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

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## Comments

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TopNewestI posted this as a question a few months ago. So feel free to go solve that. :) Parallel slabs? I think not!

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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.?

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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?

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

can you please show your process,

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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.

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i didnt use the lagrangian way, instead i took a small concentric slab, and used snells law

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

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

Note however, the theta is not the angle initially given , find it by using snells law at boundary (air-glass)

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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

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I have updated the note.

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Interesting Problem ! what is final answer ? @Akshay Bodhare

May be it can help Click here

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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.

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Hii...Q..23-14(25%-20)£22=?

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I am unable to understand

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