Refer the diagram below :
What should be the angle alpha such that the variable x is between 7mm and 7.3mm.
$\endgroup$ 13 Answers
$\begingroup$First, see the following image
From the figure,
$t=\dfrac{x}{2}$
Since,
$7<x<7.3$
$\implies 3.5<t<\dfrac{7.3}{2}$
$\implies \dfrac{2}{7.3}<\dfrac{1}{t}<\dfrac{1}{3.5}$
Also,
$t\sin\alpha=3.5$
$\implies \sin\alpha=\dfrac{3.5}{t}\in\left(\dfrac{7}{7.3},1\right)$
$\implies \alpha \in (73.52^\circ,90^\circ)$ (Approx.)
$\endgroup$ 1 $\begingroup$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?
$\endgroup$ 1 $\begingroup$
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.
