jlo2ni5x

2022-07-18

In a single slit diffraction, first minimum for green light of wavelength $5240\text{}\dot{A}$ coincides with first maximum of some other wavelength ${\lambda}^{1}$'. Find the value of ${\lambda}^{1}$.

Kaiden Weeks

Beginner2022-07-19Added 14 answers

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)

${\lambda}_{g}=5240\text{}\dot{A}\phantom{\rule{0ex}{0ex}}\frac{3}{2}{\lambda}^{1}=1(5240\dot{A})\phantom{\rule{0ex}{0ex}}{\lambda}^{1}=\frac{2\times 5240\dot{A}}{3}\phantom{\rule{0ex}{0ex}}{\lambda}^{1}=3493\dot{A}$

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

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)

${\lambda}_{g}=5240\text{}\dot{A}\phantom{\rule{0ex}{0ex}}\frac{3}{2}{\lambda}^{1}=1(5240\dot{A})\phantom{\rule{0ex}{0ex}}{\lambda}^{1}=\frac{2\times 5240\dot{A}}{3}\phantom{\rule{0ex}{0ex}}{\lambda}^{1}=3493\dot{A}$

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

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