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Initial intensity $I$. After A, $I_A = I/2$. Since $I_B$ is also $I/2$, axes of A and B are parallel ($\theta_{AB} = 0$).
Insert C at angle $\phi$ to A. Then angle between C and B is also $\phi$.
Final intensity $I' = I_A \cos^2 \phi \cos^2 \phi = \frac{I}{2} \cos^4 \phi$.
Given $\frac{I}{2} \cos^4 \phi = \frac{I}{8} \implies \cos^4 \phi = \frac{1}{4} \implies \cos^2 \phi = \frac{1}{2}$.
Thus, $\cos \phi = \frac{1}{\sqrt{2}} \implies \phi = 45^\circ$.
If $\dfrac{1}{2^{11} \cdot 5^{17}}$ is written as a terminating decimal, how many nonzero digits will it contain?
Rewrite the expression to have a denominator that is a power of 10:
$\dfrac{1}{2^{11} \cdot 5^{17}} = \dfrac{1}{2^{11} \cdot 5^{17}} \times \dfrac{2^6}{2^6} = \dfrac{2^6}{2^{17} \cdot 5^{17}} = \dfrac{64}{10^{17}}$
So the decimal is $\dfrac{64}{10^{17}} = 0.\underbrace{00\ldots0}_{15}64$
The nonzero digits are $6$ and $4$ β that is, two nonzero digits.
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