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CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 5 - MCQExams.com
CBSE
Class 12 Medical Physics
Wave Optics
Quiz 5
Water is transparent to visible light. Still it is not possible to see object at a distance in fog which consists of fine drops of water suspended in air. This is so because:
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Fine drops of water are opaque to visible light
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Most of the light is scattered away, hence the apparent opacity
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Fog affects our vision adversely
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Rays suffer total internal reflection and cannot reach the eye of the observer
Explanation
Due to scattering, most of the light is lost and is not able to reach observer's eye.
An astronomical telescope has on objective and an eyepiece of focal lengths $$10 cm$$ and $$1 cm$$ respectively. Find its tube length in normal adjustment.
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$$11\ cm$$
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$$10\ cm$$
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$$9\ cm$$
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$$1\ cm$$
Explanation
$$f_0=10 cm; \quad f_e=1 cm$$
In normal adjustment,
Magnifying power $$=\cfrac {f_0}{f_e}=\cfrac {10}{1}=10$$
Tube length $$=f_0+f_e$$
$$=(10+1)$$
$$=11\ cm$$
Light is a form of .........
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Energy
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Wave
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Both
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none
Explanation
A Light has a dual nature
Sometimes it behaves like a particle (called a photon), which explains how light travels in straight lines,
Sometimes it behaves like a wave, which explains how light bends around an object.
Magnification of an object ($$m$$), is equal to
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$$\cfrac {v+f}{f}$$
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$$\cfrac {vf}{v-f}$$
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$$\cfrac {f}{v+f}$$
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None of these
Explanation
we know,mirror formula
$$\cfrac {1}{f}=\cfrac{1}{v}+\cfrac{1}{u}, magnification(m)=\cfrac{-v}{u}$$
$$\cfrac{1}{u}=\cfrac{1}{f}-\cfrac{1}{v}$$
$$\cfrac{1}{u}=\cfrac{v-f}{fv}$$
$${u}=\cfrac{fv}{v-f}$$
$$(m)=\cfrac{-v}{u}$$,substituting $$u$$.
$$m=\cfrac{-v}{1}\times\cfrac{v-f}{fv}$$
$$m=\cfrac{f-v}{f}$$
To demonstrate the phenomena of interference, we require
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Two sources which emit radiation of same frequency
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Two sources which emit radiation of nearly same frequency
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Two sources which emit radiation of the same frequency and have a definite phase relationship
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Two sources which emit radiation of different wavelengths
Explanation
To demonstrate the phenomena of interference, we require
two sources which emit radiation of the same frequency and have a definite phase relationship.
The optical phenomenon which Newton's theory of light failed to explain is:
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Interference
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Polarization
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Diffraction
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All the above three
Explanation
Diffraction is the slight bending of light as it passes around the edge of an object. The amount of bending depends on the relative size of the wavelength of light to the size of the opening. If the opening is much larger than the light's wavelength, the bending will be almost unnoticeable. However, if the two are closer in size or equal, the amount of bending is considerable.
Who proposed wave nature of light ?
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Huygen
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Young
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Fresnel
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Maxwell
Explanation
Huygens
believed that light was made up of waves vibrating up and down perpendicular to the direction of the light travels. This is known as '
Huygens' Principle'
.
According to this theory, light waves are spherical, and these wave surfaces advance or spread out as they travel at the speed of light. This theory explains why light shining through a pin hole or slit will spread out rather than going in a straight line.
Differential refractive index is used in core to minimise loss due to
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Evanscent field
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diffraction
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polarisation
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interference
Explanation
To minimise losses due to diffraction, the refractive index of core is gradually reduce till core-cladding interface
Dichorism means
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selective absorption of unpolarised light.
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selective absorption of dispersed light.
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selective absorption of scattered light.
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selective absorption of one of the polarised component.
Explanation
Dichromism is the selective absorption of one orthogonal polarization component of an incident beam over the other. this phenomenon is due to anisotropy of the material, with one polarization component experiencing preferential absorption.
