Score 100% Reflection of Light Numericals | Mirror Formula & Magnification

[00:00:04]Hello, welcome to Hi-Tech Academy.

[00:00:08]So in the last class, we discussed the entire theoretical part of Reflection of Light.

[00:00:20]So today in class, we will discuss the numericals in Reflection of Light.

[00:00:26]Okay, one second, I'm setting up a little bit.

[00:00:27]This is the numericals in Reflection of Light.

[00:00:33]Numericals. Okay. We are going to discuss about the numericals in Reflection of Light. Okay. So before that, before doing numericals in this, we need to know what the sign convention rules are. Only if we know the sign convention rules, you can do numericals. So I request everyone to join the session Okay so I request everyone to join the session I am also adding this in the group so wait a minute like and share please like and share I request everyone to join the session I request everyone to join the session please wait for a while wait a little bit we will start the class in five minutes. So today we are going to do all the numericals. We will have a complete explanation of numericals in this class. So don't miss this class, you won't have any problem with numericals.

[00:01:43]You can do it without any problem.

[00:01:48]Okay, so I request everyone to join the session. Please join the session.

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[00:02:11]Start in five minutes and the class is still so request everyone to join the session already watching okay so please comment here or watching please comment comment okay so let's go to the topic so our topic is about Cartesian sign convention rules only if we know these sign convention rules only if we know these sign convention rules only if we know these sign convention rules only then we can solve the problem okay so what are these sign convention rules all the distances are measured from a pole so if you take this here you will see this concave mirror look here this concave mirror okay so this is a concave mirror concave mirror so this is a pole distance measured from a pole all the distances are measured from a pole wait okay so here so this is A concave mirror this is a pole this is a pole so we measure from this so all the distances are measured from pole distances are measured in the direction of incident ray are taken as positive and away from incident ray are taken as negative heights above principal axis are positive heights below principal axis are negative I will explain this to you clearly see.

[00:03:46]So look here so this is a ok so this is a concave mirror which mirror it is concave mirror so the geometric center what do we call this geometric center of a mirror the geometric center of a mirror is known as it is pole it is pole and next horizontal horizontal passing through then pole is known as principal axis what do we call this it is known as principal axis ok so principal axis this principal axis is there now example i will place one object i will place one object here i am taking an object ok i will place on object here i am placing the object here ok so wherever we place the object so before that we will find the center of Let's see what curvature is.

[00:04:42]So geometry center is not set.

[00:04:52]So here is this. So center of the curved sphere is not center of curvature.

[00:04:56]This is focus. Okay. Center of curvature and focus. I place the object beyond center of curvature.

[00:05:06]Where I place the object is. The dash is placed beyond center of curvature. When I place beyond center of curvature. When I place light, I call incident light ray. Okay. So when light is incident, this is called incident ray. Okay. When a light is incident parallel to principal axis, where does it get reflected?

[00:05:25]It gets reflected toward focus. When light is incident parallel to principal axis, where does it get reflected toward focus? When light is incident parallel to principal axis, where does it get reflected toward focus? When light is incident parallel to principal axis, where does it get reflected toward focus? If you do an incident to the focus, where will it be reflected? Parallel to the principal axis. Parallel to the principal axis. So where is the image formed? Here is the image. This is the image. Okay, this is the image.

[00:05:50]Okay, I will place it. I dash. This is the image. So if you look here, you see this. Pole. Pole. Here is the object.

[00:06:01]Pole to object. Pole to object. Pole to object. Object distance.

[00:06:08]So how does this incident happen? Incident.

[00:06:14]Reflect. Object distance. Is it away from pole? Everything that is away from pole is negative. So here is the object.

[00:06:22]Negative. Okay, where is the next image? This is the image.

[00:06:25]Look here. This is the image. So if you take the image, it is also away from pole.

[00:06:32]So image distance. Pole to image also gives you image distance denoted with this also comes minus okay pole to image comes minus okay image distance also minus okay negative means to you simply put these are this screen this screen this beyond screen leave it in that definition leave it all this screen this beyond screen distances look here distances distances on screen remember that all distances on screen should be negative.

[00:07:08]That means the object is on the screen, right?

[00:07:12]Don't write the object distance hole to object here for a minute.

[00:07:15]Okay pole to object pole to image pole to center of curvature pole to focus okay so what are all these distances these are these distances so distances if you take this mirror where do you have the distances the object the object is on the screen so it is on the screen so the minus image is also on the screen. So -v okay so center of curvature means pole to center of curvature is radius of curvature so pole to center is also on screen is on screen so minus a pole to focus ego focus focal length focal length f this is also on screen so minus a for this for this is ok same heights let's see we are done with distances now let's see heights heights heights o dash height of object i dash height of image so where is height of object is above principal axis is above principal axis see here is above principal axis. So the height above the principal axis is positive and the height of the image below the principal axis is negative. Okay, so for a concave mirror, this is what happens. But concave mirror comes like this but if you take it, he will tell you in a simple way that in concave mirror the distances which are on the screen are negative beyond the screen and positive beyond the screen should be taken positive. Okay, I told you this simply, let's take this to the concave mirror next to the convex mirror. If we take the next convex mirror, see if it is a convex mirror. Hi hello, someone is asking B.Tech coaching bro, okay, so what is the geometric center of a sorry, this is not convex, this is concave, so what is convex? The curved faces come inside.

