Question

An instructor uses light of wavelength \(633 \mathrm{nm}\) to create a diffraction pattern with a slit of width \(0.135 \mathrm{~mm} .\) How far away from the slit must the instructor place the screen in order for the full width of the central maximum to be \(5.00 \mathrm{~cm} ?\)

Short answer

Answer: The instructor must place the screen approximately 2.69 meters away from the slit for the full width of the central maximum to be 5.00 cm.

Step-by-step solution

Understand the variables given in the problem

In this problem, we are given the following information: - The wavelength of the light, λ = 633 nm - The width of the slit, a = 0.135 mm - The full width of the central maximum, W = 5.00 cm We need to determine the distance between the slit and the screen (L).

Convert units

Before we move on to calculating, let's make sure all our variables are in the same units. Here, we will convert all distances into meters: - λ = 633 nm = 633 × 10^{-9} m - a = 0.135 mm = 0.135 × 10^{-3} m - W = 5.00 cm = 5.00 × 10^{-2} m

Apply the formula for the angular width of the central maximum

The single-slit diffraction formula for the angular width of the central maximum is given by: θ = 2 arcsin(λ / (2a)) We need to first find the angle θ using the given values of λ and a. θ = 2 arcsin( (633 × 10^{-9} m) / (2 × 0.135 × 10^{-3} m) ) θ ≈ 0.0093 radians

Calculate the distance between the slit and the screen

Now that we have found the angular width (θ) of the central maximum, we can use the formula relating the angle, the distance between the slit and the screen (L) and the width of the central maximum (W): W = 2Ltan(θ / 2) We need to rearrange this formula to solve for L: L = W / (2tan(θ / 2)) Now, we can plug in the values of W and θ: L = (5.00 × 10^{-2} m) / (2tan(0.0093 / 2)) L ≈ 2.69 m So, the instructor must place the screen approximately 2.69 meters away from the slit for the full width of the central maximum to be 5.00 cm.