Challenges in optics 4

A simpler method:

As d is negligible compared to D, we can assume both the rays to be incident at an angle 30. from the figure, we need to find $\cfrac { dx }{ dt }$

let refractive index be $r=6+2t+4{ t }^{ 2 }$

using Snell's law at the first surface,

$\frac { \sin { 30° } }{ \sin { R } } =\quad r$

$\sin { R } \quad =\quad \frac { 1 }{ 2(6+2t+4{ t }^{ 2 }) }$

as this value is very small, we can consider

$R=\frac { 1 }{ 2(6+2t+4{ t }^{ 2 }) }$

at t=5,

$\frac { dR }{ dt }=-\frac { 84 }{ { 232 }^{ 2 } }=-0.00156$

Now,

$\tan { R }=\frac { x }{ D }$

$D\sec ^{ 2 }{ R } \frac { dR }{ dt }=\frac { dx }{ dt }$

As R is very small, $\sec ^{ 2 }{ R }$ =1

Ergo, $\frac { dx }{ dt }=\left| \frac { dR }{ dt } \right|=0.00156$

Upon entering the slits, the two light rays have a phase difference of $\Delta\phi = \frac{2\pi(dsin\theta)}{\lambda_0}$ . The central maximum occurs when the two rays strike the screen with 0 phase difference. The wavelength of the light in the liquid is $\lambda=\lambda_0/r$ . If the angle of the outgoing rays to the central maximum is $\alpha$ , then $0 = \Delta\phi_{tot} = \frac{2\pi(dsin\theta)}{\lambda_0}- \frac{2\pi(dsin\alpha)}{\lambda_0/r}$ , and $sin\alpha = \frac{sin\theta}{r}$ .

Differentiating, $\frac{d\alpha}{dt} = \frac{-sin\theta(dr/dt)}{r^2cos\alpha}$ .

$r(t = 5s) = 6 + (2 s^{-1})(5s) + (4 s^{-2})(5 s)^2 = 116$ ,

$\frac{dr}{dt}(t = 5s) = (2 s^{-1}) + 2(4 s^{-2})(5 s) = 42 s^{-1}$ .

Therefore $sin\alpha = sin(30)/116 << 1$ , and $cos\alpha\approx 1$ .

$\frac{d\alpha}{dt} = \frac{-sin(30)(42 s^{-1})}{(116^2)} = -0.00156 s^{-1}$ .

The speed of the central maximum is $\left|\frac{d(Dtan\alpha)}{dt}\right|\approx D\left|\frac{d\alpha}{dt}\right| = (1 m)(0.00156 s^{-1}) = 0.00156 m/s.$

Relevant wiki: Geometrical Optics

• The exact answer : 21/116^2 , the path difference irrespective of the medium there at t=0 is $$ D sin $\theta$ , where theta is the tilt provided initially , after that , using $\frac{D \lambda}{d}$ = $y_{initial}$ , we get centre maxima as $$ $\frac{D \lambda}{d*r}$ + $\frac{D}{2}$ diff. , $$ we get $v$ = $\dfrac{D\sin\theta(2bt+a)}{r^{2}}$ which is $$ 21/116^2= 0.00156064209 :)

Its easy, but very impractical