[00:09:23]We have already done this in the previous class. Okay so the geometric center of a mirror is known as the pole horizontal passing through the pole okay let's take the horizontal small. Horizontal passing through the pole is known as what it is known as it is known as the principal axis. What is this called? Okay, so what is the geometric center called?

[00:09:50]Pole Horizontal Principal Axis Here are two curved surfaces.

[00:09:54]If we draw a circle, we get this one and this one.

[00:09:58]We should take this as focal point f1, center of curvature c1, and the distance from this to the convex side is the same as this f2, this c2. Okay f2 and c2 okay c2 so let's place the object where I will place the object I will place it on the focus I will place on object atfocus so I placed the object on the focus okay so I have placed an object onfocus so it is focus on object is placed on focus so it is return yes height of object is return yes o o dash so first ray okay first ray first ray is parallel to principal axis let's say okay first ray is parallel to principal axis what is it diverging what is it diverging. Soven light is incident parallel to the principal axis what will happen to it it will get diverged where is the incident ray sorry incident ray ref reflected ray so what happened it diverged extend the diverging ray extend the diverging ray extend it extend it go where does it meet it it is extended and it is meeting towards focus means if it passes parallel where does it get reflected towards focus let's see second ray ok second second ray i will place with other color so let's see second ray incident we already know towards pole light is incident towards pole it is reflected away from pole with same angle it is reflected away from pole away from pole with Same angle same angle Let's extend this now if we extend this so where are they meeting these extended races here they are meeting beyond the screen look it is meeting beyond the screen what is this beyond the screen beyond the screen what is this screen so beyond the screen so beyond the screen so beyond the screen it is denoted with positive positive distances beyond the screen what are positive I already told you that if it is negative if it is in the screen so look here where the image fell here the image fell here the image fell here the image so the image is formed beyond the screen so the image formed beyond the screen so so let's take the image here this image so image i i dash what is this image i i dash image is formed so if you look here if you look here wait a minute okay so if you look here so object distance pole to object object distance pole to image image Distance pole to center of curvature Radius of curvature Pole to focus Focal length OK so object Distance pole to object This is the incident ray towards it, see that it is on the screen. Okay, away from the incident ray is on the screen, what do I say, what is it on the screen? I said that the distances which are in the screen are taken are negative, so this is negative, okay, and where is the image, it is beyond the screen, if it is beyond the screen, then the radius is positive, so we measure the radius like this, so this is also positive, the focal length is also positive, when the radius is positive, the focal length is also positive, and let's see the height of the image, the height of the object, h, not ho, o dash, o dash is the height of the object, i dash is the height of the image, so the height of the object is above the principal axis, so it is positive, here, so it is taken as positive, the height of the image is also above the principal axis, so this also comes out positive. Okay, so these are the sign convention rules, so tell us the logic behind this and all this. If you want to tell us in simple terms, I'll write it in simple terms. Look here for concave concave pole to weight concave pole to object pole to image pole to center of curvature pole to focal length o dash i dash height of image. So if you look here, here, we can see the weight here. You can't see it. Let's write it a little further away.

[00:14:39]Pole to object Pole to image Pole to center of curvature Pole to focus O dash I dash So pole to object Pole to image Pole to center of curvature O dash Object distance is denoted by u Image distance is denoted by v Center of curvature Pole to center of curvature Radius of curvature Pole to focus Focal length o dash Height of object i dash Height of image So if you look here, object distance is negative Image distance is only plus or minus When does plus come? That is, when the object is placed between pole and focus, the image goes beyond the screen, beyond the mirror, beyond the mirror, what happens to you Image distance Image distance comes positive, okay, so remember that one thing.

[00:15:28]Radius of curvature Negative Focal length Negative Height of object Positive Height of object Positive Height of image Plus or minus I This is the same for concave If you take convex If you take convex mirror OK Let's take convex Pole to object Pole to image Pole to center of curvature Pole to focus NextO Dash I Dash So Pole to object Object distance U Pole to image v Hole to center of curvatureR Hole to focus f o Dash Height of object i Dash Height of image So If you look here you see that all the distances Object distance is negative Image distance Positive Radius of curvature Positive Focal length Positive Height of object Positive Height of image Positive OK In mirror In a concave mirror, only the object distance is negative, everything else comes out positive. Okay, keep that in mind and do it. Okay, so sign conventions are over. What is the next next? So sign conventions are over. Let's explain it clearly. For a concave mirror, this is concave and this is convex.

[00:16:55]Okay, concave and convex. So here is the next very important topic. So what is the main topic?

[00:17:03]Important topic is the mirror formula.

[00:17:07]Mirror formula.

[00:17:09]Just so we know the formula. Object distance.

[00:17:11]This is u. And what is the image distance? What is the image distance? Okay, so this is the mirror formula.

[00:17:22]Mirror formula is over. So next, let's look at the magnification formula.

[00:17:27]Next, which formula is magnification?

