OpenStudy (agent_a):

Refractive index question.........

3 years ago
OpenStudy (agent_a):
3 years ago

OpenStudy (anonymous):

I can walk you through it if you are on... send me a message

3 years ago
OpenStudy (anonymous):

I would strongly recommend that you look at a single step at a time and actually try and reason this problem through

3 years ago
OpenStudy (anonymous):

Then once you have gotten the answer yourself you can check my to see if they agree.... This is by far the best practice since I can break down the steps for you

3 years ago
OpenStudy (anonymous):

Ok so step one is to note that in a medium the velocity (v) of light in the material is related to is frequency times the wavelength of light in the material wavelength:

3 years ago
OpenStudy (anonymous):

\[v_{m}=\nu \lambda_{m} \] Where \[ \nu \] (pronounced nu, different from v) is the frequency of the light.

3 years ago
OpenStudy (anonymous):

Now the frequency of the light is special because it remains the same when light passes from one medium to another. This is because light can be treated as electromagnetic wave. Meaning it consists of oscillating electric and magnetic fields. At an interface, the oscillations must be coupled to one another such that at no time during the crossing do the two become unsynced. If they were to become unsynced then at some time during the passage from one medium to the other, there would be a discontinuity in the wave at the interface. Meaning the electric and magnetic fields jump discontinuously from one point in the first medium to another point in the second medium and continuing on at its new different frequency. Now this does not happen in mechanical waves, nor does it take place in EM waves. What does happen though is the speed of the transmitted wave and its associated wavelength change in such a way as to preserve the frequency.

3 years ago
OpenStudy (anonymous):

Also, we can treat the air next to the glass as essentially a vacuum because the index of refraction of air is =1 up to the first 3 decimal places which is sufficient for this problem.

3 years ago
OpenStudy (anonymous):

Therefore: \[v_a \approx v_{vac} = c = \lambda \nu \ \ \ \ \&\& \ \ \ \ v_{glass}= \lambda_{glass} \nu\] \[\rightarrow \ \ \ \ \frac{\lambda}{c} = \frac{1}{\nu} = \frac{\lambda_{glass}}{v_{glass}} \\ \rightarrow \ \ \ \ \lambda_{glass}=\frac{\lambda v_{glass}}{c}\] Given the definition for index of refraction: \[ v=\frac{c}{n} \ \ \ \ or \ \ \ \ n=\frac{c}{v} \] Therefore: \[ \lambda_{glass}=\frac{\lambda \frac{c}{n_{glass}}}{c} = \frac{\lambda}{n_{glass}} \] Which is also a valid definition of the index of refraction (ratio of the wavelengths in the two mediums).

3 years ago
OpenStudy (anonymous):

Given that the wavelength of your light source in a vacuum is given as 638nm and the index of refraction of glass is 1.52; this gives: \[ \lambda_{glass}= \frac{\lambda}{n_{glass}} = \frac{638nm}{1.52} =419.73…nm \approx 420nm \]

3 years ago
OpenStudy (anonymous):

Now applying Snell’s Law to the interface between the incident medium (glass) and the transmitted medium (liquid) yields: \[ \frac{ sin \theta_{i} }{ sin \theta_{t} } = \frac{ sin \theta_{glass} }{ sin \theta_{liquid} } = \frac{ \lambda_{glass} }{ \lambda_{liquid} } = \frac{ n_{liquid} }{ n_{glass} } \]

3 years ago
OpenStudy (anonymous):

The critical angle for total internal reflectance is the angle of incidence that results in an angle of transmission of 90 degrees. If you draw the interface, you will see that the transmitted angle being 90 degree means the transmitted ray lies along the plane of the interface. Therefore: \[ n_i sin \theta_{i_{c}} = n_t sin 90^{o} = n_t \] Substituting in the known values: \[ n_t =n_{liquid_{A}} = n_i sin \theta_{i_{c}} = n_{glass} sin \theta_{c_{A}} = 1.52 * sin(52^{o}) = 1.1977…. \approx 1.20 \] Repeating this procedure for the other two liquids gives: \[ 1.06 \ for \ B \] and \[0.8998 \approx 0.9 \ for \ C \]

3 years ago
OpenStudy (anonymous):

The critical angle for total internal reflectance is the angle of incidence that results in an angle of transmission of 90 degrees. If you draw the interface, you will see that the transmitted angle being 90 degree means the transmitted ray lies along the plane of the interface. Therefore: \[ n_i sin \theta_{i_{c}} = n_t sin \theta_t = n_t sin 90^{o} = n_t \\ \rightarrow n_t =n_{liquid_{A}} = n_i sin \theta_{i_{c}} = n_{glass} sin \theta_{c_{A}} \] Now I made a mistake here earlier and foolishly took the given values to be the desired critical angle inside the glass. However if you do this you will find the index of refraction of liquid C to be less than 1 which is (given the problem at hand) unphysical. As a result, I reread the question which then caused me to realize two things. First, we didn’t need to explicitly calculate the wavelength. It wasn't, as I thought, required for the question, so I guess you can leave that out. Second and right to the point, the given angles seem to correspond to the angles of incidence WITH RESPECT TO THE INITIAL AIR-GLASS INTERFACE. This then requires one more step to solve. First note that the critical angle given is with respect to the surface normal of the air-glass interface. Thus the transmitted critical angle (now in glass) will also be given with respect to the normal. Since parallel lines cut by a transversal have equal angles, the complimentary angle to the transmitted critical angle is equal to the critical angle (with respect to the surface normal) of the glass-liquid interface. The following diagram should help:

3 years ago
OpenStudy (anonymous):

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3 years ago