 Okay, all of you got 14th, a diminished image of an object is to be obtained on a screen one meter from it. This can be achieved by placing convex lens of focal length less than 0.25 meter concave lens. Now, first of all, this is getting formed on the screen, right? So since it is getting formed on the screen, it has to be a real object, okay? So plane mirror creates a virtual object, sorry, virtual image. So since it is formed on the screen, it has to be real image. So plane mirror is not possible. Even convex mirror forms virtual image. Even this cannot be put on the screen, right? And then concave lens also will create a virtual image. So it has to be option 3 only, okay? Now let's talk about the 15th one. An eye specialist prescribes practicals having combinations of convex lens of focal and 40 in contact with the concave, the power of the combination. So you can see that similar concept was tested in J earlier also. So simply you need to write down the combined power as 1 by F1 should be in meters plus 1 by F2. Now you have to take care of the signs also. It is a concave lens, so minus of 0.25 you have to write, okay? So I think you'll get what, minus 1.5, okay? This is equal to 4 and this is 10 by 4. So yeah, okay. So let's quickly move to the next few questions. So you might be seeing that majority of the questions are straight forward from this chapter. But sometimes questions get twisted. So there is a high chance that an easy question will come from this particular chapter. I mean these two chapters. So make sure you do good amount of practice because high chance that if you do some practice you will get the answer in the exam. These two questions attempt, okay? All of you getting question 16th answer as pi as in option number 3. It's a single slit experiment. Okay, let me solve question number 16. Those who have got 16 can move to question number 17. What would monochromatic beam of light incident on a narrow slit? So this is the narrow slit incident here. Diffraction pattern is formed here in this screen, okay? At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of this slit is what? The first minimum location you should know, suppose this is distance y, okay? This is y should be equal to lambda d by a where a is the width of the single slit, okay? Similar expression comes for the maxima in the double slit experiment. But for single slit experiment, this is what the case is, okay? The maxima happens to be at a distance of lambda d by a, okay? Now we need to find the phase difference between the rays coming from the two edges of this slit, okay? Now let's connect these two rays. So one is this one and other one is this. So we need to find the phase difference between these two, okay? Now we know the standard procedure of finding the phase difference, right? You drop a perpendicular on, all right? So there is a standard way to find the, you know, phase difference or path difference. You just drop a perpendicular like this and this is the path difference. This is delta x, okay? Now in order to find delta x, what we typically do is we connect this line like that, okay? And then, you know, right? This is the angle theta, then even this angle is theta. So delta x is this distance, which is a sin of theta, okay? And if theta is very less, then delta x, I can also approximate it as a tan of theta, fine? So this is delta x. Now corresponding to this delta x, the phase difference is 2 pi by lambda into delta x, so 2 pi by lambda into delta x, which is a tan theta. Now tan theta is what? Y divided by this distance, which is d, okay? Y is what? Lambda d by a divided by d, which is tan theta, okay? Now you see that lambda get cancelled, a get cancelled and d get cancelled, okay? So phase difference is 2 pi, okay? So 16th option number 4 is correct. Many of you have said 3, right? Okay. So I guess you have tried question number 17 also, let's see how we can go about it. Now this kind of question where there is no definitive option, as in it doesn't talk about distance being 5.4 centimeter or 20 centimeter, like that. It doesn't talk about a definitive statement, but it just tries to find out the estimate of it, where exactly it will be between which and which point. This can be done best by using ray diagram, okay? Now let us see how we can utilize ray diagram to answer this question. We have concave mirror, which is placed on the horizontal table. This is the horizontal table. So we have a concave mirror, like this. This is the concave mirror and axis is vertically upwards, okay? So this is the axis, O is the pole and C is the center of curvature. So let's say C is here, okay? A point object is placed at C only. So there is an object here, okay? So the image will be where right now? Image will be at C only, okay? Because whatever ray that comes out of C will hit the mirror normally. So the ray will retrace its path, okay? Now we are filling the mirror with water. Let us say this is the water, okay? So if I fill this with water, what will happen here is that the ray that was normal to the mirror, suppose this ray, if it keeps on going, then it will hit the mirror normally and the ray will just reflect back. But what will happen? Now because of the refraction, the ray will bend towards the normal, right? So if I zoom this, I will get a situation like this. Suppose ray is coming like this, this is the normal to the water-air interface. So light will just bend towards the normal, like this, okay? And this is the path for the normal, this thing. This is the path for the, as in, this is the path where it hits the mirror normally, okay? Now it will hit like this, okay? So for the mirror, the object appears to be somewhere there at the top, beyond C, okay? Then only light will appear to bend towards the normal. So the image is slightly away from the C, okay? So that is why when image is slightly away from C, the object will come closer to the, as in between O and C, it will go, right? So it will be a real image located between point C and O, all