The decrease in intensity with increasing distance is explained by the fact that the wave is spreading out over a circular (2 dimensions) or spherical (3 dimensions) surface and thus the energy of the sound wave is being distributed over a greater surface area. ![]() Typical units for expressing the intensity of a sound wave are Watts/meter 2.Īs a sound wave carries its energy through a two-dimensional or three-dimensional medium, the intensity of the sound wave decreases with increasing distance from the source. Intensity is the energy/time/area and since the energy/time ratio is equivalent to the quantity power, intensity is simply the power/area. The greater the amplitude of vibrations of the particles of the medium, the greater the rate at which energy is transported through it, and the more intense that the sound wave is. The amount of energy that is transported past a given area of the medium per unit of time is known as the intensity of the sound wave. This relationship between energy and amplitude was discussed in more detail in a previous unit. Subsequently, the amplitude of vibration of the particles of the medium is increased, corresponding to an increased amount of energy being carried by the particles. The greater amplitude of vibration of the guitar string thus imparts more energy to the medium, causing air particles to be displaced a greater distance from their rest position. If more energy is put into the plucking of the string (that is, more work is done to displace the string a greater amount from its rest position), then the string vibrates with a greater amplitude. ![]() The amount of energy that is transferred to the medium is dependent upon the amplitude of vibrations of the guitar string. The energy that is carried by the disturbance was originally imparted to the medium by the vibrating string. The disturbance then travels from particle to particle through the medium, transporting energy as it moves. For example, a vibrating guitar string forces surrounding air molecules to be compressed and expanded, creating a pressure disturbance consisting of an alternating pattern of compressions and rarefactions. Magnification: of an image (m) is the ratio between the image and object height (hi/ho).Sound waves are introduced into a medium by the vibration of an object. The thin lens equation: 1/do + 1/di = 1/fįocal point: halfway between the mirror and the center of curvature on the principal axis.An ideal thin lens has two refracting surfaces but the lens is thin enough to assume that light rays bend only once. A thin lens is defined as one with a thickness that allows rays to refract, but does not allow dispersion and aberrations.Practice Exam 2 C/P Section Passage 10 Question 55 Section Bank C/P Section Passage 2 Question 11 Physics Question Pack Passage 8 Question 51 The lens strength can be measured in diopters and can be found by using the equation P= 1/f where f is the focal length and P is diopters of the lens. A virtual image is right side up (upright). Instead, you “see” the image because your eye projects light rays backwards. A virtual image happens when light rays do not meet at the image. A real image occurs when light rays actually intersect at the image and is inverted, or upside down. ![]() Like mirrors, the image formed by refraction may be real or virtual. The magnification m of an image is the ratio between the image and object height (hi/ho). Thin lens equation relates the object distance (do), image distance (di), and focal length (f). The treatment of a lens as a thin lens is known as the “thin lens approximation”.Ĭonverging lens: Parallel light rays entering a converging lens from the right cross at its focal point on the left (a).ĭiverging lens: Parallel light rays entering a diverging lens from the right seem to come from the focal point on the right (b). Another important characteristic of a thin lens is that light rays through its center are deflected by a negligible amount. A thin symmetrical lens has two focal points, one on either side and both at the same distance from the lens. Another way of saying this is that the lens thickness is much much smaller than the focal length of the lens. A thin lens allows rays to refract but does not allow properties such as dispersion and aberrations.Īn ideal thin lens has two refracting surfaces but the lens is thin enough to assume that light rays bend only once.
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