Step 1: Understanding the Concept:
In Young's Double Slit Experiment (YDSE), light of a specific wavelength forms an interference pattern consisting of alternate bright and dark fringes on a screen.
The distance of the \(n^{th}\) bright fringe from the central maximum is given by:
\[ y_n = \frac{n \lambda D}{d} \]
Where \(n\) is the order of the fringe, \(\lambda\) is the wavelength, \(D\) is the distance to the screen, and \(d\) is the slit separation.
When two different wavelengths \(\lambda_1\) and \(\lambda_2\) are used, they create two overlapping patterns.
A "coincidence" occurs at a point on the screen when a bright fringe from the first wavelength is at the exact same location as a bright fringe from the second wavelength.
Step 2: Key Formula or Approach:
For the first coincidence of bright fringes:
\[ y_{n1} = y_{n2} \]
\[ \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \]
Since \(D\) and \(d\) are constant for the experimental setup, the condition simplifies to:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
where \(n_1\) and \(n_2\) are the orders of the coinciding bright fringes.
Step 3: Detailed Explanation:
Given wavelengths:
\(\lambda_1 = 480 \text{ nm}\)
\(\lambda_2 = 600 \text{ nm}\)
We set up the equality:
\[ n_1 \times 480 = n_2 \times 600 \]
To find the relationship between \(n_1\) and \(n_2\), we form a ratio:
\[ \frac{n_1}{n_2} = \frac{600}{480} \]
Now, simplify the fraction by dividing both numbers by their greatest common divisor (120):
\[ \frac{600}{120} = 5 \]
\[ \frac{480}{120} = 4 \]
So, the simplified ratio is:
\[ \frac{n_1}{n_2} = \frac{5}{4} \]
The "least number" of fringes for the "first coincidence" corresponds to the smallest possible integer values of \(n_1\) and \(n_2\) that satisfy this ratio.
Therefore, \(n_1 = 5\) and \(n_2 = 4\).
This means the \(5^{th}\) bright fringe of the 480 nm wavelength coincides with the \(4^{th}\) bright fringe of the 600 nm wavelength.
Step 4: Final Answer:
The least number of bright fringes of the 480 nm light required is 5.