A disabled tanker leaks kerosene (n = 1.20) into the Persian Gulf, creating a la

Question

A disabled tanker leaks kerosene (n = 1.20) into the Persian Gulf, creating a large slick on top of the water (n = 1.30). (a) If you are looking straight down from an airplane, while the Sun is overhead, at a region of the slick where its thickness is 451 nm, for which wavelength(s) of visible light is the reflection brightest because of constructive interference? (b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest?

Explanation / Answer

(a)
Let the indices of refraction of the air, oil and water be n1,n2, and n3, respectively.

Since n1<n2 and n2<n3,

there is a phase change of from each interfaces.

The second wave also travels an additional distance of 2L (for a shift of 2n2L),

Therefore total phase difference = +2n2L.

For constructive interferes, we have m = + 2n^2 * L

for m=0,1,2…

=2n^2*L/(m-1)

=2n^2* L/m

where m=2,3,4.....

For m =1

= 2*(1.2)*(451*10^-9)/ 1

= 1082nm

For m = 2

=2*(1.2)*(451*10^-9)/2

= 541.2 nm

For m = 3

= 2*(1.2)*(451*10^-9)/3

= 360.8 nm

Only the 541.2 nm wavelength falls within the visible light range.

------------------------------

(b)

Maximum transmission into the water occurs for wavelengths for which reflection is a minimum,

i.e Destructive interference.

so (m+ ½ ) = +2n^2L

= 4n^2L/(2m+1).

For m = 1

= 4 * 1.2 * 451*10^-9/(2+1)

= 721.6 nm

For m = 2

= 4 * 1.2 * 451*10^-9/(4+1)

= 433 nm, Only the 433 nm wavelength is in the visible range.

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