# Two waves meet a Two light sources emit waves that are in phase with a wavelength of lambda = 400...

## Question:

Two waves meet a Two light sources emit waves that are in phase with a wavelength of {eq}\lambda {/eq} = 400 nm. The two waves meet at a distant point where Wave #1 has traveled 100 nm. The distance taveled by Wave #2 is variable. Which of the distances for Wave #2, listed below, would result in constructive interference?

a. 450 nm

a. 450 nm

b. 900 nm

c. 1500 nm

d. 1100 nm

e. 700 nm

## Constructive Interference:

Interference is the phenomenon in which two separate wavefronts originated from two coherent sources meet each other and form a new wavefront of higher wavelength. The condition for constructive interference is as follows.

{eq}{d_2} - {d_1} = n\lambda {/eq}

## Answer and Explanation:

**Given Data**

- The wavelength of wave is: {eq}\lambda = 400\;{\rm{nm}} {/eq}

- The distance travelled by wave 1 is: {eq}{d_1} = 100\;{\rm{nm}} {/eq}

Use relation for the constructive interference.

{eq}{d_2} - {d_1} = n\lambda {/eq}

Here, the wavelength is {eq}\lambda {/eq} , the distance travelled by the wave 1 is {eq}{d_1} {/eq} and distance travelled by wave 2 is {eq}{d_2} {/eq}.

Substitute all the values in the above equation.

{eq}\begin{align*} {d_2} - {d_1} &= n \times 400\\ {d_2} - 100 &= n \times 400\\ {d_2} &= 400n + 100 \end{align*} {/eq}

Calculate the distance travelled by wave 2 for n=1, 2, 3.

{eq}\begin{align*} {\left( {{d_2}} \right)_{n = 1}} &= 400 \times 1 + 100\\ &= 500\;{\rm{nm}}\\ {\left( {{d_2}} \right)_{n = 2}} &= 400 \times 2 + 100\\ &= 900\;{\rm{nm}}\\ {\left( {{d_2}} \right)_{n = 3}} &= 400 \times 3 + 100\\ &= 1300\;{\rm{nm}} \end{align*} {/eq}

Thus, the correct option is (b).

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from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5