All Formulas of Electricity: 1) I = Q/t 2) Q = ne 3) V = W/Q 4) V = IR 5) R = Rho x L/A 6)Resistance in Series: Rs = R1 + R2 + R3 7)Resistance in Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 (» Reciprocal of R1+R2+R3) 8)H = V × Q i) H = VIt ii) H = I²Rt iii) H = V²t/R 9) P = W/t i) P = VI ii) P = I²R iii) P = V²/R 10) E = P × t S.I. Units: Current ( I ) - Ampere Charge ( Q ) - Coulomb Time ( t ) - Second Potential Difference or Voltage ( V ) - Volt Resistance ( R ) - Ohm Ω Resistivity ( Rho ) - Ohm meter Power ( P ) - Watt Heat = Energy = Work = Joule Measuring Devices: Ammeter / Milli Ammeter - for measuring current (always connected in series) Galvanometer - for measuring small/sensitive currents (we can also get the direction of the current) Voltmeter - for measuring volts (always connected in parallel) Ohm Meter - Measures the resistivity Things to remember: electron charge = (1.6 x 10⁻¹⁹ C) 1kWh= 3.6 × 10⁶J 1Ampere = 1000mA 1 Kilowaat = 1000W 1 Horse Power = 746W i hope this list will help you! ❤ ek free ka like dete jan
h=3cm F=-12cm U=-18cm Now, using Mirror formula 1\u +1/v = 1/f, 1/-18 +1/v =1/f 1/v=1/-12+ 1/18 1/v=-3+2/36 1/v=-1/36 V=-36 Now, m=-v/u m=-(-36)/-18 m=-2 Again, m=h'\h -2=h'/3 -2×3=h' -6=h' h'=-6 Therefore, distance of the image from the mirror is 36cm. Height of the image is 6cm. Now, Answer= -36/-6. Therefore, option (b) is correct Answer. ❤❤❤❤ Please one like for it ❤❤❤ Thanks for 1k likes ♥️♥️♥️♥️♥️
00:04 Rapid revision of the Light chapter 00:41 Understanding reflection and bouncing back of light 02:00 Understanding virtual and direct images 02:39 Understanding lateral inversion and reflection in mirrors 03:59 Understanding of radius of curvature and focal length 05:03 Understanding light behavior in convex and concave mirrors 06:27 Tips and tricks to spot and use tricks in the exam 07:13 Understanding the Positioning Trick 08:51 Understanding Real and Virtual images 09:30 Convex mirror always forms virtual image 10:53 Understanding Convex and Concave Mirrors 11:36 Understanding the positioning of objects in relation to the mirror and lens 12:57 Representation of height using 'm' in Optics with formula v/u 13:39 Understanding Convex Mirror Focal Length 15:02 Understand the formula of magnification. 15:41 Understanding Snell's Law and Refractive Index 17:00 Understanding the movement of light in denser medium 17:38 Understanding the concept of lateral displacement in refraction. 18:52 Understanding Convex and Concave Lenses 19:38 Explaining the rules for optical center and mirror in geometric optics. 21:01 Focus on consistent numbering for easy referencing 21:37 Understanding the placement of lights and objects in relation to focus points 22:57 Understanding optics through lenses and mirrors 23:34 Explanation of lens formula and magnification formula. 24:52 Understanding the changes in formula and power of light. 25:29 Understanding focal length and its numerical value in meters is important for exams. 26:47 Understanding convex lens and screen positioning
Who want 90+ marks in 10th board exam Who want 90+ marks in 10th board exam Who want 90+ marks in 10th board exam Who want 90+ marks in 10th board exam
@@shreyhemavat3298 Yeh jo tum aise comment karkare dal rahe hu kya usse tumhare 90+ aa jayega Jitna time tum comment likhne mein laga rahe ho uthe padhne mein tagaoo toh I am 💯% sure ke tumhare 90 above aayege
Formulas related to reflection of light, => in reflection angle of incidence is equal to angle of relection, magnification = height of image/height of object m=hi/ho wher m is the magnification hi is the height of image ho is the height of object =>also m=-v/u formulaes related to refraction of light, => n =sini/sinr where n is the refractive index i is angle of incidence r angle of refraction =>lens formula 1/f =1/v -1/u =>power of lens,p=1/f(in m ) p=100/f(in cm) =>power of combination of lens P=P1+P2+P3... =>magnification ,m=hi/ho m=v/u => refractive index n21=speed of light in medium one /speed of light in medium 2 n21=v1/v2
h=3cm F=-12cm U=-18cm Now, using Mirror formula 1\u +1/v = 1/f, 1/-18 +1/v =1/f 1/v=1/-12+ 1/18 1/v=-3+2/36 1/v=-1/36 V=-36 Now, m=-v/u m=-(-36)/-18 m=-2 Again, m=h'\h -2=h'/3 -2×3=h' -6=h' h'=-6 Therefore, distance of the image from the mirror is 36cm. Height of the image is 6cm. Now, Answer= -36/-6. Therefore, option (b) is correct Answer.
