Week 5 PDF - Physical Optics

A screen with two slits is placed in front of a monochrome S light source. According to the Huygens principle, when light reaches these slits, it behaves like two waves in the same phase, emanating from sources S1 and S2 from here on. When these two spherical waves come together at a point on...

A screen with two slits is placed in front of a monochrome S light source. According to the Huygens principle, when light reaches these slits, it behaves like two waves in the same phase, emanating from sources S1 and S2 from here on. When these two spherical waves come together at a point on the screen, if they are in phase, they will interfere constructively and there will be maxima- bright fringes at that point. If they come in opposite phase, there will e destructive interference, and that spot remains dark. Thus, interference fringes consisting of dark and light lines are observed on the screen. The wave from a single source transforms into two waveforms in the same phase from the slits S1 and S2. In the picture, the distance between the slits is d and the screen distance is D. r1 and r2 are the paths taken by the two waves arriving at a point P at a distance y from the center O of the screen. A right triangle S1S2B is obtained from the point S1. If the screen is too far, i.e. paths r1, r2 are too large relative to distance d, the S2B side will be the difference between the two paths. In right triangle S1 S 2 B If this path difference is an integer multiple of a full wavelength (λ, 2λ, 3λ), there is constructive interference and the point P is bright. Constructive interference, Bright If this path difference equals odd multiples of half a wavelength (λ/2, 3λ/2, 5λ/2), there is destructive interference and the point P is dark. Destructive interference, Dark Let's make the formula more useful. If we draw a line from the middle point A of the slits in the figure to P and get the right triangle AOP. Since the two sides of this triangle are perpendicular to the two sides of triangle S1S2B, they are similar triangles and θ is equal. We can see rainbow colors on oil layers on soap bubbles or rainwater puddles. This effect is due to the interference of reflected light. It is a result of the interference of light reflecting from the front and back surfaces of a thin film layer as it enters our eyes. air If we examine a thin film of t thickness, some of the light falling on the surface is reflected from the front face, the other part is refracted and passes into the second medium. When the ray in the film comes to the back surface, some of it is reflected again and returns to the front surface. The part of this returning beam that comes out into the air interferes with the first reflected beam. If the film layer is denser than the medium behind: In this case, there is no phase difference in the second ray reflected from the surface. So, The phase difference between the 1st and 2nd rays is due to two reasons. ➢ the 1st reflection of 180o ➢ the 2t path difference taken during the round trip in the film environment Interference would be destructive if only the 180o phase difference in reflection were present. But this will be added to the distance difference caused by the round trip in the film environment. If the wavelength of light in air is λ, then the wavelength in a medium with a refractive index of n The phase difference is caused by 2 effects. 1. If the 2t path difference is equal to half multiples of λn, it becomes interference and light at that wavelength is observed at maximum intensity. (Maximum) 2. If the 2t path difference is equal to a multiple of λn, the interference will be destructive and the light at that wavelength will disappear, not observed. (Minimum) If the film layer is less dense than the medium behind;A phase difference of 180o will occur in the rays reflected from the front and back, both surfaces. Thus, the phase differences caused by the surface effect cancel each other and only the phase difference due to the path difference in the film medium is taken into account. In this case, the constructive and destructive interference formulas are reversed. So we use the maximum (constructive interference) formula for the minimum (destructive interference). The fact that only the blue color is reflected from one part of the thin film that reflects white light is due to the fact that all other colors are not observed in our eyes as a result of destructive interference. An oil layer with a refractive index of 1.5 is located on a puddle (n=1.33). What must be the minimum thickness of the oil layer in order for the 600nm wavelength yellow light to pass through without reflection in the white light coming from the air? Since the refractive index of the oil film is greater than the underlying water, there will be a 180o phase difference only to the 1st surface. Also, going through reflection means destructive interference. Then; If the path difference is equal to the full wavelength for M=1, it will be a multiple of the full wavelength for the other values of m. In addition, the wavelength in the oil medium At full multiples of this thickness, yellow light will not be reflected. In geometric optics, light was treated as linear propagating rays. If this rule were just right, the shadow of an obstacle placed in front of the light would have sharp edges. But in reality this is not so. If you look carefully at the shadow of a razor blade, there are second and third shadows on the edges. These shadows are called diffraction fringes. The general name of the effects caused by the deviation of light from the linear path is called diffraction. In addition to double slit interference, we can see the wave character of light even in the shadow of a single object. To understand diffraction, let's consider the single-slit diffraction experiment. In the figure, monochromatic light from a source falls on a screen at the back after passing an obstacle with only one slit on it. On the screen, a slit image is formed in the center. There are also weaker second, third ... slit images beyond the dark area on either side of the center. The slit width is important to understand the diffraction effect in this experiment. (We neglected the slit width in Young's experiment!) According to the Huygens principle, secondary waves are generated from each point of the slit on which light falls, and they propagate in the same phase. Since all these waves will come in the same phase to the central area on the screen, the central area is bright. If we examine the 1st dark fringe right next to the center, the rays coming from the various parts of the slit to this dark fringe all depart at an angle theta because the screen is so far away. If we match each point on the upper half of the slit with a point on the lower half, Since the wave coming out from the top point and the wave coming out just below the middle make destructive interference on the dark fringe on the screen, there should be a phase difference of λ/2 between them. After these two points, the two following points on the upper and lower halves of the slit should also have a phase difference of λ/2. So that λ/2 phase differs between all paired binary points. Phase difference between two matching points is (a/2)sinθ The 1. fringe condition is; dark fringe condition To find the ym distance of the fringes from the center on a screen at a distance D, dark fringe positions bright fringe positions A red light of λ1 = 700 nm and a green light of λ1 = 650 nm are sent together on a slit. Since the distance between the 5th dark fringe of the red light and the 3rd bright fringe of the green light is observed on the screen 2 m away, what is the width of the slit? Dark fringe formula for m=5 ; If the difference is 2 mm, Bright fringe formula for m=3 ; There is a limit to the clarity of the image that optical instruments such as microscope, telescope, spectroscope can give. This limit of sharpness is called the resolution or resolution of this optical instrument. The sharpness limit is due to the diffraction of light. Let's imagine two stars standing very close to each other. Let the rays emanating from these two stars pass through a slit of width a and fall on the screen. The on-screen display of each shows a distribution around a central maximum. If the central maxima of these two images overlap, they are indistinguishable. The resolution limit is considered as the boundary where the central maximum of one of the images coincides with the first minimum of the other image, as in the figure. First minimum angle; For small angles, radians can be used instead of sines. Accordingly, the smallest angle θmin that the optical instrument can distinguish The formula is valid for a rectangular slit. In many optical devices, camera lens are circular, so; Here a is the Radius of circle. The smaller the θm value, the higher the resolution of the instrument.