Light of wavelength 520 nm passes through two narrow slits 0.66 mm apart. The screen is 1.60 m away. A second source of unknown wavelength produces its second-order fringe 1.23 mm closer to the central maximum than the 520 nm light. What is the wavelength of the unknown light

λ = 266 nm

Explanation:

In this case, we need to determine first the separation of the second fringe from the central maximum. We can determine that with the following expression:

d sinθ = mλ   (1)

However, as the slits are very narrow, we can assume that sin θ ≈ θ and so θ = x/l

Replacing this in (1) we have:

d(x/l) = mλ

and solving for x:

x = mλl / d    (2)

Where:

x: separation of the 2nd fringe from the central maximum

m: order of the fringe

λ: wavelength of light

l: distance of the screen

d: distance between the slits

All the units must be in meters (m), so we can convert the units first or during the resolution. In this case, we'll do it in the resolution. Replacing the given data, we have:

x = 2 * (520 nm * 1m/10⁻⁹ nm) * 1.6 m / (0.66 mm * 1m/1000 mm)

x = 0.00252 m or just 2.52 mm

With this value, we can compute or determine the separation of the 2nd order fringe of the unknown light in the central maximum:

x₂ = 2.52 - 1.23 = 1.29 mm or 0.00129 m

Now, using (2) we can solve for λ:

λ = dx / ml   (3)

Replacing we have:

λ = (0.00129 * 0.0066) / (2 * 1.6)

λ = 2.66x10⁻⁷ m or simply 266 nm

Hope this helps

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s