# A vessel with fluids of two different refractive indices

• Vriska

## Homework Statement

both filled to height h in the vessel of length 2h. one has refractive index root 2 and the bottom fluid has refractive index n. find the apparent depth of vessel.

## Homework Equations

n = real depth/apparent depth

## The Attempt at a Solution

what these guys have done :

apparent depth = d/sqrt(2) + d/n

I don't get it, how do you add depths like these esp. when refraction happens at medium n and medium sqrt(2).?

If you were in the medium of refractive index ##\sqrt 2##, what would be the apparent depth of the other medium?

If you were in the medium of refractive index ##\sqrt 2##, what would be the apparent depth of the other medium?

if I were at the bottom of interface at distance d, the apparent depth would be d*##\sqrt 2## /n. my guess was I'd add this to the ##\frac{d} {\sqrt 2}##, that's wrong apparently

if I were at the bottom of interface at distance d, the apparent depth would be d*√2/n
Right.
my guess was I'd add this to the d/√2
No. Adapt your correct answer above to the situation in which the observer is just under the surface of the upper fluid, then compare it with an object that really is at that depth in such a fluid. How would that be seen from just above that fluid?

I
Right.

No. Adapt your correct answer above to the situation in which the observer is just under the surface of the upper fluid, then compare it with an object that really is at that depth in such a fluid. How would that be seen from just above that fluid?

just under the upper fluid? Then thered be no refraction right?

Right.

No. Adapt your correct answer above to the situation in which the observer is just under the surface of the upper fluid, then compare it with an object that really is at that depth in such a fluid. How would that be seen from just above that fluid?

okay mathematically : change in depth due to medium 2 = d- sqrt(2)d/n, due to medium 1 = d - d/sqrt(2), total change in depth is 2d - sqrt(2)d/n - d/sqrt(2) this is subtracted from total depth to get apparent depth . = sqrt(2)d/n +d/sqrt 2. wrong answer, right?

just under the upper fluid?
Just under the (upper) surface of the upper fluid.

Just under the (upper) surface of the upper fluid.

okay real depth at supper surface of upper fluid is d + d*sqrt(2)/n

so sqrt 2 = d +d*sqrt(2)/apparent depth. ad =d(n+sqrt2)/n*sqrt2. thanks! BTW, would you happen to know if the method used in the book make sense?

would you happen to know if the method used in the book make sense?
Maybe, but I cannot think of a simple argument to support it. When in doubt, draw ray diagrams.