@@TatiyaBicchuop Given, focal length =-12cm,v=-18cm ,and heigth of image=3cm after that ,we use to mirror formula ; 1/f=1/v+1/u 1/-12=1/-18+1/u 1/u=1/-12+1/18 1/u=-3+2/36 1/u=-1/36 u=-36 cm m=-v/u m=-(-18)/-36 m=-1/2 m=hi/ho -1/2=3/ho ho=-6cm
thank you so much for these revision videos, sir I have my half yearly exam tomorrow and your videos are making it a lot easier for me to revise my syllabus
Answer is 'C' h=3cm f=-12cm u=-18cm By using mirror formula, 1/v=1/f-1/u 1/v=-2+3/36 1/v=1/36 NOW RECIPROCAL V IS 36CM ---------------------------- m=-v/u m=-36/-18 m=2 NOW FOR FINDING HEIGHT m= hi/ho hi= ho x m hi=3 x 2 therefore height of image is 6 cm and image distance is 36cm
Those who are watching this for nostalgia knows that he has explained and cracked whole board paper pattern in this video. The only tip for juniors is that watch the video carefully and practice atleast 3 times to score 90+
IT'S A BIG REQUEST SIR!! Please make a video explaining the ch2 of Biology *class 9 Tissues* 🙏🏻 I'll be really grateful to you.. Please make it this week..
Thank you thank you thank you so much Prashant bhaiya Ray diagrams ki itni achi trick btaane ke liye iss video ko dekhne se pehle mujhe ray diagram smjh nhi aate the but ab mind me set ho gye hai ❤🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🔥
24:02 if someone is learning lens formula from this slide so this formula is not correct and the formula in the previous slide was correct i.e. 1/f=1/v-1/u .. I hope it helps
*AT **18:26** SO SPEED IN AIR IS DIFFERENT THAN THAT OF VACUUM* *Light in the air is 1.0003 times slower than light in a vacuum, which slows it all the way down from 299,792,458 meters per second to 299,702,547 meters per second. That's a slowdown of 89,911 meters per second, which looks like a lot but is only three ten-thousandths of the speed of light.*
h=3cm F=-12cm U=-18cm Now, using Mirror formula 1\u +1/v = 1/f, 1/-18 +1/v =1/f 1/v=1/-12+ 1/18 1/v=-3+2/36 1/v=-1/36 V=-36 Now, m=-v/u m=-(-36)/-18 m=-2 Again, m=h'\h -2=h'/3 -2×3=h' -6=h' h'=-6 therefore the correct option is (B). hence the image formed will be real and inverted ....hope the answer is correct
A rapid revision of the chapter on light covers key concepts such as reflection, properties of mirrors, and the behavior of light through lenses. It emphasizes important laws, formulas, and practical applications, including how to solve numerical problems. The session aims to clarify complex topics efficiently for Class 10 students. ##### You may be interested in these questions: - [What are the main properties of light?](#related) - [How do concave and convex mirrors differ?](#related) - [What is the significance of the focal length in lenses?](#related) Highlights: 00:04 Light is a significant chapter in physics that involves understanding its properties and behavior. This revision covers essential concepts such as reflection, plane mirrors, and spherical mirrors. -The reflection of light occurs when it bounces off a shiny surface, following specific laws that govern its behavior. Understanding these laws is crucial for solving related numerical problems. -Plane mirrors produce virtual and erect images, and their properties include lateral inversion and equal size of the image compared to the object. This understanding is vital for practical applications. -Spherical mirrors, including concave and convex types, have unique characteristics that affect how they reflect light. Recognizing their differences helps in comprehending various optical phenomena. 04:04 The focal length is the distance from the pole to the focus of a mirror, which plays a crucial role in the behavior of light rays. Understanding properties of concave and convex mirrors helps in predicting how light interacts with them. -Light rays that pass through the focus of a concave mirror will emerge parallel, illustrating the converging nature of these mirrors. This property is essential for understanding image formation. -The center of curvature for both concave and convex mirrors reflects light rays back along the same path. This fundamental property is key for various optical applications. -A practical trick for remembering the image formation positions in concave mirrors involves identifying focal points. This aids in predicting where images will be formed based on the object's location. 