Tag: explaining wave phenomena

Questions Related to explaining wave phenomena

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.

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. Bright, $5891A^o$

  2. Bright, $6891A^o$

  3. Dark, $5891A^o$

  4. Dark, $6891A^o$

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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 :

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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

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

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

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

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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 

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. 600 nm

  2. 1200 nm

  3. 60 A$^o$

  4. 120 A$^o$

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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

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $5$

  2. $4$

  3. $3$

  4. $2$

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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 :

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $1.2$

  2. $0.6$

  3. $2.4$

  4. $1.5$

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Correct Option: A

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

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. 1:4

  2. 4:0

  3. 9:4

  4. 2:3

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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 :

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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

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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:

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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

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

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

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

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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:

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $10$

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

  3. $100$

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

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Correct Option: A

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

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. increase

  2. decrease

  3. remain constant

  4. none of these

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Correct Option: B