Answer
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Hint: Interference is the superimposition of two waves which forms a resultant with amplitude of more, less or the same one. Interference produced with different waves will always either come from same source or will have nearly same frequency
Important Formulae
$\begin{gathered}
{\text{X = }}\dfrac{{\lambda {\text{D}}}}{d} \\
{\text{X = Fringe Width}} \\
{\text{D = distance between the source and the screen}} \\
\lambda {\text{ = wavelength of the light used}} \\
{\text{From above equation, we can clearly come into a conclusion that}} \\
{\text{X is directly proportional to }}\lambda \\
\end{gathered} $
Complete step by step solution
In interference, we know Fringe is the separation between two consecutive bright or dark fringe
So $\begin{gathered}
{{\text{y}}_{\text{n}}}{\text{ = }}\dfrac{{{{n\lambda D}}}}{{\text{d}}}{\text{ and }}{{\text{y}}_{n + 1}} = \dfrac{{(n - 1)}}{d}\lambda D \\
So {\text{ }}{{\text{y}}_{{n}}} - {\text{ }}{{\text{y}}_{n + 1}} = {\text{fringe width}} \\
\end{gathered} $
$\begin{gathered}
= \dfrac{{\lambda {\text{D}}}}{d} \\
\\
\end{gathered} $
Fringe width is given by
$\begin{gathered}
{\text{X = }}\dfrac{{\lambda {\text{D}}}}{d} \\
{\text{X = Fringe Width}} \\
{\text{D = distance between the source and the screen}} \\
\lambda {\text{ = wavelength of the light used}} \\
{\text{From above equation, we can clearly come into a conclusion that}} \\
{\text{X is directly proportional to }}\lambda \\
\end{gathered} $
Thus, option A is correct.
Additional Information
Interference occurs in all types of waves including it is not only limited to surface water waves, gravity waves and light waves.
Notes: If the crest of one wave meets the trough of another waves, then the amplitude of the resultant wave is the difference in the amplitude of the two waves.
Important Formulae
$\begin{gathered}
{\text{X = }}\dfrac{{\lambda {\text{D}}}}{d} \\
{\text{X = Fringe Width}} \\
{\text{D = distance between the source and the screen}} \\
\lambda {\text{ = wavelength of the light used}} \\
{\text{From above equation, we can clearly come into a conclusion that}} \\
{\text{X is directly proportional to }}\lambda \\
\end{gathered} $
Complete step by step solution
In interference, we know Fringe is the separation between two consecutive bright or dark fringe
So $\begin{gathered}
{{\text{y}}_{\text{n}}}{\text{ = }}\dfrac{{{{n\lambda D}}}}{{\text{d}}}{\text{ and }}{{\text{y}}_{n + 1}} = \dfrac{{(n - 1)}}{d}\lambda D \\
So {\text{ }}{{\text{y}}_{{n}}} - {\text{ }}{{\text{y}}_{n + 1}} = {\text{fringe width}} \\
\end{gathered} $
$\begin{gathered}
= \dfrac{{\lambda {\text{D}}}}{d} \\
\\
\end{gathered} $
Fringe width is given by
$\begin{gathered}
{\text{X = }}\dfrac{{\lambda {\text{D}}}}{d} \\
{\text{X = Fringe Width}} \\
{\text{D = distance between the source and the screen}} \\
\lambda {\text{ = wavelength of the light used}} \\
{\text{From above equation, we can clearly come into a conclusion that}} \\
{\text{X is directly proportional to }}\lambda \\
\end{gathered} $
Thus, option A is correct.
Additional Information
Interference occurs in all types of waves including it is not only limited to surface water waves, gravity waves and light waves.
Notes: If the crest of one wave meets the trough of another waves, then the amplitude of the resultant wave is the difference in the amplitude of the two waves.
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