# Find the value of lambda_2.

In a single slit diffraction, first minimum for green light of wavelength coincides with first maximum of some other wavelength ${\lambda }^{1}$'. Find the value of ${\lambda }^{1}$.
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Kaiden Weeks
Since, Between the first and second minima, there is a maxima of undetermined wavelength.
And,
Position of minima in diffraction Pattern is
$n\lambda =d\mathrm{sin}\theta$
Thus, For first minima of Green Light of wavelength ${\lambda }_{g}$ will be,
$n{\lambda }_{g}=d\mathrm{sin}{\theta }_{1}$
To Calculate
The first maxima of some other wavelength ${\lambda }_{1}$ will be at position,
$\frac{3}{2}{\lambda }^{1}=d\mathrm{sin}{\theta }_{2}$
For the first minimum of diffraction Pattern of Green light
$n{\lambda }_{g}=d\mathrm{sin}{\theta }_{1}$
For the first maxima of some other light of wavelength ${\lambda }^{1}$
$\frac{3}{2}{\lambda }^{1}=d\mathrm{sin}{\theta }_{2}$
Since, first minimum of green light coincide with the first maximum of the other light therefore,
${\theta }_{1}={\theta }_{2}\phantom{\rule{0ex}{0ex}}\text{thus}\phantom{\rule{0ex}{0ex}}\mathrm{sin}{\theta }_{1}=\mathrm{sin}{\theta }_{2}$
Therefore we can write it as,
$\frac{3}{2}{\lambda }^{1}=n{\lambda }_{g}$
Given:
n=1 (for first minima)

The Wavelength of the first maxima of some other light is ${\lambda }^{1}=3493\stackrel{˙}{A}$