Chapter 35: Q9P (page 1074)

In Fig. 35-4, assume that the two light waves, of wavelength $620 nm$ in air, are initially out of phase by $π$ rad. The indexes of refraction of the media are $n_{1} = 1.45$ and $n_{2} = 1.65$ . What are the (a) smallest and (b) second smallest value of $L$ that will put the waves exactly in phase once they pass through the two media?

Short Answer

Expert verified

(a) The smallest value of $L$ is $1.55 \times 10^{- 6} m$ .

(b) The second smallest value of L is $4.65 \times 10^{- 6} m$ .

Step by step solution

Write the given data from the question:

The wavelength in air, $λ = 620 nm$

The phase difference, $Δ ϕ = π$

The refractive index of medium 1, $n_{1} = 1.45$

The refractive index of medium 2, $n_{2} = 1.65$

Determine the formulas to calculate the smallest and second smallest value of L:

The expression to calculate the phase difference is given as follows.

$Δ ϕ = 2 π L (\frac{n_{2}}{λ} - \frac{n_{1}}{λ})$

$L = \frac{λ Δ ϕ}{2 π (n_{2} - n_{1})}$ ......(1)

The expression to calculate the second smallest value of Lis given as follows.

L = \frac{3 λ Δ ϕ}{2 π (n_{2} - n_{1})}

(a) Calculate the smallest value of L :

Calculate the smallest value of the L .

Substitute the given values into equation (1).

$\begin{array}{rcl} L & = & \frac{π \times 620 \times 10^{- 9}}{2 π (1.65 - 1.45)} \\ = & \frac{310 \times 10^{- 9}}{0.20} \\ = & 1550 \times 10^{- 9} \\ = & 1.55 \times 10^{- 6} m \end{array}$

Hence, the smallest value of $L$ is $1.55 \times 10^{- 6} m$

(b) Calculate the second smallest value of L :

Calculate the smallest value of the $L$ .

Substitute the given values into equation (1).

$\begin{array}{rcl} L & = & \frac{3 π \times 620 \times 10^{- 9}}{2 π (1.65 - 1.45)} \\ = & \frac{930 \times 10^{- 9}}{0.20} \\ = & 4650 \times 10^{- 9} \\ = & 4.65 \times 10^{- 6} m \end{array}$