In`YDSE`,how many maxima can be obtained on the screen if wavelength of light used is `200nm`and `d=700nm`:

Options

A. 12
B. 7
C. 18
D. None of these

Explanation

To solve the problem of how many maxima can be obtained on the screen in Young's Double Slit Experiment (YDSE) with the given parameters, we can follow these steps:

### Step 1: Identify the given values
- Wavelength of light, \( \lambda = 200 \, \text{nm} \)
- Distance between the slits, \( d = 700 \, \text{nm} \)

### Step 2: Understand the condition for maxima
In YDSE, the condition for constructive interference (maxima) is given by:
\[ d \sin \theta = n \lambda \]
where \( n \) is the order of the maxima.

### Step 3: Maximize the value of \( n \)
To find the maximum order of maxima, we consider the maximum value of \( \sin \theta \), which is 1. Thus, we can write:
\[ d = n \lambda \]
This leads to:
\[ n = \frac{d}{\lambda} \]

### Step 4: Substitute the values
Substituting the given values into the equation:
\[ n = \frac{700 \, \text{nm}}{200 \, \text{nm}} = 3.5 \]

### Step 5: Determine the number of maxima
Since \( n \) must be a whole number (as it represents the order of maxima), we take the integer part of \( n \):
- The maximum integer value of \( n \) is 3.

### Step 6: Calculate the total number of maxima
The total number of bright fringes (maxima) on the screen is given by the formula:
\[ \text{Total maxima} = 2n + 1 \]
Substituting \( n = 3 \):
\[ \text{Total maxima} = 2(3) + 1 = 7 \]

### Conclusion
Thus, the total number of maxima that can be obtained on the screen is **7**.

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