in this video we're going to focus on concave and convex mirrors the horizontal line is known as the principal axis this is a concave mirror and this is convex mur somewhere on the left side you have the focal point and also the center imagine if you turn this into a circle the center C is basically the center of that circle and the distance between any point of the circle and the center is known as the radius so therefore between the center of curvature and the mirror is the radius of curvature you need to know that the focal length is half of the radius of curvature the left side of the mirror is usually the front side that's where you place the object to the right this is the back side or this is behind the MER and the same is true for a convex mirror the left side is the front the right side is the back whenever you place the object in front of the mirror which is usually the case Dione is positive and whenever do is positive it's a real object if somehow the object were to be behind amiran it's really a virtual object and do is negative now the image let's say the image is over here let's just assume that it's in a position the distance between the image and the mirror is di if the image is in front of the mirror di is positive so di is positive this side if the image is behind the mirror di is negative whenever di is positive you have a real image and whenever di is negative you have a virtual image in addition the focal length F is the distance between the focal point and the mirror for a concave mirror the focal length is positive for a convex mirror the focal length is going to be negative so make sure you remember that fact because it's going to help you whenever you're solving a problem now the equation that relates the focal length with the object distance and the distance of the image is this equation 1 over F is equal to 1 over D L plus 1 over D I do is the distance between the object in the mirror the eye as you know is the distance between the image in a mirror now the next equation that we need to go over is the magnification equation magnification is equal to the ratio of the image height and the height of the object it's also equal to a negative di over do if the magnification is positive that means that the image is upright with respect to the object so let's say the object is up then the image would be in the upward direction it's upright if the magnification is negative then the image is inverted with respect to the object so this object is faced in the upward direction and the image is faced in the downward direction so we say that the image is inverted now if the absolute value of the magnification is greater than 1 the image is larger or taller than the object so the image is enlarged if the magnification is less than 1 the absolute value that is then the image is going to be reduced in size it's going to be shorter than the object so I wanted to go over the sign conventions before we work through a problem so let's say if we have a concave mirror and the focal length is 8 centimeters and we're going to place an object 24 centimeters away from the mirror so and let's say the height of the object is 4 centimeters with this information determine where the image is located how tall the images and also the magnification determine if we have a real image a virtual image if it's upright inverted enlarged or reduced so let's start with equation 1 over F is equal to 1 over D o plus 1 divided by di so for a concave mirror the focal length is positive so we're going to plug in positive 8 do is 24 now we're looking for di so whenever you have an equation with fractions and you want to solve for the missing variable you can clear away the fractions by multiplying both sides by the common denominator which is 24 di 8 and 24 can go into 24 so we can clear away every fraction so let's multiply 1 over 8 by 24 di we have to distribute the 24 di to every term in the parenthesis so 24 divided by 8 that's 3 so we have 3 di next we're going to multiply 24 di by 1 over 24 so the 24s cancel and all we have left over is just di and then we're going to multiply 24 di times 1 over di di will cancel and we'll simply get 24 so let's subtract both sides by di and let's make some space so these two variables cancel three di minus 1 di is 2 di and if we divide both sides by 2 you can see that di is 24 over 2 which means that it's 12 but it's positive 12 not negative 12 so since di is positive 12 do we have a real image or a virtual image whenever di is positive you have a real image now if di is positive this image will it form in front of the MER that's the left side or behind the mirror the back of the the back side or the right side since di is positive the image is going to form in front of the mirror that's the left side now what is the magnification magnification is negative di over do di is 12 do is 24 so the magnification is 1/2 so because the absolute value of the magnification is less than 1 the image will be reduced the image is going to be shorter than the object now is the image upright or inverted because the magnification is negative the image is inverted so if the object is facing the upward direction the image will be faced in a downward direction so now let's prove this by drawing a ray diagram but before we do that let's calculate the height of the image since the magnification is negative 1/2 you just got to multiply a negative 1/2 by H oh it's going to be a negative 2 but if you want to solve it use this equation so M is negative 1 over 2 H I we're looking for it H 0 is 4 so if you cross multiply negative 1 times 4 is negative 4 and 2 times H I that's what we have on that side then if we divide by two you can see H I is negative two it's half of H Oh so now let's see if we can draw a ray diagram so here's our concave mirror let's say this is the focal point and that's 8 centimeters from the mirror so this would be about 16 and this is 24 so let's draw the object now the first ray that we need to draw is going to be parallel to the principal axis but it's going to emanate from the object it's going to go straight to the mirror and then it's going to bounce back through the focal point now the second way that we need to draw is going to go from the object through the focal point let's do that again it's going to hit the mirror and then it's going to go that way so granted this diagram is not perfectly drawn to scale but we can see that the image is inverted and it's reduced notice that it's about half the size of the object it's shorter and the fact that the magnification is negative we see that it's inverted it's upside down now the focal point is 8 centimeters from the mirror and over here this is about 16 so we can see that the image is approximately in the right location because it needs to be about 12 centimeters from the mirror which is what we have here so it's between the focal point and the object by the way at 16 this is the center of curvature because that's twice the value of F the radius is always going to be 2f so that's all you need to do to draw the ray diagram for this particular example so whenever you have an object that's placed beyond the center of curvature the image that's forum is going to be a real image as you can see the light rays actually converge at that point it's going to be inverted because it's upside-down and it's going to be a reduce image because it's shorter than the object but now what's going to happen if we take the object and let's say if we move it on the center of curvature what type of image will form let's say the focal point is still 8 centimeters from the mirror and so this is going to be the center so let's place the object that to F from the mirror or at the center so we're going to draw the same to raise the first raise going to go