Single Slit Diffraction

For single slit diffraction, first minimum occurs at: Given: So, Thus,
Hence, the correct answer is:

Step 1: Understanding the Concept:
This refers to Fraunhofer diffraction due to a single slit. The light passing through the slit spreads out, forming a central bright maximum flanked by secondary maxima and minima. The angular width describes how much the central maximum spreads out angularly.

Step 2: Key Formula or Approach:
The angular width ( ) of the central maximum in a single slit diffraction pattern is given by:

Where:
= wavelength of light
= width of the slit

Step 3: Detailed Explanation:
The problem states that the wavelength ( ) is fixed. The separation between slit and screen ( ) is also fixed, though it only affects linear width, not angular width.
From the formula, the angular width is inversely proportional to the slit width ( ):

Let initial slit width be . Initial angular width is .
The new slit width is halved: .
The new angular width will be:




The new angular width is twice the original angular width.

Step 4: Final Answer:
The angular width becomes doubled.

Step 1: Understanding the Concept:
When monochromatic light passes through a narrow single slit, it undergoes diffraction, creating a pattern of bright and dark fringes on a screen. The dark fringes (minima) occur due to destructive interference of wavelets originating from different parts of the slit.
Step 2: Key Formula or Approach:
The general condition for minima in a single slit diffraction pattern is given by:

Where:
= width of the slit
= angular position of the minimum
= order of the minimum ( )
= wavelength of the light
Step 3: Detailed Explanation:
Given values:
Wavelength,
Slit width,
Order of minimum, (for the first minimum)
We need to find the angular position . Using the formula:


Substitute the values:



To convert this to a standard decimal:



Therefore, the angle is:

Step 4: Final Answer:
The angular position of the 1st minimum is .

Step 1: Use the relationship between the harmonics of the open and closed pipes.
For an open pipe, the frequencies of the harmonics are given by: where is the fundamental frequency and is the harmonic number. For a closed pipe, the frequencies are given by:
Step 2: Use the given frequency difference.
We are told that the frequency of the third harmonic of the closed pipe is 50Hz more than the fundamental frequency of the open pipe. This gives the equation: The third harmonic for the closed pipe is given by: So, we have:
Step 3: Solve for the fundamental frequency.
Solving for : Thus, the fundamental frequency is: