Class 12 Physics in 30 Days|Day 29|Complete Crash Course|Board 2025|Most Important Topics + PYQs#pyq

Added: 2026-05-09

[00:00:11]So we are in chapter eight. Seven chapters we completed already. So chapter in the chapter the number of problems and the spectrum electromagnetic radiations you have gamma ray xray ultraviolet visible infrared microwave radio wave application frequency and wavelength options.

[00:01:06]I think white will be fine. Distinguish between conduction current and displacement current.

[00:01:15]conduction.

[00:01:18]We have a circuit conduction current IC displacement current.

[00:01:37]Displacement current is nothing but your electric field. Discu conduction current and displacement current.

[00:01:55]Conduction current flow of electrons right I repeat conduction current is due to flow of electrons in the wire.

[00:02:07]Whereas in the displacement current right in the displacement current it is due to electric field in between the plates.

[00:02:20]Small differentiation we have magnetic field amplitude B not actually. Okay B okay B which is magnetic field. Find the electric field amplitude. Electric velocity selector concept we have studed in chapter four QV which is your lens force and then Q which is your electrostatic force without any deviation it moves in a straight line the Q and then in the Q cancel you have the remaining term to be V B equal to E electric field you know what is the value of B which is given in the question 5 10 nano repeat nano small nano which is 10^ - 9 and then V is nothing but velocity right here the velocity obviously like electric field and then magnetic field which is the combination of electromagnetic waves. Electromagnetic waves of light similar here V is nothing but 3 into 10 ^ 8 and then B is nothing but the magnetic field which is 5 10 ^ - 9 = E. So you get the electric field value.

[00:03:57]Electric field.

[00:04:01]We'll move to the next problem which is anything else.

[00:04:12]Second problem.

[00:04:14]First problem.

[00:04:20]Yeah. either >> it's there.

[00:04:36]>> Yeah, you have to. Right.

[00:04:37]>> Yeah. Okay. Thank you.

[00:04:39]>> Yeah. Yeah. Yeah. Fine. Fine.

[00:04:44]Right. Next.

[00:04:46]There are some problems also. We'll try to sort it out. But basic will come for this one in the ray optics problems from EMW in the ray optics problem I couldn't make PP because so much problems are there so based upon the repeated content I have shortlisted few things so Understand your core concept will be very clear.

[00:05:26]two thin lens L1 and L2 lens and L1 and L2 first lens and second lens focal length they have mentioned very clearly which is 24 and then - 18 cmus Okay, we'll come to that. Don't worry.

[00:05:59]Next are placed 45 cm apart.

[00:06:06]45 cm apart.

[00:06:12]An object of height 1 cm and object.

[00:06:20]Right? Or cm is placed 36 cm.

[00:06:24]Right? I repeat, is placed 36 cm in front of lens one. Right? In front of lens one.

[00:06:34]Find the position nature and then height of the image formed.

[00:06:43]You can able to understand here very clearly. You have focal convex lens and then either concave lens. Convex and then concave lens.

[00:07:09]Convex lens.

[00:07:15]Okay, we know 45m try to visualize.

[00:07:32]Okay, it will be very easy for you to write 1 cm is placed 36 cm in front of lens 36 cm.

[00:07:54]Find the position, nature and the height of the final image. Final image final.

[00:08:08]Okay. First 36 cm. U is always negative. So minus lens and object. lens.

[00:08:31]I repeat.

[00:08:38]Okay.

[00:08:43]Okay. Fine. Next.

[00:08:49]Focal length. First one. We talking about the first one height of object 1 cm given data given first lens U sorry VU first first okay First because we know f we know u 1 by v1 which is equal to 1x f plus 1x u okay I can able to find what is my v1 v1 Right.

[00:09:52]Image distance.

[00:10:07]Okay. 50 cm.

[00:10:14]No, you cannot take because 45mus 45mm I repeat that is your U2 okay Neat example 50 cm object distance.

[00:11:14]Okay. So already and then you're going to find V2.

[00:11:26]Okay. So this is the method. Okay. And this will be obviously concave.

[00:11:35]Okay. So be sure you can solve this. Okay.

[00:11:42]Hope so you understood that. Next an equilateral prism prism.

[00:11:50]Equilateral prism is made up of a material. Right?

[00:11:56]Is is made up of a material. If the angle of minimum deviation is 30°, I repeat, if the angle of minimum deviation is 30°, find the speed. Okay, find the speed.

[00:12:15]Speed of light in the prison. Obviously, find the speed of light.

[00:12:45]Find the angle of incident.

[00:12:48]Going to draw the prism.

