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Diffraction - class-XII

Description: diffraction
Number of Questions: 107
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Tags: physics wave optics optics properties of waves superposition of waves
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In a Fresnel biprism experiment, the two positions of lens give separation between the slits as $16 $cm and$9 $cm, respectively. What is the actual distance of separation?

  1. $12.5 cm$

  2. $12 cm$

  3. $13 cm$

  4. $14 cm$


Correct Option: B
Explanation:

A Fresnel Biprism is the variation on the Young's Slits experiment. The Fresnel biprism has two thin prisms which are joint at their bases to form an isosceles triangle. A single wavefront impinges on both prisms.

Separations between the slits,

$d _{1}= 16 cm$

and $d _{2}= 9 cm.$

Actual distance of separation (d)  can be computed with the formula as given:

$d = \sqrt {d _{1} \times d _{2}}\\$

$d = \sqrt {16 \times 9}\\$

$d = \sqrt {144} \\$

$d = 12 cm$

Thus actual distance will be $12 cm$.

Option B is correct.

A parallel beam of light of wavelength 6000 $\overset{0}{A}$ gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is

  1. $2 \, \times \, 10^{-3} \, rad$

  2. $3 \, \times \, 10^{-3} \, rad$

  3. $1.8 \, \times \, 10^{-3} \, rad$

  4. $6 \, \times \, 10^{-3} \, rad$


Correct Option: A
Explanation:
Here, $\lambda \, = \, 6000 \, \overset{0}{A} \, = \, 6000 \, \times \, 10^{-10} \, m \, = \, 6 \, \times \, 10^{-7} \, m \, a \, - \, 0.3 \, mm \, = \, 0.3 \, \times \, 10^{-4} \, m \, = \, 3 \, \times \, 10^{-4} \, m$

For first minima, $a sin \theta \, = \, \lambda$ where a is the slit widht 
$sin \, \theta = \, \dfrac{\lambda}{a} \, = \, \dfrac{6 \, \times \, 10^{-7}}{3 \, \times \, 10^{-4}} \, = \, 2 \, \times \, 10^{-3}$

As $sin \theta$ is very small

$\therefore \, \theta \, \cong \, sin \, \theta \, = \, 2 \, \times \, 10^{-3} \, rad$

Light of wavelength 600 nm is incident on an aperture of size 2 mm. The distance upto which light can travel such that its spread is less than the size of the aperture is:

  1. 12.13 m

  2. 6.67 m

  3. 3.33 m

  4. 2.19 m


Correct Option: B
Explanation:


Given:
 Aperture Width, $a =2mm=2\times 10^{-3}mm$
The distance upto which light can travel is Fresnel distance, $Z _F \, = \, \dfrac{a^2}{\lambda} \, = \, \dfrac{(2 \, \times \, 10^{-3})^2}{600 \, \times \, 10^{-9}} \, = \, 6.67 \, m$

For what distance is ray optics a good approximation when the aperture is 4 mm wide and the wavelength is 500 nm? 

  1. 22 m

  2. 32 m

  3. 42 m

  4. 52 m


Correct Option: B
Explanation:

Given:
 Aperture Width $a =4 $mm
Fresnel distance, $Z _F \, = \, \dfrac{a^2}{\lambda} \, = \, \dfrac{(4 \, \times \, 10^{-3})^2}{500 \, \times \, 10^{-9}}$
$\therefore \, z _p \, = \, 32 \, m$

In a Fraunhofer diffraction at single slit of width d with incident light of wavelength $5500 A^0$, the first minimum is observed at angle $30^0$. The first secondary maximum is observed at an angle $\theta$ =

  1. $sin^{-1} [\frac{1}{\sqrt{2}}]$

  2. $sin^{-1} [\frac{1}{4}]$

  3. $sin^{-1} [\frac{3}{4}]$

  4. $sin^{-1} [\frac{\sqrt{3}}{2}]$


Correct Option: C
Explanation:

Slit width = d
$\lambda = 5500 A^0 = 5.5 \times 10^{-7}m, \theta _n = 30^0$
For first secondary minima, $d sin \theta _n = \lambda$
$ d = \dfrac{\lambda}{sin \theta _n} = \dfrac{5.5 \times 10^{-7}}{sin 30^0} = 11 \times 10^{-7}$m
For the first secondary maxima, $d sin \theta _n = \dfrac{3 \lambda}{2}$
i.e. $sin \theta _n = \dfrac{3 \lambda}{2d} = \dfrac{ 3 \times 5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}} = sin \theta _n = \dfrac{3}{4} or \theta _n = sin^{-1}(3/4)$

The intensity of light at a distance $r$ from the axis of a long cylindrical source is inversely proportional to $r$.

  1. True

  2. False


Correct Option: A
Explanation:

$I=\cfrac { { I } _{ o } }{ r } \ \therefore I\propto \cfrac { 1 }{ r } $

Intensity is inversely proportional to $r$.

In diffraction occurs through a single slit then intensity of first secondary maxima become ...................  % of central maxima :-

  1. $4$ %

  2. $25$ %

  3. $75$ %

  4. $50$ %


Correct Option: C

Two slits of width $a _1$  and $a _2$ are illuminated by light of same wavelength.The first diffraction minima produced by each of them are in directions inclined at angles $\theta _1$ and $\theta _2$. The ratio of $sin\, \theta _1$ to $sin\,\theta _2$ is

  1. $\dfrac{a _1}{a _2}$

  2. $\sqrt{\dfrac{a _1}{a _2}}$

  3. $\sqrt{\dfrac{a _2}{a _1}}$

  4. $\dfrac{a _2}{a _1}$


Correct Option: A

When monochromatic light is replaced by white light in Fresnel's biprism arrangement, the central fringe is

  1. colored

  2. white

  3. dark

  4. None of these


Correct Option: B
Explanation:

When white light refracts from the biprism, all colours would form coherent sources at different positions according to their refractive indices. However the position of their central fringe would be same, and hence they would all combine at that position to again give a white fringe.

Among the Fresnel zones the operative zones contributing intensity are :

  1. last zones

  2. first few zones

  3. middle zones

  4. all the zones


Correct Option: D
Explanation:

All the fresnel zones contribute to the intensity.

The average path difference between two waves coming from third and fifth Fresnel zones of a wave front at the centre of the screen is :

  1. $\dfrac{\lambda }{2}$

  2. $2\lambda $

  3. $\lambda$

  4. $4\lambda$


Correct Option: C
Explanation:

The third and fifth fresnel zone differ in phase by $2\pi $. Thus as $2\pi $ corresponds to path difference of $\lambda $.

In Young's double slit experiment, angular width of fringes is $0.20^o$ for sodium light of wavelength $5890\overset{o}{A}$. If complete system is dipped in water, then angular width of fringes becomes.

  1. $0.15^o$

  2. $0.22^o$

  3. $0.30^o$

  4. None of these


Correct Option: A
Explanation:

b'Angular fringe width$=\theta=\displaystyle\frac{\lambda}{d}$
$\Rightarrow \theta\propto \lambda$
$\lambda _w=\displaystyle\frac{\lambda _a}{\mu _w}$
So, $\theta _w=\displaystyle\frac{\theta _{air}}{\mu _w}=\frac{0.20}{\displaystyle\frac{4}{3}}=0.15^o$'

In single slit diffraction pattern

  1. Centre fringe has negligible width than others.

  2. All fringes are of same width

  3. Central fringe does not exist

  4. None of the above


Correct Option: D
Explanation:

In single silt diffraction, the central fringe has maximum intensity and has the width double than other fringes.

