Tag: difference between interference and diffraction

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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 $(D)$