Question

An oil (μ0 = 1.45) film of thickness 280 nm floats on water (μw = 1.33). It is illuminated
by white light at normal incidence. Which colour in the visible spectrum will be most
strongly (i) reflected, and (ii) transmitted?

Answer 1

Answer on Question #50690, Physics, Optics

An oil $(\mu_0 = 1.45)$ film of thickness $280~\mathrm{nm}$ floats on water $(\mu_w = 1.33)$ . It is illuminated by white light at normal incidence. Which colour in the visible spectrum will be most strongly (i) reflected, and (ii) transmitted?

Solution:

Since $1 < 1.45$ , the light reflected from the top of the oil film undergoes phase reversal. The light reflected from the bottom undergoes no reversal because $1.45 > 1.33$ .

(i) For constructive interference, we require

$$2 \mu_ {0} d = (m + \frac {1}{2}) \lambda$$

Substituting for m gives, $m = 0$ , $\lambda_0 = 1624$ nm (infrared)

$m = 1$ $\lambda_{1} = 541$ nm (green)

$m = 2$ $\lambda_{2} = 325$ nm (ultraviolet)

Both infrared and ultraviolet light are invisible to human eye, so the dominant color in reflected light is green.

(ii) transmitted

For destructive interference - Transmission

$$2 \mu_ {0} d = m \lambda$$

Substituting for m gives, $m = 1$ , $\lambda_1 = 812$ nm (near infrared)

$m = 2$ $\lambda_{2} = 406$ nm (violet)

$m = 3$ $\lambda_{3} = 271$ nm (ultraviolet)

The dominant color visible to human eye is violet.

Therefore the color of the light in the visible spectrum most strongly transmitted is violet.

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a. Since we are talking about waves that are the most strongly reflected then we have constructive interference, we have: $2tn_{oil} = \left(m + \frac{1}{2}\right)\lambda$

Therefore: $\lambda = \frac{2tn}{m + \frac{1}{2}} = \frac{2\times 280nm\times 1.45}{m + \frac{1}{2}}.$

For $m = 0$ , $\lambda = \frac{2 \times 280nm \times 1.45}{\frac{1}{2}} = 1620nm$ (infrared - invisible)

For $m = 1$ , $\lambda = \frac{2 \times 280nm \times 1.45}{1 + \frac{1}{2}} = 542nm$ (green - visible)

For $m = 2$ , $\lambda = \frac{2 \times 280nm \times 1.45}{2 + \frac{1}{2}} = 325nm$ (ultraviolet - invisible)

Therefore the color of the light in the visible spectrum most strongly reflected is green.

b. Since we are talking about waves that are the most transmitted then we have destructive interference, we have: $2tn_{oil} = m\lambda$

Therefore: $\lambda = \frac{2tn}{m} = \frac{2\times 280nm\times 1.45}{m}$

For $m = 1$ , $\lambda = \frac{2 \times 280nm \times 1.45}{1} = 812nm$ (infrared - invisible)

For $m = 1$ , $\lambda = \frac{2 \times 280nm \times 1.45}{2} = 406nm$ (violet - visible)

For $m = 2$ , $\lambda = \frac{2 \times 280nm \times 1.45}{3} = 271nm$ (ultraviolet - invisible)

Therefore the color of the light in the visible spectrum most strongly transmitted is violet.