All particles in it have same phase

All particles have opposite phase of vibration

Few particles are in same phase, are on opposite phase

Reduction in ozone thickness in stratosphere

Thinning of ozone layer in troposphere

MCQ Questions for Class 12 Medical Physics Wave Optics Quiz 12

Wave which cannot travel in vacuum is

0%
$X-$ rays
0%
Infrasonic
0%
Ultraviolet
0%
Radiowaves

Which of the following statement is wrong

0%
Infrared photon has more energy than the photon of visible light
0%
Photographic plates are sensitive to ultraviolet rays
0%
Photographic plates can be made sensitive to infrared rays
0%
Infrared rays are invisible but can cast shadows like visible light rays

Explanation

Photographic plates uses photoelectric effect to produce the spectrum. So, if a electromagnetic ray is projected on it, it will be sensitive towards it.

Hence, its sensitive to both ultraviolet and infrared rays.

The infrared rays do cast shadows like visible rays as they are not able to pass through most opaque objects just like visible rays.

Now, we know that the energy of a electromagnetic wave is given by,

$E=\dfrac{hc}{\lambda}$

or, we can say that

$E\propto \dfrac{1}{\lambda}$

Hence, if the wavelength decreases the energy will increase. Or, the lesser the wavelength the higher the energy. Infrared photons will have higher wavelength than that of visible light. Hence, its energy should be lesser than than visible light.
So, the wrong option is (A).

In a young's double slit experiment, the central point on the screen is

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Bright
0%
Dark
0%
First bright and then dark
0%
First dark and then bright

In Young's double slit experiment if

$L$ is the distance between the slits and the screen upon which interference pattern is observed

$x$ is the average distance between the adjacent fringes and

$d$ being the slit separation. The wavelength of light is given by

0%
$\dfrac {xd}{L}$
0%
$\dfrac {xL}{d}$
0%
$\dfrac {Ld}{x}$
0%
$\dfrac {1}{Ldx}$

Explanation

We know that fringe width

$\beta =\dfrac {D \lambda}{d}$

$\therefore x=\dfrac {L \lambda}{d}\Rightarrow \lambda =\dfrac {xd}{L}$

In Young's double slit experiment the spacing between two slits is

$0.1\ mm$ . If the screen is kept at a distance of

$1.0\ m$ from the from the slits and the wavelength of light is

$5000\ \overset {o}{A}$ , then the fringe width is

0%
$1.0\ cm$
0%
$1.5\ cm$
0%
$0.5\ cm$
0%
$2.0\ cm$

Explanation

$\beta =\dfrac {\lambda D}{d}$

$=\dfrac {5000\times 10^{-10}\times 1}{0.1\times 10^{-3}}m$

$=5\times 10^{-3}m=0.5\ cm$

Wavefront means

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All particles in it have same phase
0%
All particles have opposite phase of vibration
0%
Few particles are in same phase, are on opposite phase
0%
None of these

The ozone layer absorbs

0%
Infrared radiation
0%
Ultraviolet radiation
0%
$X-$ rays
0%
$\gamma$ - rays

Explanation

Ozone layer absorbs most of the

$UV$ rays emitted by the sun.

In a Young's double slit experiment

$62$ fringes are seen in visible region for sodium light of wavelength

$5893\overset{o}{A}$ . If violet light of wavelength

$4358\overset{o}{A}$ is used in plane of sodium light, then number of fringes seen will be

0%
$54$
0%
$64$
0%
$74$
0%
$84$

Explanation

To obtain the number of fringes:

$n_1 \lambda_1=n_2 \lambda_2$

$\Rightarrow 62\times 5893 =n_3 \times 4358$

$\Rightarrow n_2=84$ .

The range of wavelength of the visible light is

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$10\ A^o$ to $100\ A^o$
0%
$4,000\ A^o$ to $8,000\ A^o$
0%
$8,000\ A^o$ to $10,000\ A^o$
0%
$10,000\ A^o$ to $15,000\ A^o$

Explanation

Wavelength of visible spectrum is

$3900\overset{o}{A} -7800\overset{o}{A}$ .

What is ozone hole

0%
Hole in the ozone layer
0%
Formation of ozone layer
0%
Thinning of ozone layer in troposphere
0%
Reduction in ozone thickness in stratosphere

Explanation

Ozone hole is depletion of ozone layer in stratosphere because of gases like

$CFC'S$ etc.

