Refractive index snell's law help

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SUMMARY

The discussion centers on the application of Snell's Law to determine whether light will be observed when transitioning from water to air through a glass plate. The user correctly identifies that total internal reflection occurs at the water-air boundary but struggles to analyze the effect of introducing a glass plate. The refractive index of glass is greater than that of water, preventing total internal reflection at the water-glass boundary. However, once the light exits the glass into air, total internal reflection can occur due to air's lower refractive index compared to glass.

PREREQUISITES
  • Understanding of Snell's Law and its formula: n1sin(θ1) = n2sin(θ2)
  • Knowledge of refractive indices of common materials (water, glass, air)
  • Familiarity with the concept of total internal reflection
  • Basic principles of light propagation and angles of incidence
NEXT STEPS
  • Study the concept of total internal reflection in detail
  • Explore the refractive indices of various materials, focusing on water, glass, and air
  • Learn how to apply Snell's Law to different boundary conditions
  • Investigate practical applications of Snell's Law in optics and fiber optics
USEFUL FOR

Students of physics, optical engineers, and anyone interested in understanding light behavior at material boundaries.

jeebs
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refractive index... snell's law... help!

Hi,

im having a LOT of trouble with what seems like a basic question. i have been stressing over this for nearly 3 hours now. can anyone help me out? here it is:

Light propagating from water into air is incident on the water surface at the angle of total internal reflection. A plane parallel glass plate is brought into contact with the surface of water.

Will any light be observed as a result of that? Justify your answer by using Snell’s law.

snell's law: n1sin \theta1 = n2sin\theta2

where n is the refractive index of the medium either side of the boundary, and theta is the angle made by the ray with the normal to the boundary either side of the boundary.

all i have managed to work out is that the light ray coming in at the reflection angle for water - air boundary would not be able to experience total internal reflection at a water - glass boundary, since the refractive index of glass is greater than water.

when the ray travels through the glass and meets the glass - air boundary, total internal reflection is possible since air has a lower refractive index than glass.

but, i am stuck here. i can't find a way to show whether the ray will be totally internally reflected or not. has anyone got any suggestions?

thanks.
 
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PS. if you haven't answered because it looks like i haven't tried hard enough or haven't given enough info then please say so, because half the time i ask for help on this site i don't even get a reply even though the question gets viewed. i want to know why.
 


you need to show some working... to both show you;ve attempted the problem & it makes it easier for people to see what you're trying to do & how to correct you

but this how i would look at it

Water to Air
nw.sin(tw) = na.sin(ta) = na.1 as at TIR so ta = 90, sin(ta) = 90

so sin(tw) = na/nw
(note na is close to 1 but we don't have to cancel so leave for now, might gain some extra insight later)

what happens when the glass is inserted, tw is the same
nw.sin(tw) = ng.sin(tg)

so we have (tg) now write the equation for the glass air interface compare them ..
 

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