Question

A primary monochromatic coherent source is used in YDSE and half of the space between the primary source and the slits is filled with a transparent liquid of refractive index \[\mu \] with respect to air, such that path of the wave \[{\text{S}}{{\text{S}}_{\text{2}}}\] is through this liquid. Everywhere else there is air and slits are sealed with equal glass plates. Find the location of central maximum.

Answer 1

Hint: First of all, we will draw the figure showing all the physical quantities involved in it. Then we will find the condition for central maxima and thereby substituting the required values in it. We will manipulate accordingly and obtain the result.

Complete step by step answer:
The double-slit experiment is a demonstration of modern physics that light and matter will show characteristics of waves and particles both classically defined; however, it demonstrates the essentially probabilistic existence of quantum mechanical phenomena.
Complete the length of the optical path protected by the two waves produced by the primary source \[\,{\text{S}}\,\] if \[\,{\text{P}}\,\] is an observing point at which the waves are the same path length, \[\,{\text{P}}\,\] is the central maximum. For better understanding, we draw a diagram, which is given below:

For the position of bright fringes, we can write:
\[\Delta z = n\lambda \]
Again, for the position of dark fringes, we can write:
\[\Delta z = \dfrac{{\left( {2n - 1} \right)n\lambda }}{2}\]
The first minima are present on either side of the central maxima.
Therefore, for the case of the central maximum, which is written as,
\[{\text{S}}{{\text{S}}_2} - {\text{S}}{{\text{S}}_{\text{1}}} = {{\text{S}}_{\text{1}}}{\text{P}} - {{\text{S}}_{\text{2}}}{\text{P}}\]
$
\Rightarrow \mu \sqrt {{{\left( {\dfrac{d}{2}} \right)}^2} + {{\left( {\dfrac{d}{2}} \right)}^2}} - \sqrt {{{\left( {\dfrac{d}{2}} \right)}^2} + {{\left( {\dfrac{d}{2}} \right)}^2}} = d\sin \theta \\
\Rightarrow \mu \sqrt {\dfrac{{2{d^2}}}{4}} - \sqrt {\dfrac{{2{d^2}}}{4}} = d\dfrac{y}{D} \\
\Rightarrow \mu \times \dfrac{d}{{\sqrt 2 }} - \dfrac{d}{{\sqrt 2 }} = d\dfrac{y}{D} \\
\Rightarrow \dfrac{1}{{\sqrt 2 }}\left( {\mu - 1} \right) = \dfrac{y}{D} \\
$
Again, we will simplify the above expression, and we get:
$
\Rightarrow y = \dfrac{D}{{\sqrt 2 }}\left( {\mu - 1} \right) \\
\Rightarrow y = \dfrac{{2000d}}{{\sqrt 2 }}\left( {\mu - 1} \right) \\
\Rightarrow y = \dfrac{{2000d \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }}\left( {\mu - 1} \right) \\
\Rightarrow y = 1000\sqrt 2 d\left( {\mu - 1} \right) \\
$

Hence, the location of the central maximum is \[1000\sqrt 2 d\left( {\mu - 1} \right)\] .

Additional information: Monochromatic coherent source: Laser light is a monochromatic source of coherent light. Coherence which means whether the phase difference is constant between the two waves is coherent. They need the same frequency in order to do this. Monochromatic which implies that there is just one light wavelength.
Refractive index: Refractive Index is a value determined from the ratio of light speed in a vacuum to that of a second density medium. The refractive index variable in the descriptive text and mathematical equations is most generally symbolized as the letter \[n\] .

Note: In this experiment, the light is passed through a double slit because two slits have two light sources that are coherent and either interfere constructively or destructively. Young used sunlight, where its own pattern is created by each wavelength, making the result more difficult to see.