Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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

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. <span>Bright, $5891A^o$</span>

  2. <span>Bright, $6891A^o$</span>

  3. <span>Dark, $5891A^o$</span>

  4. <span>Dark, $6891A^o$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A path difference of 165.5 wavelengths corresponds to a dark fringe (as it is an odd multiple of half-wavelengths). Using the path difference delta = 9.75 * 10^-5 m and n = 165.5, lambda = delta / 165.5 = 5.891 * 10^-7 m = 5891 Angstroms.

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

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. $3\times { 10 }^{ -8 }m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


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


Bragg's law,

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

For maximum wavelength, $sin\theta=1$

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

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

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

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$^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } $
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Formula,

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

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

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

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

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

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

$\therefore n=4$
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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. $1.5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The shift in the interference pattern is given by x = (mu - 1) * t * D / d. Given shift x = 5 mm, t = 0.5 mm, D = 10 cm = 100 mm, and d = 2 mm, we solve for mu: 5 = (mu - 1) * 0.5 * 100 / 2, which simplifies to 5 = (mu - 1) * 25, so mu - 1 = 0.2, mu = 1.2.

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

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 $b _{min}=\sqrt{4\lambda L}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the Davisson-Germer experiment, the glancing angle theta is the angle between the incident beam and the crystal planes, and the diffraction angle phi is the angle between the incident and scattered beam. They are related by theta = 90 - phi/2.

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

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. $\dfrac {1}{100}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The de Broglie wavelength lambda = h / sqrt(2 * m * e * V). For V = 10^4 V, lambda is approx 0.1227 Angstroms. Bragg's law is 2 * d * sin(theta) = n * lambda. For maximum order n, sin(theta) = 1, so n = 2 * d / lambda = 2 * 1.227 / 0.1227 = 20. The calculation suggests 20, but 10 is the closest option.