A transverse wave propagating along x-axis isrepresented by
$y\left (x, t \right ) = 8.0 \sin \left (0.5 \pi x - 4 \pi rt - \frac{\pi} {4}  \right )$
where $x$ is in metres and t is in seconds. The speed of the wave is:-

  1. $4 \pi$ m/s

  2. $0.5 \pi$ m/s

  3. $\frac{\pi} {4}$ m/s

  4. 8 m/s


Correct Option: D
Explanation:

$V$ = $\frac{\omega} {k}$ = $\frac{4\pi} {0.5 \pi}$ = $8ms^{-1}$

The distance between two virtual images of slits in a biprism experiment is measured using the convex-lens which is

  1. Foucault's method

  2. Huygen's method

  3. Young's method

  4. conjugate foci method


Correct Option: A

A biprism arrangement in air is immersed completely in a liquid. The fringe width

  1. remains same

  2. increases

  3. decreases

  4. none of the above


Correct Option: C
Explanation:

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

when immersed in a liquid, ${ \lambda  } _{ 1 }=\cfrac { \lambda  }{ \mu  } \ \therefore { \beta  }^{ 1 }=\cfrac { { \lambda  }^{ 1 } }{ d } =\cfrac { \lambda D }{ \mu d } =\cfrac { \beta  }{ \mu  } \ \therefore \beta \quad decreases $

A beam of natural light falls on a system of $6$ polaroids, which are arranged in succession such that each polaroid is turned through ${30}^{o}$ with respect to the preceeding one. The percentage of incident intensity that passes through the system will be

  1. $100$%

  2. $50$%

  3. $30$%

  4. $12$%


Correct Option: D
Explanation:

If $I$ is the final intensity and $I _0$ is the initial intensity then,

$I=\dfrac{I _0}{2}(\cos^2{30})^5$
$\dfrac{I _0}{I}=\dfrac{1}{2}\times \left(\dfrac{\sqrt3}{2}\right)^{10}=0.12=12 percent$

In the diffraction of light of wavelength $\lambda$ through single slit of width d, the angle between the principal maxima and first minima will be:

  1. $\lambda/d$

  2. $\lambda/2d$

  3. $\lambda/4d$

  4. $\pi/2$


Correct Option: A
Explanation:
theory of diffraction we know that minima condition is given by 
$n\lambda=d\sin \theta$
for first minima, $n=1$
$\lambda =d\sin\theta$
As $\theta\approx$ small
$\sin\theta\approx \theta$
$\lambda=d\theta$
$\theta=\dfrac{\lambda}{d}\rightarrow $ Required answer
$y=\dfrac{\lambda D}{d}$
$y=D\theta$
$\dfrac{\lambda D}{d}=D\theta$
$\theta=\dfrac{\lambda}{d}$
Required answer

A parallel beam of wavelength $\lambda =4500\mathring { A }$passes through a long slit of width $2\times 10^{-4}m$. The angular divergence for which most of the light is diffracted is (in $\times 10^{-5}$) radian)

  1. $\dfrac {2\pi}{3}$

  2. $\dfrac {5\pi}{4}$

  3. $\dfrac {3\pi}{4}$

  4. $\dfrac {\pi}{3}$


Correct Option: B

A double slit arrangement produces interference fringes for sodium light ($\lambda=5890 \mathring {A}$) that are $0.40^0$ apart. What is the angular fringe separation if the entire arrangement is immersed in water?

  1. $0.10^o$

  2. $0.20^o$

  3. $0.30^o$

  4. $0.40^o$


Correct Option: C
Explanation:

We know the distance between two slits $d=\dfrac{\lambda}{\sin\theta}$, where $\lambda$=wavelength and $\theta$=angle from the original beam.

Given wavelength of sodium light ${\lambda} _{1}=5890 \mathring { A } =5890\times{10}^{-10} m$ and ${\theta} _{1}={0.40}^{0}$
Putting the above values in the given equation we get,
$d=\dfrac{5890\times {10}^{-10}}{\sin0.40}=8.43\times {10}^{-5} m$
Also ${\lambda} _{2}=\dfrac{{\lambda} _{1}}{n}$--------(A) 
where n= refractive index of water =1.33 
Thus A becomes,
${\lambda} _{2}=\dfrac{5890\times {10}^{-10}}{1.33}=4.42\times {10}^{-7} m$
Again,
$d=\dfrac{{\lambda} _{2}}{\sin{\theta} _{2}}$
${\theta} _{2}={\sin}^{-1} (\dfrac{4.42\times {10}^{-7}}{1.33})={0.30}^{0}$ 

In a biprism experiment, the distances of a point in the focal plane of the eye-lens where the fringes are formed two optical images of the slit differ by $165.5 $wavelengths. Is the point bright or dark? If the path difference is $9.75 \times 10^{-5}m$, calculate the wavelength of light used.

  1. Bright, $5891A^o$

  2. Bright, $6891A^o$

  3. Dark, $5891A^o$

  4. Dark, $6891A^o$


Correct Option: C

If inter planar distance in a crystal is $2\times { 10 }^{ -8 }$ m then value of maximum wavelength can be diffracted is :

  1. $2\times { 10 }^{ -8 }m$

  2. $5.6\times { 10 }^{ -8 }m$

  3. $4\times { 10 }^{ -8 }m$

  4. $3\times { 10 }^{ -8 }m$


Correct Option: C
Explanation:

Given,


$d=2\times 10^{-8}m$


Bragg's law,

$\lambda=2dsin\theta$. . . . .(1)

For maximum wavelength, $sin\theta=1$

$\lambda=2d=2\times 2\times 10^{-8}$

$\lambda=4\times 10^{-8}m$

In a fresnel's bi-prism experiment , the refracting angles of the prism were 2.5$^o$ and the refracting index of the glass was 1.5 . With the single slit 10 cm from the bi-prism ,fringes were formed on a screen 1 m from the single slit . The fringe width is 0.1375 mm . The wavelength of light is 

  1. 600 nm

  2. 1200 nm

  3. 60 A$^o$

  4. 120 A$^o$


Correct Option: A
Explanation:
For a fresnel biprism having an angle of $\alpha$ and refractive index $\mu$, frindge width $\beta =\frac { \lambda D }{ 2a(\mu -1)\alpha  } \\$
Given,
$\\ \beta =0.1375mm,\quad \alpha =2.5=\dfrac { 2.5\times \pi  }{ 180 } rad,\quad D=1000mm,\quad a=\quad 100mm\quad \& \quad \mu =1.5\\$
 $\therefore \quad 0.1375=\dfrac { \lambda (mm)\times 1000 }{ 2\times 100\times (1.5-1)\times (\dfrac { 2.5\times \pi  }{ 180 } ) } \\$
$ \Rightarrow \lambda =6\times { 10 }^{ -4 }mm\quad =\quad 600nm\quad =6000\mathring { A } $

A biprism experiment was performed by using red light of wavelength $ 6500\mathring{A} $ and blue light of wavelength $ 5200\mathring{A}$. the value of n for which $ (n+1)^{th} $ blue bright band coincides with $ n^{th} $ red band is

