Okay, I'm really confused in general.
My question says:
"A solid piece of clear transparent material has an index of refraction of 1.61. If you place it into a clear transparent solution and it seems to disappear, approximately what is the index of the refraction of the solution? How do you know?"
I'm thinking of the use of snell's law to find the approximate index of refraction but I'm confused on the needed angle.
the index of the solution must be approximately the index of the material
there is no refractive boundary between the material and the solution, so the material seems invisible
To determine the approximate index of refraction of the solution, you can use Snell's law. Snell's law relates the angles of incidence and refraction to the indices of refraction of the two media.
In this case, the solid piece of clear transparent material has an index of refraction of 1.61, while the solution has an unknown index of refraction (let's call it n₂). When the solid is placed into the solution and it seems to disappear, it implies that the angle of refraction is 90 degrees.
To find the index of refraction of the solution, we can rearrange Snell's law as follows:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Here, n₁ is the index of refraction of the solid material and θ₁ is the angle of incidence (which is not given). As we don't have the angle of incidence, we can assume it to be 0 degrees, which means the light is incident perpendicularly, resulting in θ₁ = 0° and sin(θ₁) = 0.
With these assumptions, the equation becomes:
0 = n₂ * sin(θ₂)
Since sin(θ₂) cannot be zero (as the angle of refraction cannot be zero when the solid is placed in the solution), we can conclude that n₂ must be zero or very close to zero.
Therefore, the approximate index of refraction of the solution would be considered negligible or close to zero.
To find the approximate index of refraction of the solution, you can indeed use Snell's law. Snell's law relates the angles and indices of refraction of two mediums when light passes from one to another.
In this case, you have a solid piece of clear transparent material with an index of refraction of 1.61. When you place it into the clear transparent solution, it seems to disappear. This suggests that the refractive index of the solution is similar to that of the solid material, causing the light to pass through without significant reflection or refraction.
Here's how you can proceed to find the approximate index of refraction of the solution:
1. Consider the incident ray of light passing through the solid material and entering the solution. Let's call the angle between the incident ray and the normal line as angle of incidence (θ₁).
2. Use Snell's law to relate the angles and indices of refraction for the solid material and the solution:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
n₁ is the index of refraction of the solid material (1.61)
θ₁ is the angle of incidence in the solid material
n₂ is the unknown index of refraction of the solution
θ₂ is the angle of refraction in the solution (which is unknown)
3. Since the solid material and the solution are both transparent and clear, we assume the light passes through without significant refraction. In other words, the angle of refraction (θ₂) is assumed to be very close to zero degrees.
4. When θ₂ is close to zero degrees, sin(θ₂) is also close to zero. Therefore, we can approximate the equation as:
n₁ * sin(θ₁) = n₂ * 0
5. Since sin(θ₁) cannot be zero (otherwise, the light would not enter the solution), we can conclude that n₂ (index of refraction for the solution) would be approximately 0, which means the solution is most likely air.
In summary, when the solid material with an index of refraction of 1.61 seems to disappear in a clear transparent solution, it suggests that the solution has an index of refraction close to the solid material, or in this case, very close to air.