Validity of Snell's Law: Optics

Recently I working on a optics Question. Then , An Confusion is created in my mind , So please Help me to get rid out from that !

Problem : If Light ray is incident at an angle α\alpha on an glass slab which has variable refractive index which varies as : μ(y)=μ0Ky\mu (y)\quad =\quad { \mu }_{ 0 }\quad -\quad Ky. where K : is constant and refractive index of outside medium is μ0{ \mu }_{ 0 } Then Find maximum Hight at which Light ray goes up ?

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Solution: Using Snell's Law at intial state and Final State : μ0sinα=(μ0Khmax)sin90hmax=μ0(1sinα)K\\ \quad { \mu }_{ 0 }\sin { \alpha } \quad =\quad ({ \mu }_{ 0 }\quad -\quad K{ h }_{ max })\sin { 90 } \\ \\ { h }_{ max }\quad =\quad \cfrac { { \mu }_{ 0 }(1-\sin { \alpha } ) }{ K } .

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Doubt :

1)- What is the critical angle in this situation ? Can we say it as 90 degree ?

2)- How can we use snell's Law at Particular an intermediate Point ? Is always i = r ?


Expln-1 ) -Because if we use Snell's Law at the interface i.e

μsinθC=(μ0Khmax)sin90\mu \sin { { \theta }_{ C } } =({ \mu }_{ 0 }\quad -\quad K{ h }_{ max })\sin { 90 } \quad \quad .

also μ=(μ0Khmax)\mu \quad =\quad ({ \mu }_{ 0 }\quad -\quad K{ h }_{ max }). because at that point hight is almost same !

So this gives θC=90{ \theta }_{ C }\quad =\quad 90.

Expln-2) - SInce at particular point refractive index of medium is same at just below and above of the interface. So By snell's Law incident angle (i) = refracted angle (r)


I know this wrong But Please Tell me Where I'am wrong ??

Thanks a lot !!

Note by Deepanshu Gupta
4 years, 10 months ago

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To solve this problem we model the glass slab as a collection of infinitely many extremely thin glass slabs stacked together. We also assume that since the thickness of each of these slabs is so small their refractive index does not vary much and light follows a straight path inside them. Applying Snell's Law at an interface:

μ(y)sini=μ(y+dy)sinr\mu(y)\sin i=\mu(y+\text{d}y)\sin r

Thus from the equation above it is clear that rir\neq i but the value of rir\rightarrow i.

By geometry this value of rr is same as the angle of incidence at the next interface. As the angle of incidence has changed by a small amount let us call it i+dii+\text{d}i.

(μ0ky)sini=(μ0k(y+dy))sin(i+di)\therefore (\mu_{0}-ky)\sin i=(\mu_{0}-k(y+\text{d}y))\sin (i+\text{d}i)

Simplifying we obtain:

didy=kμ0kytani\dfrac{\text{d}i}{\text{d}y}=\dfrac{k}{\mu_{0}-ky}\tan i

Solving this differential equation we obtain the answer

hmax=μ0(1sinα)kh_{max}=\dfrac{\mu_{0}(1-\sin \alpha)}{k}

Karthik Kannan - 4 years, 10 months ago

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Kartik's solution is best and easy, in these type of problems light instead of travelling in straight line it travels in curve I can give that curve with another method

Suppose light is incident at origin (from 3rd quadrand) and glass is in 1st quadrant with y-axis as normal

μ0sinα=μsin(r)\mu_0 \sin \alpha = \mu sin (r)

sin(r)=μ0sinαμ\displaystyle sin (r) = \frac{\mu_0 \sin \alpha}{\mu}

We can say that

dydx=cot(r)\frac{dy}{dx} = cot (r)(slope of curve)

cot(r) can be easily found(given sin(r)) to obtain trajectory

Krishna Sharma - 4 years, 10 months ago

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@Deepanshu Gupta - you see , initially you have assumed that light is not incident at the critical angle of the base (y=0) Now as you go above the refractive index falls, but it falls extremely slowly, in any small interface,, you can see that if critical angle is reached it would mean

μ2μ1=sin(i)sin(r)μkd(y)μ=1kd(y)μ=sin(ic)sin(90)=sin(ic)\frac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } =\frac { sin(i) }{ sin(r) } \\ \\ \frac { \mu -kd(y) }{ \mu } =1-\frac { kd(y) }{ \mu } =\frac { sin({ i }_{ c }) }{ sin(90) } =sin({ i }_{ c })

clearly as dy--> 0

μ2μ1=μkd(y)μ=1kd(y)μ>1\frac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } =\frac { \mu -kd(y) }{ \mu } =1-\frac { kd(y) }{ \mu } -->\quad 1\\

so effective critical angle at any interface is 90 degrees (and it is exactly 90 degrees if the transition is absolutely smooth theoretically)

thus no internal reflection will occur till i becomes 90 degrees which is what you have used to get the maximum height, That is i gradually rises till critical angle 90 degrees

so no snells law has not been violated,, just that due to smooth gradation,,, internal reflection only occurs when ray has become paralell to the ground (which is the max height case)

Mvs Saketh - 4 years, 10 months ago

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Yeah ! Now I completely Got it ! Thanks a lot @Mvs Saketh !!

Deepanshu Gupta - 4 years, 10 months ago

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Hey depanshu can you plz guide me about some good books to excel in maths

Kundan Patil - 4 years, 10 months ago

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@Kundan Patil well , I did not use any book much because I used Brilliant.org and My Tutorial material For Maths stuff's which Helps me !

But Sometimes I refer "**Arihant New pattern IIT JEE Book" , But I'am quite uncertain about maths stuff to give you any advice , Since I didn't solve other book's ! So I'am unable to help You as much !

But I Specially Recommend You To Buy "Playing with graphs" By Amit agarwal , it is really very good book's to read !

Deepanshu Gupta - 4 years, 10 months ago

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I think we cannot determine the max height!! See the explanation

I agree with karthik and saketh that max height is reached when r=90 but after that ,total internal reflection occurs which makes the light go further upward and again the same procedure continues .So there is no max height

Ashwin Gopal - 4 years, 10 months ago

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no bro after total reflection light ray will come back downward not upward ! and finally it will come with same angle of incidence as that of initial . it's graph is symmetric about about the critical point !

Deepanshu Gupta - 4 years, 10 months ago

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where i can found a book with problems in mechanics

aris nikolaidis - 4 years, 9 months ago

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