  1. $5$

  2. $4$

  3. $3$

  4. $2$


Correct Option: B
Explanation:
Formula,

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

given that $n$ fringes of red coinciding with $n+1$ green fringes. Then,

$n\beta _r=(n+1)\beta _g$

$n\times \dfrac{6500 d}{D}=(n+1)\dfrac{5200 d}{D}$

$n\times 65=(n+1)\times 52$

$n=\dfrac{65-52}{52}$

$\therefore n=4$

A two slit youngs interference experiment is done with monochromatic light of wavelength $6000 /A$. The slits are $2 /mm$ apart. The fringes are observed on a screen placed $10 /cm$ away from the slits. Now a transparent plate of thickness $0.5 /mm$ is placed in front of one of the slits and it is found  that the interference pattem shifts by $5 /mm$. The refractive index of the transparent plate is :

  1. $1.2$

  2. $0.6$

  3. $2.4$

  4. $1.5$


Correct Option: A

The ratio of radii of Fresnel's  fourth to ninth zone is 

  1. 1:4

  2. 4:0

  3. 9:4

  4. 2:3


Correct Option: D

The box of a pin hole camera, of length $L$, has a hole of radius $a$. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $b _{min}$) when :

  1. $a=\cfrac{\lambda^{2}}{L}$ and $b _{min}=\sqrt{4\lambda L}$

  2. $a=\cfrac{\lambda^{2}}{L}$ and $b _{min}=\left(\cfrac{2\lambda^{2}}{L}\right)$

  3. $a=\sqrt{\lambda L}$ and $b _{min}=\left(\cfrac{2\lambda^{2}}{L}\right)$

  4. $a=\sqrt{\lambda L}$ and $b _{min}=\sqrt{4\lambda L}$


Correct Option: D
Explanation:

$\begin{array}{l} \sin  \theta =\dfrac { \lambda  }{ a }  \ B=2a+\dfrac { { 2L\lambda  } }{ a } .....................\left( 1 \right)  \ \dfrac { { \partial B } }{ { \partial a } } =0...........\left( 2 \right)  \ 1-\dfrac { { L\lambda  } }{ { { a^{ 2 } } } } =0 \ \Rightarrow a=\sqrt { \lambda L }  \ { B _{ \min   } }=2\sqrt { \lambda L } +2\sqrt { \lambda L }  \ by\, \, substituting\, \, for\, \, a\, \, from\, \, \left( 2 \right) \, \, in\, \, \left( 1 \right)  \ =4\sqrt { \lambda L }  \ \therefore \, \, The\, \, radius\, \, of\, \, the\, \, spot=\frac { 1 }{ 2 } 4\sqrt { \lambda L } =\sqrt { 4\lambda L }  \end{array}$

The correct relation between the angle of diffraction $\phi $ and the glancing angle $\theta $ in Davisson-Germer experiment will be:

  1. $\theta = {90^0} - \frac{\phi }{2}$

  2. $\phi = \frac{\phi }{2} - {90^0}$

  3. $\theta = {90^0} - \phi $

  4. $\phi = {90^0} - \phi $


Correct Option: A

The distance between two consecutive atoms of the crystal lattice is $1.227\overset {\circ}{A}$. The maximum order of diffraction of electrons accelerated through $10^{4}$ volt will be:

  1. $10$

  2. $\dfrac {1}{10}$

  3. $100$

  4. $\dfrac {1}{100}$


Correct Option: A

In Fresnel diffraction, if the distance between the disc and the screen is decreased, the intensity of central bright spot will 

  1. increase

  2. decrease

  3. remain constant

  4. none of these


Correct Option: B

The box of a pin hole camera, of length $L$, has a hole of radius $a$. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $b _{min}$) when

  1. $a = \sqrt {\lambda L}$ and $b _{min} = \left (\dfrac {2\lambda^{2}}{L}\right )$

  2. $a = \sqrt {\lambda L}$ and $b _{min} = \sqrt {4\lambda L}$

  3. $a = \dfrac {\lambda^{2}}{L}$ and $b _{min} = \sqrt {4\lambda L}$

  4. $a = \dfrac {\lambda^{2}}{L}$ and $b _{min} = \left (\dfrac {2\lambda^{2}}{L}\right )$


Correct Option: D

A light wave is incident normally on a glass slab of refractive index $1.5$. If $4\%$ of light gets reflected and the amplitude of the electric field of the incident light is $30\ V/m$,then the amplitude of the electric field for the wave propogating in the glass medium will be:

  1. $10\ V/m$

  2. $24\ V/m$

  3. $30\ V/m$

  4. $6\ V/m$


Correct Option: B
Explanation:

$P _{refracted}=\frac {  96}{ 100 } P _1$
$\Rightarrow K _2A^2 _t=\frac {  96}{ 100 }K _1A^2 _i$
$\Rightarrow r _2A^2 _t=\frac {  96}{ 100 }r _1A^2 _i$
$\Rightarrow A^2 _t=$ $ \frac {  96}{ 100 } \times$ $\frac { 1 }{ 3 } {2} $ $\times (30)^2$
$A _t\sqrt { \frac { 64 }{ 100 }\times(30)^2  } =24$

Light of wavelength 6328 A IS incident normally on a slit having a width of 0.2 mm. The width of the central maximum measured from minimum to minimum of diffraction pattern on a screen 9.0 meters away will be about

  1. 0.72 degrees

  2. 0.09 degrees

  3. 0.36 degrees

    1. 18 degrees

Correct Option: A

In a biprism experiment, interference bands are observed at a distance of one meter fromthe slit. A convex lens is put between the slit and the eyepiece gives two images of slit 0.7$\mathrm { cm }$ apart, the lens being 70$\mathrm { cm }$ from the eyepiece. The fringe width will be: $\left( \lambda = 6000 \mathrm { A } ^ { 9 } \right)$ 

  1. 0.3$\mathrm { mm }$

  2. 0.1$\mathrm { mm }$

  3. 0.4$\mathrm { mm }$

  4. 0.2$\mathrm { mm }$


Correct Option: A

Conditions of diffraction is

  1. $
    \frac{a}{\lambda}=1
    $

  2. $
    \frac{a}{\lambda}>>1
    $

  3. $
    \frac{a}{\lambda}<<1
    $

  4. None of these


Correct Option: C

A diffraction is obtained by using a beam of red light. What will happen if the red light is replaced by the blue light 

  1. Bands will narrower and crowd
    full together

  2. Bands become broader and further apart

  3. No change will take place

  4. Bands disappear


Correct Option: A

Direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by (a is the width of the slit)

  1. $
    \operatorname{asin} \theta=\dfrac{\lambda}{2}
    $

  2. $
    a \cos \theta=\dfrac{3 \lambda}{2}
    $

  3. $
    \operatorname{asin} \theta=\lambda
    $

  4. $
    a \sin \theta=\dfrac{3 \lambda}{2}
    $


Correct Option: A

If the whole bi-prism experiment is immersed in water then the fringe width becomes, if the refractive indices of bi-prism material and water are $1.5$ and $1.33$ respectively, 

  1. $3$ times

  2. $\displaystyle\frac{3}{4}$ times

  3. $\displaystyle\frac{4}{3}$ times

  4. $\displaystyle\frac{1}{3}$ times


Correct Option: A
Explanation:
Separation between the coherent sources when the entire setup is in air $d _{ga} = 2a( \mu _g -\mu _a)A$
Separation between the coherent sources when the entire setup is put inside water $d _{gw} = 2a( \mu _g -\mu _w)A$

where  $a=$distance  between the single slit and the biprism, A is the prism angle and $\mu _g$ is the redfractive index of biprism.