Since the objective lens merely forms an enlarged real image that is viewed by the eyepiece, the overall angular magnification M of the compound microscope is the product of the lateral magnification $$ { m }_{ 1 }$$ of the objective and the angular magnification $$ { M }_{ 2 }$$ of the eyepiece. The former is given by
$$ { m }_{ 1 }=\dfrac { { S }_{ 1 }^{ ' } }{ { S }_{ 1 } } $$
Where $$ { S }_{ 1 }and{ S }_{ 1 }^{ ' }$$ are the object and image distance for the objective lens. Ordinarily the object is very close to the focus, resulting in an image whose distance from the objective is much larger than the focal length $$ { f }_{ 1 }$$. Thus $$ { S }_{ 1 }$$ is approximately equal to $$ { f }_{ 1 }$$ and $$ { m }_{ 1 }$$ =$$ -\dfrac { { S }_{ 1 }^{ ' } }{ { f }_{ 1 } } $$, approximately. The angular magnification of the eyepiece from $$ { M }=-\dfrac { { u }^{ ' } }{ u } =\dfrac { { y }/{ f } }{ { y }/{ 25 } } =\dfrac { 25 }{ f } $$ (f in centimeters) is $$ { M }_{ 2 }=25cm/{ f }_{ 2 },$$ Where $$ { f }_{ 2 }$$ is the focal length of the eyepiece, considered as a simple lens. Hence the overall magnification M of the compound microscope is, apart from a negative sign, which is customarily ignored,
$$ { M }={ m }_{ 1 }{ M }_{ 2 }=\dfrac { \left( 25cm \right) { S }_{ 1 }^{ ' } }{ f } $$
What is the resolving power of the instrument whose magnifying power is given in the passage?
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$$ \dfrac { \mu \sin { \theta } }{0 .61\lambda } $$
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$$ \dfrac { \mu \sin { \theta } }{ 1.22\lambda } $$
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$$ \dfrac { \mu \sin { \theta } }{ \lambda } $$
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$$ \dfrac { \sin { \theta } }{ 1.22\lambda } $$
Explanation
The mentioned instrument is compound microscope and its resolving power is $$ R.P=\dfrac { 2\mu \sin { \theta } }{ 1.22\lambda } =\dfrac { \mu \sin { \theta } }{ 0.61\lambda } $$
where, $$ \mu$$ is refractive index of medium, $$ \theta$$ is the semi-vertical angle of the cone of the rays received by the objective.
A microscope is used with sodium light and its resolving power is not sufficiently large.Higher resolution will be obtained by using wavelength of
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20 micron
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2 micron
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1 micron
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400 A$$^{\circ}$$
Explanation
Since power of resolution is more for violet than for red, we conclude that resolving power is greater for light with lower wavelengths.
Wavelength of sodium light is around 589nm
Hence resolving power increases for light with lower wavelength.
Hence option D is correct.
The intensity of principal maxima in the single slit diffraction pattern is $$\displaystyle I_o$$? What will be the intensity when slit width is doubled?
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$$\displaystyle 2I_o$$
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$$\displaystyle 4I_o$$
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$$I_o$$
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$$ \dfrac{I_{0}}{2}$$
Explanation
Intensity at screen is given by-
$$I={ I }_{ 0 }\times \dfrac { { { (\sin { \dfrac { \pi a\sin { \theta } }{ \lambda } } ) }^{ 2 } } }{ { (\dfrac { \pi a\sin { \theta } }{ \lambda } ) }^{ 2 } } $$
where, a is slit size,and $$\theta $$ is angle subtended by interfering light ray on screen.
for principal(central) maxima- $$(\theta \rightarrow 0)$$
$${ I }_{ central }={ I }_{ 0 }$$ ....(independent of slit size "a")
Antinodal curves correspond to _____ interference.
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constructive
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destructive
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where intensity is less than maximum but not completely zero
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none of these
Explanation
Antinodal curves correspond to constructive interference.
In producing a pure spectrum, the incident light is passed through a narrow slit placed in the focal plane of an achromatic lens because a narrow slit
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produce less diffraction
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increase intensity
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allows only one colour at a time
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allows a more parallel beam when it passes through the lens.