[00:17:34]Magnification is denoted with m small m Let's see its formula Magnification m = hI/HO = -v/u hI/H = -v/u What does hI mean Height of object Height of object hO means hO means height of image Height of image OK So height of object is real OK So let's do problems Now we will come to it Ya Ya Ya It comes to it Clear OK Right Next This is the magnification formula The whole story is on these two formulas Now and here we will see Let's see the problems So first problem And having its focal length What is 15 cm focal length If it is a concave mirror, then it is called a concave mirror So if we go according to sign conventions If we look at sign conventions, then it is a concave mirror The Is this a concave mirror So concave If we go to the mirror sign conventions, object distance is negative, image focal length is negative, okay, object distance is focal length is negative, height of object is positive, look here, look here, focal length is negative, object distance is negative, height of object is also negative, so if we go to the problem, okay, now let's go to the problem, so here we have to take height of object plus, object distance is negative and focal length is also negative, so what do we have to find out, we have to find HI and find height of image, so what is the mirror formula that we have to use here, mirror formula, what is mirror formula, here is = sorry, sorry, sorry, sorry, sorry, sorry, 1/f = 1/v+ 1/u, so let's substitute in this, so what is 1/f? -15= 1/th is what that only has to find, we have to find out 1/u is what is -25 so if we do this 1/-15= 1/th okay 1/v + * -1/25 so let's transpose this what happens if we transpose this way and that 1/v = 1/ -25 if we go to +25 it becomes 1/25 -1/15 like that. Okay, so what should we do here? If we write this clearly, we get 1/v = 1/25 -1/15. So what is the LCM here?

[00:20:45]25 15 LCM is 25 15 LCM So 5 * 5= 25 5 * t= 15 So 5 * t= 5 5 * t= 3 is not a count, so 3 * 1= 3 * 1= 5 * 5= 25 25= LCM is 75. So here what is the LCM 75 is LCM okay so wait one second so how much is the LCM we get how much is 75 is LCM so how much is the LCM let's take 75LCM okay 25 * how much is 25 5 sorry sorry sorry 25 t= 75 minus see here is minus - 15 5 75 so therefore this can be written as 3-5 -2/75 what is 1/f = what is sorry 1/v= so i want v v = reciprocal what is -75/2 so how much is this 2 *v= 2 2 *t= 6 see once clearly see this problem here so height size of image what is given 4 cm given object distance -25 minus because it is concave mirror so its sign conventions There are minuses and the focal length is 15 cm. According to the sign convention, we take -15. We need to find the height of the image. We need to find the height of the image.

[00:22:15]And here we need to find the image distance.

[00:22:17]Okay, we need to find the height of the image and the image distance. We got the image distance. If we substitute it in the mirror formula, we get this solution. How much did we get? The image distance is 37.5 cm.

[00:22:34]How much did we get? We got 37.5 cm.

[00:22:38]What should we do now?

[00:22:40]After this, we need to find the height of the image.

[00:22:43]What do we need to find? We need to find the height of the image. So what formula do we use to find the height of the image?

[00:22:49]We will use magnification here.

[00:22:54]So what is the magnification formula m So what is the formula for m = i/h=-v/u So here you have to find m=hi i How much is h i /h 4 cm m We have given in the question itself How much is -37.5 Here -v So -that is -37.5/ / u How much is the object distance How much is the object distance in the question How much is the object distance in the question So if we check it we get 25 cm How much is 25 cm So if we substitute this 25 cm in this How much is we get Minus -25 cm So Minus minus cancels So Minus minus cancels What happens We get h i =- -37.5/ 25/4 So how can we write this simply If we send this h = this 4 Can we write 4*37.5 as -375/10? If we write -375/10 and cancel it again, we get 37.5. How much is left below this? There is 25 here. Look here. 25* 25 okay. So if we solve this, what will we get? If we cancel this, 25 * 25 25 37 goes into 25, 12 is left. There is 5 next to 12. 125 25 5= 125 okay. So 25* 15 375 comes okay. 5* 2= 10 5 t= 15 2 * 2= = 2 2= 4. So how much is left above here? hI = 2. This is minus.

[00:24:44]So there is minus. Look here. There is minus.

[00:24:45]So -3. So what is the height of the image?

[00:24:48]You got -6cm. Okay. You got -6cm.

[00:24:53]What is the image distance?

[00:24:56]37.5. How much is the height of the image? -6cm. What is the height of the image? -6cm. What is the height of the image?

[00:25:04]-37.5cm. 5cm.

[00:25:08]So the height of the image here is If it comes in negative, what is the height of the image? It is negative. If it is negative, mark it as a blind mark. If it is negative, mark it as a blind mark. If you are negative, fix it.

[00:25:20]You are the same. It is negative. It means it is real inverted and real inverted. If it is negative, it is real inverted. What is the height of the object here?

[00:25:33]We have a height of object of 4 cm. Look.

[00:25:35]What is the height of the object here? It is 4 cm. Look. It is 4 cm here. So the height of the object is 4 centimeters so 4 is greater than 6 so according to mathematically do not count as -6 we should count only the minus indicates real inverted here is greater than 6 so if we have enlarged image real inverted enlarged image so if we draw a ray diagram for real inverted enlarged image let's draw a ray diagram here so if we look at the ray diagram so this is a mirror this is a concave mirror and geometric center of a mirror is not known geometric center of a mirror is not known pole right so no it is not so big okay we call this principal axis and what do we call it geometric center geometric center we call this pole this is called principal axis its center center of Curvature is this focus okay so where is the object this object focal length is look here what is the focal length here it is 15 cm. If the focal length is 15, what will be the radius? It will be 30 cm.

[00:26:58]What will be the radius? It will be 30 cm.