08:08 Understanding image formation in mirrors is crucial for mastering optics. The nature of the image-whether real or virtual-depends significantly on the object's position relative to the mirror. -The relationship between the object's position and image type is essential. When an object is between the focus and the pole, a virtual and erect image is produced. -Concave mirrors are primarily used to enlarge images, making them valuable in applications like torches and makeup mirrors. Their ability to create real images is also noteworthy. -Conversely, convex mirrors create smaller images and are commonly used in car side mirrors and shaving mirrors. They provide a wider field of view, enhancing safety. 12:11 Understanding the rules of sign convention is essential for solving mirror-related problems in optics. Accurate application of these rules leads to correct calculations of focal length, image distance, and magnification. -The mirror formula, represented as 1/v + 1/u = 1/f, is vital for finding relationships between object distance, image distance, and focal length. This formula relies heavily on correct sign conventions. -Magnification is defined as the ratio of the height of the image to the height of the object. The formula m = -v/u helps determine whether the image is real or virtual. -Refraction describes how light bends when passing from one medium to another. Snell's Law governs this bending, establishing the relationship between angles of incidence and refraction. 16:14 The refractive index of a medium determines how much light bends when passing through it, with higher indices indicating greater bending. Understanding this concept is crucial for studying optics and light behavior. -The difference between rarer and denser media influences light behavior, as light bends towards the normal when moving from a rarer to a denser medium. This principle is essential in optics. -The formula for calculating refractive index is derived from the ratio of the speed of light in air to that in the medium. This relationship highlights the inverse nature of refractive index and light speed. -Understanding lens types, such as convex and concave, is vital for grasping how light focuses or diverges. Their unique properties determine how images are formed in optical systems. 20:16 Light passing through the optical center of a lens travels in a straight line, following three fundamental rules. Understanding these rules is essential for analyzing ray diagrams of convex and concave lenses. -The ray diagram techniques for both convex and concave lenses are similar to those used in mirrors, making the transition easier for learners. It's important to follow the same numbering method in both diagrams. -Understanding the formation of images based on object positions relative to the focal points is crucial. Various cases illustrate how images can be real or virtual depending on the object's location. -The applications of convex and concave lenses differ, with convex lenses used in magnifying glasses and concave lenses in telescopes. Each type serves specific purposes in optics. 24:20 The focal length of a convex lens is always positive, while that of a concave lens is negative. This distinction is crucial for understanding lens power and image formation. -Understanding the sign convention for focal lengths is essential. Convex lenses have positive focal lengths, while concave lenses have negative ones, affecting calculations. -The power of a lens indicates its ability to converge or diverge light. Positive power corresponds to convex lenses, and negative power to concave lenses. -When calculating power using the formula P = 1/f, ensure focal length is in meters. Converting centimeters incorrectly can lead to wrong answers.
