water waves in a ripple tank are diffracted by a narrow aperture.if the wavelength of the water waves is decreased slightly whilst keeping the aperture size the same ,which one of the following will happen?
the waves will spread out more
the waves will spread out less
the amount of diffraction will be the same
diffraction will stop
As the wavelength decreases, the spreading of the wave at the edges of the aperture also diminishes =>
the waves will spread out less
the waves will spread out less
If the wavelength of the water waves is decreased slightly while keeping the aperture size the same, the correct answer is: "the waves will spread out more."
To understand why this occurs, we need to consider the principles of diffraction and the relationship between wavelength and diffraction. Diffraction is the bending and spreading of waves as they encounter an obstacle or pass through an opening.
In this case, the narrow aperture acts as the obstacle for the water waves. When the waves pass through the aperture, they diffract or spread out. The degree to which the waves diffract depends on the relationship between the width of the aperture and the wavelength of the waves.
According to the principle of diffraction, when the wavelength of the waves decreases (while keeping the aperture size the same), the waves will spread out more. This is because the water waves now have a shorter distance to travel between the edges of the aperture, resulting in a larger angle of diffraction. As a result, the waves will exhibit more pronounced spreading or diffraction patterns.
Therefore, the correct answer is that the water waves will spread out more if the wavelength is decreased slightly while keeping the aperture size the same.