Question

A two-slit apparatus is covered with a red \((670 \mathrm{nm})\) filter. When white light is shone on the filter, on the screen beyond the two-slit apparatus, there are nine interference maxima within the 4.50 -cm-wide central diffraction maximum. When a blue \((450 \mathrm{nm})\) filter replaces the red, how many interference maxima will there be in the central diffraction maximum, and how wide will that diffraction maximum be?

Short answer

Answer: To find the number of interference maxima (N_b) and the width of the central diffraction maximum (w_b) for the blue filter, first find the slit separation 'a' using the given values for the red filter. Then, determine the angular width of the central maximum for the blue filter using the formula for diffraction by a single slit. Next, calculate the number of interference maxima within the central maximum (N_b) and the width of the central diffraction maximum (w_b) for the blue filter using the angular width and the distance between the screen and the apparatus.

Step-by-step solution

Identify the given values and the variables we need to find

The given values are: 1. Wavelength of red light, 𝜆ₐ = 670 nm 2. Width of the central diffraction maximum for red filter, wₐ = 4.50 cm 3. Number of interference maxima within the central diffraction maximum for the red filter, Nₐ = 9 The variables we need to find are: 1. Number of interference maxima within the central diffraction maximum for the blue filter, N_b 2. Width of the central diffraction maximum for the blue filter, w_b

Find the angular width of the central maximum for the red filter

The angular width of the central maximum for the red filter can be found using the formula for diffraction by a single slit: θ = (2 * m + 1) * (λ / a) Where θ is the angular width, m is the order of the maximum (m = 0 for central maximum), λ is the wavelength of light, and a is the slit width. For the red filter, we have θₐ = (2 * 0 + 1) * (𝜆ₐ / a). We need to find the value of 'a' in order to use this equation for the blue filter as well.

Find the angular separation between adjacent maxima for the red filter

The angular separation between adjacent maxima for the red filter can be found using the formula for interference by two slits: Δθ = λ / (a * sin(θ/2)) Where Δθ is the angular separation between adjacent maxima, λ is the wavelength of light, a is the slit separation, and θ is the angular width of the central maximum. Since we have Nₐ = 9 maxima within the central maximum, there are Nₐ - 1 = 8 separations between them. Therefore, angular separation between adjacent maxima for the red filter will be: Δθₐ = (θₐ) / 8

Calculate the slit separation, 'a', for the apparatus

Now, we can use the formula for interference by two slits and the angular separation for the red filter to find the slit separation, 'a': a = (𝜆ₐ) / (Δθₐ * sin(θₐ/2)) Once we have the value of 'a', we can use it to find the number of interference maxima and the width of the central diffraction maximum for the blue filter.

Find the angular width of the central maximum for the blue filter

We can now find the angular width of the central maximum for the blue filter using the formula for diffraction by a single slit and the value of 'a': θ_b = (2 * m + 1) * (λ_b / a) Where λ_b = 450 nm is the wavelength of blue light.

Calculate the number of interference maxima within the central maximum for the blue filter

Using the angular width of the central maximum for the blue filter, we can find the number of interference maxima within the central maximum: N_b = 1 + (θ_b / Δθ_b) Where Δθ_b = λ_b / (a * sin(θ_b/2)) is the angular separation between adjacent maxima for the blue filter.

Find the width of the central diffraction maximum for the blue filter

We can find the width of the central diffraction maximum for the blue filter using the angular width for the blue filter and the distance (D) between the screen and the apparatus (which is constant for both red and blue filters): w_b = D * θ_b Now, we have the number of interference maxima within the central diffraction maximum (N_b) and the width of the central diffraction maximum (w_b) for the blue filter.

Key concepts

Two-Slit Interference

When we talk about two-slit interference, we are discussing a classic physics experiment known as Young's double-slit experiment. This setup demonstrates a fundamental property of light called wave-particle duality. In this experiment, light passes through two narrow, closely spaced slits and onto a screen, creating a pattern of bright and dark fringes. The fringes are the result of constructive and destructive interference.

Constructive interference occurs when waves from different slits arrive at the same point in phase and combine to form a brighter light, or an 'interference maximum'. Destructive interference happens when waves arrive out of phase, cancelling each other out and resulting in dark fringes. The interference pattern can be altered by changing various elements such as the wavelength of light or the width between the slits, which is central to solving problems involving the interference pattern produced by different wavelengths.

Diffraction Maximum

A diffraction maximum is a bright spot in a pattern produced when light waves spread out after passing through a small opening, like a slit, and overlap with each other. The central, or zeroth-order maximum, is the brightest and widest part of the diffraction pattern.

The width of the central diffraction maximum is relevant to two-slit interference problems since the interference fringes are contained within this region. The number of bright fringes or maxima within the central diffraction maximum can be calculated using the slit width, the wavelength of light, and the angular width of the diffraction maximum. Understanding how this number changes with varying wavelengths is key to predicting the behavior of light through the double-slit apparatus.

Wavelength of Light

The wavelength of light is the distance between successive peaks of the light wave and is typically measured in nanometers (nm). Different wavelengths correspond to different colors of light in the visible spectrum, with red having longer wavelengths than blue. The wavelength affects the interference pattern because it influences the separation between maxima on the screen.

For instance, in the case of our problem, red light with a longer wavelength of 670 nm and blue light with a shorter wavelength of 450 nm will produce different interference patterns. Importantly, the number of interference maxima within the central maximum will also change with the wavelength, as the spacing between fringes becomes wider or narrower.

Angular Separation

Angular separation refers to the angle between the central maximum and the nearest interference maximum, or the angle between successive interference maxima. This angle is crucial for determining the spacing of the interference fringes on a screen located at some distance from the two slits.

In mathematical terms, the angular separation can be found using the formula \( \Delta\theta = \frac{\lambda}{a \cdot \sin(\theta/2)} \) where \( \lambda \) is the wavelength, \( a \) is the slit separation, and \( \theta \) is the angular width of the central maximum. By knowing the angular separation, one can calculate how many maxima will fit within the central maximum's angular breadth, as well as the physical distance between them on the screen.