Describe the changes in the field of view and the amount of available light when going from low to high power using the compound
1 answer:
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Answer:
λ = 5.85 x 10⁻⁷ m = 585 nm
f = 5.13 x 10¹⁴ Hz
Explanation:
We will use Young's Double Slit Experiment's Formula here:
where,
λ = wavelength = ?
Y = Fringe Spacing = 6.5 cm = 0.065 m
d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m
L = screen distance = 5 m
Therefore,
<u>λ = 5.85 x 10⁻⁷ m = 585 nm</u>
Now, the frequency can be given as:
where,
f = frequency = ?
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>f = 5.13 x 10¹⁴ Hz</u>
To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.
The altitude is,
And the velocity can be written as,
From the properties of standard atmosphere at altitude z = 20km temperature is
Velocity of sound at this altitude is
Then the Mach number
So front stagnation temperature
Therefore the temperature at its front stagnation point is 689.87K