To solve this problem, we need to find the ratio of maximum intensity to minimum intensity in Young's double slit experiment, given that the intensity of light coming from one slit is double the intensity of the other slit.
Let's denote the intensity from the first slit as \( I_1 \) and from the second slit as \( I_2 = 2I_1 \). In Young's double slit experiment, the resultant intensity \( I \) at a point on the screen where interference occurs is given by:
\(I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\delta)\)
where \( \delta \) is the phase difference between the two waves reaching that point.
The maximum intensity, \( I_{max} \), occurs when \( \cos(\delta) = 1 \):
\(I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2}\)
\(= I_1 + 2I_1 + 2\sqrt{I_1 \times 2I_1}\)
\(= 3I_1 + 2\sqrt{2}I_1\)
\(= I_1 (3 + 2\sqrt{2})\)
The minimum intensity, \( I_{min} \), occurs when \( \cos(\delta) = -1 \):
\(I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2}\)
\(= I_1 + 2I_1 - 2\sqrt{2}I_1\)
\(= 3I_1 - 2\sqrt{2}I_1\)
\(= I_1 (3 - 2\sqrt{2})\)
The ratio of maximum to minimum intensity is:
\(\frac{I_{max}}{I_{min}} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}\)
Multiplying numerator and denominator by the conjugate of the denominator:
\(\frac{(3 + 2\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}\)
\(\frac{(3 + 2\sqrt{2})^2}{9 - (2\sqrt{2})^2}\)
\(\frac{(9 + 12\sqrt{2} + 8)}{9 - 8}\)
\(\frac{17 + 12\sqrt{2}}{1}\)
Hence, the ratio is \(34\).
Therefore, the correct answer is 34.