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Diffraction at a Single Slit Video Lecture | MCAT Chemical and Physical Foundations
FAQs on Diffraction at a Single Slit Video Lecture - MCAT Chemical and Physical Foundations
| 1. What is diffraction and how does it occur? |  |
Ans. Diffraction refers to the bending or spreading of waves as they encounter an obstacle or pass through a narrow opening. It occurs when waves encounter an obstruction or opening that is comparable in size to their wavelength. This phenomenon is a result of the wavefronts bending around the edges of the obstacle or diffracting through the opening.
| 2. What is the significance of a single slit in diffraction? | |
Ans. A single slit is often used in diffraction experiments because it creates a distinct pattern of diffracted light or waves. The diffraction pattern produced by a single slit consists of a central bright band called the central maximum, flanked by alternating dark and bright bands known as the secondary maxima and minima. By studying the diffraction pattern of a single slit, scientists can further understand the nature and properties of waves.
| 3. How does the width of a single slit affect the diffraction pattern? | |
Ans. The width of a single slit plays a crucial role in determining the characteristics of the diffraction pattern. As the width of the slit decreases, the diffraction pattern becomes wider and more spread out. Conversely, a wider slit produces a narrower diffraction pattern. This relationship is described by the principle of diffraction: the narrower the slit, the greater the diffraction.
| 4. What factors influence the intensity of the diffraction pattern? | |
Ans. The intensity of the diffraction pattern depends on several factors. Firstly, the intensity of the incident wave affects the brightness of the central maximum and the secondary maxima. A more intense incident wave results in a brighter diffraction pattern. Secondly, the width of the slit affects the width and intensity of the diffraction pattern. A narrower slit produces a wider and less intense pattern. Lastly, the wavelength of the incident wave also influences the intensity of the diffraction pattern. Shorter wavelengths result in a more intense pattern.
| 5. How does the distance between the slit and the screen affect the diffraction pattern? | |
Ans. The distance between the slit and the screen affects the size and spacing of the diffraction pattern. As the distance increases, the diffraction pattern becomes more spread out and the fringes become wider. Conversely, decreasing the distance between the slit and the screen leads to a narrower and more closely spaced diffraction pattern. This relationship is governed by the principles of wave propagation and interference.
Text Transcript from Video
diffraction at a single slit let us
imagine a plane wave
Siddhant on a narrow slit a beefier
10.20 all the waves shall arrive at the
slit in the same face let the wave
proceed to Wed Oh
since all of these will arrive at O in
the same face therefore a shape image of
high intensity will be produced at O
according to the laws of ray optics or
geometrical optics we should have a
point image at O the focus of the lens l
but according to the wave theory every
point in a B will be a center of
secondary wavelets let us discuss the
net effect on the screen let us consider
a point B on the screen such that
wavelets reaching P make an angle theta
with Direction Co the wavelets start
from different parts of the slit in the
same face but they shall not arrive at
be in the same face because they will
cover unequal distance in reaching P
draw a perpendicular a n from a to the
extreme Ray at the end of the slit now
suppose the path difference be in
between the wavelets from a and B is
lambda beyond the points a and then the
secondary wavelets cover equal distances
so the only effect will be due to this
path difference lambda between the
extreme rays let the slit a B be
imagined to be divided in
to a large number of strips say 20 then
there are 10 strips in the upper region
AC and 10 strips in the local regions CD
C is a midpoint of a B of difference
between wavelets from a and C equals to
lambda by 2 the wavelets
from a and C will arrive at B with a
phase difference of Pi and thus cancel
each other's effect there
similarly the strip 1 and the strip 11
will have the path difference of lambda
by 2 and effect at P due to these will
be 0 likewise trip to and strip 12 and
so every strip in the upper half
together with its corresponding strip in
the lower half will cancel each other's
effect ultimately it follows that P will
be a dark point this will happen if BN
is equal to lambda if the width of the
slit ID D we have lambda by D equals two
sine theta does the condition of the
first minimum on the screen is that P
should have an angular separation theta
from Oh such that sine theta equals to
lambda by D if theta is very small then
sine theta equals to zero and we may
write theta equals to lambda by D a
similar pattern will be obtained on the
lower side at B thus due to diffraction
the image and the focus of the lens will
not be a point O but a
trip of finite width
you
imagine a plane wave
Siddhant on a narrow slit a beefier
10.20 all the waves shall arrive at the
slit in the same face let the wave
proceed to Wed Oh
since all of these will arrive at O in
the same face therefore a shape image of
high intensity will be produced at O
according to the laws of ray optics or
geometrical optics we should have a
point image at O the focus of the lens l
but according to the wave theory every
point in a B will be a center of
secondary wavelets let us discuss the
net effect on the screen let us consider
a point B on the screen such that
wavelets reaching P make an angle theta
with Direction Co the wavelets start
from different parts of the slit in the
same face but they shall not arrive at
be in the same face because they will
cover unequal distance in reaching P
draw a perpendicular a n from a to the
extreme Ray at the end of the slit now
suppose the path difference be in
between the wavelets from a and B is
lambda beyond the points a and then the
secondary wavelets cover equal distances
so the only effect will be due to this
path difference lambda between the
extreme rays let the slit a B be
imagined to be divided in
to a large number of strips say 20 then
there are 10 strips in the upper region
AC and 10 strips in the local regions CD
C is a midpoint of a B of difference
between wavelets from a and C equals to
lambda by 2 the wavelets
from a and C will arrive at B with a
phase difference of Pi and thus cancel
each other's effect there
similarly the strip 1 and the strip 11
will have the path difference of lambda
by 2 and effect at P due to these will
be 0 likewise trip to and strip 12 and
so every strip in the upper half
together with its corresponding strip in
the lower half will cancel each other's
effect ultimately it follows that P will
be a dark point this will happen if BN
is equal to lambda if the width of the
slit ID D we have lambda by D equals two
sine theta does the condition of the
first minimum on the screen is that P
should have an angular separation theta
from Oh such that sine theta equals to
lambda by D if theta is very small then
sine theta equals to zero and we may
write theta equals to lambda by D a
similar pattern will be obtained on the
lower side at B thus due to diffraction
the image and the focus of the lens will
not be a point O but a
trip of finite width
you
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