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,
$\left(2\times 1+1\frac{\lambda }{2}=1\times 6600\right)$
$\left(\because {\lambda }_R=6600\mathring{A}\right)$
$\lambda =\frac{6600\times 2}{3}$
$\lambda =4400\mathring{A}$