Question

The third-order bright fringe of 610-nm light is observed at an angle of 31° when the light falls on two narrow slits. How far apart are the slits?

Answer 1

Using the formula for the double-slit interference pattern, the slits' separation is calculated to be approximately 3.56 µm for a third-order bright fringe of 610-nm light observed at an angle of 31°.

Step-by-step explanation:

To calculate the separation between the two slits in a double-slit experiment, we can use the formula for the position of the bright fringes in a double-slit interference pattern, which is given by λ = d · sin(θ) / m, where λ is the wavelength of the light, δ is the separation between the slits, θ is the angle of the bright fringe and m is the order of the bright fringe.

For the third-order bright fringe (m=3) and a wavelength of 610 nm (6.10 × 10⁻¹ m), observed at an angle of 31°, the separation between the slits, d, can be calculated as follows:

δ = m · λ / sin(θ)

δ = 3 · (610 × 10⁻¹ m) / sin(31°)

δ = 3 · (610 × 10⁻¹ m) / 0.515

δ ≈ 3.56 × 10⁻¶ m or 3.56 µm

Therefore, the slits are separated by approximately 3.56 µm.

Answer 2

The distance between the slits is 3.55 μm

Step-by-step explanation:

Given that,

Order number = 3

Wave length = 610 nm

Angle = 31°

We need to calculate the distance between the slits

Using formula of distance of slit

`d\sin\theta=n\lambda`

`d=(n\lambda)/(\sin\theta)`

Where, n = order number

`\lambda`= wavelength

Put the value into the formula

`d=(3*610*10^(-9))/(\sin31)`

`d=3.55*10^(-6)\ m`

`d= 3.55\ \mu m`

Hence, The distance between the slits is 3.55 μm.