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CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 8 - MCQExams.com
CBSE
Class 12 Medical Physics
Wave Optics
Quiz 8
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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$$A_1+A_2$$
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$$|A_1+A_2|$$
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$$\sqrt{A^2_1+A^2_2-2A_1A_2 \sin ({\beta}_1 - {\beta}_2)}$$
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$$\sqrt{A^2_1+A^2_2+2A_1A_2 \cos ({\beta}_1 - {\beta}_2)}$$
Explanation
Given,Equation of waves are,
$$y_1=A_1 \sin (\omega t-{\beta}_1)$$
$$y_2=A_2 \sin (\omega t-{\beta}_2)$$
When two waves superimpose then,
$$Y=Y_1+Y_2$$
Amplitudes $$A_1$$ and $$A_2$$ are added as vectors. Angle between the two vectors $$=$$ the phase difference $$({\beta}_1-{\beta}_2)$$ between them.
$$\therefore$$ Resultant wave,
$$R=\sqrt{A^2_1+A^2_2+2A_1A_2\cos ({\beta}_1-{\beta}_2)}$$
A diffraction pattern is obtained using a beam of red light. What happens if the red light is replaced by blue light?
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Bands disappear
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No change
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Diffraction pattern becomes narrower and crowded together
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Diffraction pattern becomes broader and further apart
Explanation
The width of the fringe in diffraction pattern is given as $$\dfrac{D\lambda}{d}$$
Hence when the red light is replaced by blue light, the wavelength decreases which means that the fringe width decreases and pattern becomes narrower and crowded together.
In Fraunhofer diffraction pattern, slit width is $$0.2\ mm$$ and screen is at $$2\ m$$ away from the lens. If wavelength of light used in $$5000\overset {\circ}{A}$$ then the distance between the first minimum on either side of the central maximum is $$(\theta$$ is small and measure in radian):
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$$10^{-1}m$$
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$$10^{-2}m$$
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$$2\times 10^{-2}m$$
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$$2\times 10^{-1}m$$
Explanation
Distance between the first 2 minimum on other side $$=\dfrac{2\lambda D}{d}$$
$$=\dfrac{2\times 5\times 10^3\times 10^{-10}\times 2}{2\times 10^{-1}\times 10^{-3}}$$
$$=10\times \dfrac{10^{-7}}{10^{-4}}$$
$$=10^{-2}m$$
Newton's ring pattern in reflected system, viewed under white light consists of
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Equally spaced bright and dark bands with central dark spot
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Equally spaced bright and dark bands with central white spot
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A few coloured rings with central dark spot
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A few coloured rings with central white spot
Explanation
Rings are fringes of equal thickness. They are observed when light is reflected from a plano-convex lens of a long focal length placed in contact with a plane glass plate. A thin air film is formed between the plate and the lens. The thickness of the air film varies from zero at the point of contact to some value t. If the lens plate system is illuminated with monochromatic light falling on it normally, concentric bright and dark interference rings are observed in reflected light. These circular fringes were discovered by Newton and are called Newton’s rings. Those rings are formed with equally spaced bright and dark bands with central spot as a dark spot. Hence A is the correct answer.
An unpolarised beam of intensity $${ I }_{ 0 }$$ falls on a polaroid at an angle of $$45^0$$. The intensity of the emergent light is
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$$\dfrac { { I }_{ 0 } }{ 2 } $$
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$${ I }_{ 0 }$$
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$$\dfrac { { I }_{ 0 } }{ 4 } $$
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Zero
Explanation
$$I=I_ocos^2\omega t$$
As the angle of incidence is $$45^0$$ so $$cos^245^0= 1/2$$
Hence $$I_{av}=\dfrac{I_o}{2}$$
The ratio of resolving powers of an optical microscope for two wavelengths $$\lambda_1=4000\mathring{A}$$ and $$\lambda_2=6000\mathring{A}$$ is:
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$$16:18$$
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$$8:27$$
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$$9:4$$
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$$3:2$$
Explanation
Resolving power $$ \propto \dfrac{1}{\lambda}$$
$$\dfrac{RP_1}{RP_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{6000}{4000}$$
$$\dfrac{RP_1}{RP_2}=\dfrac{3}{2}$$
Resolving power of telescope increases when
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wavelength of light decreases
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wavelength of light increasing
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focal length of eye-piece increases
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focal length of eye-piece decreases
Explanation
Resolving power of telescope RP$$=\dfrac{a}{1.22 \lambda}$$
where $$a$$ is the diameter of objective lens and $$\lambda$$ is the wavelength of light.
$$\implies$$ RP $$\propto \dfrac{1}{\lambda}$$
Thus resolving power of telescope increases when wavelength of light is decreased.
