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प्रश्न
Explain Young’s double-slit experimental setup and obtain the equation for path difference.
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उत्तर
Young’s double-slit experiment
- Wavefronts from S1 and S2 spread out and overlapping takes place to the right side of the double slit.
- When a screen is placed at a distance of about 1 meter from the slits, alternately bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.
Equation for path difference: - Let d be the distance between the double slits S1 and S2 which act as coherent sources of wavelength λ.
- A screen is placed parallel to the double-slit at a distance D from it. The mid-point of S1 and S2 is C and the mid-point of the screen O is equidistant from S1 and S2. P is any point at a distance y from O.
- The waves from S1 and S2 meet at P either in-phase or out-of-phase depending upon the path difference between the two waves.
Young’s double-slit experimental setupδ = S2P – S1P
δ = S2P – MP = S2M -
The angular position of the point P from C is θ. ∠ OCP = θ.
- From the geometry, the angles ∠ OCP and ∠ S2S1M are equal.
∠ OCP = ∠ S2S1M = θ.
In right-angle triangle ∆ S1S2M, the path difference,
S2M = d sin θ
δ = d sin θ
If the angle θ is small, sin θ »tan θ θ
From the right angle triangle ∆ OCP,
tan θ = `"y"/"D"`
The path difference, δ = `"dy"/"d"`
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संबंधित प्रश्न
Four light waves are represented by
(i) \[y = a_1 \sin \omega t\]
(ii) \[y = a_2 \sin \left( \omega t + \epsilon \right)\]
(iii) \[y = a_1 \sin 2\omega t\]
(iv) \[y = a_2 \sin 2\left( \omega t + \epsilon \right).\]
Interference fringes may be observed due to superposition of
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
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