Yes, the ratio of the speed of light in medium 1 to the speed of light in medium 2 is equal to the refractive index of medium 2 with respect to medium 1:
h=3cm f=(-12cm) because the focal length of a concave mirror is always negative u=(-18cm) because the distance of object from the pole is always negative v=? h'=? Mirror formula= 1/v+1/u=1/f Therefore, 1/v=1/f-1/u 1/v=1/12-1/(-18) 1/v=-1/12+1/18 1/v= -3+2/36 1/v= -1/36 v×-1=36×1 v=36/-1 v= -36 To find h' , we will use the formula of magnification: m=-v/u m= -(-36)/-18 m=36/-18 m= -2 Therefore, m=h'/h=-v/u m=h'/h= -2 h'=3×-2 h'= -6cm So, the correct option is b: (-36,-2)👍 Please do it yourself first beacuse I don't know if my answer is correct or not but i am 95% sure that it is correct 💯 Okiee everyone,thanks me later byee byee 👋🎀 And if there is a mistake in my solution then please inform me ❤️
Option c is correct 💯 36,6 Mirror formula 1/v+1/u=1/f 1/v+1/-18=1/-12 1/v=1/-12-1/-18 1/v=-1/-36 v=36 Hence, The distance of the image from the mirror is 36cm Magnification formula m=-v/u m=-36/-18 m=2 Height formula m=h1/h2 2=h1/3 h1=2*3 h1=6 Hence, the height of the image is 6cm So, option c 36/6 is correct ✅ If i had done any mistake please comment ❤ Please like ❤ Thanku
All Formulas of Electricity:
1) I = Q/t
2) Q = ne
3) V = W/Q
4) V = IR
5) R = Rho x L/A
6)Resistance in Series: Rs = R1 + R2 + R3
7)Resistance in Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 (» Reciprocal of R1+R2+R3)
8)H = V × Q
i) H = VIt
ii) H = I²Rt
iii) H = V²t/R
9) P = W/t
i) P = VI
ii) P = I²R
iii) P = V²/R
10) E = P × t
S.I. Units:
Current ( I ) - Ampere
Charge ( Q ) - Coulomb
Time ( t ) - Second
Potential Difference or Voltage ( V ) - Volt
Resistance ( R ) - Ohm Ω
Resistivity ( Rho ) - Ohm meter
Power ( P ) - Watt
Heat = Energy = Work = Joule
Measuring Devices:
Ammeter / Milli Ammeter - for measuring current (always connected in series)
Galvanometer - for measuring small/sensitive currents (we can also get the direction of the current)
Voltmeter - for measuring volts (always connected in parallel)
Ohm Meter - Measures the resistivity
Things to remember:
electron charge = (1.6 x 10⁻¹⁹ C)
1kWh= 3.6 × 10⁶J
1Ampere = 1000mA
1 Kilowaat = 1000W
1 Horse Power = 746W
i hope this list will help you! ❤
ek free ka like dete jan
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Konse chapter pe comment kar rhe ho
Ye Light hai😂
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h=3cm
F=-12cm
U=-18cm
Now, using Mirror formula
1\u +1/v = 1/f,
1/-18 +1/v =1/f
1/v=1/-12+ 1/18
1/v=-3+2/36
1/v=-1/36
V=-36
Now,
m=-v/u
m=-(-36)/-18
m=-2
Again,
m=h'\h
-2=h'/3
-2×3=h'
-6=h'
h'=-6
Therefore, distance of the image from the mirror is 36cm.
Height of the image is 6cm.
Now, Answer= -36/-6.
Therefore, option (b) is correct Answer. ❤❤❤❤
Please one like for it ❤❤❤
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thanks bro me solution hi dhundh raha tha
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V = - 36 bro , but it is correct 😂😂
-1/36=1/v
Cross multipy
-v = 36
Because it is concave mirror
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7:37 damnnn bhaiyaaa lit 🔥🔥 love the trick 🗣️
Sahi mei ❤️🔥🗣️
Also give tricks for biology diagrams
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Human Eye 20 Min Rapid Revision---------Vote Now
Edit: Uploaded!
