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Goryan [66]
3 years ago
13

Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin

e and i-line radiation from a Mercury spectrum is used. Repeat this exercise for 0.30 NA and 0.36 NA lenses. Assume a k-factor k1.
Engineering
1 answer:
Vladimir79 [104]3 years ago
6 0

Answer:

a)

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

b)

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

c)

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

a)

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 =  1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 =  1.013 μm

b)

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 =  0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.243 μm

c)

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 =  0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.608 μm

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The  option that has the correct order or sequence of events are;

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3 years ago
he mean weight of a breed of yearling cattle is 11871187 pounds. Suppose that weights of all such animals can be described by th
Tamiku [17]

Question

The mean weight of a breed of yearling cattle is 1187 pounds. Suppose that weights of all such animals can be described by the Normal model ​N(1187,78). ​

a) How many standard deviations from the mean would a steer weighing 1000 pounds​ be?

b) Which would be more​ unusual, a steer weighing 1000 ​pounds, or one weighing 1250 ​pounds? ​

Answer:

a. z = -2.40

A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.

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Explanation:

a.

Let weight (in pounds) of the cattle be denoted by letter x:

z = (x - u)/ σ

Where u = mean and σ = standard deviation

u = 1187

σ = 78

x = 1000

Use z score formula to standardize the value of x:

z = (1000 - 1187)/78

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z = -2.40 ------_ Approximated

A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.

b.

x= 1250

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z = 63/78

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z = 0.81 --------- Approximated

1000 is more unusual because its contained on the extreme end from the mean

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