Using the Doppler effect equation for receding sources:

$\displaystyle f_{observed}=\frac{v}{v+v_{source}}f_{source}$

Where $\displaystyle v$ is the speed of sound in the current medium

Plugging in values:

$\displaystyle f_{observed}=\frac{340.9}{340.9+30}700$

$\displaystyle f_{observed}=643Hz$

Using the Doppler effect equation for receding sources:

$\displaystyle f_{observed}=\frac{v}{v+v_{source}}f_{source}$

Where $\displaystyle v$ is the speed of sound in the current medium

Plugging in values:

$\displaystyle f_{observed}=\frac{340.9}{340.9+45}350$

$\displaystyle f_{observed}=309Hz$

Doppler effect:

$\displaystyle \frac{\Delta \lambda}{\lambda}=\frac{v_{source}}{c}$

Where $\displaystyle \Delta \lambda$ is the change in wavelength

$\displaystyle \lambda$ is the original wavelength

$\displaystyle v_{source}$ is the velocity of the source

$\displaystyle c$ is the speed of light

Plugging in values:

$\displaystyle \frac{350}{700}=\frac{v}{3.0*10^8 \frac{m}{s}}$

Solving for $\displaystyle v$

$\displaystyle v=1.5*10^{8}\frac{m}{s}$

Doppler effect:

$\displaystyle \frac{\Delta \lambda}{\lambda}=\frac{v_{source}}{c}$

Where $\displaystyle \Delta \lambda$ is the change in wavelength

$\displaystyle \lambda$ is the original wavelength

$\displaystyle v_{source}$ is the velocity of the source

$\displaystyle c$ is the speed of light

Plugging in values:

$\displaystyle \frac{200}{600}=\frac{v}{3.0*10^8 \frac{m}{s}}$

Solving for $\displaystyle v$

$\displaystyle v=1.0*10^{8}\frac{m}{s}$

Here, we need to use the Doppler effect equation:

$\displaystyle \frac{\Delta \lambda}{\lambda_{0}}=\frac{v}{c},$

Where $\displaystyle \Delta \lambda$ refers to the wavelength difference between the two sources, $\displaystyle \lambda_{0}$ is the laboratory wavelength, $\displaystyle v$ is the speed of the source, and $\displaystyle c$ is the speed of light.

Now, let's plug in all of the values:

$\displaystyle \frac{\Delta \lambda}{\lambda_{0}}=\frac{535-500}{500}=\frac{v}{3.0*10^8}$.

$\displaystyle v=2.1*10^7\frac{m}{s}$.

Because the wavelength has been shifted to longer wavelengths (the number is larger than the rest wavelength down on Earth), we say the object is redshifted. Therefore, the source (the galaxy) is moving away from Earth at this speed. 

The process for the double-slit experiment consisted of a coherent light source aimed at a screen with a plate with two parallel slits in between them. The light traveled through each of the slits and had wave-interference (destructive and constructive interference) to produce bands of alternating light and dark along the screen. This result would not be expected if light consisted solely of particles, as was classically thought. This showed that light is a wave, because waves interfere in that manner. The light also was found to be hitting the screen at discrete points as individual particles (photons), with the alternating bands indicating the density of the particles that hit the screen. In versions of the experiment that featured detectors at the slits, the photon passed through a single slit (as would a particle) and not both (as would a wave). These outcomes demonstrate the wave-particle duality of light.

To determine the angle, use the formula for bright fringes in a two-slit experiment.

$\displaystyle dsin(\theta)=m\lambda$

Rewrite the equation so that the angle term is isolated.

$\displaystyle \theta=sin^{-1}(\frac{m\lambda}{d})$

$\displaystyle m=+1$ represents the first bright fringe.

Substitute all the givens and solve for the angle.

$\displaystyle \theta=sin^{-1}(\frac{m\lambda}{d})=sin^{-1}\left(\frac{(1)(8.5* 10^{-7}{ m})}{5* 10^{-5} { m}}\right)= 0.9741^o$

To solve for the wavelength of the light, use the formula for dark fringes of a single-slit interference.

$\displaystyle Wsin(\theta)=m\lambda$

Rewrite the equation in order to solve for lambda (wavelength). Since this is the first dark fringe, the value of $\displaystyle m=1$.

$\displaystyle \lambda=\frac{Wsin(\theta)}{m}=\frac{1.3* 10^{-4}{ m}(sin(5))}{1}=1.133* 10^{-5}{ m}$

For this question, we're presented with a scenario in which light is passing through a small slit. In doing so, the light is diffracted, meaning that it is spread out along a viewing screen. In such a case, there will be a central bright fringe, with alternating dark and bright fringes as one moves above and below the central fringe.

In this question, we're told that the size of the slit is $\displaystyle 5.0\cdot10^{-6}\:m$, and the first of these dark fringes occurs at an angle of $\displaystyle 30^{o}$ away from the slit. We're then asked to determine the wavelength of the light based on this information.

First and foremost, we'll need to use the light diffraction equation:

$\displaystyle sin(\Theta )=n\frac{\lambda }{a}$

In the above equation, $\displaystyle \lambda$ represents the wavelength of light, $\displaystyle a$ represents the length of the slit, and $\displaystyle n$ represents the location of the dark fringes (it can equal $\displaystyle 1,\:2,\:3,...$). In this case, $\displaystyle n$ is equal to one because we're considering the first dark fringe (on the top/bottom of the central fringe).

Now that we have the equation, we can isolate the wavelength term:

$\displaystyle \lambda =\frac{asin(\Theta )}{n}$

Next all we have to do is plug in the values that we know, and we can solve for the wavelength:

$\displaystyle \lambda=\frac{(5.0\cdot 10^{-6}\: m)(sin(30^{o}))}{1}$

$\displaystyle \lambda=2.5\cdot10^{-6}\:m$

Young's Double Slit Experiment is famous because it demonstrated that light, theretofore thought of as solely discrete particles called corpuscles, actually had properties of waves in addition to particles. This was demonstrated by the two light beams separated by the note card used by Young constructively and destrucively interfering with each other to form bands of light and darkness, which showed that light acts like a wave. However, the light that reached the wall was made up of discrete points, which could only occur if light also could behave like it was made of particles. Therefore, it was concluded that light can act both like a particle and a wave, known as wave-particle duality.