Here we have the light of wavelength $lambda=5.46times10^{-7}text{ m}$ passing through the slit with width $w=7.5times 10^{-1}text{ m}$ and appears on the screen at the distance $L=9.5times 10^{-5}text{ m}$. We need to find the width of the central band by using the definition of the width of the bright band in single slit diffraction.

$$
begin{align}
2cdot x&=frac{2cdot lambdacdot L}{w} \
x&=frac{lambdacdot L}{w} \
x&=frac{5.46times 10^{-7}mcdot 7.5times 10^{-1}text{ m}}{9.5times 10^{-5}m} \
&boxed{x=4.31times 10^{-3}m}
end{align}
$$

$$
x=4.31times 10^{-3}m
$$

Here we have the light of wavelength $lambda=5.89times10^{-7}text{ m}$ passing through the slit of width $w=1.1times 10^{-4}text{ m}$ and appears on the screen. The width of the central band is $2cdot x=2.6times 10^{-2}text{ m}$. We will use the definition of the width of the bright band in single slit diffraction.

$$
begin{align}
2cdot x&=frac{2cdotlambdacdot L}{w} \
L&=frac{2cdot xcdot w}{2cdotlambda} \
L&=frac{2.6times 10^{-2}mcdot 1.1times 10^{-4}m}{2cdot5.89times10^{-7}m} \
&boxed{L=2.43m}
end{align}
$$

$$
L=2.43m
$$

Here we have the light from laser of wavelength $lambda=6.328times10^{-7}text{ m}$ passing through a slit and appears on a screen. A screen is $L=1.15text{ m}$ away and the width of the central band is $2cdot x=1.5times 10^{-2}text{ m}$. We will use the definition of the width of the bright band in single slit diffraction.

$$
begin{align}
2cdot x&=frac{2cdotlambdacdot L}{w} \
w&=frac{2cdotlambdacdot L}{2cdot x} \
w&=frac{2cdot 6.328times 10^{-7}mcdot 1.15text{ m}}{1.5times 10^{-2}m} \
&boxed{w=9.7times 10^{-5}m}
end{align}
$$

$$
w=9.7times 10^{-5}m
$$

Here we have the light passes through a slit and appears on a screen $w=2.95times 10^{-5}text{ m}$. A screen is $L=6times 10^{-1}text{ m}$ away and the width of the central band is $2cdot x=2.4times 10^{-2}text{ m}$. We will use the definition of the width of the bright band in single slit diffraction.

$$
begin{align}
2cdot x&=frac{2cdotlambdacdot L}{w} \
lambda&=frac{2cdot xcdot w}{2cdot L} \
lambda&=frac{2.4times 10^{-2}mcdot 2.95times 10^{-5}m}{2cdot6times10^{-1}m} \
&boxed{lambda=5.9times 10^{-7}m}
end{align}
$$

$$
lambda=5.9times 10^{-7}m
$$

Here we have the light passes through a slit and appears on a screen $w=5times 10^{-5}text{ m}$. A screen is $L=1text{ m}$. Student uses two filters $lambda_b=4.41times10^{-7}text{ m}$ $lambda_r=6.22times10^{-7}text{ m}$. We will use the definition of the width of the bright band in single slit diffraction.

$a)$ From the definition of the bright band in single slit diffraction

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

we know that width is proportional to wavelength so if $lambda_r>lambda_b$, red light will produce the wider band

$b)$ For blue light, we have

$$
begin{align}
2cdot x_b&=frac{2cdot lambda_bcdot L}{w} \
2cdot x_b&=frac{2cdot 4.41times 10^{-7}mcdot 1text{ m}}{5times 10^{-5}m} \
&boxed{2cdot x_b=1.76times 10^{-2}m}
end{align}
$$

And for red light

$$
begin{align}
2cdot x_r&=frac{2cdot lambda_rcdot L}{w} \
2cdot x_r&=frac{2cdot 6.22times 10^{-7}mcdot 1text{ m}}{5times 10^{-5}m} \
&boxed{2cdot x_r=2.49times 10^{-2}m}
end{align}
$$

$b)$ $2cdot x_b=1.76times 10^{-2}m$ and $2cdot x=2.49times 10^{-2}m$

Here we have blue light $lambda=4.34times10^{-7}text{ m}$ passing through a diffraction grating and appears on a screen. A screen is $L=1.05text{ m}$ away and the spacing of the resulting lines is $x=5.5times 10^{-1}text{ m}$. We will use the definition of the wavelength from a diffraction grating.

