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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. - Physics

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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) \[\frac{2D\lambda}{d}\]

बेरीज
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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.

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पाठ 17: Light Waves - Exercise [पृष्ठ ३८३]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 17 Light Waves
Exercise | Q 34 | पृष्ठ ३८३

संबंधित प्रश्‍न

(i) In Young's double-slit experiment, deduce the condition for (a) constructive and (b) destructive interferences at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen.

(b) Compare the interference pattern observed in Young's double-slit experiment with single-slit diffraction pattern, pointing out three distinguishing features.


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 young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.


In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.


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?


If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.


A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1 · 4 m away. If the two slits are separated by 0·28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.


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?


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.


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.


Consider the arrangement shown in the figure. The distance D is large compared to the separation d between the slits. 

  1. Find the minimum value of d so that there is a dark fringe at O.
  2. Suppose d has this value. Find the distance x at which the next bright fringe is formed. 
  3. Find the fringe-width.

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.


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.


In Young’s double-slit experiment, using monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 x 10-2 m towards the slits, the change in the fringe width is 3 x 10-5 m. If the distance between the two slits is 10-3 m, calculate the wavelength of the light used.


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?


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).


In Young's double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. Calculate

  1. the fringe width
  2. the distance of (a) third minimum and (b) fifth maximum, from the central maximum.

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