 In this chapter, the concepts are mixed up like when the problem comes, they cannot make the problem difficult by having lot of depth in particular small concept. So what they do? They mix up like there will be multiple mirrors, there will be a lens also find together. So whenever you see a J question, it will be mix of the concept. To this question, you have a concave mirror of focal length 15 centimeter, object at 45 centimeter, height of the object is 10 centimeter or let us say 1 centimeter only. If you increase the height of object, you cannot have aperture small approximation. So height is 1 centimeter, you need to find nature of image as in where it will be and what is the size and whether it will be real inverted and all that. U is what, minus 45, focal length is what, minus 15. So what we will come out to be, minus 22.5, is this correct? What we drawn ray diagram, is this according to that or not, where is the object? Object is beyond R. So image should be between F and R, is this correct? So you should always cross check. This is V, magnification will be what? Minus of V by U, V is what, minus 22.5, object is what, U, U is minus 45, this becomes what? Minus half, magnification. So the image will be half of the object size and inverted and this is also according to the ray diagram. If it is real, it has to be inverted, now do the same thing with this type of mirror. The image will be what, virtual or real, virtual it has to be erect, so magnification has to be positive and less than 1, if you are getting otherwise it is wrong. So you can use ray diagram cases to check what you have done is right or wrong. So I am going to use mirror formula here, what is U, it is, what is F, now it is plus 15. What you get V as, plus 11.25, it has to be plus, it cannot be negative with this mirror. So V is this, magnification is what? Minus V by U, magnification has to be positive. So V is 11.25 and U is minus of 45, so it becomes positive and now you can solve it, so 1 by 4 right, 1 4th. So I will just introduce a different kind of problem, now we have done the basics. So suppose you have a concave mirror, fine, you have cut it into half, so 2 halves you have, so what you have done is you have placed the 2 halves like this, you have separated them. This is the principal axis, this was the principal axis actually right and you have moved these 2 mirrors, so that this distance become equal to delta, fine. You need to find out where the image of this object be focal length F, F is suppose 10, R will be what? R will be 20, fine. So where is the image of this object which is at 30 centimeter. Suppose this is a mirror, okay, this is a mirror, if I remove half of it, it is gone, not there. What will happen to image? Think over it, if half the mirror is gone, what happens to the image? Will it shift? It should not shift, because all the rays should converge at the same point, right, but there will be something happening, less number of rays will reach there now. Image will be slightly you know less intense, I mean the intensity of light will be lesser, that's it, that's all, okay. Now do it, yeah the horizontal distance won't change, why? See you can imagine that this is a part of complete mirror, and this is a principle axis for that, but object is not on the principle axis, where the object is here, right, but if you imagine that your object is this extended object, this much, then you can trace where is this point, okay, that is the hint. So I will just give you a hint, you can try it out later at home. See this is the object, right, what is the height of the object, delta by 2, fine, and height of the object you can say it is negative, fine. So you can find magnification for this location, and magnification times the height of object will be the height of the image, so you will be able to find out suppose magnification is let us say minus 2, and location is here, so the image will be like this, so here is your point, this point is this, so actually image is point, okay, similarly you will get a symmetrical point here for this, so there will be 2 images, okay, so if you draw the ray diagram, this was f and this was r for this, now f for this is here and r is there, okay, so this ray will go like this and come back, isn't it, similarly now r for this one is here, so this ray will go like this and come back, okay, and this ray which goes like this, what will happen to this, it becomes parallel, it passes through the focus, right, similarly this ray becomes parallel, fine, so this is the first image and this is the second image like this, okay, so this is like one of the trickiest of the problem you can see in this particular concept, so you can spend some time and try to solve it completely, right, so reflection is done, fine, we have 10 minutes, I will just give you brief about refraction, I will talk about laws of refraction, then we can study refraction in greater detail next class onwards, okay, write down refraction, what is refraction, if light changes its path as well as medium, it undergoes refraction, fine, now at times light may not change the path, but it can change the medium, even that is also refraction, but invariably whenever light changes the path, it undergoes reflection or refraction, so change of path, if it is not accompanied by change of medium, we say that it is reflection and if it changes the medium, it is refraction, now