A double slit interference experiment is set up in a chamber that can be completely evacuated with monochromatic light, an interference pattern is observed when the container is open to the air. As the container is evacuated, a careful observer will note that the interference fringes will:
A. does not change at all
B. moving slightly farther apart
C. mover slightly closer together
D. disappear completely.
Answer
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Hint: Young’s double slit experiment explains the wave nature of light using the interference of the light waves coming from two slits. The distance between the two slits is comparable to the magnitude of the wavelength.
Complete answer:
In Young's double slit experiment; the distance between the two coherent sources is comparable to the wavelength. Due to the path difference between the light coming from both the slits, interference pattern is observed at the screen which is placed away from the sources.
The path difference is given as $\Delta x=\dfrac{xd}{D}$, where $x$ is the position of the fringe from the origin, $d$ is the distance between the fringes and $D$ is the distance between the slits and source.
Then during constructive interference, $\Delta x=n\lambda$ i.e. causes the bright fringe and during destructive interference $\Delta x=(2n+1)\dfrac{\lambda}{2}$ i.e. cause the dark fringe. Where $\lambda$ is the wavelength of the coherent source.
The fringe width, the distance between two adjacent bright or dark fringes is given by $\beta=\dfrac{\lambda D}{d}$
We know that the wavelength of the light depends on the medium and so does the fringe width. Also the wavelength and speed of light is maximum in vacuum. Then we can say that when the medium is medium vacuum the will move away from each other slightly.
Hence the answer is B. Moving slightly farther apart.
Note:
Constructive interference pattern occurs when the troughs or the crests of the two coherent sources interfere. These results in addition to their amplitude, hence the fringe is bright. Similarly, Destructive interference pattern occurs when one trough and one crest of the two coherent sources interfere. These result in decreasing their amplitude, hence the fringe is dark.
Complete answer:
In Young's double slit experiment; the distance between the two coherent sources is comparable to the wavelength. Due to the path difference between the light coming from both the slits, interference pattern is observed at the screen which is placed away from the sources.
The path difference is given as $\Delta x=\dfrac{xd}{D}$, where $x$ is the position of the fringe from the origin, $d$ is the distance between the fringes and $D$ is the distance between the slits and source.
Then during constructive interference, $\Delta x=n\lambda$ i.e. causes the bright fringe and during destructive interference $\Delta x=(2n+1)\dfrac{\lambda}{2}$ i.e. cause the dark fringe. Where $\lambda$ is the wavelength of the coherent source.
The fringe width, the distance between two adjacent bright or dark fringes is given by $\beta=\dfrac{\lambda D}{d}$
We know that the wavelength of the light depends on the medium and so does the fringe width. Also the wavelength and speed of light is maximum in vacuum. Then we can say that when the medium is medium vacuum the will move away from each other slightly.
Hence the answer is B. Moving slightly farther apart.
Note:
Constructive interference pattern occurs when the troughs or the crests of the two coherent sources interfere. These results in addition to their amplitude, hence the fringe is bright. Similarly, Destructive interference pattern occurs when one trough and one crest of the two coherent sources interfere. These result in decreasing their amplitude, hence the fringe is dark.
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