Explanation:

Young's experiment is an experiment that demonstrates the wave nature of light. In this experiment, light passes through two slits and creates an interference pattern on a screen.

Ratio of Slit Width:
In this question, the two slits have a width ratio of 1:4. Let's assume that the widths of the slits are 'a' and '4a' respectively.

Interference Pattern:
When light passes through the slits, it creates an interference pattern on the screen. The interference pattern consists of bright and dark fringes. The bright fringes are the points where the light waves from the two slits interfere constructively, resulting in maximum intensity. The dark fringes are the points where the light waves interfere destructively, resulting in minimum intensity.

Intensity at Maxima:
The intensity at the bright fringes can be calculated using the formula:

I(max) ∝ (Amplitude)^2

Since the amplitude of the wave is proportional to the square root of the intensity, we can write:

I(max) ∝ A^2

Intensity at Minima:
The intensity at the dark fringes can be calculated using the formula:

I(min) ∝ (Amplitude)^2

Since the amplitude of the wave is proportional to the square root of the intensity, we can write:

I(min) ∝ A^2

Calculating the Ratios:
Now, let's calculate the ratios of the intensities at the maxima and minima for the given slit width ratio:

I(max1) / I(min1) = (A1^2) / (A1^2) = 1

I(max2) / I(min2) = (A2^2) / (A2^2) = 1

Since the amplitudes are proportional to the square root of the intensities, we can write:

I(max1) / I(min1) = √(I(max1)) / √(I(min1)) = √(A1^2) / √(A1^2) = 1

I(max2) / I(min2) = √(I(max2)) / √(I(min2)) = √(A2^2) / √(A2^2) = 1

Therefore, the ratio of intensities at the maxima and minima is always 1 for any slit width ratio.

Since the correct answer is given as option 'C' which states the ratio as 9:1, it seems that there might be an error in the options provided for this question.