[00:27:01]What is the object distance? 25cm. What we have seen so far is 25 cm, so 15 is not 25 cm, so it is inside C, so it is between F and C. So if we draw this, here is the object, right? Object is this object, okay? So object is this, so first rule is parallel to principal axis, where is it reflected, this is the focus, okay, second rule is parallel to principal axis, so where is the image? So if you look at the image, you can see where the image is. Here, let's make the image a little thicker. So the image is here, see intersecting beyond the center of curvature. So this is what is this. This is o dash o dash. Height of object. This is i i dash. What is the height of image. So the image is bigger than the object, see here, enlarged image.

[00:28:22]Okay, so let's see the next problem. Let's see the next problem.

[00:28:39]So what is the next problem? On object 6 He said cm in size so so what is the object size here 6 cm okay size of object is 6 cm so height of object is concave mirror is it not concave mirror he said so positive and it is placed at a distance what distance what object distance 20 cm means -20 cm and focal length given focal length not given what is given radius of curvature given so r = -20 cm given we know one condition r= 2f f= = focal length = r/2 hence f = 20/2 -20/2 hence what is focal length minus -10 cm comes focal length so focal length is given so what we need to find out v find out height of image. So what formula should be used? Here, the mirror formula should be used. So what is the mirror formula 1/f = 1/v+ 1/u so we need to use this and solve the problem. So look here so if we follow this we will see what is 1/f 1/eva 1/focal length how much is 10cm look here so look here 10cm is 1/-10 = 1/vp 1/ ya what is the object distance how much is -20 here so -20 so -20 so if we expand this a little bit if we do the calculation so look here 1/-10= 1/vp e -1/20 if we send this -20 here then 1/v will be -1/20 if we come here it will be +1/20 it will be -1/10 already. If we write this, 1/v = 1/20-1/10 so let's do the lcm of 20. 20 10 So 10 * 2= 20 10 * 1= 10 2 * 1= 2 * 1 So 10 * 2= 20 lcm What is the lcm of this 20 This lcm So 1/v= 20 lcm 20 * What is 20 * v = 20 See here 20 * v = 20 So 1 - 10*2= 20 So 1/v= 1-2 -1/20 Therefore v is the reciprocal -20 cm So what is the object distance? We have given the object distance as 20 cm and the image is also 2020 meters. So what should we do next, let's do magnification, that is, we need to find the height of the image. What is the formula m = h/h=-v/u hI/h = -v/u So if we substitute this, that is the only way to find the same, we have to find out the height of the object. What is the height of the object? Here, what What are the characteristics of the image? It is real and inverted. Real and inverted.

[00:32:23]Height of object is 6 centimeters. Height of image is 6 centimeters. So it is same sized image. Same sized image.

[00:32:32]Okay. So, let's do this problem.

[00:32:50]One second. Next, you can draw its ray diagram.

[00:33:19]I have already explained ray diagrams. So, its ray diagram comes at the center of curvature.

[00:33:25]Because look here, the object is 20 centimeters in radius. So, the object is near the radius of 20 centimeters. So, that's how we use the image. Okay right so if you subtract it, what is it? Ray diagram is subtracted, so what is the geometric center of a mirror?

[00:34:01]Geometric center is not a horizontal pole passing through the pole is not a principal axis. Okay, this is the pole weight. This is the pole and this is a principal axis and this is the center of curvature and this is the focus. So where is the object?

[00:34:23]Where is the object? The object is placed at the center of curvature. So first ray is parallel to the principal axis so it is reflected towards the focus. Correct exact dimension cannot be taken. That's why it came like this. So atsi means the image is also formed atsia. Okay, atsi means atsia.

[00:34:48]O dash, I dash. Okay, this is its diagram. Okay, so let's move on to the next problem.

[00:34:56]What is the next problem?

[00:35:00]Here, an object of size 3 cm is placed in front of a concave mirror at a distance of 10 cm, having a radius of curvature of 20 cm. So, what is the height of the object? What is the distance between the object and So what do we have to find out here we have to find out v we have to find out h so the mirror formula is 1/f = 1/v+ 1/u okay so if we do this if we do this so what is 1/f 1/-10= 1/v+v/ -10 u is also -10 so 1/10= 1/vp e minus -1/10 okay so if we expand this and simplify it so this -1/10 here if we transpose it 1/v it will be = -1/10 will become +1/10 -1/10 +1/10 -1/10 cancel now 1/v= 0. So what happens is 1/0 1/0 what does it mean undefined infinity undefined or infinity infinity approaches infinity image approaches infinity image remember 1/th= 0 when v= reciprocal 1/0 ok approaches infinity so then if we want to use magnification formula and find out height of image m=h/h = -th/u formula -th/u formula but here h we have to find out same find out hth how much did we give height of object how much did we give 3cm height of object how much 3cm = v how much did we get minus infinity / u how much -10 infinity / anything what is infinity / anything infinity okay infinity / anything how much infinity next this 3000 If we send it, Hi = infinity * 3, infinity * 3 is also infinity, so the image falls near infinity, the image falls near infinity, so infinity, so infinity, as I already said, means that the object is at focus, right?

[00:37:47]Look at the ray diagram, as if the object is at focus. So how does the image fall when it approaches infinity?

