Answer:
The second wavelength is 482 nm.
Explanation:
Given that,
Wavelength = 630 nm
Distance from central maxim = 0.51 m
Distance from central maxim of another wavelength = 0.39 m
We need to calculate the second wavelength
Using formula of width of fringe
[tex]\beta=\dfrac{\lambda d}{D}[/tex]
Here, d and D will be same for both wavelengths
[tex]\lambda[/tex] = wavelength
[tex]\beta [/tex] = width of fringe
The width of fringe for first wavelength
[tex]\beta_{1}=\dfrac{\lambda_{1} d}{D}[/tex]....(I)
The width of fringe for second wavelength
[tex]\beta_{2}=\dfrac{\lambda_{2} d}{D}[/tex]....(II)
Divided equation (I) by equation (II)
[tex]\dfrac{\beta_{1}}{\beta_{2}}=\dfrac{\lambda_{1}}{\lambda_{2}}[/tex]
[tex]\lambda_{2}=\dfrac{630\times10^{-9}\times0.39}{0.51}[/tex]
[tex]\lambda_{2}=4.82\times10^{-7}[/tex]
[tex]\lambda=482\ nm[/tex]
Hence, The second wavelength is 482 nm.