Quora misses you bro ;) How's your studies going? Also, tell me how are you motivating yourself everyday? And not to forget only one day left for this year to end LoL xD

@Richeek Das
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Arre... Then I know you. I have a friend named Rishav Dugar. He said the you are the smaller version of Arka Das. So I know you that you are a topper!!!

@Richeek Das
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I tried it this way. Taking the lens to be thin. I found the number of lens it will overcome until it reaches 3.925cm . Now taking the equivalent of all those lenes we ge the point where the ray would have passed if there were no further lens cutting the principal axis (focal pt) . Ebar....Using perpendicular/base.... =Tan theta(theta is angle b/w ray and principal axis). We can find out the height. If I am wrong somewhere.... Plz correct me. My answer is not matching.

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## Comments

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TopNewest@Md Zuhair

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+2348103182839...Please add me to the group.. I will be very happy

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@Md Zuhair @Rahil Sehgal @Ankit Kumar Jain - I also want to join. Please add me.

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Pls comment your contact number below so that I can add u. :)

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@Rahil Sehgal what all will you discuss in the group

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Please comment if you got it.

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Thanks...

I have added you in the group

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Thanks.

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@Md Zuhair @Rahil Sehgal - Can you please add me to the group too?

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You can comment your contact number below. Thanks.

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Would you please add me too ,my number is 8547853228

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What kind of group is it ? Formal one for physics and math problems ?

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Kind of.....

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Ohh okay!

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Quora misses you bro ;) How's your studies going? Also, tell me how are you motivating yourself everyday? And not to forget only one day left for this year to end LoL xD

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Happy New Year!

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Are you still adding students ? BTW I'm a JEE aspirant and I'm quite new to brilliant(Using it just for 1-1.5 months)

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+919903607651

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no bro... actually i deleted Whatsapp 6-7 month before :P

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Obviously. Friends with toppers like you would be a nice thing Id love to

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What do you mean by "topper" ? -_- Do you even know me ?

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your credentials say all :)

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please help me with this!

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The exact answer is \(y = hcos(\frac{x}{\sqrt(lf)})\)

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Oh so we need to find out the exact path of ray. I thought approximation would work

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Ya

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I have solved this , but I want to know if there is any better solution than what i have done ! Help me.

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