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insens350 [35]
3 years ago
13

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf

ace, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
(a) The index of refraction of the oil is 1.20. What is the minimum thickness t of the oil slick at that spot?

(b) Suppose the oil had an index of refraction of 1.50. What would the minimum thickness t be now?
Physics
1 answer:
Tasya [4]3 years ago
3 0

Answer:

<em>The correct answer is (a) 312.5nm and (b) 125nm</em>

Explanation:

<em>The first step to take is to find The minimum thickness of the slick of the oil</em>

<em>Given that,</em>

<em>(a) tmin  λ/2n </em>

<em>We substitute 750nm ( in air) for λ and 1.20 for n for the expression of  minimum thickness t of the oil slick at that spot</em>

<em>thus,</em>

<em>tmin = (750nm)/2(1.2) = 312.5nm</em>

<em>The minimum thickness of the oil slick at that spot is =312.5nm</em>

<em>(B) we find the minimum thickness t </em>

<em>The minimum thickness of the oil slick at the spot will be calculated by,</em>

<em>tmin = λ/4n</em>

<em>we then 750nm ( in air) for λ and 1.50 for n  in the expression for the minimum thickness of the slick of the oil.</em>

<em>tmin = (750nm)/4 (1.5) = 125nm</em>

<em>Therefore the minimum thickness t will now be = 125nm</em>

<em />

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[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s

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[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W

(b)the electrical power generated for a breezy winter day

T_s = T_{\infty} = -15 ^oC = 258 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k

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