In two separate set ups of the Young's double slit experiment, fringes of equal width are observed when lights of wavelength in the ratio

$1:2$ are used. If the ratio of the slit separation in the two cases is of the distance between the plane of the slits and the screen in the two set ups is

0%
$4:1$
0%
$1:1$
0%
$1:4$
0%
$2:1$

Explanation

$\beta =\dfrac {\lambda D}{d}=\dfrac {\beta_1}{\beta_2}=\left(\dfrac {D_1}{D_2}\right)\left(\dfrac {\lambda_1}{\lambda_2}\right)\left(\dfrac {d_2}{d_1}\right)$

$\Rightarrow 1=\left(\dfrac {D_1}{D_2}\right)\times \left(\dfrac {1}{2}\right)\times \left(\dfrac {1}{2}\right) \Rightarrow \dfrac {D_1}{D_2} = \dfrac {4}{1}$

Which radiation in sunlight, causes heating effect

0%
Ultraviolet
0%
Infrared
0%
Visible light
0%
All of these

Select the right option in the following

0%
Christiaan Huygens a contemporary of Newton established the wave theory of light by assuming that light waves were transverse
0%
maxwell provide the compelling theoretical evidence that light is transverse wave
0%
Thomas Young experimentally proved the wave behavior of light and Huygens assumption
0%
All the statements give above, correctly answer the question "what is light "

The penetration of light into the region of geometrical shadow is called

0%
Polarisation
0%
Interference
0%
Differection
0%
Reflection

The phase difference between incident wave and reflected wave is

$180^o$ when light ray

0%
Enters into glass from air
0%
Enters into air from glass
0%
Enters into glass from diamond
0%
Enters into water from glass

Explanation

When light reflect from denser surface phase change of

$\pi$ occurs

Following figure shows sources

$S_{1}$ and

$S_{2}$ that emits light of wavelength

$\lambda$ in all directions. the sources are exactly in phase and are seperated by a distance equal to

$1.5\lambda .$ If we start at the indicated start point abd travel along path

$1$ and

$2,$ the interference produce a maxima all along

0%
Path $1$
0%
Path $2$
0%
Any path
0%
None of these

Explanation

At any point along the path

$1,$ path difference between the waves is

$0.$
Hence maxima is obtained all along the path

$1.$
At any point along the path

$2$ , path difference is

$1.5\lambda$ which is odd multiple of

$\dfrac{\lambda}{2},$ so minima is obtained all along the path

$2.$

When a thin metal plate is placed in the path of one of the interfering beams of light

0%
Fringe width increase
0%
Fringe disappear
0%
Fringes becomes brighter
0%
Fringes becomes blurred

Which of the following is not a property of light

0%
It requires a material medium for propogation
0%
It can travel through vacuum
0%
It involves transportation of energy
0%
It has finite speed

A star emits light of

$5500 \overset{o}{A}$ wavelength. Its appears blue to an observer on the earth, it means

0%
Star is going away from the earth
0%
Star is stationary
0%
Star is coming towards earth
0%
None of the above

Explanation

Blue radiations have the wavelength around

$4600\overset{o}{A}$ . It shows that apparent wavelength is smaller than the real wavelength. It means that the star is proceeding towards earth.

A plane wavefront

$(\lambda=6 \times 10^{-7}\ m)$ falls on a slit

$0.4\ mm$ wide. A convex lens of focal length

$0.8\ m$ placed behind the slit focuses the light on a screen. What is the linear diameter of second maximum

0%
$6\ mm$
0%
$12\ mm$
0%
$3\ mm$
0%
$9\ mm$

Explanation

For secondary maxima

$d \sin \theta =\dfrac{5 \lambda}{2}$

$\Rightarrow d \theta=d.\dfrac{x}{D(\approx f)}=\dfrac{5 \lambda}{2}$

$\Rightarrow 2x=\dfrac{5 \lambda f}{d}=\dfrac{5 \times 0.8 \times 10^{-7}}{4 \times 10^{-4}}=6 \times 10^{-3}\ m=6\ mm$

If a star is moving towards the earth, then the lines are shifted towards

0%
Red
0%
Infrared
0%
Blue
0%
Green

Condition of diffraction is

0%
$\dfrac{a}{\lambda}=1$
0%
$\dfrac{a}{\lambda} > >1$
0%
$\dfrac{a}{\lambda} < <1$
0%
None of these

Explanation

For diffraction size of the obstacle must be of the order of wavelength of wave i.e.

