
In a Young’s double slit experiment, films of thickness ${t_A}$and ${t_B}$ and refractive indices ${\mu _A}$ and ${\mu _B}$ are placed in front of slits, A and B respectively. If${\mu _A}{t_A} = {\mu _B}{t_B}$, then the central maxima may:
A) Not shift
B) Shift towards A
C) Shift towards B
D) None of these
Answer
233.1k+ views
Hint: This question has more than one correct answer. When a film is introduced in front of the slits there will be a path difference. The path difference arises when the ray travels through a medium having different refractive indices. Use the formula for path difference in Young’s double-slit experiment and solve to get the answers.
Formula used:
Path difference of the ray is given by,
$\Delta x = \left( {\mu - 1} \right)t$ (Where $\Delta x$ stands for the path difference, $\mu $stands for the refractive index of the film and $t$ stands for the thickness of the film)
Complete step by step solution:
If we introduce a film having a thickness t and refractive index $\mu $between the screen and the slit, of Young’s double-slit experiment setup, the ray will have a path difference$\Delta x$.
Let ${t_A}$ and ${t_B}$ be the thickness of the films that we introduce. The films have refractive indices ${\mu _A}$and ${\mu _B}$respectively, we introduce these films between the slit and the screen.
After introducing the films, the change in path can be written as,
$\Delta x = \left( {{\mu _A} - 1} \right){t_A} - \left( {{\mu _B} - 1} \right){t_B}$
Opening the brackets, the equation will become
$\Delta x = {\mu _A}{t_A} - {t_A} - {\mu _B}{t_B} + {t_B}$……………………………..(A)
In the question, it is given that ${\mu _A}{t_A} = {\mu _B}{t_B}$
Applying this in (A), we get
$\Delta x = {t_B} - {t_A}$ (Other terms will be cancelled)
Now there can be three cases,
${t_B} = {t_A}$
In this case$\Delta x = 0$, that means there will not be any shift.
${t_B} > {t_A}$ Or ${t_A} > {t_B}$
In both cases, $\Delta x \ne 0$, that means the central maxima will shift either towards A or B.
The correct options are: Option (A), Option (B) and option (C).
Note: As there are two point charges involved in, there will be two waves in this experiment. The paths travelled by the two waves before interference will be different. This difference in their paths is called the path difference.
Formula used:
Path difference of the ray is given by,
$\Delta x = \left( {\mu - 1} \right)t$ (Where $\Delta x$ stands for the path difference, $\mu $stands for the refractive index of the film and $t$ stands for the thickness of the film)
Complete step by step solution:
If we introduce a film having a thickness t and refractive index $\mu $between the screen and the slit, of Young’s double-slit experiment setup, the ray will have a path difference$\Delta x$.
Let ${t_A}$ and ${t_B}$ be the thickness of the films that we introduce. The films have refractive indices ${\mu _A}$and ${\mu _B}$respectively, we introduce these films between the slit and the screen.
After introducing the films, the change in path can be written as,
$\Delta x = \left( {{\mu _A} - 1} \right){t_A} - \left( {{\mu _B} - 1} \right){t_B}$
Opening the brackets, the equation will become
$\Delta x = {\mu _A}{t_A} - {t_A} - {\mu _B}{t_B} + {t_B}$……………………………..(A)
In the question, it is given that ${\mu _A}{t_A} = {\mu _B}{t_B}$
Applying this in (A), we get
$\Delta x = {t_B} - {t_A}$ (Other terms will be cancelled)
Now there can be three cases,
${t_B} = {t_A}$
In this case$\Delta x = 0$, that means there will not be any shift.
${t_B} > {t_A}$ Or ${t_A} > {t_B}$
In both cases, $\Delta x \ne 0$, that means the central maxima will shift either towards A or B.
The correct options are: Option (A), Option (B) and option (C).
Note: As there are two point charges involved in, there will be two waves in this experiment. The paths travelled by the two waves before interference will be different. This difference in their paths is called the path difference.
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