Malus Law Verification Experiment by Prof Sasane S B

[@vinayakjalan5696]

Really good explanation sir.. Good to see you and this topics again..

[@manasishelar3402]

Thank you so much sir understood everything. Please keep making videos like these

[@ashishmulajkar1986]

superb explanation sir, thanks

[@SSPhysics]

Please do comment here👉

[@criminalgamer6312]

Nice explanation sir, excellent work👏👏

[@kamalpratapsinghrawat9283]

amazing explanation sir really saves my day

[@lalitakaushal45]

Helpful video 👍

[@mahsumoral4960]

How can we distinguish between the linearly polarized light, circularly polarized light, and eliptically polarized light just by looking at the intensity and angle graph?

[@karanupare3770]

Very good video sir 👍

[@tanmayraut3099]

Very nice sir😃😃😃

[@kunalj0007]

Thank you sir You saved My exam

[@vipul0710]

Thank you sir ❤️👍🏻

[@drmangeshwaghmare8766]

Nice explanation Sir

[@shardulbhosale1753]

Thank you sir

[@alphareviewsinfo7203]

Well explained!!

[@Nbeastshorts]

SUPERB

[@sohamkokane427]

C-29__SOHAM KOKANE Helpful video👍👍

[@jakirshah8329]

Nice video sir

[@niloofarkh4779]

THANK U SO MUCH

[@saritasah5731]

Thank you

[@vivekkale1684]

Excellent

[@sawanyadav4660]

Thanks sir

[@YogeshSharma-ys3hm]

What’s the use of power source here as we are getting current from lamps light only

[@SSPhysics]

Good question. Current without external voltage is too small to detect. Means almost zero for given source. I have tried that.

[@YogeshSharma-ys3hm]

@@SSPhysics got itt sir👍

[@nowitchhunt]

What is the intensity measured without the analyzer?

[@SSPhysics]

It will be constant.

[@nowitchhunt]

@@SSPhysics I understand that it will be constant but I'm wondering what the value will be relative to the maximum you get with the second polarizer in the beam. Will it be the same or higher. In other words, that value will be I0. Does the rest of your observations match the I = I0 cos^2 theta relation, or is there some adjustment needed?

[@SSPhysics]

@@nowitchhunt are you talking about maximum output intensity from analiser?

[@nowitchhunt]

@@SSPhysics Sorry if I wasn't clear. Yes I'm interested in the maximum output from the analyzer *as a percentage of the maximum intensity when the analyzer is absent*. In other words, I'm also interested in the maximum reading when the analyzer is removed.

[@SSPhysics]

Formula for Malus law itself tells this. I=Iο Cosθ Analyzer =second polariser. If polariser is perpendicular to analyzer(θ=90 )outcoming intensity(I) will be minimum(cosθ=0,I =0, ideally). When analyzer is parallel to polariser(θ=0) intensity will be maximum (cosθ=1, I=Io, ideally). Means maximum intensity coming out of analyzer is equal to incident intensity on analyzer. If you remove analyzer at all intensity will be Io.

[@AmitRaj-if6pt]

C-03

[@RiteshHarishchandre]

🙃🙃🙃🙂🙂🙂😀😀😃

[@rohitgawale3411]

C 15

[@saikatsarkar1343]

Insufficient data about graph and table...

[@SSPhysics]

This is demonstration/ practical explanation, how you can perform experiment.

[@avacdoos3101]

O.o cardiB

[@shivlingborate3169]

Nice explanation sir