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength $$400 nm$$, the first minimum is formed at an angle of $${ 30 }^{ o }$$. The direction $$\theta $$ of the first secondary maximum is given by
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$$\sin ^{ -1 }{ \dfrac { 2 }{ 3 } } $$
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$$\sin ^{ -1 }{ \dfrac { 3 }{ 4 } } $$
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$$\sin ^{ -1 }{ \dfrac { 1 }{ 4 } } $$
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$$\tan ^{ -1 }{ \dfrac { 2 }{ 3 } } $$
Explanation
Given : $$\lambda =400 nm$$
For minima in diffraction $$b \sin\theta = n\lambda$$
First minima i.e. $$n=1$$ is obtained at $$\theta = 30^o$$
$$\therefore$$ $$b \sin 30^o = 1 \times 400$$
$$\implies$$ $$b = 800 nm$$
For maxima in diffraction $$b \sin\theta = (n+\dfrac{1}{2})\lambda$$
Let first maxima i.e. $$n=0$$ is obtained at $$\theta$$.
$$\therefore$$ $$b \sin \theta = \dfrac{\lambda}{2}$$
Or
$$800 \times \sin \theta = \dfrac{400}{2}$$
Or $$\sin \theta = \dfrac{1}{4}$$
$$\implies$$ $$\theta = sin^{-1} \bigg(\dfrac{1}{4} \bigg)$$
It is difficult to observe diffraction in case of light waves, because
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Light waves can travel through vacuum
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Speed of light is more
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Light waves are transverse in nature
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Wavelength of light is small
Explanation
Because the wavelength of light wave is very small.
In an electron microscope if the potential is increased from $$20$$kV to $$80$$kV, the resolving power R of the microscope will become :
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$$\displaystyle\frac{R}{2}$$
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$$2R$$
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$$4R$$
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$$5R$$
Explanation
(b) Resolving power $$\propto \displaystyle\frac{1}{\lambda}$$ and $$\propto \sqrt{v}$$
$$\therefore R.P\propto \sqrt{v}$$
$$\Rightarrow (R.P)_2=(R.P)_1\times \sqrt{\displaystyle\frac{80kV}{20kV}}$$
$$=2(R.P)_1$$.
A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength $$6500\overset {\circ}{A}$$?
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$$4400\overset {\circ}{A}$$
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$$4100\overset {\circ}{A}$$
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$$4642.8\overset {\circ}{A}$$
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$$9100\overset {\circ}{A}$$
Explanation
$$x = \dfrac {(2n + 1)\lambda D}{2a}$$
For red light $$x = \dfrac {(4 + 1)D}{2a}\times 6500$$
For unknown wavelength of light,
$$x = \dfrac {(6 + 1)D}{2a}\times \lambda$$
Accordingly,
$$\therefore 5\times 6500 = 7\times \lambda$$
$$\Rightarrow \lambda = \dfrac {5}{7}\times 6500$$
$$= 4642.8\overset {\circ}{A}$$.
Light of wavelength $$5000\ \mathring A $$ is incident normally on a slit of width $$2.5 \times 10^{-4}\ cm $$. The angular position of second minimum from the central maximum is
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$$\sin^{-1}\left ( \dfrac {1}{5} \right )$$
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$$\sin^{-1}\left ( \dfrac {2}{5} \right )$$
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$$\left ( \dfrac {\pi}{3} \right )$$
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$$\left ( \dfrac {\pi}{6} \right )$$
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$$\left ( \dfrac {\pi}{4} \right )$$
Explanation
We know that,
Angular width of central maximum $$=\dfrac {2\lambda}{a}$$
where, $$\lambda$$ =wavelength of light,
a =width of single slit,
So, $$sin \theta =\dfrac {2\lambda}{a}$$
where $$\lambda = 5000 \mathring A = 5000 \times 10^{-10} m$$
$$a= 2.5 \times 10^{-6}\, m$$$$\Rightarrow sin \theta =\frac {2 \times 5000 \times 10^{-10}}{2.5 \times 10^{-6}}$$
$$= sin \theta =\dfrac {10000 \times 10^{-10}}{2.5 \times 10^{-6}}$$
$$= sin \theta =\dfrac {10}{25}$$
$$= sin \theta =\dfrac {2}{5}$$
$$\theta=sin^{-1}\left ( \dfrac {2}{5} \right )$$
A single slit Fraunhoffer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500 A?
