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प्रश्न
A long narrow horizontal slit is paced 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1.0 m away from the slit. Find the fringe-width if the light used has a wavelength of 700 nm.
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उत्तर
Given:-
Separation between two slits, \[d' = 2d = 2 mm = 2 \times {10}^{- 3} m..........\left(\text{as d = 1 mm}\right)\]
Wavelength of the light used,
\[\lambda = 700 nm = 700 \times {10}^{- 3} m\]
Distance between the screen and slit (D) = 1.0 m
It is a case of Lloyd's mirror experiment.
\[\text{Fringe width, }\beta = \frac{\lambda D}{d'}\]
\[ = \frac{700 \times {10}^{- 9} \times 1}{2 \times {10}^{- 3}}\]
\[ = 0 . 35 \times {10}^{- 3} m = 0 . 35\text{ mm}\]
Hence, the width of the fringe is 0.35 mm.
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संबंधित प्रश्न
State any one difference between interference of light and diffraction of light
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)
Answer in brief:
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"The fringe width ____________."
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In Young's double-slit experiment, the distance between the slits is 3 mm and the slits are 2 m away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength 480 nm and 600 run respectively. The separation on the screen between the 5th order bright fringes on the two interference patterns is ______
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