# Young double slit experiment-position of maxima and minima

In a young's double slit experimental setup, y coordinates of the central maxima and 10th maxima are 2cm and 5cm respectively. when young's double slit experimental setup is immersed in a liquid of refractive index 1.5, the corresponding y coordinates will be___________?

my attempt:-the path difference encountered in vaccum at a distance y from the centre of screen is $$\delta x=y\frac{d}{D}$$,where d is the distance between the slits and D is distance of slits from the screen.now in some other medium the path difference should change to $$\delta x''=\mu y''\frac{d}{D}$$where $\mu$ is the refractive index of the medium.for same order of maxima or minima,we must have $$y\frac{d}{D}=\mu y''\frac{d}{D}$$which gives $$y''=\frac{y}{\mu}$$.But this approach is not giving me the correct answer.Can anybody suggest where i am possibly wrong or put on a new way to approach this problem.

• what is the correct answer? – Lasper Oct 1 '16 at 9:24
• @Lasper the correct answer is 2cm for new central maximum and 4 cm for 10th maxima – Pink Oct 1 '16 at 10:06

The central maxima position is not at y=0 which means oblique incidence right? Now, for air

1. $\frac{yd}{D} - d\sin\theta = 0$ for central maxima; this gives $\sin\theta=\frac{2~\mathrm{cm}}{D}$
2. $\frac{y^\prime d}{D} - d\sin\theta = 10\lambda$; putting $y^\prime= 5~\mathrm{cm},$ you get $\lambda=\frac{3d}{10D}$

For the given medium of r.i. $1.5,$

1. Equation one doesn't change since it doesn't have to account for the medium anywhere. Hence the position of the central maxima remains the same.

2. $\frac{y'' d}{D} - d\sin\theta = \frac{10\lambda}{1.5}$; substitute the value of $\sin\theta$ and the expression for $\lambda$ from part one, you get the new position of $10^\textrm{th}$ maxima as $4~\mathrm{cm}$ please let me know if my answer is correct.

• Hi, Prasad. Phys.SE is not a homework-solving service. So, please, don't provide complete solutions t0 homework questions. – user36790 Oct 1 '16 at 19:11
• okay i did not know that. i actually worked it out for my own sake since i am giving exams. i'll be more careful – Prasad Mani Oct 2 '16 at 5:36

When immersed in the liquid the situation is the same as having the apparatus in air but changing the wavelength of light by an appropriate factor, the factor being related to the refractive index of the liquid.