the refractive index varies as

\( \mu = k*r \)

find the least distance between light ray and centre of sphere

is this approach correct?

integration of dt i.e,ds/v is minimum

using that i was able to derive

\( r_{min}^2=r_0^2* \sin \theta \)

but i am not able to derive the result by using elementary optics. i would like to hear your opinions.

Process:

Let center of sphere be the origin,then

\(x=r \cos \theta\) \(y=r \sin \theta\)

\(ds=\sqrt {1+r^2(\theta')^2} dr\),\(\theta'=\frac {d\theta}{dr}\)

\(v= \frac {c}{\mu}\) Therefore,

\(\int dt\) should be minimum

\(\int \frac {kr}{c} \sqrt {1+r^2(d\theta)^2} dr\) should be minimum

let,

\(L=\frac {kr}{c} \sqrt {1+r^2(d\theta)^2}\) Then,by Euler-Langrangian Equation,

\(\frac {\partial L}{\partial \theta}=\frac {d}{dr}(\frac {\partial L}{\partial \theta'})=0\)

Therefore,\(\frac {\partial L}{\partial \theta'}=e\),e is constant.

Therefore,\(\frac{r^3 \theta'}{\sqrt {1+r^2(\theta')^2}}=f\),f is another constant.

Now,this is valid for all r,\(\theta\),

Therefore,initial conditions will also satisfy. initially,ray is horizontal. Therefore,\(\frac {dy}{dx}=0\) Therefore,\(\theta'=\frac {\tan \theta}{r}\) substituting,for initial conditions,

\(f=r_0^2* \sin \theta\) Also,for minimum,\(\frac {1}{\theta'}=0\) Thus, \( r_{min}^2=r_0^2* \sin \theta \)

## Comments

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TopNewestInteresting Problem ! what is final answer ? @Akshay Bodhare

May be it can help Click here – Deepanshu Gupta · 2 years ago

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– Akshay Bodhare · 2 years ago

I have written the final answer.It's just that it came in one of our exams and none of the faculty was able to solve it.Also when I tried I wasn't able to solve using elementary techniques.It requires to minimize the value of an integral,the technique for which I learned from calculus of variations.Log in to reply

I posted this as a question a few months ago. So feel free to go solve that. :) Parallel slabs? I think not! – Shashwat Shukla · 2 years ago

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– Akshay Bodhare · 2 years ago

By any chance are you in FIITJEE?Also were you able to solve it when it came in CM test?Were the faculty able to solve it.?Log in to reply

– Shashwat Shukla · 2 years ago

Yes, I am from FIITJEE Kochi. And I couldn't solve it during the test... I did solve it when I got home though. And yea, our faculty also couldn't solve it. Btw, how is AITS going for you?Log in to reply

– Akshay Bodhare · 2 years ago

Not so good,the questions are really tough.The hardest one was maths in AITS PT-1.Log in to reply

– Shashwat Shukla · 2 years ago

Same here...I'm not sure if it'll help for JEE (in which questions are never this hard).Log in to reply

– Akshay Bodhare · 2 years ago

What rank did you get in PT-2Log in to reply

– Shashwat Shukla · 2 years ago

My mains rank is always better than my advanced rank. For PT-2 I think I got a rank of 63 in Mains and 183 in Advanced. What about you?Log in to reply

– Akshay Bodhare · 2 years ago

You really are good.I got mains rank 325 and advanced rank 699.Expecting a rank better than 5000 in JEE.How much do you expect?Log in to reply

– Shashwat Shukla · 2 years ago

I'm hoping for something within the first 1000. But, you never know huh? Oh, and it's really cool that you know about the Calculus of variations and the Lagrangian...I only came to know about it when I tried solving the brachistochrone problem.Log in to reply

– Akshay Bodhare · 2 years ago

Yes i too came to know it through that.Log in to reply

– Shashwat Shukla · 2 years ago

Great :D ...Well, good luck for the boards(I have my board practicals starting on the day after tomorrow). Was a pleasure corresponding with you :)Log in to reply

– Akshay Bodhare · 2 years ago

Thanks.Good luck for you too.Log in to reply

i am getting R/ sin(X) = r/ sin(x)

can you please show your process, – Mvs Saketh · 2 years ago

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– Akshay Bodhare · 2 years ago

I think that it should be R/sin(x)=r/sin(X).Also,we cannot use X directly since it is not the angle of incidence,it changes as the angle with normal of the circle changes.Log in to reply

\(\frac { sin(\theta +d\theta ) }{ sin(\theta ) } =\frac { k(r-dr) }{ kr } \)

\(\simeq \quad 1+cot(\theta )d\theta \quad =\quad 1-\frac { dr }{ r } \\ =>\quad cot(\theta )d\theta =-dr/r\\ or\quad \quad ln(\frac { sin(\theta ) }{ sin({ \theta }_{ o }) } )=-ln(\frac { r }{ R } )\\ or\quad Rsin({ \theta }_{ 0 })\quad =\quad rsin(\theta )\\ \\ when\quad ray\quad is\quad closest\quad \theta =90\\ \\ hence\\ { r }_{ min }=\quad Rsin(\theta )\)

Note however, the theta is not the angle initially given , find it by using snells law at boundary (air-glass) – Mvs Saketh · 2 years ago

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– Akshay Bodhare · 2 years ago

How did you get the first step.Log in to reply

– Mvs Saketh · 2 years ago

i have updated solutionLog in to reply

– Rajat Raj · 2 years ago

Thanks for your beautiful solutionLog in to reply

– Mvs Saketh · 2 years ago

Welcome but we do not yet know if its correct yet,, :)Log in to reply

– Akshay Bodhare · 2 years ago

I now have understood how you got the first step,but how did you conclude that minimum will occur when \(\theta\)=90?Log in to reply

– Mvs Saketh · 2 years ago

Because if the rays velocity has a component paralell to radius inward, it will move further inward, it stops moving inward, when it becomes perpendicular to radius vector, (just like an alpha particle thrown at a gold nucleus not along the radial line )Log in to reply

– Akshay Bodhare · 2 years ago

In the beginning,the light ray was horizontal,you should take that into consideration.Log in to reply

thirdly, if the ray was along the radius initially, then too my equations would hold , you may check it,, infact in ur equations i believe you have taken theta as polar angle, please see, this problem should obviously have radial symmetry , and hence the result can never explicitly depend on theta because then it would mean that it actually matters what axes you choose to define the system,,,

NOTE it, my theta is not the polar angle, it is the local angle of incidence – Mvs Saketh · 2 years ago

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– Akshay Bodhare · 2 years ago

I was trying the problem again using snell's law,you should also take into consideration the fact that the normal is changing continuously,thus the angle will not be \(\theta +d\theta\) it will be \(\theta +d\theta +d\alpha\)Log in to reply

– Akshay Bodhare · 2 years ago

I chose the axis like that to reduce the calculations.Also,please consider the thought i gave,the minimum distance should be greater than rsin\(\theta\)Log in to reply

– Akshay Bodhare · 2 years ago

let us assume that the ray never bended,then it went in a straight line,if it goes in a straight line then minimum distance = rsin\(\theta\).Please think.Log in to reply

– Mvs Saketh · 2 years ago

Yes that is indeed intriguing, please let me recheck, i am sorry for late reply, i was out of stationLog in to reply

– Akshay Bodhare · 2 years ago

I have updated the note.Log in to reply

Hii...Q..23-14(25%-20)£22=? – Ukthori Offi · 2 years ago

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– Akshay Bodhare · 2 years ago

I am unable to understandLog in to reply