$a \approx \lambda$

In the Young's double slit experiment, for which colour the fringe width is least

0%
Red
0%
Green
0%
Blue
0%
Yellow

Explanation

$\beta \propto \lambda$

Diffraction and interference of light suggest

0%
nature of light is electromagnetic
0%
Wave nature
0%
Nature is quantum
0%
Nature of light is transverse

If the separation between slit in Young's double slit experiment is reduced to

$\dfrac 13 rd$ , the fringe width becomes

$n$ times. The value of

$n$ is

0%
$3$
0%
$\dfrac 13$
0%
$9$
0%
$\dfrac 19$

Explanation

$\beta \propto \dfrac {\lambda}{d}$

$d\to \dfrac {d}{3}$

$\beta \to 3\beta$

$\therefore n=3$

In a Young's experiment, monochromatic light is used to illuminate the two slits

$A$ and

$B$ . Interference fringes are observed on a screen placed in front of the slits. Now if a thin glass plate is placed normally in the path of the beam coming from the slit

0%
The fringes will disapear
0%
The fringes width will increase
0%
The fringes width will increase
0%
There will be no change in the fringe width but the pattern shifts

A star is going away from the earth. An observer on the earth will see the wavelength of light coming from the star

0%
Decreased
0%
Increased
0%
Neither decreased nor increased
0%
Decreased or increased depending upon the velocity of the star

In order to see diffraction the thickness of the film is

0%
$100\ A^o$
0%
$10,000\ A^o$
0%
$1/ mm$
0%
$1/ cm$

In the set up shown in fig the two slits,

$S_{1}$ and

$S_{2}$ are not equidistant from the slits

$S.$ The central fringe at

$O$ is them

0%
Always bright
0%
Always dark
0%
Either dark or bright depending on the position of $S$
0%
Neither dark nor bright.

Explanation

If path difference

$\Delta =(SS+SO)-(SS+SO)=n\lambda n=0,1,2,3,...$ the central fringe at

$O$ is a bright fringe and if the path difference

$\Delta=\left(n-\dfrac{1}{2}\right)\lambda n=1,2,3,...$ the central bright fringe will be a dark fringe.

Specific rotation of sugar solution is

$0.01$ SI units.

$200\ kgm^{-3}$ of impure sugar solution is taken in a polarimeter tube of length

$0.25\ m$ and an optical rotation of

$0.4\ rad$ is observed. The percentage of purity of sugar is the sample is

0%
$80\%$
0%
$89\%$
0%
$11\%$
0%
$20\%$

Explanation

Specific rotation

$(\alpha)=\dfrac{\theta }{lc}$

$\Rightarrow c=\dfrac{\theta}{\alpha l}$

$=\dfrac{0.4}{0.01\times 0.25}=160\ kg/m^{3}$

Now percentage purity of sugar solution

$\dfrac{160}{200}\times 100=80\%$

In which of the following is the interference due to the division of wave front

0%
Young's double slit experiment
0%
Fresnel's biprism experiment
0%
Lloyd's mirror experiment
0%
Demonstration colours of thin film

Choose correct statement .
In most of the circumtances , the diffraction of sound is more possible compare to diffraction of light :

0%
Medium is necessary for sound propagation
0%
Sound waves are longitudinal while light rays are transverse
0%
Wavelength of light is very less comparable to wavelength of sound
0%
Speed of sound wave order is less than 6 compare to speed of light

Explanation

At path difference

$\dfrac{\lambda }{2}$ destructive interference will occur. So intensity will be

$l_{min}$
Let

$l_{1} = l_{2} = l_{0}$

$\therefore l_{max} = l_{1} + l_{2}2\sqrt{I_{1}I_{2}} = 1_{0} + l_{2} + 2l_{0} = 4l_{0}$
and

$l_{min} = l_{1} + l_{2} \sqrt{I_{1}I_{2}} = l_{0} + l_{0} - 2l_{0} = 0$

To show the phenomenon of interference we need two sources which emit radiation :

0%
Of equal frequency and fixed phase difference
0%
Of nearly equal frequency
0%
of equal frequency
0%
of different wavelength

The

$k$ line of singly ionised calcium has a wavelength of

$393.3\ nm$ as measured on earth. In the spectrum of one of the observed galaxies, this spectral line is located at

$401.8\ nm$ . The speed with which the galaxy is moving away from us, will be

0%
$6480\ km/s$
0%
$3240\ km/s$
0%
$4240\ km/sec$
0%
$None\ of\ these$

Explanation

$\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}\Rightarrow \dfrac{(401.8-393.3)}{393.3}=\dfrac{v}{3\times 10^{8}}$

$\Rightarrow v=6.48\times 10^{6}\ m/s=6480\ km/sec.$

According to given figure , in an experiment , electrons are passed through de-Broglie wavelength order towards a narrow's slit of width d and they are detected at D distance on a screen . The intensity distribution will be :

In young double slit experiment the maximum intensity of light is

$I_{max}$ then the intensity of light of the path difference

$\dfrac{\lambda }{2}$ will be

0%
$I_{max}$
0%
$\dfrac {Imax}{2}$
0%
$\dfrac {Imax}{4}$
0%
zero

Explanation

Sound waves are longitudinal while light rays are transverse Diffraction effect is more pronounced if the size of obstacle or aperture is of the wavelength of the waves. As the wavelength of light

$(\sim 10^{-6} m)$ is much smaller than the size of the objects around us, so diffraction of light is not easily seen. But sound waves large wavelength. They get easily diffracted by the objects around us.

