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Kazeer [188]
3 years ago
13

The Hubble Space Telescope in orbit above the Earth has a 2.4 m circular aperture. The telescope has equipment for detecting ult

raviolet light. What is the minimum angular separation between two objects that the Hubble Space Telescope can resolve in ultraviolet light of wavelength 95 nm?
Physics
1 answer:
kupik [55]3 years ago
7 0

Answer:

minimum angular separation between two objects is; θ = 4.829 x 10^(-8) rad

Explanation:

The minimum angle for resolving two adjacent point objects is proportional to the wavelength of the light and is inversely proportional to the diameter of the aperture and it's given by the formula;

θ = 1.22λ/D

Where;

θ is the resolved angle

λ is the wavelength

D is the diameter of the optics

We are given;

λ = 95nm = 95 x 10^(-9)m

D = 2.4m

Thus ;

θ = (1.22(95 x 10^(-9))/2.4

θ = 4.829 rad

Thus, minimum angular separation between two objects is; θ = 4.829 x 10^(-8) rad

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This question is incomplete

Complete Question

In the United States, a standard letter-sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for a letter-sized piece of paper is different. The international standard is based on SI units: 21.0 cm wide by 29.7 cm long.

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

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d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

Answer:

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b) 11.6929197 inches ≈ 11.7 inches

c) It's dimensions in inches for the international standard for letter - sized for paper = 8.3 wide inches by 11.7 inches long

d) The International standard letter - sized paper is longer.

Explanation:

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

21 cm =

Cross Multiply

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b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

29.7 cm =

Cross Multiply

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= 11.6929197 inches

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c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

The international standard for a letter-sized has dimensions 21.0 cm wide by 29.7 cm long.

Where

21.0cm = 8.267721 inches

≈ 8.3 inches

29.7cm = 11.6929197 inches

≈ 11.7 inches

Hence, it's dimensions in inches = 8.3 inches by 11.7 inches.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

U.S letter - sized paper = 8.5 inches wide by 11 inches long

International standard letter- sized paper = 8.3 wide inches by 11.7 inches long.

Hence, the International standard letter - sized paper is longer.

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