[00:12:57]Incident you know very well minimum deviations obvious going to find the speed of light in that particular prism. Refractive ind therefore V is nothing but C upon N suppose.

[00:13:42]Suppose glass prism example right glass prism if they have given glass prism 1.43 43 sorry 1.52 you can get the answer yes you can get the answer okay 30° sorry 60 60° Okay.

[00:14:32]The angle of prism is 60° minimum deviation 30°.

[00:14:41]Okay. So first what we have refractive index we would have calculated sin I by sin r which is equal to n_sub_2 upon n1 in Since it is very small A + D M / 2 I which is equal to refractive index 30 30 60 + 30 is nothing but 90 90 / 2 is 45° sin 45° 60° divided by 2 becomes 30° 30° value 45° Right.

[00:16:12]Sin 45 upon sin 30° which is equal to m 45° 1 by <unk>2 1 by 2 upon <unk>2 upon <unk>2.

[00:16:39]So this is your okay upon see value 3 into 10 ^ 8 and then <unk>2 which is 1.414 when you do the calculation you get the speed of light in the particular prism. Suppose the glass prism you can write since prism prism is made up of glass it's a refractive index is nothing but 1.52 make it simple they will give the mark you don't need to worry about it suppose you can find okay render the segment find find the angle of incident if the emergent just graces one phase of the prism greases, right? So, okay.

[00:18:04]Last condition is nothing but sin inverse 1 by Okay. So you can you know the value of n sin inverse the one by which is you have <unk>2 which is 1.414 414 1 / you have to use logarithm sheet since minimum R12 R1 R2 will be same minimum imum you can find your R12 right so right which is your 45° right you'll be finding your R1 R1 S I sorry sin inverse you know very well it is nothing but 45° 45° that's it you found the answer now one by root better you can go with this one by root under 45° you can write right. So pattern right find the angle of incidence if the emergent rate just grace graces one phase of the prism. Okay.

[00:20:27]Next another one. Again we are going to talk about the prism here. A prism of refractive index 1.5 and angle 75° okay 75° is immersed in a liquid 75° of liquid of refractive index root3x2 liquid root3x2 or I think uh 1 3 1.7 of course you you'll come to water so we'll discuss that but four so okay it is not water it is not water find the angle of incidence for which the ray just under goes total internal reflection so in we are talking about total Internal reflection.

[00:21:42]Sorry.

[00:22:02]Find the angle of incidence for which just the ray underos total internal reflection. Total internal reflection graces the surface and the surface.

[00:22:21]We are not talking but we said it is critical angle.

[00:22:29]question is different.

[00:22:31]So we'll discuss that.

[00:22:39]They didn't mention it is equilateral.

[00:22:42]So rightalm.

[00:22:56]Okay.

[00:23:05]1.5 <unk>2 3 <unk>2 by 4 the liquid refractive index the angle 75° is immersed. Okay. So 75° what should be the sorry uh yeah what should be the angle of incident okay they didn't mention it is minimum deviation refraction first surface surface because total internal reflection will be Again we are going to follow the same previous equation previous N1 sin I equal to N_sub_2 sin R.

[00:24:34]Right? 90° because total internal reflection liquid 3 <unk>2x 4 which is equal to sin IC 3 by 2 1.5 3x2 Not in all the cases but sometime I cancel in the two cancel and then this is nothing but 2 into <unk>2<unk>2 canceling 1<unk>2 inverse 45° we are talking about this part alone 45° find the angle of incidence for which the ray just underos the total internal reflection first so I'm going to reate sorry in the total reflection IC will be equal to your in the e actually the e 45° right.

[00:26:48]We have studied this R1 is nothing but A - R2 R2 because that is nothing but your IC 45° angle and 75°. Yeah, correct. angle of refractive index and then angle 75° 45° which is nothing but around 45 30° is 30°.

[00:27:42]Okay.

[00:27:43]30°.

[00:27:48]Simple.

[00:28:05]30°.

[00:28:14]So we can easily simplify this one because 60. So what we are going to find I again if I apply N1 sin I equal to N_sub_2 sin R first. Okay, first surface.

[00:28:50]So no need to worry about it.

[00:29:02]just okay. So the dimension 3x2 3x2 and then rin 3x4 3x I is nothing but sin inverse 3x 4 All of that right if you want you can also calculate this but so better you can stop with this also they'll give full mark okay sign in 3x4 next uh this is electro atom okay okay so we we'll skip this as of Okay, coming to this one. The focal length of the object and then IPS. Yeah, right. Especially optics in the compound microscope telescope.

[00:30:28]The focal length of the object and the eyepiece of the compound microscope the object 1 cm 2.5 cm. Okay.

