 Alright friends, so here is another question. This question is asked by one of our students as doubts. So I thought of making a video and discuss with you because this is a slightly different kind of question. So let's read this question. So as you can make out, this question is from the Ray optics chapter. So there is a cylindrical vessel over here whose diameter and heights they are equal. So it is 30 centimeter. So a point P is placed at a distance 5 centimeter from the center. So this point P location is given from the center and I is placed at a position such that the edge of the bottom is just visible. It means that if you have a straight line from the edge of this bottom, let's say this is point number one, this straight line will just pass through the top most portion of this vessel. Then only this point is just visible because if I slightly move down, then this particle or this edge will no longer be visible because there is no straight line connecting one point number one and the eye because the edge will come in between the top most edge point two. Now you need to find out up to what minimum height should the water be poured in the vessel to make the particle P visible. Right now is P visible. So if I connect point P and the eye, the vessel will come in. So the point P is not visible right now. But what will happen if you pour the water? Suppose if you pour the water like this, then because a light can bend now, there need not be a straight line connecting directly point P to the eye but light can travel in straight line in water and then it can travel in a straight line going further in air. So this situation can happen and then point P becomes visible. So we need to find out up to what height edge the water should be poured. Now let us draw this diagram little bit greater in size because we need to show a lot of angles and distances over here. So better to draw it enlarged way. So I draw a beaker or a vessel like this. Now see the location of eye is fixed. Position of eye, how will you get? If you draw a line connecting this point in the edge on this line somewhere the eye is located. Eye may be located at this point. So this is point one and let us say that you have poured the water up to this height. So this is the height of the water, we will call this as H. Now since there is a refraction going on at this point so it is better to show the normal over here. So this is the point P, let us say this is P over here. Now what should happen here is that ray of light that goes from P should refract and then travel in this line to reach the eye. So this is the path of the light from the object P. Now straight away this is an application of Snell's law isn't it? So let me call this as angle I and this as angle R. So at this point of refraction let us call this point number three. So at point three I can write down the Snell's law which is like refractive index of water into sin of I will be equal to refractive index of air into sin of R. Now what is R? R is also equal to that angle because this line is parallel to this line. So hence since this angle is 45 degree even R is equal to 45 degree and refractive index of air this is what one refractive index of water although it is not given in the question directly but I think it assumes that you know it which is 4 by 3. And angle R is also known here 45 degree. So basically what I get here is sin of angle I. So let me put the values so 4 by 3 sin of I is equal to sin of 45 which is 1 by root 2. So sin of I will come out to be 3 divided by 4 root 2. So let me call this as equation number one. Now I need to get the value of sin of I. So if I extend this normal like this this becomes a 90 degree triangle. If I say 3p and 4 3p4 is a 90 degree triangle. So if I know two of the sides I can get the sin of any of the angles in the triangle as a ratio of sides. So let me see what can be done here. So if this is edge if this is edge what do you think this length will be 1 to 4 1 to 4 length will also be edge why because this angle is 45 degree. So I know 1 to 4 distance so if I know 1 to p distance then I will get p to 4 distance. Now in order to get 1 to p distance I first draw a center line over here. Why I am drawing center line because see this 5 centimeter which is given over here that I want to use. So if let us say this is 0.5 I know 1 to 5 distance how much it is it is half of the diameter which is what 15 centimeter ok so 1 to 5 distance is 15 centimeter and I also know p to 5 distance. P to 5 distance is what p is place at a distance 5 centimeter from the center right so p to 5 is 5 centimeter ok. So I have the value of 1 to p as 1 to 5 distance minus p to 5 distance ok so this will be coming out to be 10 centimeter ok so 1 to p distance is 10 centimeter and 1 to 4 is edge. So p to 4 is what p to 4 is 10 minus h centimeter ok now I can write sine of i as what I can write sine of i as perpendicular which is p to 4 that is 10 minus h divided by the hypotenuse that is 10 minus h whole square plus h square fine. So this I get as sine of i now if I use sine of i over here I will get a quadratic question which if I solve I will get the value of h ok. So like this you have to attempt this question so I hope you have learned few things from this particular question in case you have any doubts please feel free to get in touch with us ok thank you.