Fringe width $\beta _{ga} = \dfrac{ D\lambda}{d _{ga}}$
Fringe width $\beta _{gw} = \dfrac{ D\lambda}{d _{gw}}$
$\therefore \dfrac{\beta _{ga}}{\beta _{gw}}=\dfrac{ d _{gw}}{d _{gw}}=\dfrac{ \mu _g - \mu _w}{ \mu _g - \mu _a}=\dfrac{1.5 -1.33}{1.5 -1}=\dfrac{0.17}{0.5}=\dfrac{0.17}{0.5}=\dfrac{1}{3}$
Hence, the fringe width increases 3 times.

In the Fresnel bi-prism experiment, the refractive index for the bi-prism is $\mu=3/2$ and fringe width obtained is $0.4mm$. If the whole apparatus is immersed as such in water then the fringe width will become(refractive index of water is $4/3$).

  1. $0.3mm$

  2. $0.225mm$

  3. $0.4mm$

  4. $1.2mm$


Correct Option: D
Explanation:

In a fresnel bi-prism, fringe width,$\omega _{medium}=\dfrac{D\lambda _{medium}}{d _{medium}}$

where $\lambda _{medium}=\dfrac{\lambda _{vacuum}}{\mu _{medium}}$
and $d _{medium}\propto A(\dfrac{\mu _{bi-prism}}{\mu _{medium}}-1)$
Here $\mu _{medium}=\dfrac{4}{3}$ and $\mu _{bi-prism}=\dfrac{3}{2}$
Hence $\dfrac{\omega _{water}}{\omega _{vacuum}}=3$
Hence, $\omega _{water}=1.2mm$

In bi-prism experiment, fringes are obtained by white light sources. The fringe nearest to the central fringe will be

  1. yellow

  2. green

  3. violet

  4. red


Correct Option: C
Explanation:

Position of first fringe is given by $y _{1}=\dfrac{D\lambda}{d}$

Hence the nearest fringe is one with least wavelength, which is violet.

In a bi-prism experiment fifth dark fringes are obtained at a point. If a thin transparent film is placed in the path of one of waves, then seventh bright fringe is obtained at the same point. The thickness of the film in terms of wavelength $X$ and refractive index $\mu$ will be

  1. $\displaystyle\frac{1.5\lambda}{(\mu-1)}$

  2. $\displaystyle{1.5}{(\mu-1)\lambda}$

  3. $\displaystyle{2.5}{(\mu-1)\lambda}$

  4. $\displaystyle\frac{2.5\lambda}{(\mu-1)}$


Correct Option: D
Explanation:

For 5th dark fringe to be obtained at the point, the path difference of the beams at that point =$\dfrac{9\lambda}{2}$.

For 7th bright fringe to be obtained at the same point, the new path difference of the beams at that point =$7\lambda$.
Due to insertion of the film of thickness $t$, path difference added to the beam=$t(\mu-1)$
Hence, $\dfrac{9\lambda}{2}+t(\mu-1)=7\lambda$
Hence, $t=\dfrac{2.5\lambda}{(\mu-1)}$

The refracting angle of bi-prism is

  1. $179^0$

  2. $1^0$

  3. $0.5^0$

  4. $90^0$


Correct Option: C
Explanation:
The refracting angles of the bi- prism are very small, usually about $0.5^o$

In bi-prism experiment the light source is

  1. extented

  2. narrow

  3. multichromatic

  4. all the above


Correct Option: B
Explanation:

For the sustained interference of light wave , two source must be narrow.

Which of the following formulae is incorrect in a bi-prism?

  1. $d=\sqrt{d _1d _2}$

  2. $d=2a(\mu-1)a$

  3. $d=\displaystyle\frac{D\lambda}{\beta}$

  4. $d=\displaystyle\sqrt\frac{d^2 _1}{d _2}$


Correct Option: D
Explanation:

If $d _{1}$, $d _{2}$ are the distances of two virtual sources from the central axis, the average $d$ is given by

 $d=\sqrt{d _{1}d _{2}}$.

In Fresnel's bi-prism experiment, the coherent sources are obtained by

  1. interference

  2. reflection

  3. refraction

  4. total internal reflection


Correct Option: C
Explanation:

In Fresnel bi-prism, light from the single source are made into two virtual sources on passing it through the bi-prism due to refraction in prism, so that the two sources acts as a coherent sources

Which of the following statements is correct?

  1. Diffraction is because of interference of light from same sources whereas interference is due to light form two isolated sources.

  2. Diffraction is due to interaction of light from same wave fronts whereas interference is due to interaction of two waves derived from the same source.

  3. Diffraction is due to interference of waves derived from the same source whereas interference is bending of light from the same source.

  4. Diffraction is due to reflected waves whereas interference is due to transmitted waves from a source.


Correct Option: B
Explanation:

When a same source is present, it might give different colours depending on the path difference between the interfering rays from the source.

However when two completely different sources are present, they interfere due to path difference between them and give the interference pattern. For example, Young' double slit experiment.

In a Fresnel's biprism experiment the two positions of lens separation between the slits as $16cm$ and $9cm$ respectively. The true distance of separation between the slits is 

  1. $10cm$

  2. $12cm$

  3. $14cm$

  4. $15cm$


Correct Option: B

In Young's experiment fourth bright fringe produced by light of $5000\overset{o}{A}$ superposes on the fifth bright fringe of an unknown wavelength. The unknown wavelength is _________$\overset{o}{A}$.

  1. $4000$

  2. $6000$

  3. $5000$

  4. $8000$


Correct Option: A
Explanation:
$ \beta = \dfrac{ \lambda \times D}{d} $
$4\beta _1=5\beta _2$
$4\times 5000\dfrac{D}{d}=5\times \lambda\dfrac{D}{d}$
$\lambda =4000\overset{o}{A}$.

A monochromatic plane wave of speed c and wavelength $\lambda$ is diffracted at a small aperture. The diagram illustrates successive wave fronts.
After what time will some portion of the wave front GH reach point P?

  1. $\dfrac{3\lambda}{2c}$

  2. $\dfrac{2\lambda}{c}$

  3. $\dfrac{3\lambda}{c}$

  4. $\dfrac{4\lambda}{c}$


Correct Option: C
Explanation:
The distance between two consecutive lines(wavefronts ) ia $\lambda$.
So, wavefront XY is separated by a distance of 3λ from the wavefront that has reached point P.

Distance that needs to be travelled = $3\lambda$
Speed of light = $c$

Speed =$\dfrac{Distance}{time}$
Time = $\dfrac{Distance}{Speed}$
        = $\dfrac{3\lambda}{c}$

Diffraction is a term used to describe one aspect of wave behaviour.
What does diffraction make possible?

  1. The ability to hear around corners

  2. The ability to hear high frequency and low frequency sound waves

  3. The ability to hear loud and quiet sounds

  4. The ability to hear sound through a brick wall


Correct Option: A
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.

So, diffraction makes the sound wave hear around corner.

