Wave optics part 7: Resolving power of microscope and telescope
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- Опубликовано: 19 окт 2024
- Resolving power of optical instruments
1. Angular widht
Theeta=1.22 lamda/ a
2. Resolving power of telescope
P=a/1.22lamda
3. Resolving power of microscope
P= 2mue sin beeta/1.22 lamda
4. Effect of colour on the resolving power
5. Effect of refractive index on the resolving power of microscope
6. Optical aperture
7. Numerical aperture
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By anuj Sharma anujsharma
It seems like alak sir's old board 🥺😻
Awesome explanation. Sir i watch many videos on RUclips to clear this concept but i didn't get satisfied but after watching this one i get my full concept
True
Thank you sir... Tomorrow is my physics exam.
Aur ye cheez smjh ni aa rahi thi.. 🙌🙌🙌🙌🙌
Same cheez sir thank you
After this, i am your new subscriber
Aadhe ghante se yeh topic doond rhi thii finally!! Thnkxx a lot!!
Such mein sir bahut you tube videos dekhi but apka method is superb
Thanks a lot
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Good luck for your lucky journey sir😇😇
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It just look like physics wallah set..
It's awesome
I respect you for your hardwork and passion of teaching🙌
I know once the camera ran out of battery..😅 Am i right?
😌my concept got cleared
I was struggling a lot
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I will also confusalbe . when I will find the lecture about this topick no one can understand me to clearly but watching after your lecr I am totally gain this topik and now .
Nice xplntn sir ND in a very short time period....
Osm👍
Thanks sir for making this topic video finally ab 💯 ab samaj aya
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Thx a lot...
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U r right ncert m bhot gndi Daruvation thi..🙁.. Ab smjh aa gya 👌🏻
Grt sir 🙏🙏
awsome
MaazAaa aa gya guru jii...... 👌👌👌
thanks a lot sir
Magedaar sir ...bahut achha...
Thank u sir for sharing your knowledge we love u......👍🏻👍🏻👍🏻
Nice work 😊 it really helps in understanding the concept
Sir u said resolving power is proportional to aperture..... i.e. p=a/1.22 lambda ...... So mathematically it showed that if aperture is more(distance between the object ) resolving power is more .... So how resolving power will be more for 1mm object
Aperture is not the distance between object , it is the diameter of optical instrument.
More the diameter of optical instrument (lens), more the resolving power....
Also resolving power is independent of size of object, it only depends upon size of lens , refractive index of lens, and sometimes liquid is placed before lens, so in that case it also depends upoun refractive index of liquid
@@ANUJSHARMAIci thank u sir ....got it
Just wow
thanks sir 😊
osm
Uh cleared my concept.. i was searching for the same from so long and finally i got it. Thank u so much 😊
Nice explaination
आप यूट्यूब में सबसे अच्छे पदार्थ विज्ञान के अध्यापक हो।
Well explained sir.....
Ur teaching style super cool.... Not like some serious teachers ❤
Gjb😲
Don't know why but literally I am seeing the mirror copy of alakh. Sir in this guy
Anuj Bhai bhut bhadiya
nice video but spelling of distinctly third time also wrong
Aap to phy wallah 2 h
Happy diwali sir, 😊
Mujhe laga ye physics wallah ki alakh pandiya he 😂
Same ... literatally i too
Who is here because physics wallah didn't taught this topic
Mamta i am😁
Thnks sir, ye exam me aaya tha or maine kr diya 😍😍
Great sir
Thanks sir 😊😊