Light $lambda_b=4.1times10^{-7}text{ m}$ passes through a slit of width $w=3.8times 10^{-6}text{ m}$. A screen is $L=0.29text{ m}$. We will use the definition of the width of the bright band in single slit diffraction.

The definition of the bright band in single slit diffraction is

$$
begin{align}
2cdot x=frac{2cdot lambdacdot L}{w} \
end{align}
$$

So we write

$$
begin{align}
2cdot x&=frac{2cdot lambdacdot L}{w} \
2cdot x&=frac{2cdot 4.1times 10^{-7}mcdot 0.29text{ m}}{3.8times 10^{-6}m} \
&boxed{2cdot x=6.26times 10^{-2}m}
end{align}
$$

From problem $2$ we know $lambda=4.1times10^{-7}text{ m}$, $L=0.29text{ m}$ and $2cdot x=6.26times 10^{-2}text{ m}$.

From basic geometry, we can conclude

$$
begin{align}
tantheta&=frac{x}{L} \
theta&=arctan {frac{x}{L}}
end{align}
$$

We use $x=3.13times 10^{-2}text{ m}$.

$$
begin{align}
&theta=arctan {frac{3.13times 10^{-2}text{ m}}{0.29text{ m}}} \
&boxed{theta=6.15^{circ}}
end{align}
$$

B) $6.2^circ$

In this problem we have to find the minimum separation between stars $x_{obj}=3.1text{ ly}$, the position of stars is $L_{obj}=6.2times10^{4}text{ ly}$ and the wavelength of light is $lambda=6.1times10^{-7}text{ m}$. We will use The Rayleigh criterion to solve this problem.

$$
begin{align}
{x_{obj}}=frac{1.22cdotlambdacdot L_{obj}}{D}
end{align}
$$

We need to convert light years into meters to solve the equation.

It is know that one light year is equal to $9.4605284times10^{15}text{m}$. So for $L$ wee have $L_{obj}=5.87times10^{20}text{m}$ and for $x$ wee have $x_{obj}=2.93times10^{16}text{m}$. Then we have

$$
begin{align}
&{D}=frac{1.22cdotlambdacdot L_{obj}}{x_{obj}} \
&{D}=frac{1.22cdot6.1times10^{-7}text{ m}cdot 5.87times10^{20}text{m}}{2.93times10^{16}text{m}} \
&boxed{D=1.49times10^{-2}text{m}}
end{align}
$$

C) $1.5times10^{-2}text{m}$

Here we have light $lambda=6.5times10^{-7}text{ m}$ passing through a slit $d=5.5times 10^{-5}text{ m}$. We will use the definition of the wavelength from a diffraction grating.

The definition of the wavelength from a diffraction grating is

$$
begin{align}
sintheta=frac{lambda}{d}
end{align}
$$

So we write

$$
begin{align}
sin&theta=frac{lambda}{d} \
&theta=arcsin{frac{lambda}{d}} \
&theta=arcsin{frac{6.5times10^{-7}text{ m}}{5.5times 10^{-5}text{ m}}} \
&boxed{theta=0.68^{circ}}
end{align}
$$

B) $0.68^{circ}$

A wavelength is $lambda=6.38times 10^{-7}text{ m}$. The distance between third-order bright band and the central bright band on the screen $3cdot x=7.5times 10^{-2}text{ m}$ and the distance between slits and the screen is $L=2.475text{ m}$. What we need to find is 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.

$$
begin{align}
lambda&=frac{xcdot d}{L} \
end{align}
$$

Since we have a distance for third-order band, we use $frac{3cdot x}{3}=2.5times 10^{-2}text{ m}$.

$$
begin{align}
&d=frac{lambdacdot L}{x} \
&d=frac{6.38times 10^{-7}text{ m}cdot 2.475text{ m}}{2.5times 10^{-2}text{ m}} \
&boxed{d=6.32times 10^{-5}text{ m}}
end{align}
$$

D) $6.3times 10^{-5}text{ m}$

The distance between second-order bright band and the central bright band on the screen $2cdot x=8.2times 10^{-2}text{ m}$ and the distance between slits, $d=5.3times 10^{-5}text{ m}$, and the screen is $L=4.2text{ m}$. What we need to find is wavelength of a monochromatic light.

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.

$$
begin{align}
lambda&=frac{xcdot d}{L} \
end{align}
$$

Since we have a distance for third-order band, we use $frac{2cdot x}{2}=4.1times 10^{-2}text{ m}$.

$$
begin{align}
lambda&=frac{xcdot d}{L} \
lambda&=frac{4.1times 10^{-2}text{ m}cdot 5.3times 10^{-5}text{ m}}{4.2text{ m}} \
&boxed{lambda=5.17times 10^{-5}m}
end{align}
$$

B) $5.2times 10^{-5}m$

A wavelength of red light is $lambda=6.5times 10^{-7}text{ m}$. The index of refraction is $n_f=1.41$. We will use following equation.

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

What we want is to find the minimal thickness, which is fulfilled for $m=0$. So we write

$$
begin{align}
&d=frac{1}{2}cdotleft(0+frac{1}{2} right)cdotfrac{6.5times 10^{-7}text{ m}}{1.41} \
&boxed{d=1.15times 10^{-7}m}
end{align}
$$

A) $1.2times 10^{-7}m$

A diffraction grating has $6times10^{3}frac{slit}{cm}$ which means one slit is $frac{1}{6times10^{3}}text{ cm}$ wide. So $d=1.67times10^{-6}text{ m}$. Also, we know angle $theta=20^{circ}$

Form the definition of the wavelength from a diffraction grating

$$
begin{align}
lambda=dcdotsintheta
end{align}
$$

So we write

$$
begin{align}
&lambda=1.67times10^{-6}text{ m}cdotsin20^{circ} \
&boxed{lambda=5.7times 10^{-7}text{ m}}
end{align}
$$

$$
lambda=5.7times 10^{-7}text{ m}
$$