Answer
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Hint: The difference between the paths in simple trigonometry can be written as $d\sin \theta$ , where d is the distance between the slits. The path length difference must be an integral multiple of the wavelength of the light source, or $d\sin \theta = n\lambda $ , for $n = 0,1, - 1,2, - 2....$
Complete step by step answer:
Let us assume that ${n_1}$ fringe of source of wavelength ${\lambda _1}$ coincides with the ${n_2}$ fringe of source of wavelength ${\lambda _2}$ .
${\lambda _1} = 2500\mathop A\limits^ \circ $
${\lambda _2} = 3500\mathop A\limits^ \circ $
Using the formula , $d\sin \theta = n\lambda $
We get , \[\;{n_1}{\lambda _1} = {n_2}{\lambda _2}\]
$ \Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{{\lambda _1}}}{{{\lambda _2}}}$
After substituting the values of ${\lambda _1}$ and ${\lambda _2}$ in the above equation, we get
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{2500}}{{3500}} \Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{5}{7}$
Or $\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{5k}}{{7k}}$ (where k is any integer)
By comparing the options we get ${n_1}$= 7 and ${n_2}$= 5
Hence, (b) option is the correct option.
Note: The equation $d\sin \theta = n\lambda $ (for $n = 0,1, - 1,2, - 2....$) describes constructive interference. For fixed values of d and$\lambda $ , the larger n is, the larger $\sin \theta $ is. However, the maximum value that $\sin \theta $ can have is 1, for an angle of 90 degrees (Larger angle imply that light goes backward and does not reach the screen). Similarly for destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength, or $d\sin \theta = \left( {m + \dfrac{1}{2}} \right)\lambda $ , for $m = 0,1, - 1,2, - 2.....$
Complete step by step answer:
Let us assume that ${n_1}$ fringe of source of wavelength ${\lambda _1}$ coincides with the ${n_2}$ fringe of source of wavelength ${\lambda _2}$ .
${\lambda _1} = 2500\mathop A\limits^ \circ $
${\lambda _2} = 3500\mathop A\limits^ \circ $
Using the formula , $d\sin \theta = n\lambda $
We get , \[\;{n_1}{\lambda _1} = {n_2}{\lambda _2}\]
$ \Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{{\lambda _1}}}{{{\lambda _2}}}$
After substituting the values of ${\lambda _1}$ and ${\lambda _2}$ in the above equation, we get
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{2500}}{{3500}} \Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{5}{7}$
Or $\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{5k}}{{7k}}$ (where k is any integer)
By comparing the options we get ${n_1}$= 7 and ${n_2}$= 5
Hence, (b) option is the correct option.
Note: The equation $d\sin \theta = n\lambda $ (for $n = 0,1, - 1,2, - 2....$) describes constructive interference. For fixed values of d and$\lambda $ , the larger n is, the larger $\sin \theta $ is. However, the maximum value that $\sin \theta $ can have is 1, for an angle of 90 degrees (Larger angle imply that light goes backward and does not reach the screen). Similarly for destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength, or $d\sin \theta = \left( {m + \dfrac{1}{2}} \right)\lambda $ , for $m = 0,1, - 1,2, - 2.....$
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