Off in the distance, you see the headlights of a car, but they are indistinguishable from the single headlight of a motorcycle. Assume the car's headlights are now switched from low beam to high beam so that the light intensity you receive becomes three times greater. What then happens to your ability to resolve the two light sources?

0%
It increases by a factor of $9 .$
0%
It increases by a factor of $3 .$
0%
It remains the same.
0%
It becomes one-third as good.
0%
It becomes one-ninth as good.

If the point object is moved away from mirror normal to screen then ,

0%
fringe width will increase
0%
fringe width will decrease
0%
fringe width will first increase then decrease
0%
Remain unchanged

Unpolarized light of intensity $32{\text{W}}{{\text{m}}^{{\text{ - 2}}}}$ passes through three polarizers such that the transmission axis of the last polarizer is crossed with the first. If the intensity of the emerging light is $3{\text{W}}{{\text{m}}^{{\text{ - 2}}}}$ , what is the angle between the axes of the first two polarized is.

0%
${45^0}$
0%
${60^0}$
0%
${30^0}$
0%
Zero

Light is incident normally on a diffraction grating through which the first order diffraction is seen at

$32^{\circ}$ . The second order diffraction will be seen at

0%
$48^{\circ}$
0%
$64^{\circ}$
0%
$80^{\circ}$
0%
there is no second order diffraction in this case.

Explanation

We know, that if

${\theta}_{1}$ and

${\theta}_{2}$ be the angles formed in the two orders of diffraction, then

$\dfrac{sin {\theta}_{2}}{sin {\theta}_{1}} = \dfrac{{n}_{2}}{{n}_{1}}$
Hence,

$sin {\theta}_{2} = 2 sin 32 > 1$
Hence, a second order diffraction pattern doesnt exist in this case.

If half of reflecting surface of upper part of mirror is painted black, choose correct options from the following

0%
Intensity of maxima and minima will increase
0%
Intensity at minima will increase and at maxima, intensity will decrease
0%
Intensity at minima will remain zero and at maxima, intensity will decrease
0%
Intensity at maxima remains unchanged but at minima, intensity will increase

An aperture of diameter 1.2 mm has been illuminated by monochromatic light. The diffracted light is observed on a screen. When the screen is gradually displaced towards the aperture, the centre becomes dark for the first time at a distance of 30cm from the aperture. The wavelength of light used is :

0%
6000 $A^o$
0%
4000 $A^o$
0%
8000 $A^o$
0%
3000 $A^o$

Explanation

Given that diameter =

$1.2\ mm = 0.12 \ cm$

Radius

$r = 0.06 \ cm$

$b = 30\ cm$

Here

$(b^2+r^2)=(b+\lambda)^2$

$30^2+(0.06)^2=(30+\lambda)^2$

$\displaystyle \lambda = \frac{(0.06)^2 cm^2}{2\times 30 cm}(\lambda^2 neglected)$

$=0.00006\ cm = 6000\mathring{A}$

In an electron experiment, electrons are made to pass through a narrow slit of width

$'d'$ comparable to their de Broglie wavelength. They are detected on a screen at a distance D from the slit. Which of the following graphs can be expected to represent the numbers of electron

$'N'$ detected as a function of the detector position

$'y'$ (

$y=0$ corresponds to the middle of the slit)?

Explanation

Electron beam at the edges will get diffracted away from the central line

$y=0$ whereas the beam near the central line will pass undeviated. Thus makes the diffraction pattern wider than the slit width. Hence, option C is correct.

In a Young's double slit experiment

0%
if a point source is placed symmetrically from both the slits then interference will be observed on the

screen
0%
if a point source is placed un-symmetrically from both the slits then interference will be observed on the

screen.
0%
if two slits are illuminated by two independent sources then the interference will be observed on the

screen.
0%
intensity of light at two slits should be same for better contrast

In a single slit diffraction experiment, the width of the slit is made half the original width

0%
the width of the central maxima becomes double
0%
the width of the central maxima becomes half
0%
the width of the central maxima becomes one fourth
0%
the width of the central maxima becomes four times.

