Physics

Wave Optics and Diffraction

148 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 difference between interference and diffraction explaining wave phenomena diffraction

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

Constructive interference occurs when the path difference is an integer multiple of the wavelength, i.e., n * lambda.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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

For single slit diffraction, the distance between the first dark fringes on either side is 2 * lambda * D / d. Given lambda = 600 * 10^-9 m, D = 2 m, d = 10^-3 m. Distance = 2 * (600 * 10^-9 * 2) / 10^-3 = 2.4 * 10^-3 m = 2.4 mm.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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. $6\times { 10 }^{ -3 }m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The lateral shift of the central maximum is given by (mu - 1) * t * D / d. Given mu = 1.5, t = 1.5 * 10^-6 m, D = 1 m, d = 5 * 10^-4 m. Shift = (1.5 - 1) * 1.5 * 10^-6 * 1 / (5 * 10^-4) = 0.5 * 1.5 * 10^-2 / 5 = 0.15 * 10^-2 = 1.5 * 10^-3 m.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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 ${\lambda _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Bright fringes coincide when n1 * lambda1 = n2 * lambda2. n1 * 4500 = n2 * 6000. n1 / n2 = 6000 / 4500 = 4 / 3. Thus, the 4th order of lambda1 coincides with the 3rd order of lambda2.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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. $\dfrac{\beta(\mu - 1)t}{4\lambda}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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

Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction
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. ${ \beta } _{ R }$ > ${ \beta } _{ G }$ > ${ \beta } _{ B }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $(D)$
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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.

Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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