Optics - Apparent Change in Position due to Media

[king vitamin]

Homework Statement

A plane slab of glass of thickness t and index n is inserted between an observer's eye and a point source. Show that the point source appears to be displaced to a point closer to the observer by approximately [(n-1)/n]t. Use small-angle approximations.

Homework Equations

Snell's Law and trig relations/approximations. I don't have a way to scan in the diagrams I've drawn, but a good one I've found is here: http://homepage.mac.com/cbakken/obookshelf/image033.gif .

The Attempt at a Solution

Previously I've solved for the apparent change in position of an object placed in a media of higher index of refraction which is analogous to the answer. The problem here is that both the observer and the object are outside of the media, so I can't seem grasp on to any equations relating distance of object, apparent distance of object, and slab thickness. I've been using approximations such as tan[theta]=sin[theta] so I can use Snell's Law, but since the object is located outside of the media I'm not sure how to use the refraction angle for equations. Any hints would be appreciated (I feel like there's just an approximation I'm not thinking of).

 

[king vitamin]

Nevermind, I've solved the problem (the trick was to consider another point further back from the object, where a person from inside the glass slab would see the object located).

 

[nlightner]

The apparent change in position of the point source can be explained by Snell's Law, which states that the angle of incidence equals the angle of refraction, or in this case, the angle of the light ray entering the glass slab equals the angle of the light ray exiting the slab. The small-angle approximation can be used because the angles of incidence and refraction are small enough to be considered approximately equal.

Using the diagram provided, we can see that the light ray enters the glass slab at an angle of approximately θ, and exits the slab at an angle of approximately θ'. By Snell's Law, we can write:

sinθ = n sinθ'

Since we are interested in the change in position of the point source, we can use the small-angle approximation to write:

θ ≈ tanθ = sinθ

θ' ≈ tanθ' = sinθ'

Substituting these approximations into Snell's Law, we get:

n sinθ ≈ sinθ'

n tanθ ≈ tanθ'

Since tanθ = t/L, where L is the distance between the point source and the observer's eye, and tanθ' = t/L', where L' is the apparent distance between the point source and the observer's eye, we can write:

n(t/L) ≈ (t/L')

Solving for L', we get:

L' ≈ Ln/L

Since we are interested in the change in position, we can subtract the original distance L from both sides to get:

L' - L ≈ L(n-1)/n

Therefore, the apparent change in position of the point source is approximately (n-1)/n times the original distance L.

In terms of the slab thickness t, we can write:

L' - L ≈ (n-1)t

Therefore, the point source appears to be displaced to a point closer to the observer by approximately (n-1)t.

This result is consistent with our intuition, as a higher index of refraction means that light travels slower in the medium, causing the apparent distance of the point source to decrease.