Why does the object disappear when viewed from face AD in a square glass block?

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When observing an object placed near face AB of a square glass block from face AD, the object is not visible due to total internal reflection. The discussion emphasizes that light rays from the object either reflect internally or escape through other faces, but cannot reach face AD. The key to understanding this phenomenon lies in analyzing the angles of incidence and reflection, particularly when the object is close to point A. The conversation highlights the importance of ray diagrams to visualize the light's path and confirms that total internal reflection prevents visibility from face AD. Ultimately, the object disappears from view when observed from that angle.
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Homework Statement



ABCD represents a square glass block. An object O is placed near the face AB as shown. When observed from the face AD, the object will

(a) appear midway between A and B
(b) appear nearer to A
(c) appear nearer to D
(d) not be seen at all

The answer key tells me that the answer is (d) but I don't know how to prove it.

Homework Equations



\frac{sin\ i}{sin\ r}=\frac{n_2}{n_1}

The Attempt at a Solution



I was trying to show that all the rays of light from O to AD are parallel or convergent (real image) but was unsuccessful.
 

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alexmahone said:
I was trying to show that all the rays of light from O to AD are parallel or convergent (real image) but was unsuccessful.

Hi alexmahone! :smile:

This isn't a focussing problem.

Hint: choose a typical point P on AB …

draw the ray OP … where does it go … does it hit AD …

if it does, can it get out, or is it internally reflected? :smile:
 


tiny-tim said:
Hi alexmahone! :smile:

This isn't a focussing problem.

Hint: choose a typical point P on AB …

draw the ray OP … where does it go … does it hit AD …

if it does, can it get out, or is it internally reflected? :smile:

The ray OP may escape through CD or get totally internally reflected at AD.

The minimum angle of incidence at AD is when P is very close to A.

So I only need to prove that total internal reflection occurs even when P is very close to A?
 
alexmahone said:
The ray OP may escape through CD or get totally internally reflected at AD.

The minimum angle of incidence at AD is when P is very close to A.

So I only need to prove that total internal reflection occurs even when P is very close to A?

Yup! :biggrin:
 


tiny-tim said:
Yup! :biggrin:

But what if the rays that are totally internally reflected at AD also totally internally reflect at CD, BC and AB respectively and ultimately exit through AD?

How do I exclude this possibility?
 
alexmahone said:
But what if the rays that are totally internally reflected at AD also totally internally reflect at CD, BC and AB respectively and ultimately exit through AD?

How do I exclude this possibility?

If it reflects off one side, it'll hit the next side at the complementrary (90º -) angle, so it should get out.

Even if the refractive index was so high that it couldn't get out, it would still get back to AD at the original angle, and go round again and again! :smile:
 


tiny-tim said:
If it reflects off one side, it'll hit the next side at the complementrary (90º -) angle, so it should get out.

Even if the refractive index was so high that it couldn't get out, it would still get back to AD at the original angle, and go round again and again! :smile:

Thanks so much! :)
 
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