Refer the diagram below :enter image description here

What should be the angle alpha such that the variable x is between 7mm and 7.3mm.

  • 1
    $\begingroup$ Start by drawing a vertical line straight down from the vertex of your alpha angle all the way to the bottom line, creating a right triangle with an angle alpha at the bottom right vertex. $\endgroup$ – turkeyhundt Dec 3 '14 at 20:25

First, see the following image
enter image description here

From the figure,




$\implies 3.5<t<\dfrac{7.3}{2}$

$\implies \dfrac{2}{7.3}<\dfrac{1}{t}<\dfrac{1}{3.5}$



$\implies \sin\alpha=\dfrac{3.5}{t}\in\left(\dfrac{7}{7.3},1\right)$

$\implies \alpha \in (73.52^\circ,90^\circ)$ (Approx.)

  • $\begingroup$ Nice Explanation. $\endgroup$ – Hardey Pandya Dec 3 '14 at 20:46

Here's a hint.

The thickness of one layer of the transmitting medium is $3.5$ mm. Call this $t$.

The separation of two adjacent reflected rays is the same as half of the path length inside one layer of the material, which is $d = t/\sin \alpha.$

Can you take it from there?

  • $\begingroup$ If the beam diameter is 3mm ? $\endgroup$ – user1311684 Dec 3 '14 at 20:34

enter image description here
First note that, if the length of AB increases in the fig., the alpha decreases. Now, from the fig., $AB=$ between $7$ and $7.3$, $AC=3.5+3.5=7$ and so, we have $\sin B = \frac{AC}{AB}$. Note that, angle $B$ = $\alpha$.So, $\sin B=\frac{AC}{7}$. Now, put $AC = 7$ and then put $AC = 7.3$. So, you 'll 've the values of sine B. So, you can find that two values of B i.e. of $\alpha$. Your answer will be that - the angle $\alpha$ lies between that two values.

  • $\begingroup$ Does make sense. But when AB=AC=7 then Inv Sin of 1 is 90 which cannot be the angle of alpha (two angles cannot be 90 in a triangle). This question is the simplified form of a bigger problem I'm working on. So, I'm actually satisfied and now know what I need to tweak on. (a AB distance more than 7.3 so that I have proper angle) $\endgroup$ – user1311684 Dec 3 '14 at 20:48

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