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: 600 nm

Step-by-step solution

Convert the wavelength to meters

The given wavelength is in nanometers, to work within the SI units, we need to convert it to meters. To do this, we will multiply the given value by \(10^{-9}\): \(\lambda = 600 \ nm \times 10^{-9} m/nm = 6 \times 10^{-7} m\)

Find the sine of 90 degrees

Since we are looking for the slit width where there are no minima, we want to find the slit width for which \(\theta > 90^{\circ}\). The sine of \(90^{\circ}\) is equal to 1, so we will use this value for our calculations.

Calculate the largest slit width without minima

Using the formula for the angular position of the first minimum: \(sin(\theta) = \frac{\lambda}{a}\) Since we want to find the slit width for which \(\theta > 90^{\circ}\), we will use \(sin(90^{\circ}) = 1\). Rearranging the formula to solve for \(a\): \(a = \frac{\lambda}{sin(\theta)}\) Now substitute the known values into the equation: \(a = \frac{6 \times 10^{-7} m}{1}\) \(a = 6 \times 10^{-7} m\)

Convert the slit width to nanometers

The calculated slit width is currently in meters. To convert it to nanometers, we will multiply the value by \(10^9\): \(a = 6 \times 10^{-7} m \times 10^9 nm/m = 600 nm\) The largest slit width for which there are no minima when the wavelength of the incident light on the single slit is 600 nm is 600 nm.

Key concepts

Slit Width Calculation

Understanding the calculation of slit width in single slit diffraction is crucial for understanding how light behaves when it encounters an obstacle. In the given exercise, we determine the maximum slit width that prevents the formation of minima, which occur due to destructive interference.

Let's simplify this concept: imagine your slit as a tiny opening for light waves to pass through. These waves start to spread out after passing the slit and interfere with each other. Sometimes they reinforce each other, creating bright spots, and sometimes they cancel each other out, resulting in dark spots. The dark spots, or minima, are key to calculating slit width.

To prevent minima from occurring, we need to adjust the width of the slit so that light does not spread enough to cause destructive interference. This condition is met when the sine of the angle (\theta) exceeds the value obtained from the division of the wavelength by the slit width. For no minima to occur, we need \theta to be greater than 90 degrees. Using the equation \(a = \frac{\lambda}{\sin(\theta)}\), with \(\sin(\theta) = 1\) (the sine of 90 degrees), allows us to find the slit width 'a' that aligns with our requirement of no minima.

Wavelength to Meters Conversion

The process of converting the wavelength from nanometers to meters is essential in physics to align with the International System of Units (SI), which standardizes measurements for consistency in scientific communication. In our exercise, we are given the wavelength in nanometers (nm), and we need to convert it to meters (m) to calculate the slit width appropriately.

Here's how you do it: you multiply the given wavelength by \(10^{-9}\) to convert nanometers to meters. So \(600 \ nm\) becomes \(600 \ nm \times 10^{-9} m/nm = 6 \times 10^{-7} m\).

This step may seem simple, but it is a fundamental skill necessary for any student studying wave phenomena, such as diffraction, ensuring the accuracy of all subsequent calculations and helping to establish a strong foundation in understanding wave behavior.

Diffraction Minima

Diffraction minima are the dark spots on a screen that occur when waves interfere destructively after passing through a slit. These points of destructive interference are where the waves cancel each other, leading to a minimum in light intensity. The positioning of these minima is not arbitrary; they occur at specific angles dependent on the slit width and the wavelength of the incident light.

Understanding this helps decipher why, in certain scenarios, we might prefer to eliminate these minima. A display technology, for example, would want a sharp image without such 'dead' points. In the context of our exercise, we aim to find a slit width that prevents these minima from forming, ensuring a uniform distribution of light—or at least a distribution without dark intervals.

Angular Position of First Minimum

The angular position of the first minimum is pivotal in the study of diffraction patterns because it marks the first occurrence of completely destructive interference. The angle for the first minimum helps us identify how the waves are interacting and gives a quantitative value to the behavior of light encountering a slit. In practical terms, it's the angle at which the first dark spot will appear on the screen.

In our exercise, we employed the formula \(\sin(\theta) = \frac{\lambda}{a}\) to characterize this relationship, where \(\theta\) is the angle at which the first minimum occurs, \(\lambda\) is the wavelength of the light, and 'a' is the slit width. By manipulating this formula based on our requirements, we derived the maximum slit width that would prevent the formation of any minima at all. This understanding not only helps in solving problems related to single slit diffraction but also enhances the comprehension of the wave nature of light.