(a) A tank containing methanol has walls 2.50 cm thick made of...

Question

(a) A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes the glass at a 41.3° angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. (b) The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of 20.2° from the normal, what is the refractive index of the unknown liquid?

Accepted Answer

Hi, everyone in this practice problem, we're being asked to determine an angle and also an index of refraction of an unknown medium. We will have a quartz Q V filled with turpentine oil and the refractive indices of quartz and turpentine oil are 1.54 1.47 respectively. The site thickness of the Q V is 1.25 millimeter and a laser beam strikes the Q V T site at an angle of 35 degrees with respect to the normal to the Q V surface. The reflected beam in the turpentine is theta with respect to the Q V S service. And we're being asked to determine first what that theta is and when the experiment is repeated with the same angle of incidence on the Q V side, now filled with an unknown transparent liquid. The reflected beam in the liquid forms an angle of 25 degrees with respect to the normal to the Q fat surface. We want next determine the index of refraction or N liquid of the unknown medium. And the options given are a first angle equals to 22 degrees. And second, the M liquid equals to 1.42 B, 1st 23 degrees. 2nd, 1.36 C, 1st 24 degrees and 1.34 D 1st 25 degrees and 1.38. So for light waves, we wanna in uh incorporate Snell's law of refraction, which in this case, now sla will be, and one multiplied by sign of theta, one of one medium equals to N two multiplied by sign of theta two of the second medium. I'm going to first draw the system that we have because uh from the problem statement, I think it's a little bit difficult for us to actually interpret what is going on in this case. So we'll have the um Q fat in this, in this uh problem statement here with the thickness of 1. millimeter. And then we will have first an angle uh or a laser beam coming in at an angle of 35 degrees to the normal with respect to the normal of the QVAD surface. And then that will give us or that will make a another angle of alpha which will then uh make the last angle which will be the just like. So, so the first uh the first part on the very left side is the air and air is going to be one. And then the second part is the quartz or the cuvette made out of quartz glass. So, and quartz is 1.54. And lastly on the very right side is the turpentine and the end of turpentine is 1.47. So this is going to be the system that we are working with. So in order for us to actually solve this problem, let's start with the first part and we will have to apply Snell's law to multiple boundaries on this case. So for the first part, looking at the air plus quartz boundary, we will apply Snell Slaw which is N one sine theta one equals N to sine theta two, which in this case, we will then use N of air multiplied by sine of 35 degrees equals to N of quartz multiplied by sign of alpha, N of S one multiplied by a sign of 35 degrees equals to N of quds as 1.54 multiplied by sign of alpha. We are interested to find alpha and call this, we will have the sign of alpha to be equals to 0.372. And taking the N first of this alpha will equals to sine to the power of negative one of 0.372 or alpha will equals to the arc of 0.372, which will give us an alpha value of 21.9 degrees. Now moving on to the second boundary layer now that we have alpha which is the angle of refraction with respect to the normal in the quartz alpha will also then be the angle of incidence at the quartz and turpentine boundary. So in this case, since we have an alpha here, we will also have alpha on this site. So let's move on to the second boundaries, which is going to be the quartz plus turpentine boundary. So we wanna apply snails law again which in this case, we will have N of the quartz multiplied by sine of alpha equals to the end of the turpentine multiplied by sine of theta. So the end of the courts is given to be 1.54 multiplied that by sine of alpha, which we just found, which is sine of 21.9 degrees. And the end of turpentine is 1. multiply that by sign of PETA. So um calculating this, we will then get the sign of theta to then equals to uh 1.54 multiplied by sine of 21.9 degrees divided by 1.47. And finally, after doing pretty much the same step by taking the arcs sign or the N first of sign, we will get the theta value to be 23 degrees. So the light is reflected in the turpentine at an angle of degrees with respect to the normal to the surface. So that will be the answer to the first part which is going to be the angle of the reflected beam in the turpentine which is going to be 23 degrees. So now we want to move on to the second part. So for the second part, we want to calculate the N or the index of reflection of the unknown liquid or the unknown medium. So what we want to look at is to just basically focus on the second boundary between the quartz and the turpentine or in this case, quartz and the unknown liquid. So this second boundary and neglect the first boundary here because everything on the first boundary will remain exactly the same. So the angle alpha on the boundary between the air and the quartz will still remain to be 21.9, which will mean that the angle alpha on the boundary between the quartz and the unknown liquid will also still be 21.9. So at the courts an unknown liquid boundary. So for the second part quds plus unknown at this boundary, then we will have or we will apply SNS law again where we have N of the quds multiplied by sign of alpha, which will remain the same as the first part which is 21.9 and now will equals to N of the liquid or the unknown medium multiplied by the sign of the new data, which is the apostrophe, which is going to be 25 degrees, which is given in the problem statement. So then rearranging things, we will then get the end of the liquid to then equals to the end of the quartz multiplied by sign of alpha divided by sign of data apostrophe. This will then becomes enough liquid to be equals to N of quartz is 1.54 multiply that by sine of alpha which is sine of 21.9 degrees divide that with sine of the apostrophe which is sine of degrees. And that will give us an a liquid or the index of refraction of the unknown medium to be 1.36. So the refractive index of the unknown liquid is 1.36. And then looking into the first part, the theta comes out to be 23 degrees and those two combination will give us option B to be the answer to this practice problem. So the will equals to 23 degrees and, and liquid will equal to 1.36 which will equals to or correspond to option B. So that will be all for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics, but that'll be it for this one. Thank you.

Key Concepts

Snell's Law

Refractive Index

Angle of Incidence and Refraction

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