Question

A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting off the surface. 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.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm

Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm

I only need help with part c. I put this in incase you need the information.

Part C: Now assume that the oil had a thickness of 200 nm and an index of refraction of 1.5. A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily to the diver?

Answer 1

Answer:

Explanation:

In case of oil slick a thin layer of oil is formed on water . This thin layer creates a rainbow of colour . The phenomenon is due to interference of light waves , one reflected from the upper surface of oil and the other reflected from the lower surface of the oil.

For formation of bright colour

2 μ t = ( 2n + 1 ) λ / 2

μ is refractive index of oil ,  t is thickness of oil layer  λ is wave length of light falling on the layer .

given μ = 1.2 ,   λ = 750 x 10⁻⁹   ,

2 x 1.2 t = ( 2n + 1 ) 750 x 10⁻⁹ / 2

For minimum thickness n = 0

2.4 t = 375 x 10⁻⁹

t = 156.25 n m

B ) If the refractive index of layer of medium below oil is less than that of oil , the condition of formation of colour changes

The new condition is

2 μ t = n λ

2  x 1.5 t  = 750 nm ,         n = 1 for minimum wavelength .

t = 250 nm

C ) Light mostly transmitted means dark spot is formed at that point .

For that to be observed from water side , the condition is

2 μ t = ( 2n + 1 ) λ / 2

λ = 4μ t / ( 2n + 1 )

For maximum wavelength n = 0

λ = 4μ t

= 4 x 1.5 x 200 nm

= 1200 nm .