In this problem we have two slits separated by $d=1.9times 10^{-5}text{ m}$, the distance between slits and the screen $L=0.6text{ m}$ and distance between first-order bright band and the central bright band on the screen $x=1.32times 10^{-2}text{ m}$. What we need to find is wavelength for double-slit experiment.

As we can see, this problem is an example of Young’s double slit experiment so we will find the solution by using the definition for wavelength.

$$
begin{align}
lambda&=frac{xcdot d}{L} \
lambda&=frac{1.32times 10^{-2}mcdot 1.9times 10^{-5}m}{0.6m} \
&boxed{lambda=4.18times 10^{-7}m}
end{align}
$$

$$
lambda=4.18times 10^{-7}m
$$

Here we have the light of wavelength $lambda=5.96times10^{-7}text{ m}$ passing through two slits separated by $d=1.9times 10^{-5}text{ m}$ and appears on the screen at the distance $L=0.6text{ m}$. We need to find the distance between the central band and the first-order yellow band.

As we can see, this problem is an example of Young’s double slit experiment so we will find the solution by using the definition for wavelength.

$$
begin{align}
lambda&=frac{xcdot d}{L} \
x&=frac{lambdacdot L}{d} \
x&=frac{5.96times 10^{-7}mcdot 0.6m}{1.9times 10^{-5}m} \
&boxed{x=1.88times 10^{-2}m}
end{align}
$$

$$
x=1.88times 10^{-2}m
$$

Here we have the light of wavelength $lambda=6.328times10^{-7}text{ m}$ that appears on the screen moved $L=1text{ m}$ from the slits. The first-order bright band appears at the $x=6.55times10^{-2}m$ from the central band. We need to calculate the distance between two slits.

As we can see, this problem is an example of Young’s double slit experiment so we will find the solution by using the definition for wavelength.

$$
begin{align}
lambda&=frac{xcdot d}{L} \
d&=frac{lambdacdot L}{x} \
d&=frac{6.328times10^{-7}mcdot 1m}{6.55times10^{-2}m} \
&boxed{d=9.66times 10^{-6}m}
end{align}
$$

$$
d=9.66times 10^{-6}m
$$

The light of wavelength $lambda=5.96times10^{-7}text{ m}$ appears on the screen. Slits are $d=2.25times10^{-5}text{ m}$ separated. The first-order bright band appears at the $x=2times10^{-2}text{m}$ from the central band. We need to calculate the distance between two slits.

$$
begin{align}
lambda&=frac{xcdot d}{L} \
L&=frac{xcdot d}{lambda} \
L&=frac{2times 10^{-2}mcdot 2.25times 10^{-5}m}{5.96times10^{-7}m} \
&boxed{lambda=7.55times 10^{-1}m}
end{align}
$$

$$
lambda=7.55times 10^{-1}m
$$

In problem $2$ we got the equation

$$
begin{align}
d=frac{1}{2}cdotleft(m+frac{1}{2} right)cdotfrac{lambda}{n_{oil}}
end{align}
$$

What we want here is to find the minimal thickness of the oil, $n_{oil}=1.45$ , which is fulfilled for $m=0$ for red light, $lambda=6.35times10^{-7}text{ m}$.

$$
begin{align}
d&=frac{1}{2}cdotleft(m+frac{1}{2} right)cdotfrac{lambda}{n_{oil}} \
d&=frac{1}{2}cdotleft(0+frac{1}{2} right)cdotfrac{6.35times10^{-7}text{m}}{1.45} \ &boxed{d=1.09times 10^{-7}m}
end{align}
$$

$$
d=1.09times 10^{-7}m
$$

The light ray goes through the air, $n=1$, and hits the film, $n_f=1.38$. Here the first reflection and phase inversion happens because $nn_f$, on a second reflection there is a phase inversion.

To keep yellow-green light, $lambda=5.55times10^{-7}text{ m}$, from being reflected, we use the thinnest film.

Let’s start with know equation

$$
begin{align}
d=frac{1}{2}cdotleft(m+frac{1}{2} right)cdotfrac{lambda}{n_f}
end{align}
$$

For the thinnest film, $m=0$, we have

$$
begin{align}
d&=frac{1}{2}cdotleft(0+frac{1}{2} right)cdotfrac{5.55times10^{-7}text{m}}{1.38} \
&boxed{d=1.01times 10^{-7}m}
end{align}
$$

$$
d=1.01times 10^{-7}m
$$

The light ray goes through the air, $n=1$, and hits the film, $n_f=1.45$. Here the first reflection and phase inversion happens because $nn_f$, on a second reflection, there is a phase inversion.

To keep yellow-green light, $lambda=5.55times10^{-7}text{ m}$, from being reflected, we use the thinnest film.

Let’s start with know equation

$$
begin{align}
d=frac{1}{2}cdotleft(m+frac{1}{2} right)cdotfrac{lambda}{n_f}
end{align}
$$

For the thinnest film, $m=0$, we have

$$
begin{align}
d&=frac{1}{2}cdotleft(0+frac{1}{2} right)cdotfrac{5.55times10^{-7}text{m}}{1.45} \
&boxed{d=9.57times 10^{-8}m}
end{align}
$$

$$
d=9.57times 10^{-8}m
$$

The light ray goes through the air, $n=1$, and hits the film, $n_f=1.33$. Here the first reflection and phase inversion happens because $n<n_f$.

Other situation is when ray goes through a film and hits the air. Since $n<n_f$, on the second reflection there is no phase inversion.

What we need here is the thickness of the thinnest soap film considering appearance of a black stripe. The light that is used in this problem has a wavelength $lambda=5.21times10^{-7}text{ m}$

The definition of thickness when black stripe appears is

$$
begin{align}
d=frac{1}{2}cdot mcdotfrac{lambda}{n_f}
end{align}
$$

For a thinnest film $m=0$. So, we have

$$
begin{align}
d&=frac{1}{2}cdot 1cdotfrac{5.21times10^{-7}text{ m}}{1.33} \
&boxed{d=1.96times 10^{-7}m}
end{align}
$$

$$
d=1.96times 10^{-7}m
$$

We start with an equation for the constructive interference

$$
begin{align}
d=frac{1}{2}cdotleft(m+frac{1}{2} right)cdotfrac{lambda}{n_f}
end{align}
$$

and with considering $m=0$ for the thinnest film. The light ray we have in this problem $lambda=5.21times10^{-7}text{ m}$ interfere on the soap film, $n_f=1.33$.

So, our solution is

$$
begin{align}
d&=frac{1}{2}cdotleft(0+frac{1}{2} right)cdotfrac{5.21times10^{-7}text{m}}{1.33} \
&boxed{d=9.79times 10^{-8}m}
end{align}
$$

$$
d=9.79times 10^{-8}m
$$