[00:37:51]What are its characteristics? Real inverted and it is highly magnified image. Highly magnified image is taken. This is okay. So if we look at the ray diagram of this. If we look at the ray diagram of this. This is a concave mirror and the geometric center of a mirror is known as the pole and the horizontal is known as the principal axis. This center. This center becomes the center of curvature. This becomes the focus. So where do we place the object? We place the object at the focus. The object is on the focus. So first rule is parallel to the principal axis. Reflected towards the focus.

[00:38:43]Second rule is towards the pole. So where does the reflection become parallel to the principal axis? So these two meet here. So they are parallel.

[00:38:53]Look at what is said here, these are parallel, parallel, so parallel means what I said, infinity image, infinity, let's see how it is. Extend it, if you extend it, they will not meet even beyond the screen. It means they meet on the screen itself, it means they are parallel, it means they meet near infinity, so the image is formed at infinity. Okay, incident reflection, so it meets near infinity. So that's why this is a real inverted enlarged image. Real inverted enlarged image.

[00:39:27]Okay, okay, let's go to the next problem. Right, next, object of size 5 centimeters. What is this object of size 5 centimeters?

[00:40:15]Let's make this a little bigger.

[00:40:29]What is the size of the object? 5 cm.

[00:40:33]So h = 5 cm + 5 cm. It is placed at a distance of 10 cm. The object distance is 10 cm, so what is the focal length? It is 25 cm.

[00:40:46]So what is the focal length? Find out v, find out h, and find out h. So what formula should we use here we have mirror formula so what is mirror formula 1/a = 1/v+v/u so how much is 1/a there 1/- 25= 1/vps in bracket 1/- 25 object distance sorry sorry sorry sorry 1/10 object distance how much is 1/10 object distance this is -10 ok so if we do this 1/- 25= 1/v + * -1/10 ok so this is -1/10 here what happens if we transpose this so if we transpose this becomes +1/10 +1/10 we can write 1/10 as -1/25 already is = 1/v so this 1/v = 1/10 -1/25 so what is 10 25 lcm so 10 25 lcm 5 * 2= 10 5 * 5= 25 2 * th = 2 * 5 * 1 = 1 5 * 2= 10 * 5= 50 lcm How much is 50 lcm Okay So lcm is 50 1/v= 50 lcm 10 * 5= 50 25 * 2= 50 So 1/v= 5 - 2 3/ 50 So if we do this reciprocally, what is the reciprocal of 1/v? What is the reciprocal of 1/v?

[00:42:19]v So therefore v = 50/3 v = what is 50/3 here, if we cancel this, if we cancel this, then 3 th 3 10 30 35 36 48 36 48 Let's take a different color 36 482 is 20 3c 18 e is the same 16.6 comes out if we prolong it = 16.6 centimeters 6 centimeters is in positive ok next we need to find the height of image ok for this we need to find the height of image so what is the magnification here we have so magnification m = h/h = -v/u so if we solve this simplify h / height of object how much is this how much is this height of object height of object 5cm see here 5cm given ok so 5 cm so so h = what find what place 5cm = -v -th how much is this got 16.6 got we did that already 16.6 got v value so 16.6/ object distance u how much is this object distance here 10 cm so 16.6* 6* 10 cm / -10cm minus minus cancel so If we simplify this, the weight If we simplify this, this 5 will go away hI= 5e 16.6/ / 10 10* 1.66 So how much does 5* 1.66 come out 5*v= 5c= 33 8 Something comes out Calculate 5* 1.66 5 * 1.66 So how much does 8.3 come out 8.3 3 centimeters So it comes out in positive Height of image also Height of image also came out in positive Height of image distance How much is the height of image distance We got image distance v = 16.6cm So height of image is positive and image image distance also came out positive So whenb is positive it is virtual erect and 8.3 3 Height of object and height of height of image 8.3 3 Height of object is 5cm See here 5cm is smaller than 8 means the image is bigger than the object so what happens is virtual erect and enlarged image virtual erect and enlarged image comes ok so next let's give this ray diagram so this is concave mirror and geometric center what happens geometric center of a mirror is known as it is known as pole and horizontal passing through the pole is known as principal axis this is pole and this one will be principal axis this is center of curvature and this will be the focus ok so where is the object virtual erect enlarged image means we already know where is the object between pole and focus object is here so first rule parallel to principal axis Reflected Towards Focus Okay Next Second Rule Second Rule Let's take Green Second Rule Towards Pole Okay Towards Pole Reflected Parallel to Principal Axis So Here Reflected Look Here This Is Incident Ray Wait So This Is Incident Ray This Is Reflected Ray Here This Is Incident Ray This Is Reflected Ray So Here Let's Extend The Reflected Ray Okay Reflected Ray Meet Here On Screen Meet Beyond Screen Is This Its Reflected Ray This Is This Its Reflected Ray Next What Is This Its Reflected Ray This Is Sorry This Also Extended So How Is The Image Formed Here It Formed See The Image Is Formed Here So What Will This Be Image I 8.3 cm So it fell beyond the screen, right? It fell beyond the screen, so it is a virtual erect enlarged image, right? Virtual erect enlarged image, right? Okay, so next problem, this problem you do, this is a homework problem for you, this problem you try, okay, okay, find its focal length, right? I'll tell you, look, this is a problem I'll tell you, this is another problem after this, you try, right? Height of 5 cm in size, height of object, what is 5 cm in size, is this okay? It is placed in front of a concave mirror at a distance, object distance from concave mirror is -30 cm, and from the mirror is 30 cm, what size object is 5 cm in size, 5 cm in size? It is placed in front of a concave mirror at a distance of 30 cm. He said, "From the mirror at 20 cm." Okay, so what is the image? This is 20 cm.

