Question

A laser beam of wavelength λ=632.8 nm shines at normal incidence on the reflective side of a compact disc. The tracks of tiny pits in which information is coded onto the CD are 1.60 μm apart.
For what nonzero angles of reflection (measured from the normal) will the intensity of light be maximum?

Express your answer numerically. If there is more than one answer, enter each answer separated by a comma.

Answer 1

Answer:

The intensity of light be maximum is for angles 23.3° and 52.3°.

Explanation:

Given that,

Wave length = 632.8 nm

Distance = 1.60 μm

We need to calculate the intensity of light be maximum

Using Bragg's law

$d\sin\theta=n\lambda$

$\theta=\sin^{-1}(\dfrac{n\lambda}{d})$

We need to calculate the angle for different value of n

Using Bragg's law

$\theta=\sin^{-1}(\dfrac{n\lambda}{d})$

For n₁,

Put the value into the formula

$\theta=\sin^{-1}(\dfrac{632.8\times10^{-9}}{1.60\times10^{-6}})$

$\theta=23.3^{\circ}$

For n₂,

$\theta=\sin^{-1}(\dfrac{2\times632.8\times10^{-9}}{1.60\times10^{-6}})$

$\theta=52.3^{\circ}$

Hence, The intensity of light be maximum is for angles 23.3° and 52.3°.