One astronomer has observed an apparent lengthening of the electromagnetic waves given off by a star. Which of these conclusions
2 answers:
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(A) We can solve the problem by using Ohm's law, which states:
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
and 
, therefore we can rearrange Ohm's law to find the current through the device:
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
where
is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
where
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t = 750 / 2(1.20)
t = 750 / 2.4
t = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t = 750 / 4(1.50)
t = 750 / 6
t = 125 nm
Therefore, the minimum thickness be now will be 125 nm