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$$ 9100 A^{\circ}$$
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$$4642 A^{\circ}$$
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$$4100 A^{\circ}$$
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$$4400 A^{\circ}$$
Explanation
$$\because x=\dfrac{2n+1}{2a}\lambda D$$
For red light, $$x=\dfrac{(4+1)}{2a}\times 6500 A^{\circ}\times D$$
For other light, $$x=\dfrac{(6+1)}{2a}d\times \lambda $$
$$\therefore $$ x is same for each$$\therefore 5 \times 6500=7 \lambda$$
$$\lambda =\dfrac{5}{7}\times 6500$$
$$4642 A^{\circ}$$
Two plane wavefronts of light, one incident on a thin convex lens and another on the refracting face of a thin prism. After refraction at them, the emerging wavefronts respectively become
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plane wavefront and plane wavefront
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plane wavefront and spherical wavefront
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spherical wavefront and plane wavefront
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spherical wavefront and spherical wavefront
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elliptical wavefront and spherical wavefront
Explanation
As the wave hits the lens it is the centre of the wave that meets the glass first and so this part of the wave is slowed down first (light waves move slower in glass than they do in air). This means that the outer portions of the wave 'catch up' so increasing the curvature to form a converging beam.
As the wave leaves the lens the outer portions move into the air first and so speed up first. This means that the outer portions move off more rapidly first and so the curvature of the wave is further increased so converging the light more strongly.
Hence, the emerging wavefront is spherical wavefront.
As the wave hits the prism, it is the lower half of the wave that meets the glass first and so, this part of the wave is slowed down first. This means that the upper part is moving fast and thus, the wavefront bends as it enters the prism and so, we get a plane wavefront which is slanted.
Hence, the emerging wavefront is plane wavefront.
Therefore, the answer is OPTION C.
A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 pm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will :
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Remain unstated
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Shift downward by neary two fringes
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Shift upward by nearly two fringes
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Shift downward by its fringes
Explanation
In Young's Double Slit Experiment (YDSE) be the position of central bright fringe on the screen in absence of any film, because the $$S_1O =S_2O$$.
When we introduce a film of thickness t and refractive index p, an additional path difference equal to $$(\mu t-t)=(\mu-1)t$$ is introduced. The optical path of upper bean becomes longer. For path difference on screen to be zero, path from lower slit $$S_2$$ should also be more. Thus, central bright fringe wire be located some what at O' as $$S_2O' > S_2O$$.
$$\therefore$$ The fringe pattern shifts upwards.
Now as a change in path difference of $$\lambda$$ corresponds to a change in position on the screen by $$\beta=\frac {D}{d} \lambda$$
Change in optical path difference $$\Delta y = (\mu - 1)t \times \frac {D}{d}$$
$$\Delta y= (1.5-1) ( 2\times 10^{-6}) \frac {D}{d}$$
$$\Delta y=\frac{1}{2} \times 2 \times 10^{-6} \frac {D}{d}=\frac {D}{d}\times 10^{-6} \, m$$
Fringe width $$=\frac {D}{d}\lambda =\frac {D}{d}\times 100 \times 10^{-9}m=\beta $$
Clearly, $$\frac{\Delta y}{\beta}=\frac{\frac {D}{d}\times 10^{-6}}{\frac {D}{d}\times 500 \times 10^{-9}}=2$$
$$= \frac{10^{-6}}{500 \times 10^{-9}}=\frac{10^{-6}\times 10^9}{500}=\frac{10^3}{500}$$
$$\therefore \Delta y= 2 \beta$$
The condition for obtaining secondary maxima in the diffraction pattern due to single slit is :
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$$ a \sin \theta = \dfrac {n \lambda}{2} $$
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$$ a \sin \theta = (2n - 1) \lambda / 2 $$
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$$ a \sin \theta = n \lambda $$
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$$ a \sin \theta = ( 2n -1) \lambda $$
Explanation
$$ a \sin \theta = (2n - 1) \lambda / 2 $$ is the condition for obtaining secondary maxima in the diffraction pattern due to single slit.
