Apparent Depth of a Point Source Viewed Through a Prism

[Steve Rogers]

Homework Statement

I have the following optics problem and would appreciate a detailed explanation.

A point source (P) is located a distance (\ell) from the vertical face (AB) of a right-isosceles prism (ABC). The prism has refractive index (n), and its two perpendicular sides have equal lengths (AB=BC=s). The angle at (C) is (45^\circ). An eye observes the prism from below, looking perpendicularly at face (BC), as shown in the diagram.

Question: How far below face (BC) does the virtual image of the point source appear to be?

The answer given is

\ell+\frac{s}{n}.

Relevant Equations

The apparent depth equals the actual depth divided by the refractive index.

Here is my current understanding.

Since the incident ray is perpendicular to (AB), it enters the prism without changing direction. It then strikes the hypotenuse (AC) at an angle of incidence of (45^\circ). If (n>\sqrt{2}), total internal reflection occurs, and the ray travels vertically downward toward (BC).

To locate the virtual image, I think we need to trace the emerging rays backward. Since the observation is perpendicular to (BC), the apparent depth of an object inside a medium of refractive index (n) is normally its actual depth divided by (n).

However, I'm having trouble applying this idea to the entire optical path. In particular, I don't understand why the final expression is (\ell+s/n), rather than ((\ell+s)/n).

Could someone explain how to locate the virtual image rigorously, taking the reflection at (AC) and the refraction at (BC) into account?

I'd appreciate a step-by-step explanation, especially of how the distance (\ell) contributes to the final result.

 

[kuruman]

However, I'm having trouble applying this idea to the entire optical path. In particular, I don't understand why the final expression is $(\ell+s/n)$, rather than $((\ell+s)/n).$

Figure (a) on the right shows a glass of water with a source of light at the bottom. The distance from the surface of the water ito the bottom of the glass is $d$. The eye sees the source at an apparent distance below the surface $d'=d/n$ where $n$ is the index of refraction of water.

Figure (b) shows the same glass with the same water and same source. However, the glass is elevated and the source is at its old position in air at distance $s$ below the bottom of the glass. At what distance below the surface of the of the water will the eye see the source to be? How would your answer change if distance $s$ is doubled?

 

[Steve Rogers]

View attachment 375038Figure (a) on the right shows a glass of water with a source of light at the bottom. The distance from the surface of the water ito the bottom of the glass is $d$. The eye sees the source at an apparent distance below the surface $d'=d/n$ where $n$ is the index of refraction of water.

Figure (b) shows the same glass with the same water and same source. However, the glass is elevated and the source is at its old position in air at distance $s$ below the bottom of the glass. At what distance below the surface of the of the water will the eye see the source to be? How would your answer change if distance $s$ is doubled?

Hi. My calculation gives an apparent distance of $s+\frac{d}{n}$, and if s is doubled, then the answer will be $2s+\frac{d}{n}$. But does this relate to my original question? Thank you.

 

[kuruman]

But does this relate to my original question?

Yes it does. Do you see the similarity between the answer to my question and the correct answer to your problem?

 

[kuruman]

Question: How far below face (BC) does the virtual image of the point source appear to be?

I forgot to add that the statement above is nonsensical and contradicts the familiar observation that when you look at your face in a mirror, your virtual face looks back at you behind the mirror.

A virtual image appears where light has no business being and that is why it is called "virtual". This means that the point of reflection must be between the object and the image. In this case, the eye below the mirror plane which is line AC, will see the image where light has no business being, i.e. at the extension of the vertical dotted line above the mirror line AC and, therefore, also above line BC.

 

[Merlin3189]

Hi. My calculation gives an apparent distance of $s+\frac{d}{n}$, and if s is doubled, then the answer will be $2s+\frac{d}{n}$. But does this relate to my original question? Thank you.

I wonder if the point you are missing is: the reduction in apparent distance applies only to the distance the light travels through the glass, not the s distance through air.
For air, n=1, so s/n = s. For glass n >1, so d/n is smaller than d.
In your question, only d is altered by refraction. s is not altered by refraction. So only d is multiplied by 1/n.