How Does Optical Activity Help Measure Sugar Concentration?

[chadders8903]

Homework Statement

The sugar concentration in a solution (e.g., in a urine specimen) can be measured conveniently by using the optical activity of sugar and other asymmetric molecules. In general, an optically active molecule, like sugar, will rotate the plane of polarization through an angle that is proportional to the thickness of the sample and to the concentration of the molecule. To measure the concentration of a given solution, a sample of known thickness is placed between two polarizing filters that are at right angles to each other, as shown in the figure . The intensity of light transmitted through the two filters can be compared with a calibration chart to determine the concentration.

(a) What percentage of the incident (unpolarized) light will pass through the first filter?

(b) If no sample is present, what percentage of the initial light will pass through the second filter?

(c) When a particular sample is placed between the two filters, the intersity of light emerging from the second filter is 46.0% of the incident intensity. Through what angle did the sample rotate the plane of polarization?

(d) A second sample has half the sugar concentration of the first sample. What percentage of the initial light passes through the second filter in this case?

Homework Equations

Law of Malus
I= I_o cos² (theta)
and for unpolarized light going through the first polarizer
I=.5I_o

The Attempt at a Solution

The first three problems are easy and I have gotten them correct. It is d that is giving me problems.

a. I=.5I_o
so I= 50%

b. Law of Malus
I= I_o cos² (theta)
theta is 90 and cos² 90 = 0
0%

c. This one is the Law of Malus but drawn out over a few steps. And you have to do either arccos of the answer and then subtract it from 90 because it gives you the answer of the wrong angle and you need the other one or you can just do arcsin.

so:
Law of Malus
I= I_o cos² (theta)
.46= .5 cos² (theta)
.92= cos² (theta)
sqroot (.92) = cos (theta)
.959 = cos (theta)
arccos .959 = theta
theta = 16.43
but this is the opposite angle, so
90 - 16.43 = 73.6

d.
Now for the one that I do not understand and that I keep getting wrong. And my book states that a solution containing optically active molecules that are placed between crossed polarizers, the amount of light that passes through the system gives a direct measure of the concentration of the active molecules in the solution. I do not understand how it gives the direct measure of concentration, but I will attempt a solution for this.

First I don't know how changing the concentration will affect it. Which variable is changed?

I think that it will change the 46% to 23%. So that means that

Law of Malus
I= I_o cos² (theta)
.23= .5 cos² (theta)
.46 =cos² (theta)
sqroot .46 = cos (theta)
.678 = cos (theta)
arccos .678 = theta
47.3 = theta
And again, wrong angle so minus it from 90
90 -47.3 = 42.7

then we multiply them all together for each intensity

.5 * cos² (42.7) * cos² (90-42.7) = 12.4 %.
This is off by more than 10%. I know the range of the value is between like 14% to 20%. Can someone give me a hint? I have three of the four parts correct but this last one is giving me problems.

 

[chadders8903]

I figured it out. changing the concentration by half changes the angle by half. So using the law of Malus I do:
I2= ½(I0)cos2((90- (answer to part c/2)
Solving that gives me the right answer.

 

[Redbelly98]

Welcome to Physics Forums

Glad it worked out.

As a welcoming gift, here is a copy-and-pastable θ for you.