 Gregory Emery

2022-01-11

(a) Find the angle between the first minima for the two sodium vapor lines, which have wavelengths of 589.1 and 589.6 nm, when they fall upon a single slit of width $2.00\mu m$.
(b) What is the distance between these minima if the diffraction pattern falls on a screen 1.00 m from the slit?
(c) Discuss the ease or difficulty of measuring such a distance. raefx88y

Expert

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

Expert

a) Calculation:
We use equation (1) to calculate corresponding to wavelengths values ${\lambda }_{1}$ for 589.1nm, ${\lambda }_{2}$ for 589.6nm and D for $2\mu m$,
${\mathrm{sin}\theta }_{1}=\frac{\left(1\right)\left(589.1×{10}^{-9}\right)}{2×{10}^{-3}}$
$=0.2954$
${\theta }_{1}={17.127}^{0}$
Similarly,
${\mathrm{sin}\theta }_{2}=\frac{\left(1\right)\left(589.6×{10}^{-9}\right)}{2×{10}^{-6}}$
$=0.2954$
${\theta }_{2}={17.145}^{0}$
Difference between two angles .
Conclusion:
When the two sodium vapor lines fall on a single slit of width $2.00\mu m$ the angle between the first minima is ${0.018}^{0}$.
b) Calculation:
We use equation (2) to calculate distance of first minima corresponding to the values of wavelengths ${\lambda }_{1}$ for 589.1nm, ${\lambda }_{2}$ for 589.6nm and m1.
${y}_{1}=\frac{\left(1\right)\left(589.1×{10}^{-9}\right)\left(1\right)}{2×{10}^{-6}}$
$=249.5mm$
Hence, the value of ${y}_{1}$ is 249.5mm
Similarly,
${y}_{2}=\frac{\left(1\right)\left(589.6×{10}^{-9}\right)\left(1\right)}{2×{10}^{-6}}$
$=249.8mm$
Hence, the value of ${y}_{2}$ is 249.8mm
The division among the two minima ${y}_{2}-{y}_{1}$ 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 ${0.018}^{0}$ 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.

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