Final answer:
The ratio of intensity of the dark fringe to the bright fringe in a double slit experiment with wave amplitude ratio 3:4 is calculated to be 1:49, which does not match any of the provided options.
Step-by-step explanation:
The question pertains to the topic of wave interference, specifically in the context of a double slit experiment that demonstrates the interference pattern of light or other waves. When dealing with the interference of two waves, the intensity of the resulting wave at any point is proportional to the square of the overall amplitude of the wave at that point. Given the ratio of amplitudes of the two waves as 3 to 4, to calculate the ratio of the minimum intensity (dark fringe) to the maximum intensity (bright fringe), we use the formula for constructive and destructive interference.
For constructive interference (bight fringe), the amplitudes of the waves add up, so the resultant amplitude is 3 units + 4 units = 7 units. The intensity, being proportional to the square of the amplitude, will therefore be 49 units (since 7 squared is 49). For destructive interference (dark fringe), the amplitudes subtract, giving us a resultant amplitude of 1 unit (4 units - 3 units). The intensity in this case will be 1 unit (since 1 squared is 1).
Therefore, the ratio of minimum to maximum intensity is 1:49. However, since this is not one of the provided options and based on the given question which may contain a typo with 'slides with his 4', which should likely read 'sides with ratio 4', the correct option would be 1:4, assuming 4 is the intensity of the maximum and 1 is the intensity of the minimum, contradicting the calculations above. Hence, without clear options matching the calculated result, it is not feasible to select a correct option from the provided choices.