Question

A Blu-ray player uses a blue laser that produces light with a wavelength in air of \(405 \mathrm{nm}\). If the disc is protected with polycarbonate \((n=1.58)\), determine the minimum thickness of the disc for destructive interference. Compare this value to that for CDs illuminated by infrared light.

Short answer

Answer: The minimum thickness of the polycarbonate layer required for destructive interference in a Blu-ray player is approximately 64.30 nm. In comparison, the minimum thickness for CDs illuminated by infrared light is about 123.42 nm. The Blu-ray player's polycarbonate layer has a significantly lower minimum thickness, allowing for higher data storage capacity.

Step-by-step solution

(Step 1: Understanding thin film interference)

Thin film interference happens when light waves reflect off two surfaces very close together. In our case, we have the top and bottom surfaces of the polycarbonate layer. When both reflected waves come to the detector, they interfere with each other. This interference could be constructive, where both waves add up, or destructive, where they partially or completely cancel each other. The condition for destructive interference is when the extra distance traveled by the second reflected wave is half of the wavelength (i.e., they are out of phase by half of the wavelength).

(Step 2: Formula for thin film interference-detructive condition)

For thin film interference and for destructive interference, we can use the following equation: $$ 2t = m \frac{\lambda}{2n} $$ Where: - \(t\) is the thickness of the film - \(m\) is an integer which represents the order of interference - \(\lambda\) is the wavelength of light in air - \(n\) is the refractive index of the polycarbonate layer (given as \(1.58\)) We'll use the values given in the problem and solve for \(t\) for the minimum thickness (the first-order destructive interference, with \(m = 1\)).

(Step 3: Calculate the thickness of the Blu-ray player's polycarbonate layer)

Using the values given and solving for the minimum thickness, $$ t_{Blu-ray} = \frac{1 \times \frac{405}{2 \times 1.58}}{2} $$ $$ t_{Blu-ray} \approx 64.30 \, \mathrm{nm} $$ The minimum thickness of the polycarbonate disc for destructive interference in the case of a Blu-ray player is about \(64.30 \, \mathrm{nm}\).

(Step 4: Calculate the value for CDs illuminated by infrared light.)

We need to find the minimum thickness for CDs illuminated by infrared light. Infrared light typically has a wavelength around \(780 \, \mathrm{nm}\). We'll use the same equation as before, but now with the infrared wavelength: $$ t_{CD} = \frac{1 \times \frac{780}{2 \times 1.58}}{2} $$ $$ t_{CD} \approx 123.42 \, \mathrm{nm} $$ The minimum thickness of the polycarbonate disc for destructive interference in the case of CDs illuminated by infrared light is about \(123.42 \, \mathrm{nm}\).

(Step 5: Compare the results)

Now we can compare the values obtained for both cases: - Blu-ray player's polycarbonate layer: \(64.30 \, \mathrm{nm}\) - CDs illuminated by infrared light: \(123.42 \, \mathrm{nm}\) In conclusion, the minimum thickness of the polycarbonate disc required for destructive interference in a Blu-ray player (\(\approx 64.30 \, \mathrm{nm}\)) is significantly less than that for CDs illuminated by infrared light (\(\approx 123.42 \, \mathrm{nm}\)). This difference allows Blu-ray players to read smaller and more closely spaced data, leading to higher data storage capacity.