Rapid revision of human eye 👁️💯❤️
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Rapid revision of human eye
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Yes
Option B is right [ -36, -6 ] 26:38
Option c
@@ShivakshayaSinghha option ç wrong hai
Yeah bhaya Mai i got it answer is option b
@@nbtss_799 tum ho kon aaramb ki ho kya
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1 like = 95% 2:51
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00:04 Rapid revision of the Light chapter
00:41 Understanding reflection and bouncing back of light
02:00 Understanding virtual and direct images
02:39 Understanding lateral inversion and reflection in mirrors
03:59 Understanding of radius of curvature and focal length
05:03 Understanding light behavior in convex and concave mirrors
06:27 Tips and tricks to spot and use tricks in the exam
07:13 Understanding the Positioning Trick
08:51 Understanding Real and Virtual images
09:30 Convex mirror always forms virtual image
10:53 Understanding Convex and Concave Mirrors
11:36 Understanding the positioning of objects in relation to the mirror and lens
12:57 Representation of height using 'm' in Optics with formula v/u
13:39 Understanding Convex Mirror Focal Length
15:02 Understand the formula of magnification.
15:41 Understanding Snell's Law and Refractive Index
17:00 Understanding the movement of light in denser medium
17:38 Understanding the concept of lateral displacement in refraction.
18:52 Understanding Convex and Concave Lenses
19:38 Explaining the rules for optical center and mirror in geometric optics.
21:01 Focus on consistent numbering for easy referencing
21:37 Understanding the placement of lights and objects in relation to focus points
22:57 Understanding optics through lenses and mirrors
23:34 Explanation of lens formula and magnification formula.
24:52 Understanding the changes in formula and power of light.
25:29 Understanding focal length and its numerical value in meters is important for exams.
26:47 Understanding convex lens and screen positioning
Thank you for 870 likes 🍂💫💗
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Who want 90+ marks in 10th board exam
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Who want 90+ marks in 10th board exam
Who want 90+ marks in 10th board exam
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95+ MARKS BHAI
@@shreyhemavat3298 Yeh jo tum aise comment karkare dal rahe hu kya usse tumhare 90+ aa jayega
Jitna time tum comment likhne mein laga rahe ho uthe padhne mein tagaoo toh I am 💯% sure ke tumhare 90 above aayege
Formulas related to reflection of light,
=> in reflection angle of incidence is equal to angle of relection,
magnification = height of image/height of object
m=hi/ho
wher m is the magnification
hi is the height of image
ho is the height of object
=>also m=-v/u
formulaes related to refraction of light,
=> n =sini/sinr
where n is the refractive index
i is angle of incidence
r angle of refraction
=>lens formula
1/f =1/v -1/u
=>power of lens,p=1/f(in m )
p=100/f(in cm)
=>power of combination of lens
P=P1+P2+P3...
=>magnification ,m=hi/ho
m=v/u
=> refractive index n21=speed of light in medium one /speed of light in medium 2
n21=v1/v2
8:07 thank you bhaiya mast trick h and the correct answer for the homework question is b(-36, -6)
Yah correct ans is b -36,-6
So the magnification will be negative that is real and inverted
Humlogon ne toh ek hi kitab padhi thi dost par yeh sab tune Kahan se Seekh liya mere bhai😂@@GarvJain-h7n
Ans is 36 and 6
Yeah @@GarvJain-h7n
21:43 dekho
Vote for Control and Coordination ❤
(Please Bhaiya yeh chapter revise kara do)
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h=3cm
F=-12cm
U=-18cm
Now, using Mirror formula
1\u +1/v = 1/f,
1/-18 +1/v =1/f
1/v=1/-12+ 1/18
1/v=-3+2/36
1/v=-1/36
V=-36
Now,
m=-v/u
m=-(-36)/-18
m=-2
Again,
m=h'\h
-2=h'/3
-2×3=h'
-6=h'
h'=-6
Therefore, distance of the image from the mirror is 36cm.
