In a Young's double-slit experiment, the ratio of the slit's width is 4:1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

In a Young’s double-slit experiment each slit behaves like an independent source of light. The light energy that finally reaches the screen from a slit is proportional to the geometric width of that slit, because both slits are illuminated uniformly by the same extended source.

We are told that the ratio of the widths of the two slits is 4 : 1. Therefore the individual intensities of light emerging from the two slits are also in the same ratio, i.e.

$$I_1 : I_2 = 4 : 1.$$

For any coherent source, the amplitude of the electric field is the square root of the intensity. Hence

$$a_1 = \sqrt{I_1}, \qquad a_2 = \sqrt{I_2}.$$

Taking the ratio, we have

$$\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{4}{1}} = 2.$$

Thus the individual amplitudes may be written, for convenience, as

$$a_1 = 2A, \quad a_2 = A,$$

where $$A$$ is some common amplitude factor.

The standard interference formula for the resultant intensity due to two coherent waves of amplitudes $$a_1$$ and $$a_2$$ is stated as

$$I = a_1^{\,2} + a_2^{\,2} + 2\,a_1 a_2 \cos\phi,$$

where $$\phi$$ is the phase difference between the two waves at the observation point.

For a bright fringe (maximum) the condition is $$\cos\phi = +1$$, giving

$$I_{\text{max}} = a_1^{\,2} + a_2^{\,2} + 2\,a_1 a_2 = (a_1 + a_2)^2.$$

For a dark fringe (minimum) the condition is $$\cos\phi = -1$$, giving

$$I_{\text{min}} = a_1^{\,2} + a_2^{\,2} - 2\,a_1 a_2 = (a_1 - a_2)^2.$$

Substituting $$a_1 = 2A$$ and $$a_2 = A$$ in these expressions, we get

$$I_{\text{max}} = (2A + A)^2 = (3A)^2 = 9A^{\,2},$$

$$I_{\text{min}} = (2A - A)^2 = (A)^2 = A^{\,2}.$$

Therefore the ratio of the intensity of a maximum to that of the adjacent minimum is

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{9A^{\,2}}{A^{\,2}} = 9 : 1.$$

Hence, the correct answer is Option 2.