Gregory Emery

Answered

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.

(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.

Answer & Explanation

raefx88y

Expert

2022-01-12Added 26 answers

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$\theta}_{1}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{\theta}_{2$ corresponding to wavelength $\lambda}_{1}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{\lambda}_{2$

$\mathrm{sin}\theta}_{1}=\frac{\left(1\right)\left(589.1nm\right)}{\left(2\mu m\right)$

${\mathrm{sin}\theta}_{1}=0.2945$

$\theta}_{1}={17.127}^{\circ$

$\mathrm{sin}\theta}_{2}=\frac{\left(1\right)\left(589.6nm\right)}{\left(2\mu m\right)$

${\mathrm{sin}\theta}_{1}=0.2948$

$\theta}_{1}={17.145}^{\circ$

Difference between two angles$\theta}_{2}-{\theta}_{1}\text{}is\text{}{0.018}^{\circ$ .

Solution (b)

Use equation 2 to calculate distance of first minimum corresponding to the wavelength$\lambda}_{1}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{\lambda}_{2$ .

$y}_{1}=\frac{\left(1\right)\left(589.1nm\right)\left(1m\right)}{\left(2\mu m\right)$

${y}_{1}=249.5mm$

$y}_{2}=\frac{\left(1\right)\left(589.6nm\right)\left(1m\right)}{\left(2\mu m\right)$

${y}_{1}=294.8mm$

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.

Known wavelength of sodium vapor lines:

Slit width (D) is

where D is slit width, n is order of minimum;

Step 2

Solution (a)

use equation (1) to calculate

Difference between two angles

Solution (b)

Use equation 2 to calculate distance of first minimum corresponding to the wavelength

distance between these minimum is

Solution (c)

These measurements are difficult because length and angles are too small to measure.

braodagxj

Expert

2022-01-13Added 38 answers

a) Calculation:

We use equation (1) to calculate$\theta}_{1}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{\theta}_{2$ corresponding to wavelengths values $\lambda}_{1$ for 589.1nm, $\lambda}_{2$ for 589.6nm and D for $2\mu m$ ,

$\mathrm{sin}\theta}_{1}=\frac{\left(1\right)(589.1\times {10}^{-9})}{2\times {10}^{-3}$

$=0.2954$

$\theta}_{1}={17.127}^{0$

Similarly,

$\mathrm{sin}\theta}_{2}=\frac{\left(1\right)(589.6\times {10}^{-9})}{2\times {10}^{-6}$

$=0.2954$

$\theta}_{2}={17.145}^{0$

Difference between two angles$\theta}_{2}-{\theta}_{1}\text{}is\text{}{0.018}^{0$ .

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.

$y}_{1}=\frac{\left(1\right)(589.1\times {10}^{-9})\left(1\right)}{2\times {10}^{-6}$

$=249.5mm$

Hence, the value of$y}_{1$ is 249.5mm

Similarly,

$y}_{2}=\frac{\left(1\right)(589.6\times {10}^{-9})\left(1\right)}{2\times {10}^{-6}$

$=249.8mm$

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.

We use equation (1) to calculate

Similarly,

Difference between two angles

Conclusion:

When the two sodium vapor lines fall on a single slit of width

b) Calculation:

We use equation (2) to calculate distance of first minima corresponding to the values of wavelengths

Hence, the value of

Similarly,

Hence, the value of

The division among the two minima

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

Conclusion:

These are difficult measurements, since the length and angles of these measurements are small.

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