the amount of change of path depends on what, it depends on what is the property of medium and what angle the incident light is hitting on it, fine, now angle is very easy to define, but property of medium how will you be defined, right, suppose there is a material inside, suppose there is a glass, different types of glasses, now how the glass, I mean you say that glass is having some optical property, the optical property of mirror is easy to define, you will say that focal length is the optical property of the mirror and in terms of that you have studied the entire reflection, now optical property of the medium should also be defined, because now ray is going inside the medium, fine, so optical property of medium can be anything, you can define it any which way, but the most useful way is using speed of light, because ultimately light enters the medium, right, when it enters the medium, its speed changes, fine, so if we define property of medium in terms of ratio of speeds, that we can use it further, right, if ratio of speed is this, then that will happen, if ratio of speed of something else, then something else will happen, right, so in order to quantify the property of the medium, we are using ratio of speeds, okay, so refractive index of any material is refractive index, sorry, refractive index of any material is velocity of light in air divided by velocity of light in the medium, okay, now what is velocity of light in air, 3 into transfer rate and nothing can move faster than that, fine, so velocity of medium will be less than velocity of air, right, so this will be always greater than 1, fine, so refractive index which is denoted as n or at times represented as mu also, okay, now this refractive index, I mean you can say that this is absolute refractive index, because velocity of light is absolute in air, nothing can be faster than that, right, so you can say this is absolute refractive index and absolute refractive index is actually a relative one with respect to speed of light in air, fine, now you can also find out relative refractive index, refractive index of first medium relative to second medium, this you can write like this, velocity of light in the second medium divided by velocity of light in the first medium, this can be less than 1 or greater than 1, getting it, so if n1 is absolute refractive index of medium 1, how can I write it as vA by v1 and n2 can be written as vA by v2, right, hence n12, how it can be written as n1 by n2, right, so this is n1 by n2, find out on this, right, so we have quantified a property of medium, this defines what medium it is and how it will behave, fine, write down laws of refraction, first, first law is actually Snell's law, Snell's law says that refractive index multiplied by sine of angle in that medium and that angle is what, angle between the ray and the normal is a constant, fine, so I can say that n1 into sine of i1 is n2 into sine of i2, you know, you can keep on writing like this, this is what the Snell's law is, in our textbook it is written like this, you know, n1 sine i is equal to n2 sine r, fine, so when you draw the diagram, this is medium 1 represented as n1 and this is n2, suppose there is another medium n3, fine, so there is no limit, you can have multiple mediums, so this is n4, fine, so it says that if a light comes like this, this is what, i1, so n1 into sine of i1 is equal to, this is i2, n2 into sine of i2 and that is also equal to, this will be i2 only, fine, this is i3, it is equal to n3 into sine i3, fine, here also refraction will happen, this is i4 is equal to n4 into sine i4, simple, right, so but then if it is only one interface, then you can say n1 sine i is equal to n2 sine r, rather than writing i2 you can say refraction angle, okay, so this is how the Snell's law is, now look at it carefully, if n1 is less, if n1 is less and n2 is more, tell me angle i will be greater than angle r, angle i will be less than angle r, what it will be, n1 is less than n2, now tell me, i and r cannot be more than 90 degrees, between 0 to 90 degrees sine is increasing function, right, if n1 is less, sine i has to be more and sine r has to be less, so r has to be less than i and n2 is more, so light should bend towards the normal to make r less, right, so when you go from lower refractive index to higher refractive index, light will bend towards the normal, fine, if you go from higher refractive index to lower refractive index, it will bend away from the normal, fine, this is the normal direction, it has bended away from the normal, from the from it normal course, so n2 has to be less than n1, okay, we also say that for medium which has higher refractive index, we call them denser medium and lower as rarer medium, okay, denser or rarer does not mean mass density, okay, because there are materials which are mass density wise lighter, but optically denser, okay, terpentine oil and water, they are like one such example, fine, so this is the first law of refraction, Snell's law, second law, write down, second law says similar thing, incident ray normal and what refracted ray now, all lie on the same plane, fine, so these are the two laws of refraction, these law we will use extensively apart from those approximation where angles are small, that I will state again, so we will use these laws extensively to study entire refraction, okay, so next class we will continue from here.