The resolving power of a microscope is
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Inversely proportional to numerical aperture
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Directly proportional to wavelength
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Directly proportional to square of the wavelength
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Directly proportional to numerical aperture
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Independent of numerical aperture
Explanation
Resolving power of a microscope $$=\dfrac { 2\mu \sin { \theta } }{ 1.22\lambda } $$
i.e., resolving power of a microscope is directly proportional to numerical aperture.
A slit of width a is illuminated by white light. For red light ($$\displaystyle \lambda =6200\overset { \circ }{ A } $$), the first minima is obtained at a diffraction angle of $$\displaystyle { 30 }^{ \circ }$$. then the value of a is
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$$\displaystyle 3250\overset { \circ }{ A } $$
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$$\displaystyle 6.5\times { 10 }^{ -4 }mm$$
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$$\displaystyle 1.24$$ micron
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$$\displaystyle 2.6\times { 10 }^{ -4 }cm$$
Explanation
For first minima
$$\displaystyle a\sin { \theta } =\lambda $$
$$\displaystyle \therefore \quad a=\frac { \lambda }{ { \sin { \theta } }_{ 1 } } =\frac { 6200\times { 10 }^{ -10 } }{ { \sin { 30 } }^{ \circ } } =1.24\times { 10 }^{ -6 }m$$
$$\displaystyle =1.24u$$
The condition for diffraction of $$mth$$ order minima is
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$$d\sin \theta_{m} = m\lambda, m = 1, 2, 3, ....$$
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$$d\sin \theta_{m} = \dfrac {m\lambda,}{2} m = 1, 2, 3, ....$$
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$$d\sin \theta_{m} = (m + 1) \dfrac {\lambda}{2}, m = 1, 2, 3, ....$$
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$$d\sin \theta_{m} = (m - 1) \dfrac {\lambda}{2}, m = 1, 2, 3, ....$$
Explanation
For obtaining $$mth$$ secondary minima at a point on screen, path difference between the diffracted waves, $$\triangle = d\sin \theta_{m} = \pm m\lambda$$ where, $$m = 1, 2, 3, .....$$
In Young's double slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength $$\lambda$$. In another experiment with the same set up the two slits are same of equal amplitude of wavelength $$\lambda$$ but are incoherent. The ratio of intensity of light at the mid point of the screen in the first to the second case is?
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$$4:1$$
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$$2:1$$
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$$1:1$$
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$$1:2$$
Explanation
When sources are coherent, interference will take place and the intensity at mid - point of the screen will be $$ (a+a)^{2} = 4A^{2}$$. In case of incoherent sources, no interferencer will taake places and the intensity at mid - point will be $$A^{2}+A^{2}=2A$$. Hence the required ratio wil be $$2 : 1 $$
What will be the angle of diffraction for the first order maximum due to Fraunhofer diffraction by a single slit of width $$0.50\ mm$$, using light of wavelength $$500\ nm$$?
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$$1\times 10^{-3}rad$$
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$$3\times 10^{-3}rad$$
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$$1.5\times 10^{-4}rad$$
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$$1.5\times 10^{-3}rad$$
Explanation
Using $$d\sin \theta = n\lambda$$, for $$n = 1$$
$$\sin \theta = \dfrac {\lambda}{d} = \dfrac {500\times 10^{-9}}{0.5\times 10^{-3}} = 10^{-3} rad$$.
When a red glass is heated in dark room it will seem.
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Black
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Green
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Yellow
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Red
Consider superposition of waves coming from three light source (slits) $$A,B,C$$ as shown in the figure. Given that
$$B{ P }_{ 0 }-A{ P }_{ 0 }=\cfrac { \lambda }{ 3 } ;d=\sqrt { \cfrac { 2\lambda D }{ 3 } } $$
the ratio of intensity at $${P}_{0}$$ compared to intensity due to individual slit is
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$$1.5$$
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$$2$$
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$$2.5$$
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$$3$$
If numerical aperture of a microscope is increased then its
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resolving power remains constant
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resolving power becomes zero
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limit of resolution is decreased
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limit of resolution is increased
Explanation
On increasing the numerical aperture of the microscope,more light rays are gathered that were diffracted at larger angles and hence its limit of resolution increases.
