In a single slit diffraction experiment the first minimum for red light of wavelength \[6600{{A}^{0}}\] coincides with the first maximum for other light of wavelength \[\lambda \]. The value of \[\lambda \] is:
\[\begin{align}
  & A.2200{{A}^{0}} \\
 & B.3300{{A}^{0}} \\
 & C.4400{{A}^{0}} \\
 & D.5000{{A}^{0}} \\
\end{align}\]

Hint: Whenever there is a coinciding condition of maxima and minima, the path difference must be the same. This will help us to solve the question.

Formula used:
\[a\sin \theta =n\lambda \]

Where:
a = size of a single slit
\[\theta =\] angular path difference
n = integer to give the position of minima
\[\lambda =\]wavelength of incident light

Complete step by step answer:
Diffraction: The phenomenon of bending of light around corners of an obstacle or aperture in the path of light is called diffraction.

For the condition of minima

\[a\sin \theta =n\lambda \]

Where:

\[n=\pm 1,\pm 2,\pm 3...\]

First minima of red light is at n=1

\[a\sin \theta ={{\lambda }_{red}}\]

\[\sin \theta =\dfrac{{{\lambda }_{red}}}{a}\]

Maxima of unknown wavelength lies between the first and second minima.

First minima:

\[a\sin \theta =\lambda \] …….(1)

Second minima

\[a\sin \theta =2\lambda \] …….(2)

Therefore first maxima will lie in between two consecutive minima;

Adding (1) and (2)
\[2a\sin \theta =3\lambda \]
\[a\sin \theta =\dfrac{3}{2}\lambda \]
\[\sin \theta =\dfrac{3\lambda }{2a}\]

And we know that \[\sin \theta =\dfrac{{{\lambda }_{red}}}{a}\], ( angle is same because two wavelengths are coinciding)

\[\begin{align}
  & \dfrac{{{\lambda }_{red}}}{a}=\dfrac{3\lambda }{2a} \\
 & \lambda =\dfrac{2{{\lambda }_{red}}}{3} \\
 & \lambda =\dfrac{2\times 6600{{A}^{0}}}{3} \\
 & \\
\end{align}\]

Answer is option (C). \[4400{{A}^{0}}\]

Note: There is an alternate option to solve this question:
In diffraction,
Path difference for minima is given by \[\Delta x=\dfrac{n\lambda a}{D}\]
Where D is the distance between slit and screen.

Therefor path difference of red light for first minima (n=1) is \[\Delta x=\dfrac{{{\lambda }_{red}}a}{D}\]

And,

Path difference for maxima is given by \[\Delta x=\left( n+\dfrac{1}{2} \right)\dfrac{{{\lambda }_{{}}}a}{D}\]

Therefore, path difference of unknown light for first maxima (n=1)

\[\begin{align}
  & \Delta x=\left( 1+\dfrac{1}{2} \right)\dfrac{{{\lambda }_{{}}}a}{D} \\
 & \Delta x=\left( \dfrac{3}{2} \right)\dfrac{{{\lambda }_{{}}}a}{D} \\
\end{align}\]

And for coinciding conditions the path difference of two lights must be the same.

\[\begin{align}
  & \dfrac{{{\lambda }_{red}}a}{D}=\dfrac{3\lambda a}{2D} \\
 & \lambda =\dfrac{2{{\lambda }_{red}}}{3} \\
 & \lambda =\dfrac{2\times 6600{{A}^{0}}}{3} \\
 & \\
\end{align}\]