Advertisements
Advertisements
प्रश्न
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) \[\frac{2D\lambda}{d}\]
Advertisements
उत्तर
Given:-
Fours slits S1, S2, S3 and S4.
The separation between slits S3 and S4 can be changed.
Point P is the common perpendicular bisector of S1S2 and S3S4.

(a) For \[z = \frac{\lambda D}{d}\]
The position of the slits from the central point of the first screen is given by \[y = {OS}_3 = {OS}_4 = \frac{z}{2} = \frac{\lambda D}{2d}\]
The corresponding path difference in wave fronts reaching S3 is given by \[∆ x = \frac{yd}{D} = \frac{\lambda D}{2d} \times \frac{d}{D} = \frac{\lambda}{2}\]
Similarly at S4, path difference, \[∆ x = \frac{yd}{D} = \frac{\lambda D}{2d} \times \frac{d}{D} = \frac{\lambda}{2}\]
i.e. dark fringes are formed at S3 and S4.
So, the intensity of light at S3 and S4 is zero. Hence, the intensity at P is also zero.
(b) For \[z = \frac{3\lambda D}{2d}\]
The position of the slits from the central point of the first screen is given by \[y = {OS}_3 = {OS}_4 = \frac{z}{2} = \frac{3\lambda D}{4d}\]
The corresponding path difference in wave fronts reaching S3 is given by \[∆ x = \frac{yd}{D} = \frac{3\lambda D}{4d} \times \frac{d}{D} = \frac{3\lambda}{4}\]
Similarly at S4, path difference,
\[∆ x = \frac{yd}{D} = \frac{3\lambda D}{4d} \times \frac{d}{D} = \frac{3\lambda}{4}\]
Hence, the intensity at P is I.
(c) For \[z = \frac{2\lambda D}{d}\]
The position of the slits from the central point of the first screen is given by \[y = {OS}_3 = {OS}_4 = \frac{z}{2} = \frac{2\lambda D}{2d}\]
The corresponding path difference in wave fronts reaching S3 is given by \[∆ x = \frac{yd}{D} = \frac{2\lambda D}{2d} \times \frac{d}{D} = \lambda\]
Similarly at S4, path difference, \[∆ x = \frac{yd}{D} = \frac{2\lambda D}{2d} \times \frac{d}{D} = \lambda\]
Hence, the intensity at P is 2I.
APPEARS IN
संबंधित प्रश्न
In Young' s experiment the ratio of intensity at the maxima and minima . in the interference pattern is 36 : 16. What is the ratio of the widths of the two slits?
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
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?
In Young’s experiment, the ratio of intensity at the maxima and minima in an interference
pattern is 36 : 9. What will be the ratio of the intensities of two interfering waves?
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?
In a Young's double slit experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away?
A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle \[\theta = \sin^{- 1} \left( \frac{\lambda}{2d} \right)\] with the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.
White coherent light (400 nm-700 nm) is sent through the slits of a Young's double slit experiment (see the following figure). The separation between the slits is 0⋅5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1⋅0 mm away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?

In Young’s double slit experiment, what is the effect on fringe pattern if the slits are brought closer to each other?
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
The Young's double slit experiment is performed with blue and with green light of wavelengths 4360Å and 5460Å respectively. If x is the distance of 4th maxima from the central one, then:
In Young's double slit experiment, the minimum amplitude is obtained when the phase difference of super-imposing waves is: (where n = 1, 2, 3, ...)
In Young's double slit experiment shown in figure S1 and S2 are coherent sources and S is the screen having a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits. Which wavelength passing through the hole has strong intensity?

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.

In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is `1/8`th of the wavelength. The ratio of intensity at this point to that at the centre of a bright fringe is close to ______.
ASSERTION (A): In an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
REASON (R): Fringe width is proportional to (d/D).
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?
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, 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.
