How Deep Does the Acrylic Bottom Appear Under Water?

Click For Summary
SUMMARY

The discussion focuses on calculating the apparent depth of an acrylic block submerged under water, with specific refractive indices of water (n=1.33) and acrylic (n=1.5). The relationship between real depth and apparent depth is established using Snell's Law and the equation (n1/p) - (n2/q) = (n1-n2)/r. Participants suggest using ray diagrams to visualize the refraction of light as it passes from acrylic to water, aiding in determining the perceived position of the acrylic's bottom surface.

PREREQUISITES
  • Understanding of Snell's Law and refractive indices
  • Familiarity with ray diagrams in optics
  • Basic knowledge of light refraction principles
  • Ability to manipulate equations involving refractive indices
NEXT STEPS
  • Study Snell's Law in detail to understand light behavior at interfaces
  • Learn how to construct ray diagrams for optical systems
  • Explore the concept of apparent depth in different mediums
  • Investigate the effects of varying refractive indices on light paths
USEFUL FOR

Students and educators in physics, particularly those focused on optics, as well as anyone interested in understanding the principles of light refraction and its applications in real-world scenarios.

feliperrr
Messages
1
Reaction score
0

Homework Statement


A container with a layer of water (n=1.33) of 5 cm thick is over a block of acrylic (n=1.5) of 3 cm thick. An observer watches (perpendicularly from above) the lowest surface of the acrylic. What distance does it (the bottom surface) seems to be from the top of the water?

Homework Equations


(n1/p) - (n2/q) = (n1-n2)/r

The Attempt at a Solution


(n1/p) - (n2/q) = 0

q = n2p/n1
 
Last edited:
Physics news on Phys.org
feliperrr said:

Homework Statement


A container with a layer of water (n=1.33) of 5 cm thick is over a block of acrylic (n=1.5) of 3 cm thick. An observer watches (perpendicularly from above) the lowest surface of the acrylic. What distance does it (the bottom surface) seems to be from the top of the water?

Homework Equations


(n1/p) - (n2/q) = (n1-n2)/r

The Attempt at a Solution


(n1/p) - (n2/q) = 0

q = n2p/n1

i think you can use the relation ship between real depth of an abject and apparent depth of the same object related to ratio of the refractive indices of the medium.
as the bottom is being viewed vertically so one can take paraxial rays close to the vertical and extend the emergent ray below to find the apparent position of the image.
what will be depth of the acrylic slab when viewed from water?
and what will be the depth of this image when viewed from above -the water surface?
 
I would rather suggest to use Snell's law and a ray diagram :wink:

Draw such a diagram with rays from a point on the bottom surface of the block.
These rays are refracted when they pass from the block into the water (ignore the glass)
and again as they exit the water surface. These rays will then indicate where the virtual
image of the point is located if you extend the rays backwards.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
8K
  • · Replies 6 ·
Replies
6
Views
12K