
In interference pattern, using two coherent sources of light, the fringe width is
A) Directly proportional to wavelength.
B) Inversely proportional to square of the wavelength.
C) Inversely proportional to wavelength.
D) Directly proportional to square of the wavelength.
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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