Variation of a graph requiring a proportionality to hold

[aspodkfpo]

Homework Statement

https://www.asi.edu.au/wp-content/uploads/2015/03/2010_Physiscs_solutions.pdf
Q12 A)

Pg 11

Relevant Equations

Ben’s method requires that the voltage output be directly
proportional to the intensity, which it is not.

Do not understand the statement:
Ben’s method requires that the voltage output be directly proportional to the intensity, which it is not.

https://www.asi.edu.au/wp-content/uploads/2015/03/2010_Physiscs_solutions.pdf

My thoughts are that by I= Io cos(theta)^2

We can relate voltage to theta which is then related to I by a costheta^2. Thus the constant value of 73mV that she reduces the voltages by will just result in a shift of the graph which doesn't really change anything. There seems to be something wrong in my logic, but I think their statement also doesn't seem compeltely right.

 

[Tom.G]

...the constant value of 73mV that she reduces the voltages by...

First, HE (Ben) subtracted the 73mV, Charlotte darkened the room.

Second, I agree the defined procedure is not ideal. Part b) seems to use I₀ for the calculations, which is not really defined. The graph shows I_θ, not I₀. Since Ben subtracted 73mV, I took that to mean use the voltage reading obtained in the experiment. But there is no value given for any subtraction by Charlotte, which would normally indicate either no subtraction was done, or Zero was subtracted. But it couldn't be Zero if I₀ was Zero. That would make the Output Ratio, defined as multiples of I₀, rather ridiculous.

Also, it does not account for the nonlinearity of output voltage vs incident radiation; compensation using the COS² graph would be needed for accurate results.

The given answer for part b) was obviously not written by a Test Engineer! (at least not one that is still employed )

Cheers,
Tom

 

[aspodkfpo]

First, HE (Ben) subtracted the 73mV, Charlotte darkened the room.

Second, I agree the defined procedure is not ideal. Part b) seems to use I₀ for the calculations, which is not really defined. The graph shows I_θ, not I₀. Since Ben subtracted 73mV, I took that to mean use the voltage reading obtained in the experiment. But there is no value given for any subtraction by Charlotte, which would normally indicate either no subtraction was done, or Zero was subtracted. But it couldn't be Zero if I₀ was Zero. That would make the Output Ratio, defined as multiples of I₀, rather ridiculous.

Also, it does not account for the nonlinearity of output voltage vs incident radiation; compensation using the COS² graph would be needed for accurate results.

The given answer for part b) was obviously not written by a Test Engineer! (at least not one that is still employed )

Cheers,
Tom

Please explain why "Ben’s method requires that the voltage output be directly proportional to the intensity, which it is not."

For instance if I had a graph of y = cos (x-79), that would be the same as a graph of y= cos(x) shifted horizontally. I don't see why subtracting 79 mV is invalid.

 

[haruspex]

The way they set it up, they have defined I₀ to be the greatest intensity they will create. So all other intensities they create will be less, and so can be expressed as I₀cos²(θ) for some theta. This allow s them to use the graph.
But equally, we could go to the trouble of creating a new graph from it that plots v against I/I₀.
Suppose that can be represented as a function f, $v =f(I/I_0)$. Ben's method is to measure ambient light, giving $v_A=f(I_A/I_0)$ and bulb+ambient, giving $v_{A+B}=f(I_{A+B}/I_0)$, then taking the difference: $v_{A+B}-v_A=f(I_{A+B}/I_0)-f(I_A/I_0)$. He needs this to approximate $f(I_{A+B}/I_0-I_A/I_0)$, i.e. he needs f to be linear.
But if f is linear then the printed graph should look like a piece of a sine curve. It certainly doesn't at the right hand end, where the gradient tends to minus infinity, but it would take a bit of work to check more generally.
So I do wonder whether the examiner has confused linearity of f with linearity of the printed graph. Given the blunder rate of these papers, that is a distinct possibility.

 

[Tom.G]

Please explain why "Ben’s method requires that the voltage output be directly proportional to the intensity, which it is not."

Let's start with "...which it is not."
If you plot on the original given graph, the function
230*COS²θ
(230 being the maximum point originally plotted),
you will see that the curves do not match.

The above formula models the transmission thru crossed polarizers. Since the two curves don't match, that shows the sensor output is not linear with intensity. If the sensor were linear over the range, the curves would overlay each other.
Therefore the sensor correction factor (73mV in the problem statement) varies over the intensity range in a nonlinear fashion, you can't get them to line up with a simple shift.

Cheers,
Tom