Tag: diffraction

Questions Related to diffraction

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

difference between interference and diffraction explaining wave phenomena diffraction

  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

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Correct Option: D

In Young's double slit experiment if monochromatic light used is replaced by white light, then

difference between interference and diffraction explaining wave phenomena diffraction

  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.

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Correct Option: D
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.

On a rainy day, small oil films on water show brilliant colours. This is due to

difference between interference and diffraction explaining wave phenomena diffraction

  1. dispersion

  2. interference

  3. diffraction

  4. polarization

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Correct Option: B
Explanation:

Interference 

Interference is a phenomenon in which two waves superimpose to form a resultant wave of greater, lower or same amplitude. 

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

difference between interference and diffraction explaining wave phenomena diffraction

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: B
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 $

In $YDSE$ how many maxima can be obtained on the screen if wavelength of light used is $200\ nm$ and $d = 700\ nm$.

difference between interference and diffraction explaining wave phenomena diffraction

  1. $12$

  2. $7$

  3. $18$

  4. $None\ of\ these$

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Correct Option: B
Explanation:

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

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

difference between interference and diffraction explaining wave phenomena diffraction

  1. $\dfrac { 1 }{ \sqrt { 2 } } $

  2. $\dfrac { \sqrt { 3 } }{ 2 } $

  3. $\dfrac { 1 }{ 2 } $

  4. $\dfrac { 3 }{ 4 } $

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Correct Option: D
Explanation:

For maximum intensity $\phi =0$

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

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

 

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

difference between interference and diffraction explaining wave phenomena diffraction

  1. $\frac{M}{2}$

  2. $\frac{M}{4}$

  3. $\frac{M}{8}$

  4. $\frac{M}{16}$

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Correct Option: B
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 $\phi = \dfrac{2 \pi}{3} rad$
$I _R' = I + I + 2 \sqrt{I \times I} cos \dfrac{2 \pi}{3} = I = \dfrac{M}{4}$

How can resolving power of the instrument be increased?

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

  1. use UV light

  2. immerse in oil

  3. use IR light

  4. use one more lens.

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Correct Option: A,B
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.

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

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

  1. dispersive power

  2. magnifying power

  3. resolving power

  4. none of these

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Correct Option: C
Explanation:

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

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 $)

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

  1. $95 cm$

  2. $100 cm$

  3. $105 cm$

  4. $500 cm$

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Correct Option: C
Explanation:

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