[00:48:47]This is not from the mirror. Forms on image.

[00:48:49]This is forms on image. Image from the mirror. Okay, correct me. He said, "From the mirror." If it is a mirror, where did the image fall? From the mirror means that it fell from the pole. Image from the mirror means that it fell here. Towards the mirror means that it is a virtual image.

[00:49:15]Since the image from the mirror is a real image, minus 20 should be taken.

[00:49:17]Remember this. From the mirror.

[00:49:19]From the mirror.

[00:49:21]Suppose he said, "From the mirror." It is negative, which means that it is on the screen. And towards the mirror. Suppose he said, " Towards the mirror." It is positive, which is here. The image is here. The image is positive. What is the fifth? It is a positive and virtual image. Is it okay? So what do you want to find out?

[00:49:39]Here you want to find out the focal length.

[00:49:41]Just find out the focal length.

[00:49:43]So we know the formula. 1/f = 1/v+ 1/u 1/v+ 1/u So 1 /f = 1/-20 +1/-30 Okay. So if we do this, 1/f = 1/-20 -1/30 So if we look at this, then what happens? This is 1/f = 20 30.

[00:50:12]We have to do the LCM of 20 30. So 10*2= 10 * t= 10*2 = 30= 10 * 20 * t= 60. So there is a minus.

[00:50:22]So 20t= 60 - 30 2= 60. So 1/ 1/a = -5/ 60. So therefore. What is the reciprocal of f= f= 60 - 60/5 5th 5 how much is left and what is left is 10 left 52= 10th f=-2 centimeters f= -2 centimeters So this is the focal length of a mirror What is the focal length of a mirror Okay so let's do the next problem This is a very important problem There are chances of it being asked in the exam Important problem This is okay Very important problem There are chances of it being asked in the important Four marks Okay we have done five problems Now we will only do it on concave mirror Next let's do it on convex mirror Okay let's do this problem too This is also a very important problem This is also a very important problem Let's do this too Okay Write the image of on Object formed by concave mirror has magnification is -1 Before telling this problem, we should know the concept of magnification. Let us see the concept of magnification. Let us see. So, let us see. So, let us tell you a little logic. Logic is easy. Logic is okay. Logic for magnification.

[00:52:04]See here. Magnification. In the exam, you are asked m = 2.75. Right asks the nature and characteristics of the image. He says nature and characteristics of the image. He asks the image. So what do you say? You are confused.

[00:52:26]How is it? He says m = -1. He says m = -1.75. He says m = -0.75. He says 75. So you are confused. Instead of being confused, let me show you simple logic. See here. So this is the logic of magnification. So see here. So that is m.

[00:52:49]Nature of image and mirror. Okay.

[00:52:57]Nature of image mirror. So that is =p1 = -1 >p1 > -1 <1 < -1. Okay. Less than +1 < than -1. Okay. So a little logic. See here. Very important.

[00:53:24]Magnification. Plus. If it is seen. Plus. If it is seen blindly. Virtually erect. That is virtually erect. Image.

[00:53:33]Plus. If it is seen blindly.

[00:53:36]Virtually erect. Write virtually erect. Okay. Erected. Is it straight. Is it inverted? So if there is a plus, it is virtual erect, right. If it is equal to plus, it is virtual erect, right. If it is minus, it is blind. If it is minus, it is blind. What should we write as blind? Real inverted, real inverted, right? Real inverted, right? This is also real inverted, right? Real inverted. So what is mirror here? Mirror what is mirror? Let's do it now. Not now. Wait. Let's do it like a line table. We are drawing a table. This table is very important. Okay. So, right. So, virtual erect, real inverted, virtual erect, real inverted. We know how.

[00:55:11]If the magnification is plus, it is virtual erect. If minus, it is okay. If minus, it is real inverted. If minus, it is real inverted. Now let's see. If =th, it is the same sized image. If =th, what is the same sized image? If =th, what is the same sized image?

[00:55:34]So =th, this is =way, this is also the same sized image?

[00:55:42]Next greater Danva Gedanva means +1.75 + 2.75 + 3.75. Anything can be. So it is greater than one, plus indicates virtual erect, okay, plus indicates virtual erect, minus indicates real inverted, if it is greater than one, what is it? If it is greater than one, it is an enlarged image, what is it? The enlarged image here is greater than one, so enlarged here is another one, what is it? The following is also greater than one.

[00:56:12]Greater than is this a greater than symbol so it is enlarged image enlarged image ok less than is < is 0.75 + 0.75 -0.75 75 which means there is nothing i.e. it should be less than one if less then it is diminished image this is also diminished ok it is also diminished image so a little logic will tell you another thing see if you get a virtual erect same sized image in the mirror then this is a plane mirror remember plane mirror is ok plane mirror and real images all mirror ok next virtual erect next virtual erect diminished virtual erect diminished think what if you get virtual erect diminished What is a convex mirror?

[00:57:15]Convex Virtual Erected Diminished Only Convex means that convex mirrors form a virtual erect diminished image in any position. Okay, so this is the concept of mirror magnification.

