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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestInteresting Problem ! what is final answer ? @Akshay Bodhare

May be it can help Click here

Log in to reply

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

can you please show your process,

Log in to reply

I have updated the note.

Log in to reply

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

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)

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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!

Log in to reply

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

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Hii...Q..23-14(25%-20)£22=?

Log in to reply

I am unable to understand

Log in to reply