Question

A soap bubble (n = 1.33) is 555 nm thick.What is the third longest (m = 3)wavelength IN NANOMETERS that givesa MINIMUM (destructive interference) atthat point?(Hint: If you leave the wavelength in nm,the answer will be in nm. No conversionnecessary.)(Unit = nm)

Answer 1

We are given the following information.

t = 555 nm

n = 1.33

m = 3

We are asked to find the third longest wavelength that gives a minimum destructive interference at that point.

Recall that the destructive interference is given by

$2nt=m\lambda$

Re-arranging the above equation for wavelength.

$\lambda=\frac{2nt}{m}$

Substitute the known values.

$\lambda=\frac{2\cdot1.33\cdot555}{3}=492.1\;nm$

Therefore, the third longest wavelength that gives a minimum destructive interference at that point is 492.1 nm.