The beautiful iridescent (like a rainbow) colors on the wings of a tropical or morpho butterfly are due to

$A$ thin oil film of refractive index $1.2$ floats on the surface of water $(\mu = \frac{4}{3})$. When light of wavelength $\lambda = 9.6 \times 10^{-7} \ m$ falls normally on the film from air,it appears dark when seen normally. What is the minimum change in its thickness for which it will appear bright in normally reflected light by the same light?

In a biprism experiment, the fifth dark fringe is obtained at a point. $A$ thin transparent film of refractive index '$\mu$' is placed in one of the interfering paths. Now, the $7^{th}$ bright fringe is obtained at the same point. If '$\lambda$' is the wavelength of light used, the thickness of the film is equal to:

White light reflected from a soap film (Refractive Index $\mu = 1.5$) has a maxima at $600 \ nm$ and a minima at $450 \ nm$ with no minimum in between. Then the thickness of the film in units of $10^{-7} \ m$ is:

Calculate the minimum thickness of a soap film $(n=1.33)$ that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is $532 \,nm$. (in $\,nm$)

Light is incident on a glass plate $(\mu = 1.5)$ at an angle of $60^o$. A dark fringe is observed corresponding to a wavelength of $6000 \, \mathring{A}$. If the thickness of the glass plate is $1.2 \times 10^{-3} \, mm$, find the order of the interference fringe.