[00:30:49]Find the tube length of the microscope for obtaining the magnification of 300.

[00:31:00]Magnification of 300 in the length.

[00:31:04]Suppose length of the telescope F all just you're going to add both 3.5 will be your length of the telescope.

[00:31:32]length magnification is nothing but Hdash by H or L by F 1 + D near D infinite going to do the product of your magnification right which is nothing but L by F not I'm just writing the condition for both okay either the FE and then L by F not D by FE correct you can use either equation to find the lengths it can be anything okay so instead like better I can go with the second equation which is 25 f therefore your L is nothing but M * of F 300 F 1 F 2.5 D 25 cm right so everything is in cm. So cime represent the better.

[00:33:03]So that will be your length of the uh in the component microscope and the tube length. Okay.

[00:33:12]Fine.

[00:33:14]A ray of light is incident right? A ray of light is incident at an angle of 60° on one face of a prism.

[00:33:36]Incident angle 60 of the angle. Yeah. One face of the prism of a angle 30°.

[00:33:47]A ray of light is incident at an angle of 60° on one face of a prism of angle face of a prism of angle 30°.

[00:33:59]30° emerging out of the prism makes an angle of 30°.

[00:34:32]What is actually going to happen here with the incident?

[00:34:38]Okay.

[00:34:44]same calculate the refractive index of the prism. refractive index of the prism.

[00:34:59]Last sin a + d m by 2 upon sin a by 2.

[00:35:19]refractive ind right you can simply substitute the value value you'll get the answer okay direct value you can directly substitute direct okay refractive index suppose glass you no need to worry since the refractive index of the glass 1.52 you can write directly right you can write directly in the minimum deviation minimum deviation R1 R1 + R2 minimum condition I + EU 60° E so this is what you're going to follow minimum deviation I don't want to make it too big complicated.

[00:36:56]N1 sin I N2 Sin R sin I by sin R 60° 60° 30° 3 Right 30.

[00:37:34]So this is the way you have to solve.

[00:37:38]Oh, the angle sorry number this angle you cannot cancel sin 60° 60° upon 2 <unk>3 by 2 in two cancel okay So right at the sin 60° sin 30° go for this an equilateral prism of a prism has a refractive index 1.6 6 in air.

[00:38:30]Calculate the angle of minimum deviation. Minimum deviation d when kept in a medium of refractive index 4 <unk>3 by2 angle because it is equilateral angle of prism 60° 1.6 6. Okay. Calculate the minimum deviation of the prism when it is kept in a medium medium 4 <unk>3 sorry roo<unk>2 upon 5 different.

[00:39:12]We going to compare.

[00:39:15]We are going to compare when you take the ratio N2 upon N_sub_2 by N1 N1 by N_sub_2 whichever way you like.

[00:39:33]So obviously you will have a refractive index. So for air and medium so you will have your refract index for that particular medium you have your glass and the prism plus on the medium prism and then medium. You know the refractive index term sin I sin rul you have your r right sin r. So R1 you know that is a = r1 + r2. So the minimum deviation 2= a minimum s a + d m upon 2 / sin a by 2.

[00:40:44]Okay.

[00:40:46]Minimum deviation refractive index glass 1.6 upon 4 <unk>2.6 into 5 and then 8 by 42.

[00:41:13]Ah <unk>2 which is equal to the sign yeah sin angle of prism equal so 60 + dm you don't know upon 2 divided by sin a 60 by 2 which is 30° sin 30° 1 by 2 yeah 2 <unk>2 I'll go with this <unk>2 sin a + the 60° with another 1x2 Okay.

[00:42:16]This becomes 2<unk>2 sin 60° plus dm.

[00:42:29]Okay.

[00:42:33]So it is common for both. So like uh yeah it is common this 1.414 you know that and then you have this sign 60° and then in the dm.

[00:42:56]So we can also negate angle is very small minimum deviation.

[00:43:03]So since we neglect that term you have this right this and then so you're going to split okay or sin 60 plus dm Right. Again 30° that becomes even more easier.

[00:43:45]I can also go in that pattern same. So finally you can able to find the minimum deviations. I repeat direct condition since they mentioned minimum deviation.

[00:44:15]Angle of prism a 60° 60° which is 60 by 2 is 1 by 2 30° the equation equation. So just substitute you'll get the value of dm. Okay. So this is your minimum deviation in terms of angle.

[00:45:02]Next array of light passes through an equilateral glass prism. Okay. A ray of light passes through equilateral 60° such that it goes a minimum deviation again when the inc when the angle of incident is 3x4 times the angle of prism. So angle of prism When the angle of incidence is 3x4 times the angle of prism okay 3x4 times determine the refractive index of the class prism and the speed of light at the prism already minimum right angle.