[00:57:29]This is how magnification is written. If we know this concept, see here. So here see magnification. So here see question.

[00:57:39]Magnification is minus. Minus indicates real inverted =. It means same sized image. Okay, next magnification is +1. -1.75.

[00:57:56]This is so minus indicates real inverted.

[00:58:02]1.75 is bigger than one. So it is enlarged image.

[00:58:09]Next, see this.

[00:58:11]See this. Magnification is how magnification is. It is in minus. It is in minus so this is also real inverted and 0.75 75 is less than one so this is a diminished image diminished image okay so he will give it like this so next let's take another question in which that magnification magnification m = p3.75 75 is plus if it is virtual erect plus if it is virtual erect and 3.75 greater than two so if it is greater than two then it is enlarged image if it is smaller than two then what is the enlarged image ok m value is 0.75 is plus if it is virtual erect virtual erect less than one 0.5 5 is less than one so it is diminished image so this is the magnification concept.

[00:59:20]So what is the question here?

[00:59:24]What is the question here? An image of height The image of object formed by the concave mirror is m = -1. Okay, what is the object distance minus the concave mirror, so it becomes -60. He said, "Find the image distance and characteristics of image and its focal length for the concave mirror." So, what is the magnification formula? We have -v/u. So what is it? We have m. What is it? -1 = -a -60/ u. Okay, so what happens? This is -1= - * - +60/u.

[01:00:01]So, all these things come and go. So, u = 60/ -1. Therefore, u = -60 cm. What is the value of u? It will be -60 cm.

[01:00:15]We have u. So we have u. So what should we find here? He said, "Find the image distance." So simply remember that when m = -1, what is m = here, when it is -1, this is the real inverted same sized image, same sized image, so wherever the object is, the image will also be the same.

[01:00:45]The image will also be taken by ATCA. So what should we do then what is the simple answer to this is this object is on what is the object on what is the object on this so the object is on the radius i.e. its radius is also -60cm radius is -60 then r= 2f what is the focal length r/2 means 60/2 -60/2 so what is f -30cm f -30 cm so there are chances of asking such questions. Okay, so look at this, it is also very important.

[01:01:31]Next, I will tell you a problem about a convex mirror. The rest of the problems will be homework. The rest of the problems will be homework.

[01:01:35]Okay, right, weight, weight for one. So what is the question here?

[01:01:59]What is the question on an object of size 4 cm? What is the height of the object? 4 cm is placed in front of a convex mirror at a distance of 25 cm. Okay, and having its focal length, what is the focal length? 15 cm. So, find the nature and characteristics of the image. We need to find the height of the image.

[01:02:17]Here, however, we need to know the sign conventions of a convex mirror. If we also go into the sign conventions of a convex mirror, we will find the distance of the object. So this is the sign conventions we did in convex one object distance is negative all the rest are positive height of object height of image all positive so we can make it simple this is ok so if we make it simple let us do so problem so mirror formula is 1/f = 1/v+ 1/u so 1/f = 1/v + * -1/25 -1/25 so what will this be so 1 this -1/25 if we transpose here it will become +1/25 so 1/v = -1/25 will become +1/25 so it will be like that. So 25 15 we have to take the LCM of 5 * 5 = 25 5 * t = 15 5 * t = 5 3 * 1 = 1 5 * 5 = 25 25 * t = 75 So what is the LCM of 1/v = LCM of 75 25 75 15 5 75 This is a plus, it is not a minus So what happens then 1/v = 8/75 What happens 1/v = 8/75 So what happens to v Here you have v = 75/8 8 * t = 8 8 = 64 8 9 723 30 8 t = 24 So there is no need to do it yellow We just need to get one number after the point So how much v we got + 9.3 cm There is nothing so as it is said plus it got v We have got v What do we need to find out next We need to find out h I need to find out Height of image v Now we need to find out the height of the image. So to find out the height of the image, what formula should be used? We need to use the magnification formula. So magnification = hI/ hV=-v/u So if we substitute here, we have to find out h in this same by hO What is the height of object Here the height of object is 4 cm What is 4 cm Object distance is 25 cm and image distance is 9.3 We already got it So here h/h 4 cm = -v -9.3/u 3/u - 25cm Okay So if we do this, minus minus cancels out So if we transpose this here 4 nihI = 4* 9.3/ 25 So we can write this hI = 4* 9.3 ni 93/10 93/10 means 9.3 3 becomes * 25 Okay So if we cancel this in the table 25 25 25 25 = 75 375 That is how much more The remaining 80 98 18 is the remaining 25 25 120 25 7 145 25 8 25* 10= 250 Okay 25 4 100 25C 150 25 7 comes 3.7 Something So we do n't need to do the eject click, it's enough to do it. So how much does it come? So hI= 4E 3.7 So 3.7*4 How much does it come? 74 28 43 12 134 So 14.8 So 14.8/10 So hI= 1.48 Okay Wait Wait Wait What is the weight? Did something small happen? What didn't happen?

[01:06:34]1.48 centimeters So how much is the height of the image?

[01:06:37]We got 1.48 centimeters So how much is the image distance?

[01:06:44]9.3 centimeters So how much is the image distance?

[01:06:49]9.3 3sm It came in plus It came in plus. What is the actual height of the object? The height of the object is 4 cm. Look here, it is 4 cm. But what is the height of the image? We get 1.48.