Hence, answer is option-(D).
If the wavelength of light used is $$6000\mathring { A } $$. The angular resolution of telescope of objective lens having diameter $$10cm$$ is ______ rad
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$$7.52\times { 10 }^{ -6 }$$
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$$6.10\times { 10 }^{ -6 }$$
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$$6.55\times { 10 }^{ -6 }$$
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$$7.32\times { 10 }^{ -6 }$$
Explanation
Limit of resolution $$\sin { \theta } =\theta =\cfrac { 1.22\lambda }{ D } $$
putting the values
$$\theta=\dfrac{1.22\times6000\times10^{-10}}{0.1}$$
$$\theta=7.32\times10^{-6}$$
Option (D) is correct.
In which of the following cases do we obtain a plane wave front?
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Light emitted by a point source in an isotropic medium
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Light emerging from a convex lens when a point source is placed at its focus
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Light of the sun reaching the earth
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Light diverging from a slit.
Explanation
Isotropic medium is medium which doesn't depend on direction of light.
In this medium speed of light is constant and wave travel from source to infinity than it make plane wave front.
In option A given light emitted by the point source in an isotropic medium that means direction of light is constant and it can travel for a long distance. If light will travel for long distance than it will form plane wave front.
Hence,
correct option is A.
Two coherent point sources of sound wave $$S_1$$ and $$S_2$$ produce sound of same frequency 50 Hz and wavelength 2 cm with amplitude 2 x $$(10)^-$$$$^3$$ m. Each circular arc represents a wavefront at a particular time and is separated from next arc by a distance 1 cm. Both the sound waves propagate through the medium and interfere with each other. Read paragraph carefully and answer the following questions. [r = 1 cm]
The point (s) where constructive interference occurs
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G only
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P and A
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G and F
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T and U
In Young's double slit experiment shows in figure, $$S_1$$ and $$S_2$$ are coherent sources and S is the screen having a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits. Which wavelength passing through the hole has the strongest intensity?
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400 nm
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700 nm
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500 nm
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667 nm
Explanation
From youngs double slit experiment we have
path difference $$(\Delta x)=\dfrac{y\ d}{D}$$
$$y=1\ mm, d=0.5\ mm\ \ \ D=50\ cm =500\ mm$$
So,
$$\Delta x=\dfrac{1\times 0.5}{500}\ mm$$
$$\Delta x=\dfrac{0.5}{500}\times 10^{6}\ nm$$
$$\Delta x=1000\ nm$$
For the strongest intensity at $$H$$ there should be constructive interference
$$\Rightarrow \Delta x=n\lambda$$
where
$$\lambda \Rightarrow$$ wavelength of light
$$n\rightarrow$$ integer $$(1, 2, 3,....)$$
$$1000=n\lambda$$
For $$n=1$$
$$\lambda =1000\ nm$$
For $$n=2$$
$$\lambda=\dfrac{1000}{2}=500\ nm$$
For $$n=3$$
$$\lambda=\dfrac{1000}{3}=333.3\ nm$$
So,
For the visible light $$\lambda =500\ nm$$
Direction :
The question has a paragraph followed by two statements, Statement-1 and Statement-Of the given four alternatives after the statements, choose the one that describes the statements.
A thin air film is formed by putting the convex surface of a piano-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.
Statement 1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of $$\pi$$.
Statement 2 : The centre of the interference pattern is dark.
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Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1.
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Both statements 1 and 2 are true but statement 2 is not the correct explanation of statement 1.
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Statement 1 is true but statement 2 is false.
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Both statements 1 and 2 are false.
Explanation
When the light is reflected from a denser medium, a phase change of $$\pi$$ occurs in the reflected wave. Hence statement-1 is true.
Since the phase difference between the two reflected waves at the centre of interference pattern is $$\pi$$. Hence the centre of the interference pattern is dark. Hence statement-2 is true.
The limit of resolution of an optical instrument is the smallest angle that two points on an object have to subtend at the eye so that they are.