Two wave-fronts are emitted from coherent sources of path difference between them is $2.1$ micron. Phase difference between the wave-fronts at that point is $7.692\ \pi$. Wavelength of light emitted by sources will be :

  1. $5386\mathring{A}$

  2. $5400\mathring{A}$

  3. $5460\mathring{A}$

  4. $5892\mathring{A}$


Correct Option: C
Explanation:
Given:
$\Delta x=2.1 \times 10^{-6}\\\Delta \phi=7.6922$

$\Phi =\dfrac{2\pi \Delta x}{\lambda}$

$\lambda=\dfrac{2\pi\times 2.1\times 10^{-6}}{7.692\pi}$

$\lambda=5460\times 10^{-10}m$

A screen is at a distance of $1\ m$ away from the aperture. If light of wavelength $500\ nm$ falls on an aperture, then area of first HPZ and radius of third HPZ are :

  1. $1.57\ mm^2, 1.22\ mm$

  2. $1.22\ mm^2, 1.57\ mm$

  3. $1.65\ mm^2, 2.79\ mm$

  4. $2.63\ mm^2, 0.22\ mm$


Correct Option: A

The sodium yellow doublet has wavelengths $5890\mathring{A}$ and $'\lambda' \mathring{A}$ and resolving power of a grating to resolve these lines is $982$, then value of $\lambda$ is :

  1. $5896\mathring{A}$

  2. $5880\mathring{A}$

  3. $5869\mathring{A}$

  4. $5876\mathring{A}$


Correct Option: A

The phenomenon of diffraction can be treated as interference phenomenon if the number of coherent sources is 

  1. one

  2. two

  3. zero

  4. infinity


Correct Option: D
Explanation:

The single slit diffraction pattern acts as an envelope for the multiple slit interference patterns.

Diffraction on a single slit is equivalent to interference of light from infinite number of coherent sources contained in slit.

In Fresnel's biprism expt., a mica sheet of refractive index 1.5 and thickness 6 $\times$ 10$^{-6}$m is placed in the path of one of interfering beams as a result of which the central fringe gets shifted through 5 fringe widths. The wavelength of light used is

  1. 6000 $\overset{o}{A}$

  2. 8000 $\overset{o}{A}$

  3. 4000 $\overset{o}{A}$

  4. 2000 $\overset{o}{A}$


Correct Option: A
Explanation:

Where n is equivalent number of fringe by which the centre fringe is shifted due to mica sheet
$\displaystyle\lambda = \frac{(\mu - 1)t}{n} = \frac{(1.5 - 1) 6 \times 10^{-6}}{5} = 6 \times 10^{-7} m = 6000 \overset{o}{A}$

An aperture of size a is illuminated by a parallel beam of light of wavelength $\lambda$. The distance at which ray optics has a good approximation is

  1. $\dfrac {a^{2}}{\lambda}$

  2. $\dfrac {\lambda}{a^{2}}$

  3. $\dfrac {\lambda}{a}$

  4. $\dfrac {\lambda^{2}}{a}$

  5. $a^{2}\lambda$


Correct Option: A
Explanation:

An aperture of size a is illuminated by a parallel beam of light of wavelength $\lambda$. The distance at which ray optics has a good approximation is $\dfrac {a^{2}}{\lambda}$. This is the Fresnel distance.

Diffraction gratings provide much brighter interference pattern since more light passes through them compared with double slits.

  1. True

  2. False


Correct Option: A
Explanation:

The statement is true.

That is diffraction gratings provide much brighter interference pattern since more light passes through them compared with double slits.
One difference between the interference of many slits (diffraction grating) and double-slit (Young's Experiment) is that a diffraction grating makes a number of principle maxima along with lower intensity maxima in between.  The principal maxima occur on both sides of the central maximum for which a formula similar to Young's formula holds true.

In biprism experiment, the distance of 20th bright band from the centre of the interference pattern is 8 mm. The distance of the 30th bright band is

  1. 4 mm

  2. 8 mm

  3. 12 mm

  4. 16 mm


Correct Option: C
Explanation:

Given

$20\cfrac { \lambda D }{ d } =8\ =\cfrac { \lambda D }{ d } =\cfrac { 8 }{ 20 } \ \therefore  { 30 }^{ th }$ bright,
$=30\times \cfrac { \lambda D }{ d } \ =30\times \cfrac { 8 }{ 20 } \quad =12mm$

In a biprism experiment, the fringe width obtained on the screen is $6\ mm$ from the slits which are $1.5\ m$ away from each other. Keeping the setting unchanged if the eye-piece is moved $20\ cm$ towards the biprism, find the change in fringe width.

  1. $0.90\ mm$

  2. $0.86\ mm$

  3. $0.80\ mm$

  4. $0.53\ mm$


Correct Option: A

In a Fresnel's bi-prism experiment, the fringe of width $0.05mm$ is observed on a screen at a distance of $1.5m$ from the source . When a convex lens is placed between the source and the screen, for two positions of the lens image of interfering sources are produced on the screen. The separation between the two images being $0.04$ and $0.01mm$, respectively. The wavelength of light used is

  1. $6.67nm$

  2. $0.667nm$

  3. $667nm$

  4. $667A^o$


Correct Option: B
Explanation:

Distance between slits, $d=\sqrt{d _{1}d _{2}}=0..02mm$

Fringe width, $\beta=\dfrac{D\lambda}{d}$
hence $\lambda=\dfrac{\beta d}{D}=0.667nm$

When a certain photosensistive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is-${ V } _{ 0 }/2.$ When the surface frequency v/2, the stopping potential is -${ V } _{ 0 }.$ The threshold frequency for photoelectric emission is

  1. $\dfrac { 5v }{ 3 } $

  2. $2v$

  3. $\dfrac { 4 }{ 3 } v$

  4. $\dfrac { 3v }{ 2 } $


Correct Option: A

A thin water of $9.9\mu$ m thickness and refractive index $1.5$ is introduced in front of one of the slits in a Young's double slit experiment. By how many fringe width does the fringe pattern shift? (Given wavelength of the wave is $5.5\times 10^{-7}$m)

  1. $11$

  2. $9$

  3. $3$

  4. $8$


Correct Option: B

Monochromatic light of wavelength $400\ nm$ and $560\ nm$ are incident simultaneously and normally on double slits apparatus whose slits separation is $0.1\ mm$ and screen distance is $1\ m$. Distance between areas of total darkness will be:

  1. $4\ mm$

  2. $5.6\ mm$

  3. $14\ mm$

  4. $28\ mm$


Correct Option: D
Explanation:

For $\lambda =400$mm

fringe width
$\beta =\cfrac { \lambda D }{ d } \ =\cfrac { 400\times 1\times 10\times 1000 }{ 1\times { 10 }^{ 6 } } $
$ =4$mm
For $\lambda =560$mm
$\beta =\cfrac { \lambda D }{ d } =5.6$mm
$LCM$ of both fringe width will give combined fringe width which is also the distance between $2$ minima .
$LCM 4,5,6=28$mm

In, which of the following the interference is produced by the division of wave front? 

  1. Young's double slit experiment

  2. Fresnel's bi-prism

  3. Lloyd's mirror

  4. Colours of thin film


Correct Option: A
Explanation:

In Young's Double Slit Experiment, wavefront from a single sources divides when passed through the two slits. These two wavefronts interfere after emerging out of the slits.

