# The laser in a CD player has a vacuum wavelength of 780 nm. The plastic coating over the pits has...

## Question:

The laser in a CD player has a vacuum wavelength of 780 nm. The plastic coating over the pits has an index of refraction of 1.3. What is the thickness of the pits on the CD?

A : {eq}7.8 \times 10^{-7} {/eq} m

B : {eq}1.5 \times 10^{-7} {/eq} m

C : {eq}6.0 \times 10^{-7} {/eq} m

D : {eq}3.9 \times 10^{-7} {/eq} m

E : {eq}3.0 \times 10^{-7} {/eq} m

## Interference:

Interference is the interaction of waves superimposing with each other forming another wave of different or same amplitude. There are two types of interference namely Constructive and Destructive Interference. Constructive interference is when in-phase waves superimpose each other forming a wave with a greater amplitude. Destructive interference is when out-of-phase waves superimpose forming a wave of reduced amplitude.

## Answer and Explanation:

{eq}\lambda_{vacuum} = 78- \times 10^{-9} \ m {/eq}

{eq}n = 1.3 {/eq}

Given:

We can compute for the thickness of the pits on the CD given by,

{eq}2t = \frac {1}{2} \lambda_{coating} {/eq}

Furthermore, the wavelength of the coating in terms of the wavelength of the vacuum is given by,

{eq}\lambda_{coating} = \frac {\lambda_{vacuum}}{n} {/eq}

Solving for the thickness t,

{eq}t = \frac {\lambda_{vacuum}}{4n} = \frac {780 \times 10^{-9} \ m}{4 * 1.3} {/eq}

{eq}t = \frac {780 \times 10^{-9} \ m}{4 * 1.3} = 1.5 \times 10^{-7} \ m {/eq} 