[00:46:41]Determine the refractive index. Okay.

[00:46:45]Determine the refractive index of the prism. Okay. The glass you can write.

[00:46:55]So angle of yeah correct. You can write simply since it is a glass the refractive index of glass is 1.52 okay they'll give the mark no need to worry about it they will give the mark the speed= c upon vantelity V= C upon C of the glass which is 1.52 divide you will get the value.

[00:47:44]Next, a sun shine recorder globe of 30 cm radius. Sorry diameter diameter which is 15. Okay. Sunshine recorder globe.

[00:48:06]It's it's not like normal one globe 30 cm diameter and then it is made up of glass refractive index 1.52 exact 1.5 array enters the globe right array enters the globe parallel to its axis.

[00:48:36]upon the sunshine recorder globe. So overall it is nothing but 30 ray enters the globe in the globe to the principal axis.

[00:49:02]Calculate the position from the center of the sphere.

[00:49:07]Center of the sphere. Calculate the position from the center of the sphere where the ray crosses the axis. Obviously the glass either like righter rarer to denser towards denser to rarer away the normal. Okay.

[00:49:43]Sorry.

[00:49:53]You know that very well.

[00:50:06]Infinite, right?

[00:50:10]We do know the exact position.

[00:50:14]We don't know but globe like normal round.

[00:50:24]Okay. Sperical in this first surface I will change the color.

[00:50:34]The first surface radius of curvature.

[00:50:38]Second what we are talking.

[00:50:58]Yes.

[00:51:02]Single surface N1 U N_sub_2 N1 by Radius of curvature right we don't know But you know it is rad because 30 here you can able to find your V. I repeat First surface n_sub_2 v n_sub_1 u n_sub_2 nsub_1 by r surface.

[00:52:38]I repeat 1.52. Okay.

[00:52:50]1.5 in the U value right in the U value and then in the V value we don't know but already you cannot take directly So radi I repeat 30 minus in the V will be your U renderance concept Okay, you're going to find your Vgative You'll get the final image like second surface Okay, great.

[00:54:36]A point source of light. Yes.

[00:54:40]Right. We don't know what but a point source of light. Yes. Is placed at the bottom of the vessel. at the bottom of the vessel containing a liquid of refractive index 5 by3 containing liquid of refractive index 5.3.

[00:55:06]A person is viewing the source from just above the surface and the surface.

[00:55:20]There is an opaque disc of radius 1 cm.

[00:55:32]The light refractive indic The center of the disc, I repeat, the center of the disc lies vertically above the surface.

[00:56:07]Vertically above the surface above above the surface. Sorry. source.

[00:56:15]The liquid from the vessel is gradually drained, right? And the liquid through the tap. What is the maximum height of the liquid for which the source cannot be seen at all?

[00:56:42]I think for a certain distance learning.

[00:57:13]Okay.

[00:57:31]There is a refraction apparent depth. Real depth concept related book problem.

[00:57:58]detail.

[00:58:04]Okay.

[00:58:12]Radius of illumination or window radius.

[00:58:22]Okay.

[00:58:24]Radius.

[00:58:30]Okay.

[00:58:33]Direct substitution.

[00:58:40]What is the maximum height of the liquid for which the height?

[00:59:05]Because alternative interior angles are equal.

[00:59:11]90° N1 sin IC= N2 Rin IC I 1 by refractive index refractive index hypotenus Right. Obviously you'll be using your tan because right angle triangle opposite adjacent. Right. So tan you need your this side height opposite by adjacent. You can also go with that pattern. Right.

[01:00:18]I don't know the radius.

[01:00:27]Opposite hypotenus hypotenus right hypotenus by hypotenus opposite angle is Okay.

[01:01:05]Which is your r² + h² root hypoten which is equal to 3x f 3x 5.

[01:01:18]If I do squaring because I need to remove the square root term r² = sorry divid r² + h² 9 by 25 you can able to find hub okay you can take reciprocal reciprocal I I'll go with R² H² by R² = 25 upon 9 - 9 by 9. Okay.

[01:02:04]H² = R² * 25 - 9 by 9 all radius they didn't do they mentioned anything no I think nothing travel radius 1 cm they have given very clearly I substitute this r you'll get the value Okay. And then in the root.

[01:02:38]So this is your uh problem. We'll discuss some other problems from this chapter and also we'll move to the other chapter problems in the next uh session.

[01:02:51]Uh thank you so much for joining the class. Thank you. Thank you so much.