The light of one single color is called as:

  1. Monochromatic light

  2. Bi-chromatic light

  3. Tri-chromatic light

  4. Poly-chromatic light


Correct Option: A
Explanation:

Monochromatic light 

A light which is composed of a single colour is called monochromatic light. It consists of single wavelength. 
Eg : LASER 

Monochromatic light beam passed through two narrow slits form a pattern of bright and dark lines on a screen. The phenomena causing this pattern is: 
  1. Refraction

  2. Reflection

  3. Polarization

  4. Interference

  5. Intensity


Correct Option: D
Explanation:

The waves pass through the double-slit and interfere with each other, creating nodes and antinodes on the screen, which we see as alternating bright and dark lines. 

State whether given statement is True or False
In Young's double slit experiment, the two coherent sources are real.

  1. True

  2. False


Correct Option: B
Explanation:
False.
Young's double slit experiment was done to demonstrate interference. The essential and the most sufficient condition for interference is that there must be a constant phase difference between the two waves emitted by the two sources (real or virtual). Such sources, having a constant phase difference and same frequency, are called coherent sources. Therefore, the sources used should be coherent, real or virtual does not matter. 

In Young's experiment, the source is red light of wavelength $7\times 10^{-7}m$. When a thin glass plate of refractive index $1.5$ at this wavelength is put is the path of one of the interfering beam, the central bright fringe shifts by $10^{-3} m$ to the position previously occupied by the $5^{th}$ bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength $5\times 10^{-7}m$ the central fringe shifts to a position initially occupied by the $6^{th}$ bright fringe due to red light. Find the refractive index of glass for the green light. Also estimate the change in fringe width due to the change in wavelength.

  1. $7\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.

  2. $8\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.

  3. $9\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.

  4. $17\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.


Correct Option: A

If the intensity of the waves observed by two coherent sources is $I$. Then the intensity if resultant wave in constructive interference will be:-

  1. $2 I$

  2. $4 I$

  3. $I$

  4. None of the above


Correct Option: A

In Young's experiment two coherent sources are placed $0.90\ mm$ apart and fringe are observed one metre away. If it produces second dark fringes at a distance of $1\ mm$ from central fringe., the wavelength of monochromatic light is used would be 

  1. $60\ \times 10^{-4}\ cm$

  2. $10\ \times 10^{-4}\ cm$

  3. $10\ \times 10^{-5}\ cm$

  4. $6\ \times 10^{-5}\ cm$


Correct Option: D
Explanation:
Formula,

$x=(2n-1)\dfrac{\lambda D}{2d}$

$\Rightarrow \lambda =\dfrac{2xd}{(2n-1)D}$

$=\dfrac{2\times 10^{-3}\times 0.9\times 10^{-3}}{(2\times 2-1)\times 1}$

$=6 \times 10^{-5}cm$

The yellow light source in Young's double- slit experiment is replaced by a monochromatic red light source of same intensity. Then the fringe width of the interference pattern in comparison with that of the previous pattern will 

  1. increase

  2. decrease

  3. remain unchanged

  4. vanish


Correct Option: A

In a Young's double slit experiment, $d = 1\ mm,$ $\lambda = 6000 \overset {0}{A}$ & $D = 1\ m.$ The slits produce same intensity on the screen. The minimum distance between two points on the screen having $75\%$ intensity of the maximum intensity is:

  1. $0.45\ mm$

  2. $0.40\ mm$

  3. $0.30\ mm$

  4. $0.20\ mm$


Correct Option: B

Direction:
The question has a paragraph followed by two statements, Statement-1 and Statement-2. 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 plane-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 wave change of $\pi $
Statement-2: The centre of the interference pattern is dark

  1. Statement-1 is true, Statement-2 is false.

  2. Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1

  3. Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1

  4. Statement-1 is false, Statement-2 is true


Correct Option: C
Explanation:

As the center of the pattern is dark, we know that the phase difference is $(2n+1)\pi$. Furthermore, we also know that for the center of the interface pattern, there is no path difference. So the difference can only be due to reflection which is $\pi$. Therefore, both the assertion and the reason are correct and the reason is explanatory,

The distance of n bright fringe to the $(n+1)^{th}$ dark fringe in Young's experiments is equal to : 

  1. $\frac{n\lambda D}{d}$

  2. $\frac{n\lambda D}{2d}$

  3. $\frac{\lambda D}{2d}$

  4. $\frac{\lambda D}{d}$


Correct Option: A

In a young's bouble slit experiment ,$d=1mm,\lambda  = 6000\mathop A\limits^0 $ and $D=1m$(where d,$\lambda$ and D have unit meaning). Each of slit individually produces same intensity on the screen. The minimum distance between two points  on the screen having $75$% intensity of the maximum intensity is:

  1. $0.45mm$

  2. $0.40mm$

  3. $0.30mm$

  4. $0.20mm$


Correct Option: B

 In Young's experiment , the separation between 5th maxima and 3rd minima is how many times as that of fringr width?

  1. 5 time

  2. 3 times

  3. 2 times

  4. 2.5times


Correct Option: C
Explanation:
For interference pattern, in double slit experiment maxima will be found at a distance y from the central maximum,
$y=n\lambda D/d$
$D=$distance between slit and screen
$d=$ separation of the slits
For $n=5$,
$y=5\lambda D/d$
For minima
$y=(2n+1)\lambda D/2d$
For $n=3$,
$y=7\lambda D/d$
Difference between maxima and minima$=7\lambda D/d-5\lambda D/d=2\lambda D/d$
Now, fridge width is given by $x=\lambda D/d$
Thus, the given difference is $2$ times the fringe width.

Monochromatic light of wavelength 580 mm incident on a slit of width 0.30 mm. The screen 2 m from the slit. the width of the center maximum is 

  1. $3.35\times 10^{-3}m$

  2. $2.25\times 10^{-3}m$

  3. $6.20\times 10^{-3}m$

  4. $7.7\times 10^{-3}m$


Correct Option: D
Explanation:

$d=580nm=580\times {10}^{-9}m$

$d=0.3mm=0.0003m$
width of the cental maxima $=\cfrac{2\lambda D}{d}$
$=\cfrac{2\times 580\times {10}^{-9}\times 2}{0.0003}$
$=0.0077m$

Monochromatic green light of wavelength 5 $\times 10^{-7}$ m illuminates a pair of slits 1 mm apart. The separation of bright lines in the interference pattern formed on a screen 2 m away is :

  1. 0.25 mm

  2. 0.1 mm

  3. 1.0 mm

  4. 0.01 mm


Correct Option: C
Explanation:

Given,


$\lambda=5\times 10^{-7}m$


$d=1mm=0.001m$

$D=2m$

The separation between two bright fringe is called fringe width.

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

$B=\dfrac{2\times 5\times 10^{-7}}{10^{-3}}$

$B=1mm$

The correct option is C.

A metal plate is placed 2 m away from a monochromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as large as 10 atomic diameters (${10^{ - 9}}\;m$) in radius, calculate how long it will take for such a 'target' to 'soak off' % eV of energy for its emission from the metal?

  1. 1.5 hr

  2. 2.5 hr

  3. 3.5 hr

  4. 4.5 hr


Correct Option: C

To make the central fringe at the centre O, a mica sheet of refractive index $1.5$ is introduced. Choose the correct statements (s).

