Question

What is the largest slit width for which there are no minima when the wavelength of the incident light on the single slit is $600.\mathrm{nm}?$

Short answer

Answer: The largest slit width for which there are no minima when the wavelength of the incident light is 600 nm is 600 nm.

Step-by-step solution

Understand the Concept of Single-Slit Diffraction

Single-slit diffraction occurs when light waves pass through a narrow slit and then diffract, or bend, around the edges of the slit. This bending results in a pattern of light and dark areas on a screen behind the slit due to constructive and destructive interference of the light waves.

Know the Condition for Destructive Interference

In the single-slit diffraction pattern, destructive interference happens when the path length difference between two waves arriving at the screen is an odd multiple of half the wavelength (λ/2, 3λ/2, 5λ/2, etc). For the first minimum, the path length difference between the waves from the top and bottom edges of the slit should be λ/2.

Derive the Slit Width Formula

Let's consider a single slit of width 'a' and let 'θ' be the angle with respect to the central maximum where the first minimum occurs. According to the geometry of the diffraction pattern, the path length difference is (a/2)sin(θ). Now, equating the path length difference to λ/2. (a/2)sin(θ) = λ/2 Now, rearrange the equation to get the formula for the slit width, a. a = λ/sin(θ)

Calculate the Largest Slit Width for No Minima

If there are no minima, it means that the angle θ should be larger than 90 degrees for the first minimum, but this is not possible since the maximum value of θ for destructive interference is 90 degrees. So we should analyze the situation for θ = 90 degrees. a = (600 * 10^{-9}m) / sin(90) a = 600 nm The largest slit width for which there are no minima when the wavelength of the incident light is 600 nm is 600 nm.