Explanation
If the slit is wide, different part of the slit produces different spectra, which overlap resulting in impure spectrum. Therefore narrow slit is used as it allows more parallel beam of light when it passes through the lens.
The slits in a Young's double- slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is $$I_o$$. If one of the slit is closed, the intensity at this point will be_______
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$$I_o$$
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$$\displaystyle \frac{I_o}{4}$$
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$$\displaystyle \frac{I_o}{2}$$
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$$4I_o$$
Explanation
for interference of two waves,
$${ I }_{ total }={ I }_{ 1 }+{ I }_{ 2 }+2\sqrt { { I }_{ 1 }{ I }_{ 2 } } \cos { \theta }$$
for equal intensities and central fringe,
$${ I }_{ total }={ 4I }_{ 1 }={ I }_{ 0 }$$ ..... given.
$$\therefore { \quad I }_{ 1 }=\dfrac { { I }_{ 0 } }{ 4 } $$
Hence when one slit is closed the intensity will be $$\dfrac { { I }_{ 0 } }{ 4 } $$
Four light waves are represented by
$$(i)y=a_1\sin\omega t$$ $$(ii)y=a_2\sin(\omega t+\varepsilon)$$
$$(iii)y=a_1\sin2\omega t$$ $$(iv)y=a_2\sin2(\omega t+\varepsilon)$$
Interference fringes may be observed due to superposition of
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$$(i)$$ and $$(ii)$$
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$$(i)$$ and $$(iii)$$
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$$(ii)$$ and $$(iv)$$
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$$(iii)$$ and $$(iv)$$
Explanation
For superposition, frequency of waves should be same
therefore superposition of (i) with (ii) and (iii) with (iv) will happen
In a diffraction(single slit experiment), the slit is exposed by white light. The fringe surrounding the central fringe is
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Red
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Yellow
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Violet
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Green
Explanation
From the condition, $$a \sin \theta=n\lambda$$
the cental spot is bright since $$n=0$$ and it is satisfied for all wavelengths.
The next angle will be for $$\lambda_{violet}$$
Red will have the maximum angle.
The path difference between two wave fronts emitted by coherent sources of wavelength $$5640 \ \mathring{A}$$ is $$2.1$$ microns. The phase difference between the wave fronts at that point is :
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$$7.692$$
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$$7.692\pi $$
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$$\displaystyle \frac{7.692}{\pi}$$
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$$\displaystyle \frac{7.692}{3\pi}$$
Explanation
phase difference is $$\dfrac{d}{\lambda} \times 2\pi =\dfrac{2\pi *2.1}{0.564}=7.692\pi $$
option $$B$$ is correct
Two coherent points sources $$S_1$$ and $$S_2$$ vibrating in phase emit light of wavelength $$\lambda$$. The separation between them is $$2\lambda$$. The light is collected on a screen $$\sum$$ placed at a distance $$D>>\lambda$$ from the slit $$S_1$$ as shown, in the fig. Find the minimum distance so that intensity at $$P$$ is equal to intensity at $$O$$.
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$$D$$
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$$D_{\displaystyle/3}$$
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$$\sqrt{3}D$$
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$$D_{\displaystyle/2}$$
Explanation
$$S_1S_2\cos\theta=n\lambda$$
$$2\lambda\cos\theta=\lambda$$
$$\theta=60^0$$
$$x=D\tan 60^0=\sqrt{3}D$$.
Huygen's principle of secondary wavelets may be used to
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find the velocity of light in vacuum
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explain the practical behaviour of light
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find the new position of a wave front
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explain Snell's law
Explanation
Huygen's principle states that every point on a wave-front may be considered a source of secondary spherical wavelets which spread out in the forward direction at the speed of light. The new wave-front is the tangential surface to all of these secondary wavelets.
Hence option C is correct.
Also this can be used to find the expression of snells's law $$\mu_1 sini=\mu_2sinr$$
Hence option D is correct.
Which statement is true?
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The secondary wavelets cause interference in YDSE
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The secondary wavelets cause diffraction in single slit experiment.
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If the colliminating and focusing lens are used secondary wavelets do not exist.
