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प्रश्न
- Assertion (A): In Young's double slit experiment all fringes are of equal width.
- Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).
विकल्प
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A).
Assertion (A) is true and Reason (R) is false.
Assertion (A) is false and Reason (R) is also false.
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उत्तर
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
Bright and dark fringes are produced as light travels through a slit in Young's double-slit experiment. Both the black and brilliant fringes are the same width as YDSE.
Statement 2 explanation: White light is used in Young's double-slit studies, and it is directed through the slit. Therefore, only brilliant and dark fringes may be seen utilising this source.
We know that fringe width is given by,
Width = `(lambdaD)/d`
Where,
λ = Wavelength of source used.
D = Distance between screen and slit.
d = Distance between slits.
As fringe width depends on all these factors so the fringe width remains constant.
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संबंधित प्रश्न
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.
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 double slit S1 − S2 is illuminated by a coherent light of wavelength \[\lambda.\] The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen ∑ is placed behind the double slit at a distance D2 from it (see the following figure). The screen ∑ receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.
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 using monochromatic light of wavelength 600 nm, 5th bright fringe is at a distance of 0·48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate:
(i) Distance between the two slits.
(ii) Fringe width, i.e. fringe separation.
Draw the intensity distribution as function of phase angle when diffraction of light takes place through coherently illuminated single slit.
The force required to double the length of a steel wire of area 1 cm2, if its Young's modulus Y= 2 × 1011/m2 is:
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is ______.
In an interference experiment, a third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain the fifth bright fringe at the same point?
In Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.