from the object to the mirror and then it's going to bounce towards the focal point the second ring is going to go from the object through the focal point it's going to touch the mirror it's going to bounce back parallel to the principal axis so as you can see we have another image that's inverted but where it really should be if I drew this correctly it should be over here the image should be at the exact position as the object but inverted with the same size so it's not enlarged or reduced let's see if I can draw this better maybe this line is that we want that's bad my drawn is not to scale but it's supposed to be something like that if it's on the center it's supposed to be the same size as the object so you get a real image that's inverted now what's going to happen if we place the object between the center and the focus so we're going to draw the same race the first ray it's going to go straight to the mirror and pass through the focal point the second ring it's going to go through let's do that again just going to go through here and bounce back parallel to the principal axis so the image is between the center and the focal point but it's still inverted but notice that it appears to be a little bit larger than the object so it's enlarged but it still liberal image because the light rays they converge at that point but now what's going to happen if we place the object on the focal point so what's going to happen in this case if we try to apply the same method of drawing a very diagram as we did before we can draw the first ray but the second ray we can't draw because typically it would go from the object to the focal point but the the object is on a focal point so we can't draw the second ring so let's draw this ray backwards and say if it passes through the focal point and bounces back towards the object let's say if we have another ray that passes through the focal point and bounces this way these light rays they don't converge it turns out that if you place the object on the focal point you really don't get an image and if you use the equation it appears as if the image is that infinity for example the focal length is still 8 but the object is also 8 centimeters from the mirror so if you subtract both sides by 1/8 these will cancel so therefore 0 is 1 over di the only way this could be true is if a DI is infinity which means the image is so far that it doesn't exist now what's going to happen if we place the object between the focal point and the mirror let's say if we put it here what's going to happen so we're going to draw the first ray the same way it's going to pass to the object bounce remember through the focal point now we need to draw another ring and the second ray is going to go through the focal point and it's going to pass to the object and it's going to interact with the mirror and then it's going to bounce straight now notice that the light rays they don't converge anywhere the only time the intersect is at the object which is where they should intersect but notice what's going to happen if we extend these lines let's extend a blue line if we extend the yellow line notice that these two they intersect at this point by the way just in case you're wondering the Ray that we need to extend is the reflective ring which is the Ray with the arrow so we have to extend this ray to the other side and also this one the one of the arrow to the other side as you can see if we place the object between the focal point and the mirror we get an image that enlarged it's taller than the object it's also upright it's in the same direction as the object is faced in the upward direction and is it a real image or a virtual image this image is a virtual image for one reason it's formed behind the mirror keep in mind the left side is the front of the MER that's where the object is and the right side is in the back or behind the mirror so therefore di is negative behind the mirror and so we have a virtual image as you can see the light rays do not actually converge behind the mirror you can see that with the dashed lines they appear to converge but they do not actually converge whenever the light rays appear to converge as indicated with the dotted lines or the dashed lines do you have a virtual image whenever the light rays actually converge at a point to form an image the image is a real image and it's going to be in front of the mirror it's going to be on the left side as opposed to the back of the MER so just make sure you know those facts so now let's move on to our next example and that is the concave mirror now let's say that the focal length is 6 centimeters and the object is placed 3 centimeters in front of the mirror where is the image located and what is the magnification so let's use the same equation now keep in mind we have a convex mirror instead of a concave mirror so for convex mirror the focal length is negative so we need to multiply both sides of the equation by 6 di to clear away all three fractions negative 1 6 times 6 di the sixths will cancel and so it would simply be negative di 6 divided by 3 is 2 and for the last two these will cancel and we'll simply get 6 so let's subtract both sides by 2 di negative di o- 2 di is negative 3 di so 6 divided by negative 3 is negative 2 so the fact that di is negative means that we have a virtual image that it's going to form behind the mirror now let's calculate the magnification which is negative di divided by do di is negative to do streets so the magnification is positive two thirds two negative signs will convert into a positive sign so because the magnification is positive that means that the image is upright so if the object is facing upward the image is also faced in the upward direction but notice that the magnification is less than one so the image is reduced it's going to be shorter than the object the now let's draw a picture so first let's draw the principal axis and then the convex mirror so the focal point is 6 centimeters from the mirror so over here is the center and the object it's 3 centimeters from the mirror so should be somewhere over here let's make the object a little bit taller so the first ray that we're going to draw is going to be parallel to the principal axis is going to go from the object to the mirror and then if you draw a dashed line you can see that the Ray is going to bounce this way the mon line is not straight so let's do that again so that's better now we need to draw another ring we need to draw a line that passes through the object to the center so another library let me put it in green just going to go this direction and this is where the image is going to form so as you can see it's between the mirror and the focal point but it's behind the mirror keep in mind where the object is is the front of the MER so the right side is behind the mirror so because it's behind the mirror di is negative and so we have a virtual image and a modification we said was positive 2/3 it's positive so it's upright the image is in the same direction as the object but it's 2/3 the height of the object so as you can see it's shorter than the object its reduced and so now you have two ways to figure out if the image is going to be real a virtual upright or inverted enlarge or reduce you can simply use the equations keep in mind if di is negative it's a virtual image and it's behind the mirror and if M is positive its upright and if the absolute value of M is greater than 1 its enlarged as long as you remember those key details and the fact that the focal length is negative for a convex mirror but positive for concave Myr you can figure out anything you need even without drawing a diagram but it helps to draw the ray diagram if you remember how to draw it so that is it for this video thanks for watching and have a great day