In order for diffraction to happen a wave must “bump into” an obstacle or a slit whose dimensions are comparable to its’ wavelength. Since the wavelength of light is of the order of $10^{-7}text{ m}$ and the wavelength of sound is of the order of a few dozens of centimeter it is much easier to achieve more noticeable diffraction of sound than of light.

The smaller the angle of diffraction, the less blurring and so the greater the resolution. Since the angle of diffraction is obtained from

$$
sintheta=1.22frac{lambda}{D}
$$
we see that by increasing $D$ we decrease the sine and so the angle of diffraction (since the sine is an increasing function in the range of $0-90^circ$) which enhances camera’s resolution.

$textbf{Given values:}$

$$
begin{align*}
m &= 1 \
lambda &= 676 text{ nm} \
W &= 7.64 ~ mu text{m}
end{align*}
$$

In order to calculate the angle to the first dark fringe above the central fringe, we need to apply the expression for dark fringes in single-slit interference. The value of the integer, $m$ , is equal to 1 since it is the first dark fringe.

$$
begin{align*}
W sin theta &= m lambda \
sin theta &= dfrac{m lambda}{W} \
theta &= {sin}^{-1} left( dfrac{m lambda}{W} right) \
&= {sin}^{-1} left[ dfrac{1 (676 cdot 10^{-9} text{ m})}{7.64 cdot 10^{-6} text{ m}} right]
end{align*}
$$

$$
{boxed{theta = 5.078 text{textdegree}}}
$$

$$
{theta = 5.078 text{textdegree}}
$$

$textbf{Given values:}$

$$
begin{align*}
m &= 1 \
W &= 2.7 ~ mu text{m} \
theta &= 12 text{textdegree}
end{align*}
$$

The wavelength of the light that passes through a given slit width to the first dark fringe with integer equal to 1 can be obtain by applying the expression for dark fringes in single-slit interference :

$$
begin{align*}
W sin theta &= m lambda \
lambda &= dfrac{W sin theta}{m} \
&= dfrac{2.7 cdot 10^{-6} text{ m} (sin 12 text{textdegree})}{1}
end{align*}
$$

$$
{boxed{lambda = 561.4 text{ nm}}}
$$

$$
{lambda = 561.4 text{ nm}}
$$

$textbf{Given values:}$

$$
begin{align*}
m &= 1 \
lambda &= 592 text{ nm} \
theta &= 21 text{textdegree}
end{align*}
$$

The slit width, $W$ , can be obtain by applying the relation for dark fringes in single-slit interference :

$$
begin{align*}
W sin theta &= m lambda \
W &= dfrac{m lambda}{sin theta} \
&= dfrac{1 (592 text{ nm})}{sin 21 text{textdegree}}
end{align*}
$$

$$
{boxed{W = 1651.93 text{ nm}}}
$$

$$
{W = 1651.93 text{ nm}}
$$

$textbf{Given values:}$

$$
begin{align*}
m &= 1 \
lambda_{He} &= 632.8 text{ nm} \
theta &= 17 text{textdegree} \
lambda_Y &= 591 text{ nm}
end{align*}
$$

$textbf{(a)}$ The width of the slit, $W$ , produced by the passing light from a helium-neon laser can be obtain by applying the relation for dark fringes in single-slit interference :

$$
begin{align*}
W sin theta &= m lambda_{He} \
W &= dfrac{m lambda_{He}}{sin theta} \
&= dfrac{1 (632.8 text{ nm})}{sin 17 text{textdegree}}
end{align*}
$$

$$
{boxed{W = 2164.37 text{ nm}}}
$$

$textbf{(b)}$ If a monochromatic yellow light of wavelength $591 text{ nm}$ is used instead of the helium-neon laser, the angle to the first dark fringe will be $textbf{less than}$ to $17 text{textdegree}$. From the expression for dark fringes in single-slit interference, the angle is directly proportional with the wavelength therefore, the smaller the wavelength is the lesser the angle will be.

$textbf{(a)}$ ${W = 2164.37 text{ nm}}$

$textbf{(b)}$ $textbf{less than}$ to $17 text{textdegree}$

$textbf{Given values:}$

$$
begin{align*}
lambda &= 690 text{ nm} \
D &= 5.1 text{ mm}
end{align*}
$$

The minimum angle of separation between the two point sources of light can be calculated using the expression for the diffraction pattern of a circular opening :

$$
begin{align*}
sin theta &= 1.22 ~ dfrac{lambda}{D} \
theta &= {sin}^{-1} left( 1.22 ~ dfrac{lambda}{D} right) \
&= {sin}^{-1} left( 1.22 times dfrac{690 cdot 10^{-9} text{ m}}{0.0051 text{ m}}right)
end{align*}
$$

$$
{boxed{theta = 0.00946 text{textdegree}}}
$$

$$
{theta = 0.00946 text{textdegree}}
$$