Refractive index and depth of object in water.

[heartOFphysic]

Question:
A Fsherman sees a Fish in a river at an apparent depth below the surface of
the water of 0.75 m. Given that the refractive index of water is 1.33, is the
true depth of the Fish below the water's surface:
A 0.75 m?
B Less than 0.75 m?
C 1 m?
D More than 1 m?


My proposal:

I am not really sure about this at all. All I can think of is "if the refractive index of water is 1.33 then the depth could be 1.33 times that of what it seems to be" (however I CAN'T prove this, it's just a hunch)

So 0.75 X 1.33 is roughly 1, so is the answer 1m?

 

[NascentOxygen]

Hunches won't help you, but a neat diagram is sure to.

Draw a light ray from the fish and entering the eye of the observer. Make sure you show the correct way it bends as it passes from water into air.

 

[PeterO]

Question:
A Fsherman sees a Fish in a river at an apparent depth below the surface of
the water of 0.75 m. Given that the refractive index of water is 1.33, is the
true depth of the Fish below the water's surface:
A 0.75 m?
B Less than 0.75 m?
C 1 m?
D More than 1 m?


My proposal:

I am not really sure about this at all. All I can think of is "if the refractive index of water is 1.33 then the depth could be 1.33 times that of what it seems to be" (however I CAN'T prove this, it's just a hunch)

So 0.75 X 1.33 is roughly 1, so is the answer 1m?

If you have ever looked into a swimming pool while the surface is smooth [you have to get there early before everyone jumps in] you would realize that the water appears different depths at different angle of observation.
Looking straight down it appears deepest - but even then shallower that it actually is.
I think I have seen somewhere that the apparent depth is 1/n times the real depth - referring to the apparent depth when looking directly into the medium - angle of incidence = 0 degrees.

 

[tiny-tim]

hi heartOFphysic!

yes, do as NascentOxygen suggests …

Hunches won't help you, but a neat diagram is sure to.

Draw a light ray from the fish and entering the eye of the observer. Make sure you show the correct way it bends as it passes from water into air.

i'm a fish, and i know!

 

[heartOFphysic]

I think I have seen somewhere that the apparent depth is 1/n times the real depth - referring to the apparent depth when looking directly into the medium - angle of incidence = 0 degrees.

so it's (0.75)/ (1/1.333) = 0.99975


thus 1m?

 

[NascentOxygen]

Can you work out the diagram (a vertical cross-section through the water) using the basic formula, Snell's Law? http://en.wikipedia.org/wiki/Refractive_index

That will be more useful to helping you understand what's involved here than guessing, and hoping for the best.

 

[emailanmol]

Hey,

This problem can be solved only for small angles.
Everyone above has pointed put some really useful tips.

The best way for you to start is to first derive a general formulae.(you will need a diagram.you can take help of textbooks because its not as simple as it sounds).
(Hint: You have some idea regarding the resulting formulae .Just go through the derivation again )


From there you will obtain a result which will be valid only for small angles of observation.
Plug in values and find the answer

 

[tiny-tim]

This problem can be solved only for small angles.

and small fish

 

[emailanmol]

Lol tiny-tim.

I have read many posts of yours and I got to admit,
'Not only are your posts to the point, they also tend to create an ambience of fun and happiness which helps the OP a lot'.

Cheers
Anmol