(a) Find the angle between the first minima for the two sodium vapor lines, which...

Step 1
Known wavelength of sodium vapor lines:
${\displaystyle {\lambda }_1:589.1nm}$
${\displaystyle {\lambda }_2:589.6nm}$
Slit width (D) is ${\displaystyle 2\mu m}$
${\displaystyle D\sin \theta =n\lambda }$ (1)
${\displaystyle y=\frac{n\lambda L}{D}}$ (2)
where D is slit width, n is order of minimum; ${\displaystyle \lambda }$ is wavelength of light, ${\displaystyle \theta }$ is angle at which minimum is present, y is distance of minimum from center maximum, and L is distance between slit and screen
Step 2
Solution (a)
use equation (1) to calculate ${\displaystyle {\theta }_1\text{and}{\theta }_2}$ corresponding to wavelength ${\displaystyle {\lambda }_1\text{and}{\lambda }_2}$
${\displaystyle {\sin \theta }_1=\frac{\left(1\right)\left(589.1nm\right)}{\left(2\mu m\right)}}$
${\displaystyle {\sin \theta }_1=0.2945}$
${\displaystyle {\theta }_1={17.127}^{\circ }}$
${\displaystyle {\sin \theta }_2=\frac{\left(1\right)\left(589.6nm\right)}{\left(2\mu m\right)}}$
${\displaystyle {\sin \theta }_1=0.2948}$
${\displaystyle {\theta }_1={17.145}^{\circ }}$
Difference between two angles ${\displaystyle {\theta }_2-{\theta }_{1}is{0.018}^{\circ }}$.
Solution (b)
Use equation 2 to calculate distance of first minimum corresponding to the wavelength ${\displaystyle {\lambda }_1\text{and}{\lambda }_2}$.
${\displaystyle y_1=\frac{\left(1\right)\left(589.1nm\right)\left(1m\right)}{\left(2\mu m\right)}}$
${\displaystyle y_1=249.5mm}$
${\displaystyle y_2=\frac{\left(1\right)\left(589.6nm\right)\left(1m\right)}{\left(2\mu m\right)}}$
${\displaystyle y_1=294.8mm}$
distance between these minimum is ${\displaystyle y_2-y_1=0.3mm}$
Solution (c)
These measurements are difficult because length and angles are too small to measure.

a) Calculation:
We use equation (1) to calculate ${\displaystyle {\theta }_1\text{and}{\theta }_2}$ corresponding to wavelengths values ${\displaystyle {\lambda }_1}$ for 589.1nm, ${\displaystyle {\lambda }_2}$ for 589.6nm and D for ${\displaystyle 2\mu m}$,
${\displaystyle {\sin \theta }_1=\frac{\left(1\right)(589.1\times {10}^{-9})}{2\times {10}^{-3}}}$
${\displaystyle =0.2954}$
${\displaystyle {\theta }_1={17.127}^0}$
Similarly,
${\displaystyle {\sin \theta }_2=\frac{\left(1\right)(589.6\times {10}^{-9})}{2\times {10}^{-6}}}$
${\displaystyle =0.2954}$
${\displaystyle {\theta }_2={17.145}^0}$
Difference between two angles ${\displaystyle {\theta }_2-{\theta }_{1}is{0.018}^0}$.
Conclusion:
When the two sodium vapor lines fall on a single slit of width ${\displaystyle 2.00\mu m}$ the angle between the first minima is ${\displaystyle {0.018}^0}$.
b) Calculation:
We use equation (2) to calculate distance of first minima corresponding to the values of wavelengths ${\displaystyle {\lambda }_1}$ for 589.1nm, ${\displaystyle {\lambda }_2}$ for 589.6nm and m1.
${\displaystyle y_1=\frac{\left(1\right)(589.1\times {10}^{-9})\left(1\right)}{2\times {10}^{-6}}}$
${\displaystyle =249.5mm}$
Hence, the value of ${\displaystyle y_1}$ is 249.5mm
Similarly,
${\displaystyle y_2=\frac{\left(1\right)(589.6\times {10}^{-9})\left(1\right)}{2\times {10}^{-6}}}$
${\displaystyle =249.8mm}$
Hence, the value of ${\displaystyle y_2}$ is 249.8mm
The division among the two minima ${\displaystyle y_2-y_1}$ is 0.3mm.
Conclusion:
There is a distance from the two minima of 0.3mm.
c) It is seen from sub part (a) and (b) that the angle is ${\displaystyle {0.018}^0}$ and distance between two minima is 0.3mm respectively which is very small. Such small cannot be measured precisely. Hence, these are difficult measurements.
Conclusion:
These are difficult measurements, since the length and angles of these measurements are small.