Question

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 $\mathring{A}$ , then wavelength of first maximum will be :

• A. 3300 $\mathring{A}$
• B. 4400 $\mathring{A}$
• C. 5500 $\mathring{A}$
• D. 6600 $\mathring{A}$

Answer 1

Answer: (B)

In a single slit experiment, For diffraction maxima,
$a\,\sin\,\theta=\left(2n+1 \frac{\lambda}{2}\right)$
and for diffraction minima, $a\,\sin\,\theta =n\lambda$
According to question,
$\left(2\times1+1 \frac{\lambda}{2}=1\times6600\right)$
$\left(\because \lambda_{R}=6600\,\mathring{A}\right)$
$\lambda=\frac{6600\times2}{3}$
$\lambda=4400\,\mathring{A}$