  1. The thickness of sheet is $2\left( {\sqrt 2 - 1} \right)d$ in front of ${S _1}$

  2. The thickness of sheet is $\left( {\sqrt 2 - 1} \right)d$ in front of ${S _2}$

  3. The thickness of sheet is $2\sqrt 2 d$ in front of ${S _1}$

  4. The thickness of sheet is $\left( {2\sqrt 2 - 1} \right)d$ in front of ${S _1}$


Correct Option: A

Monochromatic green light of wavelength $5x{ 10 }^{ -7 }$ m illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed on a screen 2m away is 

  1. 0.25 mm

  2. 0.1 mm

  3. 1.0 mm

  4. 0.01 MM


Correct Option: C

Each of the four pairs of light waves arrives at a certain point on a screen. The waves have the same wavelength. At the arrival point, their amplitudes and phase differences are :
$2 \mathrm { a } _ { 0 } , 6 \mathrm { a } _ { 0 } $ and $\pi$ rad
$3 \mathrm { a } _ { 0 } , 5 \mathrm { a } _ { 0 } $ and $\pi$ rad
$9 \mathrm { a } _ { 0 } , 7 \mathrm { a } _ { 0 } $ and $3\pi$ rad
$2 \mathrm { a } _ { 0 } , 2\mathrm { a } _ { 0 } $ and $0$
The pair/s which has greatest intensity is /are :

  1. $I$

  2. $II$

  3. $II, III$

  4. $I, IV$


Correct Option: D

In Young's double slit. experiment, distance between two sources is 0.1 mm. The distance of screen from the sources is 20 cm. Wavelength of light used is 5460 k Then angular position of first dark fringe is 

  1. $0.20 ^ { \circ }$

  2. $0.32 ^ { \circ }$

  3. $0.08 ^ { \circ }$

  4. $0.16 ^ { \circ }$


Correct Option: D

The waves of $600 \mu m$ wave length are incident normally on a slit of $1.2\ mm$ width. The value of diffraction angle corresponding to the first minima will be (in radian):

  1. $\dfrac{\pi}{2}$

  2. $\dfrac{\pi}{6}$

  3. $\dfrac{\pi}{5}$

  4. $\dfrac{\pi}{4}$


Correct Option: B
Explanation:

Given,

$\lambda=600\mu m$
$d=1.2mm$
condition for minima 
$dsin\theta=m\lambda$
for $m=1$
$sin\theta=\dfrac{m\lambda}{d}=\dfrac{\lambda}{d}$

$sin\theta=\dfrac{600\times 10^{-6}}{1.2\times 10^{-3}}=0.5$

$\theta=sin^{-1}(0.5)=\dfrac{\pi}{6}$
The correct option is B.

In the case of interference, The maximum and minimum intensities are in the ratio $16:9$. Then

  1. The maximum and minimum amplitude will be in the ratio 9:5

  2. The intensities of the individual waves will be in the ratio 4:3.

  3. The amplitudes of the individual waves will be in the ratio 4:1

  4. None of the above is true


Correct Option: C

In young's double slits experiments , the distance between the slits is 1 mm and that between slit and screen is 1 meter and 10th fringe is 5 mm away from the central bright fringe, then wavelength of light used will be

  1. $ 8000 A^0 $

  2. $ 7000 A^0 $

  3. $ 6000 A^0 $

  4. $ 5000 A^0 $


Correct Option: A

Using monochromatic light of wavelength $\lambda $  an experimentalist sets up the Young's double sit experiment in three ways as shown, If the observes they $y=\beta $  the wavelenght of light used is 

  1. 520 nm

  2. 540 nm

  3. 560 nm

  4. 580 nm


Correct Option: A

In young's double slit experiment, when two light waves form third minimum intensity, they have 

  1. phase difference of 3$\pi$

  2. phase difference of $\dfrac { 5 \pi } { 2 }$

  3. path difference of 3$\lambda$

  4. path difference of $\dfrac { 5 \lambda } { 2 }$


Correct Option: B

For constructive interference to take place between two monochromatic light waves of wavelength $  \lambda  $ , the path difference should be

  1. $
    (2 n-1) \frac{\lambda}{4}
    $

  2. $
    (2 n-1) \frac{\lambda}{2}
    $

  3. $
    n \lambda
    $

  4. $
    (2 n+1) \frac{\lambda}{4}
    $


Correct Option: C

A young double slit experiment uses a monochromatic source The shape of the interference fringes formed on a screen is

  1. circle

  2. hyperbola

  3. parabola

  4. straight line


Correct Option: D

A slit of width $d$ is illuminated by white light. For red light ($\lambda = 6500 $), the first minima is obtained at $\theta = 30$. Then, the value of $d$ will be

  1. $3250 \mathrm { A }$

  2. $6.5 \times 10 ^ { - 4 } \mathrm { mm }$

  3. $1.24$ microns

  4. $2.6 \times 10 ^ { - 4 } \mathrm { cm }$


Correct Option: C

A beam of light of $  \lambda=600 n m  $ from a distant source falls on a single slit $1$ $ \mathrm{mm}  $ wide and the resulting diffraction pattern is observed on a screen $2$ $ \mathrm{m}  $ away. The distance between first dark fringes on either side of the central bright fringe is

  1. $
    1.2 \mathrm{cm}
    $

  2. $
    1.2 \mathrm{mm}
    $

  3. $
    2.4 \mathrm{cm}
    $

  4. $
    2.4 \mathrm{mm}
    $


Correct Option: D

A concave mirror having radius of curvature 40cm is placed in front of an illuminated source at a distance of 30cm. If find the location of the image.

  1. $4F$

  2. $6F$

  3. $3F$

  4. None of these


Correct Option: C

A monochromatic light of $\lambda =5000{ A }^{ \circ  }$ is incident on the slits separated by a distance $5\times { 10 }^{ -4 }m.$ The interference pattern is seen on a screen placed at a distance 1 m from the slits. A thin glass plate of thickness $1.5\times { 10 }^{ -6 }m$ and refractive index $\mu =1.5$ is placed  between one of the slits and the screen. The lateral shift of the central maximum is 

  1. $1.5\times { 10 }^{ -3 }m$

  2. $3\times { 10 }^{ -3 }m$

  3. $4.5\times { 10 }^{ -3 }m$

  4. $6\times { 10 }^{ -3 }m$


Correct Option: A

Two wavelengths of light of wavelength ${\lambda _1} = 4500\mathop {\text{A}}\limits^{\text{o}} $
 and ${\lambda _2} = 6000\mathop {\text{A}}\limits^{\text{o}} $ are sent through a Young's double slit apparatus simultaneously then

  1. no interference pattern will be formed

  2. the third order bright fringe of ${\lambda _1}$ will coincide with the fourth order bright fringe of ${\lambda _2}$

  3. the third order bright fringe of ${\lambda _2}$ will coincide with fourth order bright fringe of ${\lambda _1}$

  4. the fringes of wavelength ${\lambda _1}$ will be wider than the fringes of wavelength ${\lambda _2}$


Correct Option: C

In $YDSE$, slab of thickness $t$ and refractive index $\mu$ is placed in front of any slit. Then displacement of central maximum in terms of fringe width when light of wavelength $\lambda$ is incident on system is 

  1. $\dfrac{\beta(\mu - 1)t}{2\lambda}$

  2. $\dfrac{\beta(\mu - 1)t}{\lambda}$

  3. $\dfrac{\beta(\mu - 1)t}{3\lambda}$

  4. $\dfrac{\beta(\mu - 1)t}{4\lambda}$


Correct Option: B
Explanation:

Displacement of central maximum $(y)$
$=\beta(\mu - 1)t = \dfrac{dy}{D}$
$y = \dfrac{\lambda D(\mu -1)t}{\lambda d}$            $\left(\beta = \dfrac{\lambda D}{d}\right)$
$sy = \dfrac{\beta (\mu - 1)t}{\lambda}$

