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

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Comments

Interesting Problem ! what is final answer ? @Akshay Bodhare

May be it can help Click here

Deepanshu Gupta - 6 years, 5 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.

Akshay Bodhare - 6 years, 5 months ago

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

can you please show your process,

Mvs Saketh - 6 years, 5 months ago

I have updated the note.

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.

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)

@Mvs Saketh – How did you get the first step.

@Akshay Bodhare – i have updated solution

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

@Akshay Bodhare – 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 )

@Mvs Saketh – In the beginning,the light ray was horizontal,you should take that into consideration.

@Akshay Bodhare – 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

@Mvs Saketh – 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.

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

Mvs Saketh - 6 years, 4 months ago

@Mvs Saketh – 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 $\theta$

@Mvs Saketh – 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 $\theta +d\theta +d\alpha$

Akshay Bodhare - 6 years, 4 months ago

@Mvs Saketh – Thanks for your beautiful solution

Rajat Raj - 6 years, 4 months ago

@Rajat Raj – Welcome but we do not yet know if its correct yet,, :)

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

Shashwat Shukla - 6 years, 4 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.?

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?

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

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

@Shashwat Shukla – What rank did you get in PT-2

@Akshay Bodhare – 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?

@Shashwat Shukla – 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?

@Akshay Bodhare – 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.

@Shashwat Shukla – Yes i too came to know it through that.

@Akshay Bodhare – 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 :)

@Shashwat Shukla – Thanks.Good luck for you too.

Hii...Q..23-14(25%-20)£22=?

Ukthori Offi - 6 years, 5 months ago

I am unable to understand

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$