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Question: The distances of a point on the screen from two slits in a biprism experiment is \(1.8 \times {10^{ ...

The distances of a point on the screen from two slits in a biprism experiment is 1.8×10−5m1.8 \times {10^{ - 5}}m and 1.23×10−5m1.23 \times {10^{ - 5}}m. If wavelength of light used is 6000Ao6000\mathop A\limits^o , then fringe formed at that point is
A. 10th10th bright
B. 10th10th dark
C. 9th9thbright
D. 9th9thdark

Explanation

Solution

This is a question of interference. First calculate the path difference. Then, use the path difference and see whether it fits in the formula of maxima or in the formula of minima. Finally calculate the position of the minima or the position of maxima, whichever is applicable.

Complete step by step answer:
Path difference is the difference in lengths of path given. We need to subtract the two path lengths given in question. Here given the path lengths are 1.23×10−5m1.23 \times {10^{ - 5}}m and 1.8×10−5m1.8 \times {10^{ - 5}}m
The path difference here is Δx=0.57×10−5m\Delta x = 0.57 \times {10^{ - 5}}m
For maxima or bright fringe, Δx=nλ\Delta x = n\lambda
For minima or dark fringe, Δx=2n−12λ\Delta x = \dfrac{{2n - 1}}{2}\lambda
Given wavelength of light used is 6000Ao6000\mathop A\limits^o
For bright fringe,
0.57×10−5m=nλ n=0.57×10−56000×10−10=9.5  0.57 \times {10^{ - 5}}m = n\lambda \\\ n = \dfrac{{0.57 \times {{10}^{ - 5}}}}{{6000 \times {{10}^{ - 10}}}} = 9.5 \\\
Which is not a whole number. Thus no bright fringe can be obtained from the given path length and wavelength
For dark fringe,

0.57×10−5m=2n−12λ n=12(2×0.57×10−56000×10−10+1) n=10  0.57 \times {10^{ - 5}}m = \dfrac{{2n - 1}}{2}\lambda \\\ n = \dfrac{1}{2}\left( {\dfrac{{2 \times 0.57 \times {{10}^{ - 5}}}}{{6000 \times {{10}^{ - 10}}}} + 1} \right) \\\ n = 10 \\\

Thus the 10th10th dark fringe is obtained.

So, the correct answer is “Option B”.

Note:
Students should know that sometimes the path difference can be directly calculated and directly checked for maxima and minima. Also avoid making mistakes in doing big calculations. Sometimes you need to approximate while doing calculations. Since, Δx=2n−12λ\Delta x = \dfrac{{2n - 1}}{2}\lambda we used thus the value of nn starts from 11 . If we use Δx=2n+12λ\Delta x = \dfrac{{2n + 1}}{2}\lambda then the value of nn will start from 00 . So we need to add one to the calculated value of nn . Students generally make mistakes in this.