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The wavelength of light is a fundamental property that greatly influences the behavior of light waves. In the context of single slit diffraction, the wavelength determines how much the light spreads out or bends when it encounters an obstacle like a slit. The light's wavelength is often represented by the Greek letter lambda (\( \lambda \)).
The exercise specifies that the wavelength is \(680\ \mathrm{~nm}\). This measurement is in nanometers, which is a common unit for wavelengths in the visible spectrum. To put it in perspective:

• 1 nanometer (nm) is one-billionth of a meter (\(1\ \mathrm{~nm} = 10^{-9}\ \mathrm{~m}\)).
• The visible spectrum ranges roughly from about 380 nm to about 750 nm.
• 680 nm corresponds to a reddish-orange color of light.

Wavelength is crucial for determining the diffraction pattern seen in such problems. When light with a specific wavelength passes through a slit, it forms patterns of dark and light regions on a screen, based on the interference of the light waves.

A diffraction pattern is a series of dark and light bands formed as light waves spread out after passing through a slit or around an object. This phenomenon results from the constructive and destructive interference of light waves. Given the exercise, when light with a wavelength of 680 nm passes through a single slit, it creates a diffraction pattern on the screen 50.0 cm away. Let's break down the essential elements of a diffraction pattern:

• The pattern consists of alternating bright and dark areas called fringes.
• Bright fringes (maxima) happen where the light waves add together constructively.
• Dark fringes (minima) occur due to destructive interference, where waves cancel out.

The exercise gives the distance between the first and third minima in the pattern as 3.00 mm. By applying the formula for the location of the minima, you can determine the width of the slit. Each order of the minima corresponds to a specific angle where light waves interfere destructively. The relationship among the minima, the wavelength, and the distance to the screen helps find the slit width.

The width of the slit is a crucial parameter in experiments involving diffraction. It affects how much the light waves spread out to create the diffraction pattern seen on the screen. The width, typically represented as \( a \), is determined using the diffraction formula.
To find the width of the slit in this exercise, use the following steps based on the given formula \( y_m = \frac{m\lambda L}{a} \):

• Know that the positions of the first and third minima are represented by \( y_1 \) and \( y_3 \).
• The given distance between \( y_1 \) and \( y_3 \) is 3.00 mm, so you have \( y_3 - y_1 = 3.00\ \mathrm{~mm} \).
• Use the relationship \( 2\lambda L / a = 3.00\ \mathrm{~mm} \) from the problem to find \( a \).
• Substitute known values: \( \lambda = 680\ \mathrm{~nm} \), \( L = 50.0\ \mathrm{~cm} \), and convert these into consistent units.

After solving the equation, you will obtain the width of the slit, which helps explain how the diffraction pattern is formed. A narrower slit generally causes more spreading, while a wider slit results in less diffraction.