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CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 15 - MCQExams.com
CBSE
Class 12 Medical Physics
Wave Optics
Quiz 15
In a Young's double slit experiment sources of equal intensities are used. Distance between slits is $$d$$ and wavelength of light used is $$\lambda (\lambda <<d)$$, Angular separation of nearest points on either side of central maximum where intensities become half of the maximum value is
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$$\dfrac{\lambda}{d}$$
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$$\dfrac{\lambda}{2d}$$
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$$\dfrac{\lambda}{4d}$$
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$$\dfrac{\lambda}{6d}$$
The YDSE apparatus is as shown in figure below. The condition for point $$P$$ to be a dark fringe is $$(\lambda =$$ wavelength of light waves) :
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$$(l_{1} - l_{3}) + (l_{2} - l_{4}) = n\lambda$$
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$$(l_{1} - l_{2}) + (l_{3} - l_{4}) = n\lambda$$
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$$(l_{1} + l_{3}) - (l_{2} + l_{4}) = \dfrac {(2n - 1)\lambda}{2}$$
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$$(l_{1} + l_{2}) - (l_{2} + l_{3}) = \dfrac {(2n - 1)\lambda}{2}$$
Unpolarised light beam of intensity $$I_{0}$$ is incident on polaroid $$P_{1}$$. The three polaroids are arranged in such a way that transmission axis of $$P_{1}$$ and $$P_{3}$$ are perpendicular to each other. Angle between the transmission axis of $$P_{2}$$ and $$P_{3}$$ is $$60^{\circ}$$. The intensity of the beam coming out from $$P_{3}$$ will be
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$$\dfrac {I_{0}}{2}$$
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$$\dfrac {3I_{0}}{8}$$
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$$\dfrac {3I_{0}}{32}$$
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$$\dfrac {3I_{0}}{64}$$
Explanation
$$I_{2} = I_{1}\cos^{2} 30^{\circ} = \dfrac {I_{0}}{2}\times \dfrac {3}{4} = \dfrac {3I_{0}}{8}$$
$$\therefore I_{3} = I_{2}\cos^{2} 60^{\circ} = \dfrac {3I_{0}}{8}\times \dfrac {1}{4} = \dfrac {3I_{0}}{32}$$.
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 $$S_{2}$$. If central bright fringe is obtained on screen at $$O$$.
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$$(\mu - 1)t = d\sin \theta$$
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$$(\mu - 1) t = d\cos \theta$$
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$$(\mu - 1)t + d\sin \theta = 0$$
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$$\dfrac {t}{\mu - 1} = \dfrac {d}{\sin \theta}$$
If the plate is heated so that it temperature rises by $$10^{\circ}C$$, then how many fringes will cross a particular point on the screen? (Neglect the thermal expansion of plate):
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$$1000$$
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$$10^{-4}$$
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$$5000$$
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$$\dfrac {1}{3000}$$
One of the two slits in YDSE is painted over, so that it transmits only light waves having intensity half of the intensity of the light waves through the other slit. As a result of this
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fringe pattern disappears
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bright fringes become brighten and dark one becomes darker
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dark and bright fringes get fainter
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dark fringes get brighter and bright fringes get darker
Explanation
We know that,
$$I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}cos\triangle\phi$$
Here $$I_{1}=I_{0}$$
$$I_{2}=I_{0}/2$$
For maximum intensity,$$cos \triangle \phi=1$$
For minimum intensity, $$cos \triangle \phi=-1$$
Before painting,
$$I=I_{2}=I_{0}, I_{max}=4I_{0}, I_{min}=0$$
After painting,
$$I_{1}=I_{0},I_{max}=\left(\dfrac{3+2\sqrt{2}}{2}\right)I_{0} < 4I_{0}$$
$$I_{2}=\dfrac{I_{0}}{2}, I_{min}= \left(\dfrac{3-2\sqrt{2}}{2}\right)I_{0} > 0$$
Hence,
dark fringes get brighter and bright fringes get darker.
The maximum intensity in Young's double-slit experiment is $$I_{0}$$. Distance between the slits is $$d=5\lambda$$, where $$\lambda$$ is the wavelength of monochromatic light is used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $$D=d$$ ?