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secondary wavelets travel in a straight line
Explanation
In diffraction experiment as there is only single slit when the secondary wavelet reaches the slit it gets diffracted and thus cause pattern on screen , where as in YDSE wave forms gets interfaced on one another to form pattern on screen .
Two independent monochromatic sodium lamps can not produce interference because
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The frequencies of the two sources are different.
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The phase difference between the two sources changes will respect to time.
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The two sources become coherent.
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The amplitude of two sources is different.
Explanation
Answer is A
Two sources of light are said to be coherent if they emit light which always has a constant phase difference between them. It means that the two sources must emit radiations of the same wavelength. The two independent sources cannot be coherent because of the fact that independent sources cannot maintain a constant phase difference between them. For experimental purposes, two virtual sources obtained from a single parent source can act as coherent. In such case all the random phase changes occurring in the parent source are repeated in the virtual sources also, thus maintaining a constant phase difference between them. Since the wavelength of light waves is extremely small, the two sources must be narrow and must also be close to each other.
Interference can never be obtained with two independent sources of light, such as two bulbs or two candles, due to the fact that any two independent beams of light are always incoherent.
Light waves travel in a vacuum, along the $$X-$$axis. Which of the following may represent the wave fronts?
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$$x=c$$
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$$y=c$$
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$$z=c$$
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$$x+y+z=c$$
Explanation
Given the direction of propagation is $$ \hat{i} $$
Wave fronts will be planes $$ \perp \hat{i} $$
In the given options, only plane $$ x=c $$ is $$ \perp \hat{i} $$
In coherent sources it is necessary that their
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amplitudes are same
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wavelengths are same
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initial phase remains constant
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None of these
Explanation
For sources to be coherent it is necessary that their initial phase difference remains the same or constant.
Two coherent sources with intensity ratio $$\beta$$ produce interference. The fringe visibility will be
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$$\displaystyle \frac{2\sqrt\beta}{1+\beta}$$
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$$\displaystyle 2\beta$$
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$$\displaystyle \frac{2}{1+\beta}$$
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$$\displaystyle \frac{\beta}{1+\beta}$$
Explanation
Fringe visibility is defined as: $$Visibility = \dfrac{ 2\sqrt{I_1I_2}}{I_1 + I_2}$$
Given, $$ \dfrac{I_1}{I_2} = \beta$$
$$ \therefore visibility = \dfrac{2 I_2 \sqrt{\beta} }{I_2( 1 + \beta)}= \dfrac{ 2 \sqrt{\beta} }{1 + \beta }$$
Lattice constant of a crystal is $$3 \times 10^{-8} cm$$ and glance angle of X-ray is $$30^\circ$$ for first order diffraction, then the value of $$\lambda$$ will be:-
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$$6 \times 10^{-8} cm$$
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$$3 \times 10^{-8} cm$$
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$$1.5 \times 10^{-8} cm$$
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$$10^{-8} cm$$
What does the term point to correspondence in the paragraph refer to ?
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Waves having constant amplitude
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Waves having constant phase relation
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Waves having same frequency
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Wave having same amplitude,frequency and constant phase relation
Explanation
Broad sources may be treated to be a collection of many point sources. These sources have a constant phase relation, that is, the light from any two points picked up form the collection have a constant phase difference. This phase difference would exist even in reflected as well as transmitted wave, even when frequency might change.
The correct relation between the time interval $$'\partial\ '$$ and phase difference $$'\delta\ '$$ is
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$$\displaystyle\partial=\frac{T}{2\pi}\delta$$
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$$\displaystyle\partial=\frac{2\pi}{T}\delta$$
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$$\displaystyle\partial=2\pi \delta$$
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$$\displaystyle\partial=\frac{\delta}{2\pi}$$
Explanation
Time interval $$T$$ generates p
hase difference of $$2\pi$$
So time difference of 1 generates phase difference $$\dfrac {2\pi}{T}$$
Time difference of $$\displaystyle \partial$$ generates phase difference $$\delta = \dfrac{2\pi}{T} \partial$$
=> $$\partial = \delta \dfrac { T} { 2\pi}$$
Interference pattern can be produced by two identical sources. Here the identical sources mean that
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their size is same
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their wavelength is same
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the intensity of light emitted by them is same
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the emplitudes of light waves emitted by them are same
Explanation
For observing interference the term identical source means that their wavelength are the same ( i.e., they are coherent).
