Answer
Verified
333.9k+ views
Hint:Try to visualize the diagram and use different approximations to solve the question. General geometry can be used. The observations of Young’s double-slit experiment will give a perspective about the occurrence of interference. Coherency is the key to implement the experiment successfully.
Complete step by step answer:
(a) When two light sources produce waves having the same frequency and have a constant phase difference with time, they are called to be coherent. The two conditions for two light sources to be coherent are as follows:
-The coherent sources of light must have originated from a single source and must be monochromatic (that is they have a single wavelength) in nature.
-The path difference between the light waves from the two sources must be small.
(b) Let, a monochromatic source of light with wavelength $\lambda $ is used to illuminate a narrow slit, $S$ therefore producing two coherent sources of light through the two slits ${S_1}$ and${S_2}$.
Now, the light waves from the slits ${S_1}$ and${S_2}$ superimpose with each other and reach point $P$ as shown in the figure and have a path difference of${S_2}P - {S_1}P$.
Let $d = $ the separation between the slits, $D = $ the distance between the screen and the slits, $x = $ the distance between the centre of the screen to point $P$ and $\theta = $ the angle between $MP$ and $MO$. Now, draw a perpendicular,${S_1}N$ on line ${S_2}P$ such that distances $P{S_1}$ and $PN$ are equal. Therefore, the path difference is${S_2}N$. Let, the point $P$ be at a distance $x$ from $O$. Therefore,
$PE = x - \dfrac{d}{2}$ and $PF = x + \dfrac{d}{2}$ .
Consider the right-angled triangles in the figure, we get:
\[ \Rightarrow {\left( {{S_2}P} \right)^2} - {\left( {{S_1}P} \right)^2} = \left[ {{D^2} + {{\left( {x + \dfrac{d}{2}} \right)}^2}} \right] - \left[ {{D^2} + {{\left( {x - \dfrac{d}{2}} \right)}^2}} \right]\]
Simplify the equation:
$ \Rightarrow \,{\left( {{S_2}P} \right)^2} - {\left( {{S_1}P} \right)^2} = 2xd$
$ \Rightarrow \left( {{S_2}P + {S_1}P} \right)\left( {{S_2}P - {S_1}P} \right) = 2xd$
\[\, \Rightarrow {S_2}P - {S_1}P = \dfrac{{2xd}}{{{S_2}P + {S_1}P}}\]
Approximate that,
$ \Rightarrow {S_2}P \cong {S_1}P \cong D$
So, Path difference $ = {S_2}P - {S_1}P = \dfrac{{xd}}{D}$
The condition for bright fringe at point P on the screen is:
$ \Rightarrow \,\,{S_2}P - {S_1}P = m\lambda $
$ \Rightarrow \,\dfrac{{xd}}{D} = m\lambda $
The condition for dark fringe at point P on the screen is:
$ \Rightarrow \,\,{S_2}P - {S_1}P = \left( {2m + 1} \right)\dfrac{\lambda }{2}$
$ \Rightarrow \,\dfrac{{xd}}{D} = \left( {2m + 1} \right)\dfrac{\lambda }{2}$
Where $m$ is the order of the fringe and $\lambda $ is the wavelength.
In order to find the fringe width, $\beta $ subtract the distance of the two consecutive bright or dark fringes.The fringe of ${m^{th}}$ order will be seen at:
${x_m} = \dfrac{{m\lambda D}}{d}$
The fringe of ${\left( {m + 1} \right)^{th}}$ order will be seen at:
${x_{m + 1}} = \dfrac{{\left( {m + 1} \right)\lambda D}}{d}$
So, the fringe width,$\beta $ is given by:
$ \Rightarrow \,\beta = {x_{m + 1}} - {x_m}$
$ \therefore \beta = \dfrac{{\lambda D}}{d}$
(c) If $s$ is the size of the source and $b$ its distance from the plane of the two slits, then the criterion for the interference fringes to occur is:
$\dfrac{s}{b} < \dfrac{\lambda }{d}$
Where $d = $ the separation between the slits and $\lambda = $ wavelength of the monochromatic light.Also, the size of the source and the distance between the slits should be small.
Note: The fringe width for both the bright and dark fringes will be the same. This is because air is present between the slits and the screen and thus acting as a parallel film. Therefore, the light will not deviate from its original path and produce fringes of equal widths. If a Plano-convex lens is used to produce interference then circular fringes will occur as the air film will be of variable thickness.