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$$\dfrac{I_{0}}{2}$$
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$$\dfrac{3}{4}I_{0}$$
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$$I_{0}$$
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$$\dfrac{I_{0}}{4}$$
Explanation
Path difference, $$\triangle x= \dfrac{yd}{D}$$
Here, $$y = \dfrac {5\lambda}{2}$$
and $$D= 10d = 50 \lambda$$ $$(as \quad d \quad = \quad 5\lambda)$$
So,
$$\triangle x= \left(\dfrac{5\lambda}{2}\right)\left(\dfrac{5\lambda}{50\lambda}\right)= \dfrac{\lambda}{4} $$
Corresponding phase difference will be
$$\phi= \left(\dfrac{2\pi}{\lambda}\right)(\triangle x)= \left(\dfrac{2\pi}{\pi}\right)\left(\dfrac{\lambda}{4}\right)= \dfrac{\pi}{2}$$
or $$\dfrac{\phi}{2}=\dfrac{\pi}{4}$$
$$ \therefore I= I_{0}cos^{2}\left(\dfrac{\phi}{2}\right)$$
$$=I_{0}cos^{2}\left(\dfrac{\pi}{4}\right)= \dfrac{I_{0}}{2}$$
The path difference between two interfering waves at a point on the screen is $$\lambda/6$$. The ratio of intensity at this point and that at the central bright fringe will be (assume that intensity due to each slit in same)
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$$0.853$$
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$$8.53$$
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$$0.75$$
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$$7.5$$
Explanation
We know that at path difference = $$\dfrac{\lambda}{6}$$, phase difference = $$\dfrac{\pi}{3}$$
and, $$I=I_{0}+I_{0}+2I_{0}cos\dfrac{\lambda}{3}= 3I_{0}$$
So, the required ratio is $$\dfrac{3I_{0}}{4I_{0}}=0.75$$
In YDSE, if a bichromatic light having wavelengths $$\lambda_{1}$$ and $$\lambda_{2}$$ is used, then the maxima due to both lights will overlap at a certain distance $$y$$ of from the central maxima. Take separation between slits as $$d$$ and distance between screen and slit as $$D$$. Then, the value $$y$$ will be
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$$\left({\lambda_{1}+\lambda_{2}}{2D}\right)d$$
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$$\dfrac{\lambda_{1}-\lambda_{2}}{2D} \times 2d$$
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LCM of $$\dfrac{\lambda_{1}D}{d}$$ and $$\dfrac{\lambda_{2}D}{d}$$
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HCF of $$\dfrac{\lambda_{1}D}{d}$$ and $$\dfrac{\lambda_{2}D}{d}$$
Explanation
We know that the distance of nth maxima from the central maxima is given by:-
$$y = \dfrac{n \lambda D}{d}$$
Let $$n_{1}^{th}$$ maxima corresponding to $$\lambda_{1}$$ wavelength be overlapping with $$n_{2}^{th}$$ maxima corresponding to $$\lambda_{2}$$ wavelength. Then the required distance,
$$y=\dfrac{n_{1}\lambda_{1}D}{d}=\dfrac{n_{2}\lambda_{2}D}{d}$$
Here, $$n_1$$ and $$n_2$$ are changing while $$\dfrac{ \lambda_1 D }{d} $$ and
$$\dfrac{ \lambda_2 D }{d} $$ are constant.
Hence,
$$y$$=LCM of $$\dfrac{\lambda_{1}D}{d}$$ and $$\dfrac{\lambda_{2}D}{d}$$
When the object is self-luminous, the resolving power of a microscope is given by the expression
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$$\dfrac{2\mu \sin \theta }{1.22\lambda}$$
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$$\dfrac{\mu \sin \theta }{1\lambda}$$
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$$\dfrac{2\mu \cos \theta }{1.22}$$
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$$\dfrac{2\mu }{\lambda}$$
Coherent sources are characterized by the same
[KCET 1993]
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Phase and phase velocity
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Wavelenght, amplitude and phase velocity
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Wavelenght, amplitude and frequency
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Wavelenght and phase
In a Young's double slit experiment with a source of light of wavelength $$6320\overset{o}{A}$$, the first minima will occur when
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Path difference is $$9480\overset{o}{A}$$
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Phase difference is $$2\pi $$ radian
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Path difference is $$6320\overset{o}{A}$$
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Phase difference is $$\pi $$ radian
Explanation
For maxima path difference $$\Delta = n\lambda$$
So for $$n=1, \Delta =\lambda =6320\overset{o}{A}$$
Now, $$\triangle\phi$$=$$\dfrac{\triangle(x)×2\pi}{\lambda}$$
So, $$\triangle\phi=2\pi$$
In Young's double slit experiment the source $$S$$ and the two slits $$A$$ and $$B$$ are vertical with slit $$A$$ above slit $$B$$. The fringes are observed on a vertical screen $$K$$. The optical path length from $$S$$ to $$B$$ is increased very slightly ( by introducing a transparent material of higher refractive index) and the optical path length from $$S$$ to $$A$$ is not changed, as a result the fringe system on $$K$$ moves
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Vertically downward slightly
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Vertically upwards slightly
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Horizontally, slightly to the left
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Horizontally, slightly to the right
A $$20\ cm$$ length of a certain solution causes right-handed rotation of $$38^{o}$$. A $$30\ cm$$ lemgth of another solution causes left-handed rotation of $$24^{o}$$. The optical rotation caused by $$30\ cm$$ length of a mixture of the above solution is the volume ration $$1:2$$ is
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Left handed rotation of $$14^{o}$$
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Right handed rotation of $$14^{o}$$
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Left handed rotation of $$3^{o}$$
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Right handed rotation of $$3^{o}$$
Explanation
As $$\theta \propto l$$
Volume ratio $$1:2$$ in a tube of length $$30\ cm$$ means $$10\ cm$$ length of first solution and $$20\ cm$$ length of second solution.
Rotation produced by $$10\ cm$$ length of first solution
$$\theta_{1}=\dfrac{38^{0}}{20}\times 10=19^{0}$$
Rotation produced by $$20\ cm$$ length of second solution
$$\theta_{2}=-\dfrac{24^{0}}{30}\times 20=-16^{0}$$
$$\therefore $$ Total rotation produced $$=19^{0}-16^{0}=3$$
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