The definition of the wavelength from a diffraction grating is

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

So we write

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

But we don’t know $theta$.

From basic geometry we can conclude

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

So we write

$$
begin{align}
d&=frac{lambda}{sin{left(arctan {frac{x}{L}}right)}} \
d&=frac{4.34times10^{-7}text{ m}}{sin{left(arctan {frac{5.5times 10^{-1}text{ m}}{1.05text{ m}}}right)}} \
&boxed{d=9.35times 10^{-7}m}
end{align}
$$

$$
d=9.35times 10^{-7}m
$$

Here we have violet light $lambda=4.21times10^{-7}text{ m}$ passing through a diffraction grating slits, $d=8.6times 10^{-7}text{ m}$ separated, and appears on a screen. A screen is $L=8times 10^{-1}text{ m}$ away from the slits. We will use the definition of the wavelength from a diffraction grating and basic geometry.

The definition of the wavelength from a diffraction grating is

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

From basic geometry, we can conclude

$$
begin{align}
tantheta=frac{x}{L}
end{align}
$$

We need to find the separation of the lines in the diffraction pattern, $x$.

We will do that by combining (1)

$$
begin{align}
theta=arcsin{frac{lambda}{d}}
end{align}
$$

and (2).

$$
begin{align}
x&=Lcdottantheta \
x&=Lcdottan{left( arcsin{frac{lambda}{d}} right)} \
x&=8times 10^{-1}text{ m}cdottan{left( arcsin{frac{4.21times10^{-7}text{ m}}{8.6times 10^{-7}text{ m}}} right)} \
&boxed{x=4.49times 10^{-1}m}
end{align}
$$

$$
x=4.49times 10^{-1}m
$$

Here blue light produces dots on the wall that is $L=6.5times 10^{-1}text{ m}$ away from slits, $x=5.8times 10^{-1}text{ m}$ separated. From $Example Problem 3$ we know $d=7.41times 10^{-7}text{ m}$. To find a wavelength of blue light we will use the definition of the wavelength from a diffraction grating and basic geometry.

The definition of the wavelength from a diffraction grating is

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

and from basic geometry we can conclude

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

$$
begin{align}
lambda&=dcdotsin{left(arctan {frac{x}{L}} right)} \
lambda&=7.41times 10^{-7}text{ m}cdotsin{left(arctan {frac{5.8times 10^{-1}text{ m}}{6.5times 10^{-1}text{ m}}} right)} \
&boxed{lambda=4.93times 10^{-7}m}
end{align}
$$

$$
lambda=4.93times 10^{-7}m
$$

Here we have light $lambda=6.32times10^{-7}text{ m}$ passing through a diffraction grating and appears on a screen. A screen is $L=5.5times 10^{-1}text{ m}$ away and the spacing of the resulting lines is $x=5.6times 10^{-2}text{ m}$. We will use the definition of the wavelength from a diffraction grating.

The definition of the wavelength from a diffraction grating is

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

and from basic geometry we can conclude

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

$$
begin{align}
d&=frac{lambda}{sin{left(arctan {frac{x}{L}}right)}} \
d&=frac{6.32times10^{-7}text{ m}}{sin{left(arctan {frac{5.6times 10^{-2}text{ m}}{5.5times10^{-1}text{ m}}}right)}} \
d&=6.24times 10^{-6}m \
d&=6.24times 10^{-4}cm
end{align}
$$

This means our solution is

$$
begin{align}
frac{1 slit}{d}&=frac{1 slit}{6.24times 10^{-4}cm} \
&boxed{frac{1 slit}{d}=1.6times 10^{3}frac{slit}{cm}}
end{align}
$$

$$
frac{1 slit}{d}=1.6times 10^{3}frac{slit}{cm}
$$