Tag: diffraction

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Questions Related to diffraction

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

If a torch is used in place of monochromatic light in Young's experiment what will happens?

  1. Fringe will appear for a moment then, it will disappear

  2. Fringes will occur as from monochromatic light

  3. Only bright fringes will appear

  4. No fringes will appear
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A torch emits white light, which is a mixture of many wavelengths. Each wavelength produces its own interference pattern, and these patterns overlap, washing out the fringes.

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

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

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In a Young's double slit experiment the intensity at a point where the path difference is $ \dfrac { \lambda  }{ 6 } $ ($\lambda $ being the wavelength of the light used) is $I$. if ${ I } _{ 0 }$ denotes the maximum intensity, is equal to

  1. $\dfrac { 1 }{ \sqrt { 2 } } $
  2. $\dfrac { \sqrt { 3 } }{ 2 } $
  3. $\dfrac { 1 }{ 2 } $
  4. $\dfrac { 3 }{ 4 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For maximum intensity $\phi =0$

Assume that ${ I } _{ 1 }={ I } _{ 2 }={ I } _{ 3 }$(source intensity)

$\therefore { I } _{ 0 }={ I } _{ 3 }+{ I } _{ 2 }+2\sqrt { { I } _{ 2 }{ I } _{ 3 } } \\ { I } _{ 0 }={ 4I } _{ 3 }\longrightarrow 1\\ \Delta x=\cfrac { \lambda  }{ 6 } \\ \therefore \Delta \phi =\cfrac { 2\pi  }{ \lambda  } \Delta x\quad =\cfrac { 2x }{ \lambda  } \times \cfrac { \lambda  }{ 6 } \\ \quad \quad \quad \quad =\cfrac { \pi  }{ 3 } \\ \therefore I={ I } _{ s }+{ I } _{ s }+2\sqrt { { I } _{ s }{ I } _{ s } } \cos { \cfrac { \pi  }{ 3 }  } \\ \quad \quad =3{ I } _{ 3 }\\ I=\cfrac { 3 }{ 4 } { I } _{ 0 }$

 

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In Young's double slit experiment using monochromatic light of wavelengths $\lambda$, the intensity of light at a point on the screen with path difference $\lambda$ is M units. The intensity of light at a point where path difference is $\lambda/3$ is then

  1. $\frac{M}{2}$
  2. $\frac{M}{4}$
  3. $\frac{M}{8}$
  4. $\frac{M}{16}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resultant intensity
$I _R = I _1 + I _2 + 2\sqrt{I _1 I _2} cos \theta$
If path difference = $\lambda$ phase difference $2 \pi$
$\therefore I _R = I + I + 2 \sqrt{I \times I} cos 2 \pi = 4 I = M$
If path difference $\dfrac{\lambda}{3}$, phase difference $\phi = \dfrac{2 \pi}{3} rad$
$I _R' = I + I + 2 \sqrt{I \times I} cos \dfrac{2 \pi}{3} = I = \dfrac{M}{4}$

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

How can resolving power of the instrument be increased?

  1. use UV light

  2. immerse in oil

  3. use IR light

  4. use one more lens.
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Resolving power for the instrument is found to be $\dfrac{\mu\sin\theta}{0.61\lambda}$ , UV light has short wavelength, hence higher resolving power. Oil is optically denser than air, that is, its $\mu$ is greater than that of air. Thus immersing in oil would increase the resolving power.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

The ability of an optical instruments to show the images of two adjacent point objects as separate is called :

  1. dispersive power

  2. magnifying power

  3. resolving power

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

By definition, resolving power of an optical instrument is its ability to show two closely adjacent point (closely spaced) as distinct as possible.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Two lenses of focal lengths $+ 100 cm$ and $+ 5 cm$ are used to prepare an astronomical telescope. The minimum tube length will be : (final image is at $\displaystyle \infty $)

  1. $95 cm$
  2. $100 cm$
  3. $105 cm$
  4. $500 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

the length of telescope =focal  length  of  object $(-f _0)$ +focal  length  of  eyepiece  $(f _e)$$=100+5=105cm$