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Unresolved
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Well resolved
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Just resolved
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None of these
Explanation
Limit of resolution of an optical instrument is
the minimum angle
that two points on an object have to subtend at the eye so that they are just resolved. (C)
What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident on the film is $$\text{750 nm}$$? Assume that the refractive index for the film is $$\mu \, = \, 1.33.$$
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$$\text{282 nm}$$
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$$\text{70.5 nm}$$
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$$\text{141 nm}$$
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$$\text{387 nm}$$
Explanation
Given:
The wavelength of the light incident on the soap film is $$750\ nm$$.
The refractive index of the film is $$1.33$$.
To find:
The minimum thickness of the soap film required for constructive interference.
The thickness of the film for the constructive interference is given by:
$$2t=\left(m+\dfrac12\right)\lambda'$$
The new wavelength in the medium is given as:
$$\lambda'=\dfrac{\lambda}{\mu}$$
So, the minimum thickness of the film $$'t'$$ will be at $$m=0$$ :
$$2t_{min}=\left(0+\dfrac12\right)\dfrac{750\times10^{-9}}{1.33}$$
$$t_{min}=\dfrac{750\times10^{-9}}{4\times1.33}$$
$$=1.41\times10^{-7}\ or\ 141\ nm$$
Two light waves superimposing at the mid - point of the screen are coming from coherent sources of light with phase difference $$3\pi$$ rad . Their amplitude at the given 1 cm each . The resultant amplitude at the given point will be,
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5m
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3m
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2m
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zero
Explanation
Resultant amplitude,
A= $$\sqrt{A_{1}^{2}+A_{2}^{2}+2A_1A_2\ cos\ \phi}$$
Here, $$A_1=A_2=1cm, ϕ=3π \ rad, ϕ=3π\ rad $$
$$A_R=0$$
Coherent sources for studies in interference of light are obtained from.
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Two sources derived from a single source of light having a constant phase difference
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Two independent sources of light having a varying phase difference
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Two independent sources of light having a constant phase difference
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None of the above
Consider sunlight incident on a slit of width $$104\ A^o$$. The image seen through the slit shall
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be a fine sharp slit white in colour at the Centre
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a bright slit white at the centre diffusing to zero intensities at the edges
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a bright slit white at the centre diffusing to regions of different colours
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only be a diffused slit white in colour
Explanation
The wavelength of the visible light lies in the range $$4000\ A^o-7000\ A^o $$ whereas the width of the slit is about $$104\ A^o$$. Since the width of the slit is comparable to that of wavelength. Diffraction occurs with maxima at centre.
The maxima consist of all the colour of lights present in the wave. Therefore, At the centre all colours appear i.e, mixing of colours from the white patch at the centre
Choose the Correct answer from alternative given.
The idea of secondary wavelets for the propagation of a wave was first given by:
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Newton
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Huygens
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Maxwell
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Fresnel
Explanation
According to the Huygens theory, all the points on a wavefront act as source for a secondary wavefront or a wavelet which spread outward from the point and move at the speed of light.
In the case of the waves from two coherent sources $$S_1$$ and $$S_2$$ , there will be constructive interference at an arbitrary point P, the path difference $$S_1P - S_2P$$ is then
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$$[ n + \dfrac{1}{2}] \lambda$$
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$$ n \lambda$$
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$$[ n - \dfrac{1}{2}] \lambda$$
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$$ \dfrac{\lambda}{2}$$
Explanation
constructive interference occurs when the path differences $$(S_1P-s_2P)$$ is an integral multiple of $$\lambda$$ or $$(S_1P-s_2P)=n\lambda$$
Where n=0,1 ,2 3,........
In Young's double slit eperiment two disturbances arriving at a point P have phase difference of $$\dfrac{\pi}{3}$$ . The intensity of this point expressed as a fraction of maximum intensity $$I_0$$ is then
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$$\dfrac{3}{2}I_0$$
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$$\dfrac{1}{2}I_0$$
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$$\dfrac{4}{3}I_0$$
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$$\dfrac{3}{4}I_0$$
Explanation
The resultant intensity ,$$ I=I_o cos^2 \dfrac{\phi}{2}$$
Here, $$I_0$$ is the maximum intensity and $$\phi$$=$$\dfrac{\pi}{3}$$
$$\therefore I=I_0 cos^2$$
$$I=I_0 cos^2 \dfrac{\pi}{3 \times 2}=\dfrac{3}{4}I_0$$
Which of the following properties of laser beam can be used to measure long distances?
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It is very intense.
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It is highly monochromatic.
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It is an unidirectional beam of light.
