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प्रश्न
White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe \[\left( \lambda = 400\text{ nm} \right)\] which overlaps with a red fringe \[\left( \lambda = 700\text{ nm} \right).\]
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उत्तर
Let the separation between the slits be d and distance between screen from the slits be D.
Suppose, the mth bright fringe of violet light overlaps with the nth bright fringe of red light.
Now, the position of the mth bright fringe of violet light, yv = \[\frac{m \lambda_v D}{d}\]
Position of the nth bright fringe of red light, yr = \[\frac{n \lambda_r D}{d}\]
For overlapping, yv = yr
So, as per the question,
\[\frac{m \times 400 \times D}{d} = \frac{n \times 700 \times D}{d}\]
\[ \Rightarrow \frac{m}{n} = \frac{7}{4}\]
Therefore, the 7th bright fringe of violet light overlaps with the 4th bright fringe of red light.
It can also be seen that the 14th violet fringe will overlap with the 8th red fringe.
Because,
\[\frac{m}{n} = \frac{7}{4} = \frac{14}{8}\]
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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.
What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters same?
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- Assertion (A): In Young's double slit experiment all fringes are of equal width.
- Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).
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- The angular separation of the fringes.
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