The position of bright fringes Y_m on screen in double slit experiment is given by following expressión
[tex]y_m={m\lambdaD}{d}[/tex]
We can solve for d, clearing the variables, so
[tex]d=\frac{m\lambda D}{y_m}[/tex]
So making the substitution for λ= 594nm, D=3m, y_1=4.84 and m=1, we have
[tex]d=\frac{594*10^{-9}*3*1}{4.84*10^{-3}}\\d=3.7*10^{-4}m[/tex]
Through this value we can find the position of dark fringes y_m on screen. The following expressión can approach better,
[tex]y_m=(m+\frac{1}{2})\frac{\lambda D}{d}[/tex]
To solve the wavelength m is equal to 0,
[tex]y_0=\frac{\lambda D}{2d}[/tex]
clearing for \lambda
[tex]\lambda = \frac{2y_0 d}{D}[/tex]
Making the substitution for [tex]y_0=4.84mm, d=3.7*10^{-4}, D=3m[/tex]
[tex]\lambda=\frac{2(4.84*10^{-3})(3.7*10^{-4})}{3}\\\lambda=1.19 \mu m[/tex]
So we have that the wavelength required is [tex]1.19 \mu m[/tex]
The wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen [tex]\lambda=1.19\ \mu m[/tex]
The wavelength of any wave is defined as the distance between two max adjacent amplitudes, or the distance between two successive troughs or crest.
The position of bright fringes Y_m on screen in double slit experiment is given by following expressión
[tex]y_m=md[/tex]
We can solve for d, clearing the variables, so
[tex]d=\dfrac{m\lambda D}{y_m}[/tex]
So making the substitution for λ= 594nm, D=3m, y_1=4.84 and m=1, we have
[tex]d=\dfrac{594\times 10^{-9}\times 3\times1}{4.84\times 10^{-3}}[/tex]
[tex]d=3.7\times 10^{-4}\ m[/tex]
Through this value we can find the position of dark fringes y_m on screen. The following expressión can approach better,
[tex]y_m=(m+\dfrac{1}{2})\dfrac{\lambda D}{d}[/tex]
To solve the wavelength m is equal to 0,
[tex]y_o=\dfrac{\lambda D}{2d}[/tex]
clearing for [tex]\lambda[/tex]
[tex]\lambda=\dfrac{2y_od}{D}[/tex]
Making the substitution for
[tex]y_o=4.84\ mm \ d=3.7\times 10^{-4}\ D=3\ m[/tex]
[tex]\lambda=\dfrac{2(4.84\times 10^{-3}(3.7\times 10^{-4})}{3}[/tex]
[tex]\lambda= 1.19\ \mu m[/tex]
So we have that the wavelength required is [tex]\lambda=1.19\ \mu m[/tex]
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