Physics

Wave Optics and Diffraction

163 Questions

Wave optics explores light phenomena like interference and diffraction. This hub features problems on Young's double slit experiment, Fresnel biprism, and single slit diffraction patterns. These topics regularly appear in physics sections of competitive exams.

Young's double slitFresnel biprismSingle slit diffractionConstructive interferenceFringe width calculations

Wave Optics and Diffraction Questions

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 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 $b _{min} = \left (\dfrac {2\lambda^{2}}{L}\right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The angular width of the central maximum is given by 2 * lambda / a. Substituting lambda = 6328 angstroms and a = 0.2 mm yields the angular width in radians, which can then be converted to degrees to find approximately 0.36 degrees.

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

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

Conditions of diffraction is

  1. $
    \frac{a}{\lambda}=1
    $
  2. $
    \frac{a}{\lambda}>>1
    $
  3. $
    \frac{a}{\lambda}<<1
    $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Diffraction occurs significantly when the size of the aperture or obstacle is comparable to or smaller than the wavelength of the incident wave. Thus, a/lambda << 1 is the condition for significant diffraction.

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

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}
    $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In single slit diffraction, the secondary maxima occur approximately where the path difference is an odd multiple of lambda / 2, given by a sin(theta) = (2n + 1)lambda / 2. For the first secondary maximum, n = 1, which gives a sin(theta) = 3lambda / 2.

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

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

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. $\displaystyle\frac{2.5\lambda}{(\mu-1)}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $t=\dfrac{2.5\lambda}{(\mu-1)}$

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

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. $d=\displaystyle\sqrt\frac{d^2 _1}{d _2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

 $d=\sqrt{d _{1}d _{2}}$.