In Lloyd's single mirror method we have
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Both sources virtual
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One source virtual and one real
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Both sources real
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None of these
Explanation
By reflection we create a virtual source from real source and light from both of these sources are used to create interference pattern in Lloyd's single mirror method.
The equations of waves emitted $$S_1,S_2,S_3$$ and $$S_4$$ are respectively $$y_1=20\sin(100\pi t), y_2=20\sin(200\pi t), y_3=20\cos(100\pi t)$$ and $$y_4=20\cos(100\pi t)$$. The phenomenon of interference will be produced by
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$$y_1$$ and $$y_2$$
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$$y_2$$ and $$y_3$$
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$$y_1$$ and $$y_3$$
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Interference will not possible
Explanation
To set up a stable and clear interference pattern, two conditions must be met:
The sources of the waves must be coherent, which means they emit identical waves with a constant phase difference.
And the waves should be monochromatic - they should be of a single wavelength.
Since the frequency of $$y_1$$ and $$y_2$$ is same, they form interference pattern.
Are interference of light and production of beats in sound identical?
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Yes, both increase or decrease the intensity.
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No, interference occurs in space regime and beats occur in time regime.
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No,light waves are em waves and sound waves are mechanical.
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Yes because both correspond to enforcement.
Explanation
No, because even though both occur due to constructive and destructive interference between the waves, but light waves are em waves and sound waves are mechanical.
The transverse nature of light waves is verified by
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reflection of light
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polarisation of light
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refraction of light
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interference of light
Explanation
Phenomenon of polarisation helps in establishing the fact that light waves are transverse in nature, otherwise it was believed that they are longitudinal in nature like sound waves.
All particles of a wave front vibrate
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in same phase
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in opposite phase
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up and down
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left and right
Explanation
Wave front by definition is the locus of points having same phase.
Answer. A) in same phase
A parallel beam of light $$\lambda=5000A^o$$ falls normally on a single narrow slit of width $$0.001mm$$. The light is focused by a convex lens on a screen placed in the focal plane. The first minimum will be formed for the angle of diffraction are equal to:
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$$0^0$$
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$$15^0$$
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$$30^0$$
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$$50^0$$
Explanation
Let the angle be $$\theta$$
$$dsin\theta =\lambda$$
$$sin\theta = \dfrac{500nm}{0.001mm}=0.5$$
$$\theta=30^0$$
option $$C$$ is correct .
At what maximum width $$\delta_{max}$$, of the slit are the interference fringes on the screen observed still sharp?
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$$42\mu m$$
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$$36\mu m$$
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$$64\mu m$$
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none of these
Explanation
$$\displaystyle\frac{b\delta_{max}}{r}=\frac{\beta}{2}$$
$$\Rightarrow \delta_{max}=\displaystyle\frac{\beta r}{2b}=\displaystyle{10^{-3}\times 1.1\times (.1)}{2\times 1.3}=\displaystyle 42\mu m$$
The ratio of slit widths in Young's double slit experiment is $$4:9$$. The ratio of maximum and minimum intensities will be
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$$169:25$$
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$$81:16$$
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$$13:5$$
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$$25:1$$
Explanation
The intensity ratio will be same as slit width ratio
$$I_{1} : I_{2} = 4:9$$
$$\dfrac {I_{max}} {I_{min}} =( \dfrac {\sqrt {I_{2}} + \sqrt {I_{1}}} {\sqrt{I_2} - \sqrt{I_{1}}})^{2}$$ = $$(\dfrac {3+2} {3-2})^2$$ = 25:1
Answer. D) 25:1
Diffraction of sound is very easy, to observe in day-to-day life. This is not so with light. This is so because
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$$\lambda_S>\lambda_L$$
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$$\lambda_S<\lambda_L$$
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light waves are transverse and sound waves are longitudinal
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$$\lambda_S=\lambda_L$$
Explanation
A wavelength of light $$ { \lambda }_{ L }$$ is comparable to the size of the obstacles while the wavelength of sound $$ { \lambda }_{ S }$$ is much greater than that of light and henceforth to that of the obstacle.
