In Young's double slit experiment, if the widths of the slits are in the ratio `4:9`, the ratio of the intensity at maxima to the intensity at minima will be

Options

A. `169:25`
B. `81:16`
C. `25:1`
D. `9:4`

Explanation

To solve the problem of finding the ratio of the intensity at maxima to the intensity at minima in Young's double slit experiment given the widths of the slits in the ratio of 4:9, we can follow these steps:

### Step 1: Understand the relationship between slit width and intensity
The intensity of light from each slit is directly proportional to the square of the amplitude of the light wave produced by that slit. The amplitude is also proportional to the width of the slit. Thus, if the widths of the slits are in the ratio \( w_1:w_2 = 4:9 \), we can express the intensities as:
\[ I_1 : I_2 = w_1^2 : w_2^2 = 4^2 : 9^2 = 16 : 81 \]

### Step 2: Calculate the amplitudes
Since the intensity is proportional to the square of the amplitude, we can find the ratio of the amplitudes:
\[ A_1 : A_2 = \sqrt{I_1} : \sqrt{I_2} = \sqrt{16} : \sqrt{81} = 4 : 9 \]

### Step 3: Write the formulas for maximum and minimum intensity
The formula for the maximum intensity \( I_{max} \) and minimum intensity \( I_{min} \) in terms of amplitudes is given by:
\[ I_{max} = (A_1 + A_2)^2 \]
\[ I_{min} = (A_1 - A_2)^2 \]

### Step 4: Substitute the values of amplitudes
Substituting \( A_1 = 4 \) and \( A_2 = 9 \):
\[ I_{max} = (4 + 9)^2 = 13^2 = 169 \]
\[ I_{min} = (4 - 9)^2 = (-5)^2 = 25 \]

### Step 5: Calculate the ratio of intensities
Now, we can find the ratio of the intensity at maxima to the intensity at minima:
\[ \frac{I_{max}}{I_{min}} = \frac{169}{25} \]

### Final Answer
Thus, the ratio of the intensity at maxima to the intensity at minima is:
\[ \frac{I_{max}}{I_{min}} = \frac{169}{25} \]

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