The distance between the second-order fringes for these two wavelengths is 0.3 × 10—³ m.
The wavelength of first light is,
[tex]λ _{1} = 720 \: nm = 720 \times 10 ^{ - 9} \: m[/tex]
The wavelength of the second light is,
[tex]λ _{1} = 660 \: nm = 660 \times 10 ^{ - 9} \: m[/tex]
The separation is d.
d = 0.61 mm
[tex]d = 0.61 \times 10 ^{ - 9} \: m[/tex]
The distance of the screen is D.
D = 1 m
For wavelength 1 the distance of the second fringe is,
The number of fringes is 2.
[tex]x _{n} = \frac{nλD}{d} [/tex]
[tex]x _{2} = \frac{2 \times 720 \times 10 ^{ - 9} \times 1.5}{0.61 \times 10 ^{ - 3} } [/tex]
[tex] = 3.5 \times 10 ^{ - 3} [/tex]
For wavelength 2 the distance of the second fringe is,
[tex]x ^{'} _{n} = \frac{nλD}{d} [/tex]
[tex]x ^{'} _{2} = \frac{2 \times 660 \times 10 ^{ - 9} \times 1.5}{0.61 \times 10 ^{ - 3} } [/tex]
[tex]x ^{'} _{2} = 3.2\times 10 ^{ - 3} [/tex]
The distance between the second-order fringes for these two wavelengths on a screen 1.5 m away is,
[tex]∆x = x _{2} - x ^{'} _{2} [/tex]
[tex] = (3.5 \times 10 ^{ - 3} - 3.2 \times 10 ^{ - 3} )[/tex]
[tex] = 0.3 \times 10 ^{ - 3} \: m[/tex]
Therefore, the distance between the second-order fringes for these two wavelengths is 0.3 × 10—³ m.
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