Give the relationship between the wavelength of light and its angle of deviation, when it is passed through a prism.
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Let, A = angle of prism D = angle of deviation N = refractive index $\lambda\text{vaccum}=$ wavelength in vacuum $\lambda\text{medium}=$ wavelength in medium We knowa$\text{n}=\frac{\text{sin}(\frac{\text{A}+\text{D}}{2})}{\text{sin}(\frac{\text{A}}{2})}$
$\frac{\text{n}\lambda\text{vaccum}}{\lambda\text{medium}}$
So,$\lambda\text{vacuum}=\lambda\text{medium}\frac{\text{sin}\big(\frac{\text{A}}{2}\big)}{\text{sin}\big(\frac{\text{A+D}}{2}\big)}$
$\frac{\text{n}\lambda\text{vaccum}}{\lambda\text{medium}}$
So,$\lambda\text{vacuum}=\lambda\text{medium}\frac{\text{sin}\big(\frac{\text{A}}{2}\big)}{\text{sin}\big(\frac{\text{A+D}}{2}\big)}$
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