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All of these
Explanation
A laser is a source of very intense, monochromatic and undirectional beam of light. These properties of a laser light can be exploited to measure long distances.
Light of wavelength $$600$$ is incident on a single slit. The first minimum of the diffraction pattern is obtained pattern is obtained at a distance of 4 from the center. The distance between the screen and the slit is 2m. What is the width of slit?
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0.2 m
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0.3 m
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0.5 m
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0.6 m
In Young's double-slit experiment, the angular width of a fringe formed on a distant screen is $$1^o$$. The slit separation is 0.01 mm. The wavelength of the light is
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0.174 nm
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$$0.174 \, \overset{0}{A}$$
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$$0.174 \, \mu m$$
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$$0.174 \, \times \, 10^{-4} \, m$$
Explanation
Angular fringe width, $$\theta \, = \, \dfrac{\lambda}{d}$$
$$\therefore \, \lambda \, = \, \theta d$$
Here, $$d = \, 0.01 mm \, = \, 10^{-5} \, m, \, \theta \, = \, 1^{\circ} \, = \, \dfrac{\pi}{180} \, radians$$
$$\lambda \, = \, \dfrac{\pi}{180} \, \times \, 10^{-5} \, = \, 1.74 \, \times \, 10^{-7} \, m$$
$$= \, 0.174 \, \times \, 10^{-6} \, m \, = \, 0.174 \mu m$$
Young's expt. the ratio of intensity at maxima and minima in the interference pattern is The 25 :The ratio of slit width will be
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4 : 1
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2 : 1
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16 : 1
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8 : 1
Explanation
$$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \left( {\dfrac{{25}}{9}} \right) = \left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)$$
$$\dfrac{5}{3} = \left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)$$
$$5{a_1} - 5{a_2} = 3{a_1} + 3{a_2}$$
$$2{a_1} = 8{a_2}$$
$$\dfrac{{{a_1}}}{{{a_2}}} = 4$$
$$\dfrac{{{a_1}}}{{{a_2}}} = \sqrt {\dfrac{{{\omega _1}}}{{{\omega _2}}}} = \dfrac{4}{1}$$
$$\left( {\dfrac{{{\omega _1}}}{{{\omega _2}}}} \right) = \dfrac{{16}}{1}$$
The human eye has an approximate angular resolution of $$\phi = 5.8 \times 10^{-4}$$rad and typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots?
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14.5 cm
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20.5 cm
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29.5 cm
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28 cm
Explanation
Here, angular resolution of human eye,
$$\phi \, = \, 5.8 \, \times \, 10^{-4} \, red$$
The linear distance between two successive dots in a typical photo printer is $$l \, = \, \dfrac{2.54}{300} \, cm \, = \, 0.84 \, \times \, 10^{-2} \, cm.$$
At a distance of z cm, the gap distance l will subtend an angle
$$\phi \, = \, \dfrac{l}{z} \, \therefore \, z \, = \, \dfrac{l}{\phi} \, = \, \dfrac{0.84 \, \times \, 10^{-2} \, cm}{5.8 \, \times \, 10^{-4}} \, = \, 14.5 \, cm$$
Transverse nature of light was confirmed by the phenomenon of the
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refraction of light
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diffraction of light
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dispersion of light
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polarization of light.
Explanation
The phenomenon of polarization confirms that light is a transverse wave because for polarization, the light should have different components oscillating in the different planes and a transverse wave has the oscillations perpendicular to the direction of propagation of the wave.
A parallel beam of sodium light of wavelength 5890 $$\overset{0}{A}$$ is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is $$60^o$$. The smallest thickness of the plate which will make it dark by reflection:
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3926 $$\overset{0}{A}$$
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4353 $$\overset{0}{A}$$
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1396 $$\overset{0}{A}$$
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1921 $$\overset{0}{A}$$
Explanation
Let the thickness of the Glass slab is $$t$$
The condition for minimum thickness corresponding to a dark band is $$2\mu t \,cosr = \lambda$$
$$ \therefore t= \dfrac{\lambda}{2 \mu cos \,r} = \dfrac{5890 \times 10^{-10}}{2 \times 1.5 \times cos 60^o} $$
$$t=3926 \times 10^{-10} m = 3926 {\overset{o}A} $$
Four identical monochromatic sources A, B, C, D as shown in the figure produce waves of the same wavelength $$\lambda$$ and are coherent. Two receiver $$R_1$$ and $$R_2$$ are at great but equal distances from B. Which of the two receivers picks up the larger signal when B is turned off?
