Two coherent light sources each of wavelength - are separated by a distance 3- The total number of minima formed on line AB which runs from - to - is-

The correct option is C 6The correct answer is option(C).highest order minima, n_min=( dλ 12)=3λλ 12= 52 Total number of minima formed = 2n_min+1=2× ...

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Standard XII

Physics

Introduction to YDSE

Two coherent ...

Question

Two coherent light sources each of wavelength $λ$ are separated by a distance $3 λ$ . The total number of minima formed on line AB which runs from $- \infty$ to $+ \infty$ is:

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Solution

The correct option is C 6
The correct answer is option(C).
highest order minima, $n_{m i n} = (\frac{d}{λ} - \frac{1}{2}) = \frac{3 λ}{λ} - \frac{1}{2} = \frac{5}{2}$
Total number of minima formed = $2 n_{m i n} + 1 = 2 \times \frac{5}{2} + 1 = 6$

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Two coherent light sources each of wavelength λ are separated by a distance 3λ. The total number of minima formed on line AB which runs from −∞ to +∞ is:

The correct option is C 6The correct answer is option(C).highest order minima, nmin=(dλ−12)=3λλ−12=52Total number of minima formed = 2nmin+1=2×52+1=6

Two coherent light sources each of wavelength $λ$ are separated by a distance $3 λ$ . The total number of minima formed on line AB which runs from $- \infty$ to $+ \infty$ is:

The correct option is C 6
The correct answer is option(C).
highest order minima, $n_{m i n} = (\frac{d}{λ} - \frac{1}{2}) = \frac{3 λ}{λ} - \frac{1}{2} = \frac{5}{2}$
Total number of minima formed = $2 n_{m i n} + 1 = 2 \times \frac{5}{2} + 1 = 6$