An un-publicized beam of intensity $$2a^2$$ passes through a thin Polaroid. Assuming zero absorption in the Polaroid the intensity of emergent planes polarized light is
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$$2a^2$$
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$$a^2$$
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$$\sqrt2a^2$$
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$$\dfrac {\ a^2}{2}$$
Explanation
When an upolarized beam passes through a polaroid, the intensity of the beam halves. When this polarized beam passes through a polaroid again, the beam's intensity varies by Malus' Law.
Thus the intensity becomes $$\dfrac{2a^2}{2}=a^2$$
Two waves
$$y_1=A_1\sin(\omega t-\beta_1)$$ and
$$y_2=A_2\sin(\omega t-\beta_2)$$
superimpose to form a resultant wave whose amplitude is
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$$\displaystyle\sqrt{A^2_1+A^2_2+2A_1A_2\cos(\beta_1-\beta_2)}$$
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$$\displaystyle\sqrt{A^2_1+A^2_2+2A_1A_2\sin(\beta_1-\beta_2)}$$
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$$A_1+A_2$$
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$$|A_1+A_1|$$
Explanation
The phase difference the two waves $$y_1$$ and $$y_2$$, $$\delta= |\beta_1 - \beta _2|$$
Amplitude of the wave $$y_1$$ and $$y_2$$ is $$A_1$$ and $$A_2$$ respectively.
Thus resultant amplitude $$R= \sqrt{A_1^2 + A_2^2 + 2A_1A_2 cos\delta}$$
$$\implies R= \sqrt{A_1^2 + A_2^2 + 2A_1A_2 cos|\beta_1 - \beta_2|}$$
The shift of the interference pattern on the screen when the slit is displayed by $$Sl=1mm$$ along the arc of radius $$r$$ with centre at $$0$$.
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$$4mm$$
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$$6mm$$
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$$10mm$$
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$$13mm$$
Explanation
The mirror will rotate by $$\dfrac{\bigtriangleup l}{r}$$ or rotation of reflected ray $$=\dfrac{\bigtriangleup l}{r}$$ shift in fringe magnitude
$$=\displaystyle\frac{b\bigtriangleup l}{r}$$ $$=1.3\times (10)^2=13mm$$
A person wishes to distinguish between two pillars located at a distance of 11 km. What should be the minimum distance between these pillars (resolving power of normal human eye is 1')?
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1 m
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3.2 m
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0.5 m
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5 m
Explanation
Resolving power is given by the distance between two objects to be distinguished per unit distance of objects from the object distinguishing them.
Hence,$$\theta=\dfrac{d}{D}$$
Hence,$$d=\theta D=\dfrac{1}{60}\times \dfrac{\pi}{180}\times 110000=3.2m$$
Two identical lights sources $$S_1$$ and $$S_2$$ emit the light of same wavelength $$\lambda$$. These light rays will exhibit interference if
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their phase difference remain constant.
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their phase difference is distributed randomly.
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their light intensities remain constant.
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their light intensities change continuously.
Explanation
For interference to take place the light sources need to be either in phase or have a constant phase difference. In case the phase difference keeps changing the interference pattern will keep on changing, as a result of interference pattern will be observed.
Answer. A) Thier phase difference remain constant.
What happens to the interference pattern if the two slits in Young's experiment are illuminated by two independent sources such as two sodium lamps $$S$$ and $$S'$$ as shown in figure.
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Two sets of interference fringes overlap.
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No fringes are observed.
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The intensity of the bright fringes is doubled.
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The intensity of the bright fringes becomes four times.
Explanation
No interference pattern i.e., no fringes would be observed because the two sources are non-coherent.
If one of the two slits of Young's double-slit experiment is painted so that it transmits half the light intensity as the second slit, then
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fringe system will altogether disappear
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bright fringes will become brighter and the dark fringes will become darker
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both dark and bright fringes will become darker
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dark fringes will become less dark and bright fringes will become less bright.
