In Young's double slit experiment, assume intensity of light on screen due to each source alone is `I_(0) and K_(1)` is equal to difference of maximum and minimum intensity. Now intensity of one source is made `(I_(0))/(4) and K_(2)` is again difference of maximum and minimum intensity . Then `(K_(1))/(K_(2))=`

Options

A. 4
B. 3
C. `3/4`
D. `2`

Answer

To solve the problem, we need to determine the ratio \( \frac{K_1}{K_2} \) based on the given conditions in Young's double slit experiment. Let's break down the solution step by step.

### Step 1: Understanding Intensity in Young's Double Slit Experiment
In Young's double slit experiment, the intensity \( I \) at a point on the screen due to two coherent sources can be expressed as:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
where \( I_1 \) and \( I_2 \) are the intensities from the two sources, and \( \phi \) is the phase difference between the waves from the two sources.

### Step 2: Finding Maximum and Minimum Intensities
The maximum intensity \( I_{\text{max}} \) occurs when \( \cos \phi = 1 \):
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
The minimum intensity \( I_{\text{min}} \) occurs when \( \cos \phi = -1 \):
\[ I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2} \]

### Step 3: Calculating \( K_1 \)
The difference between maximum and minimum intensity is given by:
\[ K_1 = I_{\text{max}} - I_{\text{min}} = 4\sqrt{I_1 I_2} \]
Assuming \( I_1 = I_0 \) and \( I_2 = I_0 \), we have:
\[ K_1 = 4\sqrt{I_0 I_0} = 4I_0 \]

### Step 4: Modifying One Source's Intensity
Now, we change the intensity of one source to \( \frac{I_0}{4} \). Let’s denote the new intensities as:
- \( I_1 = I_0 \)
- \( I_2 = \frac{I_0}{4} \)

### Step 5: Calculating \( K_2 \)
Now we find \( K_2 \) using the new intensities:
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} = I_0 + \frac{I_0}{4} + 2\sqrt{I_0 \cdot \frac{I_0}{4}} = I_0 + \frac{I_0}{4} + 2\sqrt{\frac{I_0^2}{4}} = I_0 + \frac{I_0}{4} + I_0 = \frac{9I_0}{4} \]
\[ I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2} = I_0 + \frac{I_0}{4} - 2\sqrt{\frac{I_0^2}{4}} = I_0 + \frac{I_0}{4} - I_0 = \frac{I_0}{4} \]
Thus,
\[ K_2 = I_{\text{max}} - I_{\text{min}} = \frac{9I_0}{4} - \frac{I_0}{4} = \frac{8I_0}{4} = 2I_0 \]

### Step 6: Finding the Ratio \( \frac{K_1}{K_2} \)
Now we can find the ratio:
\[ \frac{K_1}{K_2} = \frac{4I_0}{2I_0} = 2 \]

### Final Answer
Thus, the ratio \( \frac{K_1}{K_2} \) is:
\[ \frac{K_1}{K_2} = 2 \]