Height of the image is 6cm.
Now, Answer= -36/-6.
Therefore, option (b) is correct Answer.
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This week my exam please sir 😊
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Option B is always correct and whoever likes this video will get 92% in boards 2024-25🎉🎉💯💯💯
@@rudduxd6266 It's B
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Acids , base and salt rapid revision 👍🏻
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Next week is my exam ... Plz ASAP🥲
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Ye lo tissue ch
@@ronakrawal07tissue ch ruclips.net/video/ccNmX3zcbss/видео.htmlsi=jAnkz0LbCjfffi90
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Tissue ch
7:22 remarkable trick 👍
Lecture starts at 0:40
8:07 best trick to remember ray diagrams..... Thanku so much bhaiya..
Kisi aur teacher ka trick bataen
@@k-poplover6612 kiska?
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LearnwithVY channel par ye tricks pehle se hai
@@Alokchaudhary8430-n1g Yes
option B is the right answer ( -36,-6)
Bhai mera option c aya
Mera bhi bhai 😅😅@@TatiyaBicchuop
@@TatiyaBicchuop
Given, focal length =-12cm,v=-18cm ,and heigth of image=3cm
after that ,we use to mirror formula ;
1/f=1/v+1/u
1/-12=1/-18+1/u
1/u=1/-12+1/18
1/u=-3+2/36
1/u=-1/36
u=-36 cm
m=-v/u
m=-(-18)/-36
m=-1/2
m=hi/ho
-1/2=3/ho
ho=-6cm
m= -2 not -1/2
Because
m=-v/u
v= -36 , u = -18
v=-(-36)/-18
v=2/-1
v=-2
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26:40 the correct answer is b) -36,-6❤
Write answer
@@Rupadsul *Right
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mere ghr pr jitne phone h utne phone se like kr denge
Next chapter control and coordination plz 🥺
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thank you so much for these revision videos, sir I have my half yearly exam tomorrow and your videos are making it a lot easier for me to revise my syllabus
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Thank you very much prasant bhaiya itne saare topics ko short revision se padha de rahe..itne easy way se padhane k liye thank you 😊
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Homework:
27:09------> -36,-6 (option B)
bhai solution btao ge ek barr
@@gurtejsingh2549
Bilkul..
Height of object=3cm.(Given)
F= -12 (Given)
u= -18
v= ?
Height of image= ?
..
Mirror formula --> 1/f=1/v+1/u
-1/12 = 1/v - 1/18
1/v = -1/36
v= -36.
m=-v/u
= 36/(-18)
= -2.
m= height of image/ height of object
-2= height of image/ 3(given)
Height of image= -2×3
= -6.
Answers------> -36,-6(option B)
Hope it helped 👍🏻
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15:10 - Refraction
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Control coordination rapid revision
Answer is 'C'
h=3cm
f=-12cm
u=-18cm
By using mirror formula,
1/v=1/f-1/u
1/v=-2+3/36
1/v=1/36
NOW RECIPROCAL
V IS 36CM
----------------------------
m=-v/u
m=-36/-18
m=2
NOW FOR FINDING HEIGHT
m= hi/ho
hi= ho x m
hi=3 x 2
therefore height of image is 6 cm and image distance is 36cm
galat hai sarthak tera ans it will be 1/v = 1/f +1/u and now input the values
Bhai u (-) ka h
And answer is (-36,-6)
Acid, base and salt rapid revision
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24:02 if someone is learning lens formula from this slide so this formula is not correct and the formula in the previous slide was correct i.e. 1/f=1/v-1/u .. I hope it helps
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-36,-6 (b)
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answer is option = (B) thank you sir for your help
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Pagal
*AT **18:26** SO SPEED IN AIR IS DIFFERENT THAN THAT OF VACUUM*
*Light in the air is 1.0003 times slower than light in a vacuum, which slows it all the way down from 299,792,458 meters per second to 299,702,547 meters per second. That's a slowdown of 89,911 meters per second, which looks like a lot but is only three ten-thousandths of the speed of light.*
h=3cm
F=-12cm
U=-18cm
Now, using Mirror formula
1\u +1/v = 1/f,
1/-18 +1/v =1/f
1/v=1/-12+ 1/18
1/v=-3+2/36
1/v=-1/36
V=-36
Now,
m=-v/u
m=-(-36)/-18
m=-2
Again,
m=h'\h
-2=h'/3
-2×3=h'
-6=h'
h'=-6
therefore the correct option is (B).
