beta = \frac{590 \times {10}^{- 9} \times 1}{12 \times {10}^{- 4}}\]
\[= 4 . 9 \times {10}^{- 4} m\]
(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
\[∆ x = \left( \mu_m - 1 \right) t - \left( \mu_p - 1 \right) t\]
\[= \left( \mu_m - \mu_p \right) t\]
\[ = \left( 1 . 58 - 1 . 55 \right) \times \left( 0 . 5 \right) \left( {10}^{- 3} \right)\]
\[ = \left( 0 . 015 \right) \times {10}^{- 3} m\]
∴ Number of fringes shifted, n = \[\frac{∆ x}{\lambda}\]
\[\Rightarrow n = \frac{0 . 015 \times {10}^{- 3}}{590 \times {10}^{- 9}} = 25 . 43\]
∴ 25 fringes and 0.43th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,
\[x = \left( 0 . 43 \right) \times 4 . 91 \times {10}^{- 4}...........\left( \because \beta = 4 . 91 \times {10}^{- 4} m \right)\]
\[= 0 . 021 cm\]
On the other side,
\[x' = \left( 1 - 0 . 43 \right) \times 4 . 91 \times {10}^{- 4} \]
\[ = 0 . 028 cm\]
APPEARS IN
संबंधित प्रश्न
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
Find the intensity at a point on a screen in Young's double slit experiment where the interfering waves have a path difference of (i) λ/6, and (ii) λ/2.
How does an unpolarized light incident on a polaroid get polarized? Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium.
Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third polaroid ‘C’ be placed between them so that the intensity of polarized light transmitted by polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A?
What is the effect on the interference fringes to a Young’s double slit experiment when
(i) the separation between the two slits is decreased?
(ii) the width of a source slit is increased?
(iii) the monochromatic source is replaced by a source of white light?
Justify your answer in each case.
The intensity at the central maxima in Young’s double slit experimental set-up is I0. Show that the intensity at a point where the path difference is λ/3 is I0/4.
If the separation between the slits in a Young's double slit experiment is increased, what happens to the fringe-width? If the separation is increased too much, will the fringe pattern remain detectable?
White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe \[\left( \lambda = 400\text{ nm} \right)\] which overlaps with a red fringe \[\left( \lambda = 700\text{ nm} \right).\]
In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength \[\lambda.\] Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum.
In Young's double-slit experiment, the two slits are separated by a distance of 1.5 mm, and the screen is placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths of 650 nm and 520 nm is used to obtain interference fringes.
Find the distance of the third bright fringe for λ = 520 nm on the screen from the central maximum.
In Young's double slit experiment the slits are 0.589 mm apart and the interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe is at a distance of 7.5 mm from the dark fringe which is second from the center of the fringe pattern. Find the wavelength of the light used.
Two slits in Young's interference experiment have width in the ratio 1 : 2. The ratio of intensity at the maxima and minima in their interference is ______.
Two balls are projected at an angle θ and (90° − θ) to the horizontal with the same speed. The ratio of their maximum vertical heights is:
Why is the diffraction of sound waves more evident in daily experience than that of light wave?
Consider a two-slit interference arrangement (Figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O.
How will the interference pattern in Young's double-slit experiment be affected if the screen is moved away from the plane of the slits?
A beam of light consisting of two wavelengths 600 nm and 500 nm is used in Young's double slit experiment. The silt separation is 1.0 mm and the screen is kept 0.60 m away from the plane of the slits. Calculate:
- the distance of the second bright fringe from the central maximum for wavelength 500 nm, and
- the least distance from the central maximum where the bright fringes due to both wavelengths coincide.
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
In Young's double-slit experiment, the separation between the two slits is d and the distance of the screen from the slits is 1000 d. If the first minima fall at a distance d from the central maximum, obtain the relation between d and λ.
