Answer:
0.82 mm
Explanation:
The formula for calculation an [tex]n^{th}[/tex] bright fringe from the central maxima is given as:
[tex]y_n=\frac{n \lambda D}{d}[/tex]
so for the distance of the second-order fringe when wavelength [tex]\lambda_1[/tex] = 745-nm can be calculated as:
[tex]y_2 = \frac{n \lambda_1 D}{d}[/tex]
where;
n = 2
[tex]\lambda_1[/tex] = 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
[tex]y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]
[tex]y_2[/tex] = 0.00276 m
[tex]y_2[/tex] = 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength [tex]\lambda_2[/tex] = 660-nm is as follows:
[tex]y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]
[tex]y^'}_2[/tex] = 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
[tex]\delta y = y_2-y^{'}_2[/tex]
[tex]\delta y[/tex] = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
[tex]\delta y[/tex] = 10 ⁻³ (2.76 - 1.94)
[tex]\delta y[/tex] = 10 ⁻³ (0.82)
[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m
[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m [tex](\frac{1.0mm}{10^{-3}m} )[/tex]
[tex]\delta y[/tex] = 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
