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A Double Slit S1 − S2 is Illuminated by a Coherent Light of Wavelength λ . the Slits Are Separated by a Distance D. - Physics

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प्रश्न

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.

योग
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उत्तर

Given:-

Separation between the two slits = d

Wavelength of the coherent light =λ

Distance between the slit and mirror is D1.

Distance between the slit and screen is D2.

Therefore,

apparent distance of the screen from the slits,

\[D = 2 D_1 + D_2 \]

Fringe width, \[\beta = \frac{\lambda D}{d} = \frac{\left( 2 D_1 + D_2 \right) \lambda}{d}\]

Hence, the required fringe width is \[\frac{\left( 2 D_1 + D_2 \right)  \lambda}{d}.\]

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अध्याय 17: Light Waves - Exercise [पृष्ठ ३८२]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 17 Light Waves
Exercise | Q 24 | पृष्ठ ३८२

संबंधित प्रश्न

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Estimate the number of fringes obtained in Young's double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to single slit.


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How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?


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  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, \[\lambda = 500\text{ nm, d = 1.0 mm and D = 1.0 m.}\] Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.


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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).


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`β = (λ"D")/"d"`

Where the terms have their usual meaning.


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