hence the image formed will be real and inverted ....hope the answer is correct
Thanks
You are genius bro ...............
Actually I can't understand the question but understand your answer 😂😂😹🤧
A rapid revision of the chapter on light covers key concepts such as reflection, properties of mirrors, and the behavior of light through lenses. It emphasizes important laws, formulas, and practical applications, including how to solve numerical problems. The session aims to clarify complex topics efficiently for Class 10 students.
##### You may be interested in these questions:
- [What are the main properties of light?](#related)
- [How do concave and convex mirrors differ?](#related)
- [What is the significance of the focal length in lenses?](#related)
Highlights:
00:04 Light is a significant chapter in physics that involves understanding its properties and behavior. This revision covers essential concepts such as reflection, plane mirrors, and spherical mirrors.
-The reflection of light occurs when it bounces off a shiny surface, following specific laws that govern its behavior. Understanding these laws is crucial for solving related numerical problems.
-Plane mirrors produce virtual and erect images, and their properties include lateral inversion and equal size of the image compared to the object. This understanding is vital for practical applications.
-Spherical mirrors, including concave and convex types, have unique characteristics that affect how they reflect light. Recognizing their differences helps in comprehending various optical phenomena.
04:04 The focal length is the distance from the pole to the focus of a mirror, which plays a crucial role in the behavior of light rays. Understanding properties of concave and convex mirrors helps in predicting how light interacts with them.
-Light rays that pass through the focus of a concave mirror will emerge parallel, illustrating the converging nature of these mirrors. This property is essential for understanding image formation.
-The center of curvature for both concave and convex mirrors reflects light rays back along the same path. This fundamental property is key for various optical applications.
-A practical trick for remembering the image formation positions in concave mirrors involves identifying focal points. This aids in predicting where images will be formed based on the object's location.
08:08 Understanding image formation in mirrors is crucial for mastering optics. The nature of the image-whether real or virtual-depends significantly on the object's position relative to the mirror.
-The relationship between the object's position and image type is essential. When an object is between the focus and the pole, a virtual and erect image is produced.
-Concave mirrors are primarily used to enlarge images, making them valuable in applications like torches and makeup mirrors. Their ability to create real images is also noteworthy.
-Conversely, convex mirrors create smaller images and are commonly used in car side mirrors and shaving mirrors. They provide a wider field of view, enhancing safety.
12:11 Understanding the rules of sign convention is essential for solving mirror-related problems in optics. Accurate application of these rules leads to correct calculations of focal length, image distance, and magnification.
-The mirror formula, represented as 1/v + 1/u = 1/f, is vital for finding relationships between object distance, image distance, and focal length. This formula relies heavily on correct sign conventions.
-Magnification is defined as the ratio of the height of the image to the height of the object. The formula m = -v/u helps determine whether the image is real or virtual.
-Refraction describes how light bends when passing from one medium to another. Snell's Law governs this bending, establishing the relationship between angles of incidence and refraction.
16:14 The refractive index of a medium determines how much light bends when passing through it, with higher indices indicating greater bending. Understanding this concept is crucial for studying optics and light behavior.
-The difference between rarer and denser media influences light behavior, as light bends towards the normal when moving from a rarer to a denser medium. This principle is essential in optics.
-The formula for calculating refractive index is derived from the ratio of the speed of light in air to that in the medium. This relationship highlights the inverse nature of refractive index and light speed.