Explanation
as the intensity from second slit decreases the bright fringe becomes darker due to constructive interference and dark fringe becomes bright due to destructive interference
option $$D$$ is correct
Light waves travel in vacuum along the y-axis. Which of the following may represent the wavefront?
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x$$=$$constant
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y$$=$$constant
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z$$=$$constant
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x+y+z$$=$$constant
Explanation
SYNOPSIS WAVE FRONT:
A continuous locus of all the points which are ; in the same phase or state of vibration, is called wavefront. A Point source of light produces a spherical wavefront. A linear source of light produces a cylindrical wave.
If the wave is traveling in vacuum along the y-axis, the wavefront can be a surface perpendicular to the direction of the wave i.e. x-z plane which is $$y=constant$$.
A monochromatic beam of light falls on Young's double slit experiment apparatus as shown in figure. A thin sheet of glass is inserted .in front of lower slit $$S_{2}$$.
The central bright fringe can be obtained :
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At $$O$$
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Above $$O$$
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Below $$O$$
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Anywhere depending on angle $$\theta$$, thickness of plate $$t$$, and refractive index of glass $$\mu$$
Explanation
Because of beam inclination means light reaches $$ { S }_{ 1 }$$ before it reaches $$ { S }_{ 2 }$$, hence an induced time lag before the slits.
It means that, without any glass sheet, the central fringe would be above point O because of this inclination
But due to presence of glass sheet, it can be anywhere depending on angle θ, thickness of plate t, and refractive index of glass μ
A beam of light of wavelength $$600\ nm$$ from a distance source falls on a single slit $$1\ mm$$ wide and a resulting diffraction pattern is observed on a screen $$2\ m$$ away. The distance between the first dark fringes on either side of central bright fringe is
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$$1.2 cm$$
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$$1.2 mm$$
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$$2.4 cm$$
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$$2.4 mm$$
Explanation
for a dark fringe to form
$$\dfrac{dy}{D}=\lambda$$
$$y=\dfrac{D\lambda}{d}=\dfrac{2\times 600 \times 10^{-9}}{10^{-3}}=1.2mm$$
distance between the first dark fringes on either side of central bright fringe is $$2\times y=2.4mm$$
option $$D$$ is correct
In a YDSE with identical slits, the intensity of the central bright fringe is $${I}_{0}$$.If one of the slits is covered, the intensity at the same point is
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$$2{I}_{0}$$
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$$ { I }_{ 0 }$$
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$${ I }_{ 0 }/2$$
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$${ { I }_{ 0 } }/{ 4 } $$
Explanation
Let intensity due to ea
ch slit on the central bright fringe be $$I_1$$.
Thus intensity at the point of bright fringe due to interference of the light from two slits is $$I_0=I_1+I_1+2\sqrt{I_1.I_1}\cos\phi$$
For central bright fringe, $$\phi=0$$
$$\implies I_0=4I_1$$
$$\implies I_1=\dfrac{I_0}{4}$$
The wavefront of a light beam is given by the equation $$x+2y+3z=c$$ (where $$c$$ is arbitrary constant of light). What is the angle made by the light with the y-axis is
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$$\cos ^{ -1 }{ \cfrac { 1 }{ \sqrt { 14 } } } $$
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$$\sin ^{ -1 }{ \cfrac { 2 }{ \sqrt { 14 } } } $$
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$$\cos ^{ -1 }{ \cfrac { 2 }{ \sqrt { 14 } } } $$
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$$\sin ^{ -1 }{ \cfrac { 3 }{ \sqrt { 14 } } } $$
Explanation
the wavefront equations must be of form $$ax+by+cz=1$$ where $$a^{2}+b^{2}+c^{2}=1$$ (a,b,c are cosines of angles made with x,y,z axes respectively)
applying this to the given equation ,
$$1^{2}+2^{2}+3^{2}=c^{2}$$
$$c^{2}=14$$
$$c=\sqrt{14}$$
angle made with y-axis is $$cos ^{-1}\dfrac{2}{\sqrt{14}}$$
option $$C$$ is correct.
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Practice Class 12 Medical Physics Quiz Questions and Answers
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