In an experiment of single slit diffraction pattern, first minimum...

In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is

6600 Å

, then wavelength of first maximum will be:

Options

A $3300 Å$
B $4400 Å$
C $5500 Å$
D $6600 Å$

Correct answer

B. $4400 Å$

Step-by-step solution

In a single slit experiment, For diffraction maxima,

a \sin \theta=\left(2 n+1 \frac{Å}{2}\right. and for diffraction minima,

\mathrm{a} \sin \theta=\mathrm{n} \lambda

According to question,

\begin{aligned}&\left(2 \times1+1 \frac{\lambda}{2}=1 \times6600\right. \\ &\left(\because \lambda_{\mathrm{R}}=6600 Å\right) \\ &\lambda=\frac{6600 \times 2}{3}\\ &\lambda=4400 Å \end{aligned}

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