so this video is all about ray diagrams so let's start with a spherical mirror particularly a concave mirror let's draw a horizontal line now this horizontal line is known as the principal axis and let's say this is the focal point the distance between the focal point and the mirror is known as the focal length and let's place the object outside of the focal point so let's put it there and so the distance between the object and the mirror that's d-o h-o is the height of the object so let's draw a ray from the object to the mirror i think i went too far let's do that again and then from the mirror let's draw a line to the focal point now let's draw another ray from the object through the focal point to the mirror and then it's going to bounce back and go in this direction so where the rays intersect that is the location of the image and so what type of image do we have would you say this is a real image or a virtual image now because the light rays actually converge at this point we have a real image and the distance between the image and the mirror is known as di for a spherical mirror d i is positive when the image is on the left and it's negative when it's on the other side and whenever di is positive you have a real image now the image is inverted as you can see it's in the opposite direction of the object the object is facing up the image is facing in the downward direction now notice that the height of the image is greater than the height of the object if that's the case then the magnification the absolute value is greater than one so we have an enlarged image because it's bigger than the object if the magnification is positive then you have an upright image if the magnification is negative you have an inverted image so the magnification has to be negative in this case because the image is inverted now let's look at another example using a concave mirror so let's say this is the focal point and this time we're going to place the object inside the focal point that is between the focal point and the mirror so the first phrase that i'm going to draw is going to go from the object to the mirror and then that ray is going to bounce back towards the focal point and then i'm going to draw an extension in this general direction now the center of curvature is twice it's located at twice the focal length so let's say the focal length is 5 the radius of curvature will be 10 and that's where the center of curvature is located so what i'm going to do is draw a ray that connects the object with the center of curvature and then past the mirror it's going to go in this general direction now granted my lines are not perfectly straight but at some point these two they will intersect and so if you place the object between the mirror and the focal point you're going to get an upright image it's upright because it's in the same direction as the object is facing upward and we can see that the image is enlarged in size it's bigger than the object and this image is a virtual image for one thing it's on the other side so di is negative also notice that the light rays appear to converge here but they do not actually converge the actual light rays are the solid lines if you see like a dashed line that's where the light rays appear to converge so that means that we have a virtual image as opposed to a real image now for this example let's use a convex mirror as opposed to a concave mirror and so that's going to be the focal point and here is the center of curvature and let's place the object on that side so we're going to draw the first ray from the object to the mirror and then it's going to bend towards the focal point now the second ring is going to go from the object to the center of curvature so notice that the rays appear to converge at this point and so what we have is a virtual image the image is upright and notice that it's reduced in size the image is smaller than the object so this is the height of the image and this is the height of the object this is d o and this is d i since d i is on the opposite side for mirror it's going to be negative which in the case that we have a virtual image since h i is less than h o we have a reduced image and here's some equations that you want to keep in mind that you can apply for mirrors or lenses one over f is equal to one over d o plus one over g i now this equation is referred to as the mirror equation for mirrors and it's also known as the thin lens equation for convergent lens and divergence so this is a concave mirror and the focal length is positive and the other spherical mirror that we talked about is the convex mirror where the focal length is negative because the focal length is negative d i will always be negative and so therefore a convex mirror will always produce a virtual image now some other formulas they need to know is that the magnification is equal to h i divided by h o so the image height is the magnification times the height of the object now the magnification is also equal to negative d i over d o and if you need to calculate the power of the lens it's equal to one over f and this is measured in diopters which is meters to the minus one so if you have the focal length in centimeters you need to convert it to meters now let's talk about lens so this is a convergent lens it's very big so let me draw a smaller version and here this is a divergent lens the focal length for a divergent lens will be negative but for a convergent lens the focal length is positive and so a divergent lens will always produce a virtual image for a convergent lens it can produce a real image or a virtual image depending on where the object is located relative to the focal length so let's start with a divergent lens so let's say this is the focal point let's place the object here so the first ray we're going to draw is going to go from the object to the center of the diverge lens and then this is going to go in that direction now let's draw the second ray from the object through the center of the lens so the image is located here the rays appear to converge here but they don't so therefore what we have is a virtual image it's reduced in size but it's upright now let's go over an example using a convergence so let's say this is the focal length or rather the focal point now let's put the object beyond the focal point so let's draw a line from the object to the center of the lens and then it's going to bounce towards the focal point the second ray is going to go from the object to the focal point until it reaches the center of the lens where it's going to change direction and then we can draw a third array from the object to the center and as we can see they're going to converge in this general region so the image is located here so in this example we have a real image because the rays they actually converge at this point it's inverted it's upside down but it's enlarged it's bigger than the object or at least it looks bigger and so those are some examples of how you can locate the image using ray diagrams and so that's it for this video thanks for watching you