Explanation

We have

$w \propto \dfrac {1}{d}$ , where

$d$ is the width of the slit.
If

$d' = \dfrac {d}{2}$ then

$w' \propto 2w$

Microwaves of frequency

$3\times 10^{4}\ MHz$ and ultrasonic waves of wavelength

$1\ cm$ are passed through a slit of width

$2\ cm$ . Then

0%
diffraction will occur only in microwaves
0%
diffraction will occur only in ultrasonic wave
0%
diffraction will occur in both but diffraction pattern will be different
0%
diffraction will occur in both and diffraction patterns will be identical

Explanation

We have:

$C = \nu \lambda$

The wavelength of the microwaves is given by:

$\lambda_m = \dfrac {3 \times 10^8}{3 \times 10^4 \times 10^6} = 10^{-2}\ m = 1\ cm$

The wavelength of the microwave is

$1\ cm$ . Also the wavelength of ultrasonic wave is

$1\ cm$ . Both the waves have wavelengths less than the slit width. So, diffraction will occur for both the waves.

Since the two wave systems have the same wavelength, their diffraction patterns will be the same for the same slit.

In an electron microscope the accelerating voltage is increased from 20 kV to 80 kV, the resolving power of the microscope will change from R to

0%
$2 R$
0%
$\dfrac{R}{2}$
0%
$4R$
0%
$3R$

Explanation

Electron microscope is a microscope that can magnify very small details with high resolving power due to the use of electrons as the source of illumination. Since the wavelength of electrons are 100,000 times shorter than visible light the electron microscopes have greater resolving power
We have the Abbe's formula as resolution limit

$d=\dfrac{0.61\lambda}{NA}$ (NA is the numerical aperture)

The resolving power increases when d, the minimum distance that can be seen between two points in the image, decreases. Thus, according to the formula the resolving power is inversely proportional to the wavelength.

Resolving Power

$\propto \dfrac{1}{\lambda}$

A higher voltage will give the electrons a higher speed. Thus the electrons will have a smaller de Broglie wavelength according to the equation,

$\lambda=h/mv$

$\lambda\propto\dfrac{1}{\sqrt V}$

Thus we get Resolving power

$\propto \sqrt{V}$

$\implies\dfrac{R}{R'} = \sqrt{\dfrac{20}{80}}$

Thus,

$R' = 2R$

Interference pattern is observed at 'P' due to superimposition of two waves coming out from a source 'S' as shown in the figure. The value of 'l' for which maxima is obtained at 'P' is (R is perfect reflecting surface)

0%
$l=\dfrac {2n\lambda}{\sqrt 3-1}$
0%
$l=\dfrac {(2n-1)\lambda}{2(\sqrt 3-1)}$
0%
$l=\dfrac {(2n-1)\lambda \sqrt 3}{4(2-\sqrt 3)}$
0%
$l=\dfrac {(2n-1)\lambda}{\sqrt 3-1}$

Explanation

From the figure straight path SP=

$2l$

Reflected path SP =

$2l\sec 30^{\circ}$

So path difference is

$2l(\sec 30^{\circ}-1)$

Also the ray, when reflected by the mirror, suffers a phase change of

$\pi$

So the total difference in phase is

$2l(\sec 30^{\circ}-1)\times \dfrac{2\pi}{\lambda} + \pi$

For constructive interference

$2l(\sec 30^{\circ}-1)\times \dfrac{2\pi}{\lambda} + \pi = 2n\pi$

Solving this, we get

$l=\dfrac{(2n-1)\lambda \sqrt{3}}{4(2-\sqrt{3})}$

Find fringe width and number of possible maxima on the screen

$E$ :

0%
$1.1mm, 8$
0%
$1.1mm, 9$
0%
$0.8mm,8$
0%
$0.9mm,9$

Explanation

$D=b+r=1.4m$

$d={ S }_{ 1 }{ S }_{ 2 }=2r\delta =\dfrac { 2\times 0.1\times 1 }{ 5\times 57 } rad$

$\beta =\dfrac { \lambda D }{ d } =\dfrac { 550\times { 10 }^{ -9 }\times 1.4\times 5\times 57 }{ 2\times 0.1\times 1 } =1.1mm$

Number of possible maxima

$=\dfrac { 2b\alpha }{ \beta } =\dfrac { 2\times 1.3\times 0.2 }{ 1.1\times { 10 }^{ -3 }\times 5\times 57 } =8.3$

$n=8.3+1\approx 9$

The shape of the nodal curves is

0%
Elliptical
0%
Parabolic
0%
Hyperbolic
0%
Circular

CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 12 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Wave Optics Quiz 12 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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