# I need help with a varying refractive index problem 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, $r_{min}^2=r_0^2* \sin \theta$ Note by Akshay Bodhare
5 years, 10 months ago

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Interesting Problem ! what is final answer ? @Akshay Bodhare

- 5 years, 9 months 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.

- 5 years, 9 months ago

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

- 5 years, 9 months ago

I have updated the note.

- 5 years, 9 months 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.

- 5 years, 9 months ago

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)

- 5 years, 9 months ago

How did you get the first step.

- 5 years, 9 months ago

i have updated solution

- 5 years, 9 months ago

I now have understood how you got the first step,but how did you conclude that minimum will occur when $\theta$=90?

- 5 years, 9 months 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 )

- 5 years, 9 months ago

In the beginning,the light ray was horizontal,you should take that into consideration.

- 5 years, 9 months ago

that is not at all a problem, for two reasons, one is that my theta is not the polar angle as yours, mine is the angle the ray makes with the local radial line which also happens to be the normal which was not 0 initially, secondly, that is not the actual angle at r=R , because i only care about the angles inside the glass ball, not outside,,

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

- 5 years, 9 months 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.

- 5 years, 9 months ago

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

- 5 years, 9 months 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$\theta$

- 5 years, 9 months 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 $\theta +d\theta +d\alpha$

- 5 years, 9 months ago

- 5 years, 9 months ago

Welcome but we do not yet know if its correct yet,, :)

- 5 years, 9 months ago

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

- 5 years, 9 months 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.?

- 5 years, 9 months 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?

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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

- 5 years, 9 months ago

What rank did you get in PT-2

- 5 years, 9 months 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?

- 5 years, 9 months 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?

- 5 years, 9 months 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.

- 5 years, 9 months ago

Yes i too came to know it through that.

- 5 years, 9 months 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 :)

- 5 years, 9 months ago

Thanks.Good luck for you too.

- 5 years, 9 months ago

Hii...Q..23-14(25%-20)£22=?

- 5 years, 10 months ago

I am unable to understand

- 5 years, 10 months ago