
The distances of interference points on a screen from two slits are $18\mu m$ and $12.3\mu m\,$. If the wavelength of light used is $6\times {{10}^{-7}}m$ then the number of dark or bright fringe formed there will be-
(A) 8th dark
(B) 9th bright
(C) 10th dark
(D) 11th dark
Answer
126.3k+ views
Hint: In present day material science, the double slit experiment is a demonstration that light and matter can show attributes of both traditionally characterized waves and particles. It also shows the essentially probabilistic nature of quantum mechanical phenomena. With this theory being clear, we can derive and/or apply equations and determine the various numerical aspects that can be solved in this topic.
Complete step-by step answer:
Path difference between the two light waves can be written as, $\Delta x=1.8\times {{10}^{-5}}-1.23\times {{10}^{-5}}=0.57\times {{10}^{-5}}m$
From the above equation and the given data, we can determine that,
Wavelength of the light used $\lambda =6000A{}^\circ =6000\times {{10}^{-10}}m$
Using, $\Delta x=m\lambda \Rightarrow m=\dfrac{\Delta x}{\lambda }$
$\therefore m=\dfrac{0.57\times {{10}^{-5}}}{6000\times {{10}^{-10}}}=9.5$
As m does not come out to be an integral, a dark fringe must be formed at that point according to the condition for dark fringes.
$\therefore$Using the condition for dark fringes, we can determine that, $m=n-\dfrac{1}{2}=9.5\Rightarrow n=10$
Thus the 10th dark fringe is formed at that point according to the condition for dark fringes.
Hence, the correct answer is Option B.
Note: We must know that, on the off chance that light comprises carefully of ordinary or classical particles, and these particles were fired in an orderly fashion through a slit and permitted to strike a screen on the opposite side, we would hope to see a pattern according to the size and state of the slit. Nonetheless, when this "single-slit test" is practically performed, the example on the screen is a diffraction design in which the light is spread out. The littler the slit, the more noteworthy the point of spread.
Complete step-by step answer:
Path difference between the two light waves can be written as, $\Delta x=1.8\times {{10}^{-5}}-1.23\times {{10}^{-5}}=0.57\times {{10}^{-5}}m$
From the above equation and the given data, we can determine that,
Wavelength of the light used $\lambda =6000A{}^\circ =6000\times {{10}^{-10}}m$
Using, $\Delta x=m\lambda \Rightarrow m=\dfrac{\Delta x}{\lambda }$
$\therefore m=\dfrac{0.57\times {{10}^{-5}}}{6000\times {{10}^{-10}}}=9.5$
As m does not come out to be an integral, a dark fringe must be formed at that point according to the condition for dark fringes.
$\therefore$Using the condition for dark fringes, we can determine that, $m=n-\dfrac{1}{2}=9.5\Rightarrow n=10$
Thus the 10th dark fringe is formed at that point according to the condition for dark fringes.
Hence, the correct answer is Option B.
Note: We must know that, on the off chance that light comprises carefully of ordinary or classical particles, and these particles were fired in an orderly fashion through a slit and permitted to strike a screen on the opposite side, we would hope to see a pattern according to the size and state of the slit. Nonetheless, when this "single-slit test" is practically performed, the example on the screen is a diffraction design in which the light is spread out. The littler the slit, the more noteworthy the point of spread.
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