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$$R_1$$
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$$R_2$$
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$$R_1$$ and $$R_2$$
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None of these
Explanation
Let the signal picked up at $$R_2$$ from $$B$$ be $$yB = a_1 \cos wt$$.
The path difference between signal at D and that at $$B$$ is $$\dfrac{\lambda}{2}$$
$$:yD = -a_1\cos wt$$
The path difference between signal at $$A$$ and that at $$B$$ is
$$\sqrt{d^2 + \left(\dfrac{\lambda^2}{2}\right)} - d = d\left(1 + \dfrac{\lambda^2}{4d^2}\right)^{1/2} - d \simeq \dfrac{1\lambda^2}{8d^2}$$
As $$d > > \lambda$$, therefore this path difference $$\to 0$$
and phase differnece $$= \dfrac{2\pi}{\lambda} \left(\dfrac{1}{8}\dfrac{\lambda^2}{d^2}\right) \to 0$$
Hence, $$yA = a_1\cos (wt - \phi)$$
similarly, $$yC = a_1 \cos(wt - \phi)$$
$$\therefore $$ Signal picked up by $$R_2$$ is
$$yA + yB + yC + yD = y = 2a_1\cos (wt- \phi)$$
$$\therefore |y|^2 = 4a_1^2 \cos^2(wt - \phi)$$
$$\therefore < I > = 2a_1^2$$
Thus, $$R_1$$ picked up the larger signal.
Yellow light is used in a single slit diffraction experiment with slit width of $$0.6\ mm$$. If yellow light is replaced by X-rays, then the observed pattern will reveal.
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That the central maximum is narrower
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More number of fringes
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Less number of fringes
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No diffraction pattern
Explanation
No diffraction pattern will be observed because
$$(i)$$ X-rays cannot be seen through naked eye.
$$(ii)$$ If X-rays are not of comparable size to diffract for a slit width of $$0.6$$mm.
When two waves with same frequency and constant phase difference interfere,
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there is a gain of energy
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there is a loss of energy
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the energy is distributed and the distribution changes with time
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the energy is redistributed and the distribution remains constant in time
Explanation
All the frequency does not change the energy will be redistributed in the regions of constructive interference and distribution remains constant in time.
In an interference arrangement similar to Young's double-slit experiment, the slits $$S_{1}$$ and $$S_{2}$$ are illuminated with coherent microwave sources, each of frequency $$10^{6}Hz$$. The source are synchronized to have zero phase difference. The slits are separated by a distance $$d = 150.0\ m$$. The intensity $$I(\theta)$$ is measured as a function of $$\theta$$, where $$\theta$$ is defined as shown. If $$I_{0}$$ is the maximum intensity, then $$I(\theta)$$ for $$0\leq \theta \ 90^{\circ}$$ is given by
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$$I(\theta) = \dfrac{I_{0}}{2}$$ for $$\theta = 30^{\circ}$$
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$$I(\theta) = \dfrac{I_{0}}{4}$$ for $$\theta = 90^{\circ}$$
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$$I(\theta) = I_{0}$$ for $$\theta = 0^{\circ}$$
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$$I(\theta)$$ is constant for all values of $$\theta$$
A point sources $$S$$ emitting light of wavelength $$600\ nm$$ is placed at a very small height $$h$$ above a flat reflecting surface $$AB$$ (see figure). The intensity of the reflected light is $$36\%$$ of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance $$D$$ from it.What is the shape of the interference fringes on the screen?
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Circular.
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helical
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eliptical
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spiral
Explanation
Shape of interference fringes is circular due to the symmetry of circular wave bounding from plane number.
For minima to take place between two monochromatic light waves of wavelength $$\lambda ,$$ the path difference should be
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$$n\lambda $$
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$$\left( {2n - 1} \right)\dfrac{\lambda }{4}$$
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$$\left( {2n - 1} \right)\dfrac{\lambda }{2}$$
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$$\left( {2n - 1} \right)\lambda $$
Explanation
For minima,
Path diff $$ = \left( {2n - 1} \right)\frac{\lambda }{2}$$
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Practice Class 12 Medical Physics Quiz Questions and Answers
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