Question

In a double-slit experiment the distance between slits is 5.6 mm and the slits are 0.81 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 m

Answer 1

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 $y$ of the bright fringe can be determined from

$y =\frac{m\lambda D}{d}$

Where $m$ is the order

$\lambda$ is the wavelength

$D$ is the distance from the screen

$d$ is the distance between the slits

From the question,

$d$ = 5.6 mm = 5.6 × 10⁻³ m

$D$ = 0.81 m

m = 3

For the first interference pattern, $\lambda$ = 410 nm = 410 × 10⁻⁹ m

∴ $y =\frac{m\lambda D}{d}$ becomes

$y =\frac{3 \times 410 \times 10^{-9} \times 0.81}{5.6 \times 10^{-3} }$

$y = 1.78 \times 10^{-4}$ m

For the second interference pattern, $\lambda$ = 660 nm = 660 × 10⁻⁹ m

∴ $y =\frac{3 \times 660 \times 10^{-9} \times 0.81}{5.6 \times 10^{-3} }$

$y = 2.86 \times 10^{-4}$ 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.