-Understanding lens types, such as convex and concave, is vital for grasping how light focuses or diverges. Their unique properties determine how images are formed in optical systems.
20:16 Light passing through the optical center of a lens travels in a straight line, following three fundamental rules. Understanding these rules is essential for analyzing ray diagrams of convex and concave lenses.
-The ray diagram techniques for both convex and concave lenses are similar to those used in mirrors, making the transition easier for learners. It's important to follow the same numbering method in both diagrams.
-Understanding the formation of images based on object positions relative to the focal points is crucial. Various cases illustrate how images can be real or virtual depending on the object's location.
-The applications of convex and concave lenses differ, with convex lenses used in magnifying glasses and concave lenses in telescopes. Each type serves specific purposes in optics.
24:20 The focal length of a convex lens is always positive, while that of a concave lens is negative. This distinction is crucial for understanding lens power and image formation.
-Understanding the sign convention for focal lengths is essential. Convex lenses have positive focal lengths, while concave lenses have negative ones, affecting calculations.
-The power of a lens indicates its ability to converge or diverge light. Positive power corresponds to convex lenses, and negative power to concave lenses.
-When calculating power using the formula P = 1/f, ensure focal length is in meters. Converting centimeters incorrectly can lead to wrong answers.
Hi
Yes, the ratio of the speed of light in medium 1 to the speed of light in medium 2 is equal to the refractive index of medium 2 with respect to medium 1:
f=12
hi=3
u=18
now using mirror formula
1/u+1/v=1/f
then v= -36
now using m= -v/u
then m=-2
now using height formula m=hi/ho
then ho 6
= -36,-6
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Ans= -36cm and - 6cm
Best lecture❤❤❤❤❤❤❤❤❤. I am very happy sir thank u very much ❤❤
-36,-6 is the answer 🎉🎉
h=3cm
f=(-12cm) because the focal length of a concave mirror is always negative
u=(-18cm) because the distance of object from the pole is always negative
v=?
h'=?
Mirror formula= 1/v+1/u=1/f
Therefore, 1/v=1/f-1/u
1/v=1/12-1/(-18)
1/v=-1/12+1/18
1/v= -3+2/36
1/v= -1/36
v×-1=36×1
v=36/-1
v= -36
To find h' , we will use the formula of magnification:
m=-v/u
m= -(-36)/-18
m=36/-18
m= -2
Therefore, m=h'/h=-v/u
m=h'/h= -2
h'=3×-2
h'= -6cm
So, the correct option is b:
(-36,-2)👍
Please do it yourself first beacuse I don't know if my answer is correct or not but i am 95% sure that it is correct 💯
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And if there is a mistake in my solution then please inform me ❤️
Targeted class- 10th...
Real class- no need to tell... 😂😂
9th kw ho tum?
Option c is correct 💯
36,6
Mirror formula
1/v+1/u=1/f
1/v+1/-18=1/-12
1/v=1/-12-1/-18
1/v=-1/-36
v=36
Hence, The distance of the image from the mirror is 36cm
Magnification formula
m=-v/u
m=-36/-18
m=2
Height formula
m=h1/h2
2=h1/3
h1=2*3
h1=6
Hence, the height of the image is 6cm
So, option c 36/6 is correct ✅
If i had done any mistake please comment ❤
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Thanku
rapid revision of human eye❤❤❤😊
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#acid base and salt jldi rapid revision
26:38 option B is correct sir
H=3cm
F=13cm
U=18cm
1/v + 1/u=1/f
1/f= 1/v-1/u
1/12=1/v-1/18
v=-36
Magnification formula
m=-v/u
m=-36/18
m=-2
h'=3×-2
=-6
Ans:-b ) -36,-6
Next revision on control and coordination
Sir please chemistry ch 2. Acid bases and salts
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Magnetic effect k v revision karwa dijiye bhaiya 😢
Homework question ka answer option b) -36, -6
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Lecture start at 0:40
Refraction and reflection 1:40
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