Light of wavelength $ 5000 \mathring { A }  $ passes through a slit of width 6.5 cm and forms a difference pattern with a lens of focal length 40 cm, held close to the slit.The distance between the first minimum and the first secondary maximum is

  1. $ 2 \times 10^{-6} m $

  2. $ 2 \times 10^{-4} m $

  3. $ 4 \times 10^{-6} m $

  4. $ 4 \times 10^{-5} m $


Correct Option: B

In Young's double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of a refractive index $1.6$ and thickness $1.964$ microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the plane of slits and the screen is doubled. It is found that the the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. The wavelength of the light will be

  1. $3000\overset {\circ}{A}$

  2. $4850\overset {\circ}{A}$

  3. $5892\overset {\circ}{A}$

  4. None of these


Correct Option: A
Youngs double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are ${ \beta  } _{ G }$, ${ \beta  } _{ R }$, ${ \beta  } _{ B }$ and respectively. Then,
  1. ${ \beta } _{ G }$ > ${ \beta } _{ B }$ > ${ \beta } _{ R }$

  2. ${ \beta } _{ B }$ > ${ \beta } _{ G }$ >${ \beta } _{ R }$

  3. ${ \beta } _{ R }$ > ${ \beta } _{ B }$ > ${ \beta } _{ G }$

  4. ${ \beta } _{ R }$ > ${ \beta } _{ G }$ > ${ \beta } _{ B }$


Correct Option: D
Explanation:
Fring width, $\beta =\frac{\lambda D}{d}$  and hence $\beta \propto \lambda $
since $\lambda _{red} > \lambda _{green}  > \lambda _{blue}$
So $\beta _{red} > \beta _{green } > \beta _{blue}$
So, the correct option will be $(D)$

The Young's modulus of a wire is determined by the apparatus known as

  1. Kater's pendulum apparatus

  2. Poiseuille's apparatus

  3. Maxwell's needle

  4. Searle's apparatus


Correct Option: D
Explanation:

It is the Searle's apparatus which is used to determine the Young's modulus of a wire.

If a torch is used in place of monochromatic light in Young's experiment what will happens?

  1. Fringe will appear for a moment then, it will disappear

  2. Fringes will occur as from monochromatic light

  3. Only bright fringes will appear

  4. No fringes will appear


Correct Option: D

In Young's double slit experiment if monochromatic light used is replaced by white light, then

  1. all bright fringes become white.

  2. all bright fringes have colors between violet and red.

  3. no fringes are observed.

  4. only central fringe is white, all other fringes are colored.


Correct Option: D
Explanation:

Thus if we use white light in place of monochromatic light the central fringe is white, containing on either side a few coloured fringes (in order VIBGYOR) and the remaining screen appears uniformly illuminated.

On a rainy day, small oil films on water show brilliant colours. This is due to

  1. dispersion

  2. interference

  3. diffraction

  4. polarization


Correct Option: B
Explanation:

Interference 

Interference is a phenomenon in which two waves superimpose to form a resultant wave of greater, lower or same amplitude. 

In Young's double slit experiment, the wavelength of red light $7500\overset {\circ}{A}$ and that of blue light is $5000\overset {\circ}{A}$. The value of $n$ for which $n^{th}$ bright band due to red light coincides with $(n + 1)^{th}$ bright band due to blue light, is

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: B
Explanation:

For bright 

distance $y=n\lambda \cfrac { D }{ d } \ \therefore \quad n=7500\cfrac { D }{ d } =(n+1)5000\cfrac { D }{ d } \ \therefore \quad \cfrac { n }{ n+1 } =\cfrac { 2 }{ 3 } \ \therefore \quad n=2\quad $

In $YDSE$ how many maxima can be obtained on the screen if wavelength of light used is $200\ nm$ and $d = 700\ nm$.

  1. $12$

  2. $7$

  3. $18$

  4. $None\ of\ these$


Correct Option: B
Explanation:

$Given\quad that\quad \lambda =200nm,\quad d=700nm\ \theta =\frac { \lambda  }{ d } \ We\quad know\quad that\quad maximum\quad angle\quad can\quad be\quad { 90 }^{ 0 }.\ No\quad of\quad rings\quad that\quad can\quad be\quad obtained\quad say\quad n,\ (n)=\frac { sin90 }{ sin\theta  } =\frac { 1 }{ sin\theta  } \ If\quad \theta \quad is\quad so\quad small\quad then-\ n=\frac { 1 }{ \theta  } \ =\frac { d }{ \lambda  } =\frac { 7 }{ 2 } \ No\quad of\quad maxima\quad =2n\ \quad \quad \quad =2\times \frac { 7 }{ 2 } \ \quad \quad \quad =7$

In a Young's double slit experiment the intensity at a point where the path difference is $ \dfrac { \lambda  }{ 6 } $ ($\lambda $ being the wavelength of the light used) is $I$. if ${ I } _{ 0 }$ denotes the maximum intensity, is equal to

  1. $\dfrac { 1 }{ \sqrt { 2 } } $

  2. $\dfrac { \sqrt { 3 } }{ 2 } $

  3. $\dfrac { 1 }{ 2 } $

  4. $\dfrac { 3 }{ 4 } $


Correct Option: D
Explanation:

For maximum intensity $\phi =0$

Assume that ${ I } _{ 1 }={ I } _{ 2 }={ I } _{ 3 }$(source intensity)

$\therefore { I } _{ 0 }={ I } _{ 3 }+{ I } _{ 2 }+2\sqrt { { I } _{ 2 }{ I } _{ 3 } } \\ { I } _{ 0 }={ 4I } _{ 3 }\longrightarrow 1\\ \Delta x=\cfrac { \lambda  }{ 6 } \\ \therefore \Delta \phi =\cfrac { 2\pi  }{ \lambda  } \Delta x\quad =\cfrac { 2x }{ \lambda  } \times \cfrac { \lambda  }{ 6 } \\ \quad \quad \quad \quad =\cfrac { \pi  }{ 3 } \\ \therefore I={ I } _{ s }+{ I } _{ s }+2\sqrt { { I } _{ s }{ I } _{ s } } \cos { \cfrac { \pi  }{ 3 }  } \\ \quad \quad =3{ I } _{ 3 }\\ I=\cfrac { 3 }{ 4 } { I } _{ 0 }$

 

In Young's double slit experiment using monochromatic light of wavelengths $\lambda$, the intensity of light at a point on the screen with path difference $\lambda$ is M units. The intensity of light at a point where path difference is $\lambda/3$ is then

  1. $\frac{M}{2}$

  2. $\frac{M}{4}$

  3. $\frac{M}{8}$

  4. $\frac{M}{16}$


Correct Option: B
Explanation:

Resultant intensity
$I _R = I _1 + I _2 + 2\sqrt{I _1 I _2} cos \theta$
If path difference = $\lambda$ phase difference $2 \pi$
$\therefore I _R = I + I + 2 \sqrt{I \times I} cos 2 \pi = 4 I = M$
If path difference $\dfrac{\lambda}{3}$, phase difference $\phi = \dfrac{2 \pi}{3} rad$
$I _R' = I + I + 2 \sqrt{I \times I} cos \dfrac{2 \pi}{3} = I = \dfrac{M}{4}$

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