right? So this actually will be solved best if you are able to analyze the ray diagram, okay? Because there is no definitive statement over here. You just have to just, you know, evaluate the options, okay? All of you clear about this? Any doubt, type in yes or no, all right? See anyways, this is getting recorded, so you can always refer this any moment. These two questions start solving, okay? So what is the answer for 18th? Again it is a theoretical question, right? It's a theoretical one. All right, should we solve now? Fine. Question number 18. Once of all, yellow light will be like of the order of 500 nanometers, right? And this is a wavelength of yellow light. And wavelength of X-ray is like, I think, 1, 2, 100 Armstrong, okay? Or whatever it may be, you know that the wavelength of X-ray is very, very less than the wavelength of yellow light. A frequency of X-ray is very high compared to the frequency of yellow light, okay? Now in the young, sorry, in the single set experiment or in fact in the double set experiment also, both of the cases, the assumption is that the size of slit is comparable to the wavelength, okay? Now if you use X-ray, the wavelength is so less that for X-ray it is not a small hole. It's not a small hole where you can say that it's a single set experiment, right? It's not a narrow opening. For X-ray, it is like, you know, a huge opening. So there will be no fringes that will be observed. So that is the option number 4 is correct over here, okay? Because light will bend from the edges or diffraction will happen only when wavelength is comparable to the size of the hole, okay? Now to the 19th one, 19th, what do you think is the answer, okay? Now this is a thin slice of the cylinder, thin slice that cut out of the cylinder. So this cylinder actually, it's a part of cylinder that comes out of your screen, okay? So when you look at this ray, okay? And if you look at all the rays, suppose are hitting along the line which line is coming out of your screen, then all of these line will be similar, right? They will encounter same amount of path difference. They will have an, you know, refraction over here, like this it will go and then it will simply bend away from the normal and then reflection will happen. So all the lights which are in this line, the line which comes out of your screen will encounter this scenario, fine? So there is a line symmetry over here, fine? So there is a line symmetry. It is not a circular symmetry. That is why it will be straight, the fringes will be straight, okay? So these are, I think, these are not numerical. So you just have to know actually some concepts to solve these theoretical ones. But it is very easy to go wrong in these theoretical questions because, you know, you can think in any direction. So it becomes tricky at times, whereas numerical, you will get a definitive answer. So all these ones, oh, you are saying option D, fringe spacing increases as we go outwards. See outwards means you are coming out of the screen, okay? If it is straight, then there is no question of spacing when you are coming out of the screen because along the same screen, same along the one straight line, there is only one fringe, okay? In last option, it was not clearly written whether it is a straight or a circular fringe they are talking about when they talk about spacing, okay? Good that you have answered the second one initially. In a compound microscope, the intermediate image, we know intermediate image is inverted, real and magnified. So this is just knowledge-based question. If you have studied your NCRD properly, then you will get it immediately. Question number 20, are you able to read it properly because the font size became very less? Should I upload it again? So 20th question is on the lens maker formula. So we have this formula, 1 by F is equal to mu 2 by mu 1 minus 1 multiplied by 1 by R1 minus 1 by R2, right? Stress, okay, okay, fine. Now it is fine. Stress, are you watching on mobile phone? Anyways, it's fine. So we have a concave lens which has effective index 1.5. So basically, mu 2 is refractive index of the material, which is 1.5, okay? So same radius of curvatures were there, all right, in the concave lens. So it was like this. This is the concave lens, right? So basically for this, radius of curvature if you take minus R, and for that side you have to take it as plus R, okay? So R1 is minus R and R2 is plus R, okay? So when you immerse in a medium of refractive index 1.75, then what will happen? Let's say 1.75 minus 1, 1 divided by minus R, minus of 1 divided by R, okay? So this will become equal to 1.5 minus 1.75 to minus 2 by R, okay? So this is 1 by F, fine. So you are getting focal length as positive. So F is greater than 0. So focal length is positive means what? It means that when parallel rays come and hit this lens, it will get converged forward. Then only you will say that focal length is positive, right? So it becomes a converging lens, okay? So concave usually is diverging. But when the medium's refractive index becomes more than the material's refractive index, the same diverging lens becomes or it will act as a converging lens, okay? So it becomes a converging lens. So option 3 and 4, they are out of question, okay? All of you have answered third for this. F focal length is 1.75 divided by 0.25 into R divided by 2, okay? So this is 4 into 0.875 times R, which is 3.5 times R, okay? So question 20, option 1 is correct, okay? Any doubt, guys, on these two questions, anything you want to ask, how do you know R1 is minus R? See, R is geometrical property. It is a concave lens, right? So this side should be like this, okay? So central curvature will be on the left-hand side, okay? So if central curvature on the left-hand side, you are measuring distance from here. You are moving against the incident ray, right? So for concave surface, center of curvature is this side, right? So that is the reason why radius of curvature of this side is negative. We are using sine convention.