Respuesta :
Answer:
The separation in meters on the screen between the bright fringes of the two interference patterns is 1.08 × 10⁻⁴ m
Explanation:
Here is the complete question:
In a double-slit experiment, the distance between slits is 5.6 mm and the slits are 0.18 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 410 nm, and the other due to light of wavelength 660 nm. What is the separation in meters on the screen between the third-order (m = 3) bright fringes of the two interference patterns?
Explanation:
For a bright fringe, the distance [tex]y[/tex] of the bright fringe can be determined from
[tex]y =\frac{m\lambda D}{d}[/tex]
Where [tex]m[/tex] is the order
[tex]\lambda[/tex] is the wavelength
[tex]D[/tex] is the distance from the screen
[tex]d[/tex] is the distance between the slits
From the question,
[tex]d[/tex] = 5.6 mm = 5.6 × 10⁻³ m
[tex]D[/tex] = 0.81 m
m = 3
For the first interference pattern, [tex]\lambda[/tex] = 410 nm = 410 × 10⁻⁹ m
∴ [tex]y =\frac{m\lambda D}{d}[/tex] becomes
[tex]y =\frac{3 \times 410 \times 10^{-9} \times 0.81}{5.6 \times 10^{-3} }[/tex]
[tex]y = 1.78 \times 10^{-4}[/tex] m
For the second interference pattern, [tex]\lambda[/tex] = 660 nm = 660 × 10⁻⁹ m
∴ [tex]y =\frac{3 \times 660 \times 10^{-9} \times 0.81}{5.6 \times 10^{-3} }[/tex]
[tex]y = 2.86 \times 10^{-4}[/tex] m
Now, the separation between the bright fringes of the two interference patterns will be the difference in the distances of the bright fringes for the two interference patterns. That is
2.86 × 10⁻⁴ m - 1.78 × 10⁻⁴ m = 1.08 × 10⁻⁴ m
Hence, the separation in meters on the screen between the bright fringes of the two interference patterns is 1.08 × 10⁻⁴ m.
