Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency 1. 72 x 1015 hz?
1 answer:
The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3× / 1.72×
or approximately 1.74×
m.
The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:
f = c / λ
where, f = frequency of light
c = speed of light
λ = wavelength of light
Given data = f = 1.72×Hz
Therefore, λ = 3× / 1.72×
λ = 1.74×m
The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3× / 1.72×
or approximately 1.74×
m.
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second option
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Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop, IR = E- V
IR = 12 V - 5 V = 7 V
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Answer:
1.3 x 10⁻⁴ m
Explanation:
= wavelength of the light = 450 nm = 450 x 10⁻⁹ m
n = order of the bright fringe = 1
θ = angle = 0.2°
d = separation between the slits
For bright fringe, Using the equation
d Sinθ = n
Inserting the values
d Sin0.2° = (1) (450 x 10⁻⁹)
d (0.003491) = (450 x 10⁻⁹)
d = 1.3 x 10⁻⁴ m