# Homework Help: Easy one

1. Nov 25, 2005

### meadow

i just need a little help with a relatively easy lens question

A small fish is cast into the center of a glass sphere of radius R=5 cm and n=1.5. Where will an observer see the fish? Where will the observer see a decorative pattern painted on the back side of a sphere?

So, I thought I could use the formula to find the image distance i=n-n1/R; and i subtracted 1.0 (n of air) from 1.5, then divided by the radius to find the distance of the image of the fish. For the background pattern, I used 10 cm as the value of "R", because it is the length of the diameter away from the observer.
Am I right in my thinking? I feel like I am just plugging and chugging, rather than really understanding the concept...(so i am not sure it i am even evaluating correctly)
Thank you.

2. Nov 25, 2005

### lightgrav

Doesn't the image distance (measured from what?)
depend on the location of the object?

You COULD do this with Snell's Law, if you have too much time ...

3. Nov 25, 2005

### meadow

I thought because the fish was placed in the center of the sphere, the distance of the object would be the distance of the radius. I don't, however, know the distance of the observer. Is there a way to explain where the observer would see the fish and the pattern without that distance? How should I approach this problem?

4. Nov 25, 2005

### lightgrav

( i = n-n/R is obviously incorrect : a distance [meter] is not equal 1/R [1/m] )

If you don't like Snell ...
$$\frac{n_g}{d_{ob}} - \frac{n_a}{d_{im}} = \frac{n_g - n_a}{R}$$
should've been derived in your textbook ("single-surface refraction"):
distances are measured from the front of the glass, closest to person.

draw 2 rays from the fish ... are they deflected at the edge of the glass?

draw 2 rays from the back ... are they deflected at the front of the glass?

Last edited: Nov 25, 2005
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