[01:07:06]1.48 means that the height of the object is greater than the height of the image. Image is smaller Image is smaller image But remember one thing here Height of image is if in plus If in minus Real inverted If in plus Virtual Erected Virtual Erected Image Virtual Erected Image and Virtual Erected Height of image is smaller than height of object So diminished diminished image Virtual Erected diminished image So if we look at this ray diagram we should take convex mirror Okay So geometric center of a mirror is not its pole and horizontal passing through the pole is not its principal axis This is pole and this is principal axis From this we should take this curved sphere So its center is c1 Its focus is f1 This is f2 From this How much distance is it? Okay, that's the distance, take f2, that's the distance, take c2, so look at where the object is.

[01:08:25]Here, what is the focal length? The focal length is 15 cm. I'll write it down. I'll tell you as I write it. Look, what is the focal length? The focal length is 15 cm. The object is 25 cm. What is the radius?

[01:08:40]When the focal length is 15, what is the radius? It is 30 cm. That means the object is 30 x 25, sorry, 30 x 15. That means the object is 30 x 25. Sorry, the object is 15 x 15. So the object is placed between the pole and focus. Sorry, the focus and center of curvature. So this is the height of the object. So the first rule is parallel to the principal axis. Write this color. What? The first rule is parallel to Where does the principal axis get reflected? This is the focus. Second rule.

[01:09:29]Where does the focus get reflected? Parallel to the principal axis get reflected.

[01:09:31]Oh, sorry, sorry, sorry, sorry, so where should I place the object?

[01:09:43]I want to place the object here. This is the screen, right? Okay, so this is the screen. So I'm placing the object here. Where will the object be? Between pole and focus.

[01:09:50]So, the object placed here is the first rule.

[01:09:54]First rule. Parallel to the principal axis get reflected. Diverge. Diverge. So, we have to extend the diverging ray. So, where did it get extended? This is the focus.

[01:10:09]Where did it get extended?

[01:10:12]This is the focus. So, this is the dash. The height of the object. Okay, so, this is the incident. Ray this is reflected ray second rule what is it towards pole towards pole second rule towards pole incident oh sorry so towards pole incident so where is it reflected this this is also away from pole with same angle away from pole is reflected with same angle so let's extend this now reflected ray this incident ray this is its reflected ray this is its incident ray this is its reflected ray this so if we extend this so where does the image fall here here intersected image person this intersected here here intersected image so here intersected so where does the image fall you so beyond mirror beyond mirror this is image where image fell beyond mirror fell.

[01:11:24]I I dash image this is the screen this is the screen this is beyond mirror or beyond screen beyond screen or beyond mirror image is beyond screen so that's why it's virtual virtual image and above principal axis see this is the principal axis if this is the principal axis if this is the principal axis then what is the image so what is the image so so this is the erect image above the principal axis if the erect image size of object see now height of object is big height of image is small see here here see height of image height of image is small here so small so it is diminished image is an image diminished image so like this like this numericals like this we have to do a problem okay and okay so this is a problem next wait wait this Okay, this is a problem.

[01:12:40]Okay, look, first, second, third, fourth, fifth, sixth.

[01:13:00]Put all these problems in. So, concave, convex, same questions, but in the place of concave, that's all. I put convex in place. Take this as a reference and put all these problems in. Okay, I'll put it slowly. You take a screenshot. It's easy. This is the third problem. This is the second problem. This is the fourth problem. This is the fifth problem.

[01:13:18]This is the sixth problem.

[01:13:20]So, take these problems as a reference. Take a reference and do it like how concave was done.

[01:13:26]Give me the answers in the comments. Okay, so today we completed this class. So I think this class is completely understood. So this is very important. From this class, we will most likely get a long answer question. Okay, so a long answer question.

[01:13:59]Okay, a LAQ will be asked. There are chances of asking a LAQ. Let's go. Okay, there are chances of asking a long answer numerical LAQ of six marks. R Itla Table R Question Okay Asking and Hypothesis Question Asking and Hypothesis Hypothesis Question He asks about magnification. Even though he talks about the concept of magnification, does magnification equal 2.75 75? He asks about the nature of image. So there are chances that questions like this will arise. What does a Ray diagram say? He will give you a problem and just ask you to draw a Ray diagram. So you all stay through it. So those who haven't watched this live, those who haven't watched this live, watch it completely. Okay, so I request everyone to like, share and subscribe. Please do like, share and subscribe. Please do like, share and subscribe. And whoever your friends are, mainly government school students.

[01:14:49]Okay, whoever you are, I will have a class completely for all of you. Mine will be daily. I will tell you this time from 5:30 to 6:30. So don't miss this class at all. So, today the physics topics are over, right? Tomorrow there won't be a class. Maybe there won't be a class.

[01:15:04]Next class will come and we will start from real numbers in mathematics or at least from trigonometry.

[01:15:13]So stay tuned daily. Please stay tuned daily regularly. So please do. Like Share and Subscribe Thanks for watching my channel Okay Thanks for your support Okay So please do Like Share and Subscribe So I am going to end the stream So I request everyone to watch complete video Watch this entire video. Don't skip at all.

[01:15:41]Okay, so don't skip it at all, watch the entire video.

[01:15:47]Each and everything is important.

[01:15:49]Many important concepts are mentioned in this. So very important. So you don't miss it at all.

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