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A random variable $Y$ follows a normal distribution with mean $200$ and standard deviation $10$. Compare:
Quantity A: The probability that $Y > 220$
Quantity B: $\dfrac{1}{6}$
The value $220$ is $\dfrac{220 - 200}{10} = 2$ standard deviations above the mean.
For a standard normal distribution, the probability that $Z > 2$ is approximately $0.0228$ (about $2.28\%$).
Quantity B $= \dfrac{1}{6} \approx 0.1667$.
Since $0.0228 < 0.1667$, Quantity A is less than Quantity B.
Therefore Quantity B is greater.
Spot size $b \approx \text{geometric spread} + \text{diffraction spread} = a + \frac{L\lambda}{a}$.
To minimize $b$, differentiate with respect to $a$ and set to zero:
$\frac{db}{da} = 1 - \frac{L\lambda}{a^2} = 0 \implies a = \sqrt{\lambda L}$.
Substituting back: $b_{min} = \sqrt{\lambda L} + \frac{L\lambda}{\sqrt{\lambda L}} = 2\sqrt{\lambda L} = \sqrt{4\lambda L}$.
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