Refraction at a spherical surface

1. Nov 22, 2012

Kurokari

1. The problem statement, all variables and given/known data

A speck of dust is 3cm from the center of a glass sphere with radius of 5cm. If the glass sphere is placed in a tank of glycerin with a refractive index of 1.47, find the image distance, as viewed along the diameter through the speck of dust from the far side.
Refractive index of glass is 1.50

2. Relevant equations

(n1/u) + (n2/v) = |n2-n1|/r

3. The attempt at a solution

The solution is actually given, but I want to confirm what I understand from the solution given is absolutely correct.

The object distance is given that u = 5 + 3 = 8cm
This means that the speck of dust is the glass sphere, and since the question says we're looking from the far side, it means we're looking at the other side, that would explain the 5 + 3 = 8 cm.

In addition, since the we're looking that way, the light source and the object are on the same side of the spherical surface, does this mean object distance is positive?

Lastly, the n1 given was 1.50, which means n1 is the index of refraction of where your object is, and n2 the index of refraction of glycerin?

2. Nov 22, 2012

Simon Bridge

Yep - "other side" begs the question "other side from what?" The only marker you have is the dust position.
The light "source" is the object. When is it not?
I always get confused by that too :( Consider, the glass is acting as a lens - sort-of. As light coming from the dust is concerned, it travels through a bunch of glass, hits a spherical interface (the "lens" position) and then travels through some glycerin.
How would you normally figure the object distance for that?

Alternatively you can just re-derive the equation geometrically and use whatever signs you like.

If the object is inside glass then why wouldn't the refractive index where the object is be the refractive index of glass?