In a Young's double slit experiment, the separation between the two slits is d and the wavelength of the light is `lamda`. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice (s).

Options

A. If `d=lambda`, the screen will contain only on e maximum.
B. If `lambda lt d lt 2 lambda`, at least one more maximum(besides the central maximum) will be observed on the screen
C. If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit2, the intensites of the observed dark and bright firnges will increase
D. If the intensity of light fallin on slit 2 is increased so that is becomes equal to that of slit 1, th eintensites of the observed dark and bright fringes will increase

Answer

To solve the problem regarding the Young's double slit experiment where the intensity of light falling on slit 1 is four times that falling on slit 2, we will analyze the situation step by step.

### Step 1: Understand the Intensity Relationship
Let the intensity of light falling on slit 2 be \( I \). Therefore, the intensity of light falling on slit 1 is \( I_1 = 4I \).

### Step 2: Calculate Maximum and Minimum Intensities
The maximum intensity \( I_{max} \) and minimum intensity \( I_{min} \) in the interference pattern can be calculated using the formula:
- \( I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \)
- \( I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \)

Substituting the values:
- \( I_{max} = (\sqrt{4I} + \sqrt{I})^2 = (2\sqrt{I} + \sqrt{I})^2 = (3\sqrt{I})^2 = 9I \)
- \( I_{min} = (\sqrt{4I} - \sqrt{I})^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I \)

### Step 3: Analyze the Conditions for Maxima and Minima
From the calculations:
- The maximum intensity observed on the screen is \( 9I \).
- The minimum intensity observed on the screen is \( I \).

### Step 4: Consider the Effects of Equalizing Intensities
If the intensity of light falling on slit 1 is reduced to equal that of slit 2, then both slits will have intensity \( I \). The new maximum and minimum intensities will be:
- \( I_{max} = (\sqrt{I} + \sqrt{I})^2 = (2\sqrt{I})^2 = 4I \)
- \( I_{min} = (\sqrt{I} - \sqrt{I})^2 = 0 \)

### Step 5: Conclusion on the Statements
1. If \( d = \lambda \), the screen will contain only one maxima (True).
2. If \( d \) is between \( \lambda \) and \( 2\lambda \), there will be multiple maxima (True).
3. If the intensity of the light falling on slit 1 is reduced to that of slit 2, the intensity of observed dark and bright fringes will increase (True).
4. If the intensity of the light falling on slit 2 is increased to that of slit 1, the intensity of observed dark and bright fringes will also increase (True).

### Final Answer
All statements regarding the Young's double slit experiment are correct.