Complete step by step answer:
(a) When two light sources produce waves having the same frequency and have a constant phase difference with time, they are called to be coherent. The two conditions for two light sources to be coherent are as follows:
-The coherent sources of light must have originated from a single source and must be monochromatic (that is they have a single wavelength) in nature.
-The path difference between the light waves from the two sources must be small.
(b) Let, a monochromatic source of light with wavelength $\lambda $ is used to illuminate a narrow slit, $S$ therefore producing two coherent sources of light through the two slits ${S_1}$ and${S_2}$.
Now, the light waves from the slits ${S_1}$ and${S_2}$ superimpose with each other and reach point $P$ as shown in the figure and have a path difference of${S_2}P - {S_1}P$.
Let $d = $ the separation between the slits, $D = $ the distance between the screen and the slits, $x = $ the distance between the centre of the screen to point $P$ and $\theta = $ the angle between $MP$ and $MO$. Now, draw a perpendicular,${S_1}N$ on line ${S_2}P$ such that distances $P{S_1}$ and $PN$ are equal. Therefore, the path difference is${S_2}N$. Let, the point $P$ be at a distance $x$ from $O$. Therefore,
$PE = x - \dfrac{d}{2}$ and $PF = x + \dfrac{d}{2}$ .
Consider the right-angled triangles in the figure, we get:
\[ \Rightarrow {\left( {{S_2}P} \right)^2} - {\left( {{S_1}P} \right)^2} = \left[ {{D^2} + {{\left( {x + \dfrac{d}{2}} \right)}^2}} \right] - \left[ {{D^2} + {{\left( {x - \dfrac{d}{2}} \right)}^2}} \right]\]
Simplify the equation:
$ \Rightarrow \,{\left( {{S_2}P} \right)^2} - {\left( {{S_1}P} \right)^2} = 2xd$
$ \Rightarrow \left( {{S_2}P + {S_1}P} \right)\left( {{S_2}P - {S_1}P} \right) = 2xd$
\[\, \Rightarrow {S_2}P - {S_1}P = \dfrac{{2xd}}{{{S_2}P + {S_1}P}}\]
Approximate that,
$ \Rightarrow {S_2}P \cong {S_1}P \cong D$
So, Path difference $ = {S_2}P - {S_1}P = \dfrac{{xd}}{D}$
The condition for bright fringe at point P on the screen is:
$ \Rightarrow \,\,{S_2}P - {S_1}P = m\lambda $
$ \Rightarrow \,\dfrac{{xd}}{D} = m\lambda $
The condition for dark fringe at point P on the screen is:
$ \Rightarrow \,\,{S_2}P - {S_1}P = \left( {2m + 1} \right)\dfrac{\lambda }{2}$
$ \Rightarrow \,\dfrac{{xd}}{D} = \left( {2m + 1} \right)\dfrac{\lambda }{2}$
Where $m$ is the order of the fringe and $\lambda $ is the wavelength.
In order to find the fringe width, $\beta $ subtract the distance of the two consecutive bright or dark fringes.The fringe of ${m^{th}}$ order will be seen at:
${x_m} = \dfrac{{m\lambda D}}{d}$
The fringe of ${\left( {m + 1} \right)^{th}}$ order will be seen at:
${x_{m + 1}} = \dfrac{{\left( {m + 1} \right)\lambda D}}{d}$
So, the fringe width,$\beta $ is given by:
$ \Rightarrow \,\beta = {x_{m + 1}} - {x_m}$
$ \therefore \beta = \dfrac{{\lambda D}}{d}$
(c) If $s$ is the size of the source and $b$ its distance from the plane of the two slits, then the criterion for the interference fringes to occur is:
$\dfrac{s}{b} < \dfrac{\lambda }{d}$
Where $d = $ the separation between the slits and $\lambda = $ wavelength of the monochromatic light.Also, the size of the source and the distance between the slits should be small.
Note: The fringe width for both the bright and dark fringes will be the same. This is because air is present between the slits and the screen and thus acting as a parallel film. Therefore, the light will not deviate from its original path and produce fringes of equal widths. If a Plano-convex lens is used to produce interference then circular fringes will occur as the air film will be of variable thickness.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is October 161905 regarded as an important day class 10 social science CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which river is known as Ganga of the south A Krishna class 11 social science CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE