Final answer:
In a Young's double-slit experiment, the intensity of light at a point on the screen is directly proportional to the square of the amplitude, dependent on the color of light only in that color indicates wavelength, and characterized by interference patterns creating bright and dark fringes. Option A is correct.
Step-by-step explanation:
The student's question pertains to the Young's double-slit experiment, which is used to demonstrate the interference patterns of light, thus providing evidence of its wave-like properties. In this context, the intensity of light at a point on the screen is:
Directly proportional to the square of the amplitude of the light wave. The intensity, I, is proportional to the square of the amplitude, A, so I ∝ A². This means that as the amplitude of the wave doubles, the intensity will quadruple.
Dependent on the color of the light used. Different colors of light have different wavelengths. Even though color affects wavelength, the intensity depends primarily on the amplitude and the interference pattern.
Exhibition of interference patterns. When monochromatic light passes through two slits, it produces an interference pattern consisting of bright and dark fringes on the screen due to constructive and destructive interference, respectively.
The intensity of light at a point does not generally decrease with increasing wavelength. The claim that intensity decreases with increasing wavelength is not true in the context of the Young's double-slit experiment. The intensity is influenced by the amplitude and the interference pattern, rather than the wavelength of the light.
Thus, the correct statements about the intensity of light in a Young's double-slit experiment are its direct proportionality to the square of the amplitude, its dependence on the color of light to the extent that color reflects the wavelength, and its patterned distribution due to interference.