Tag: explaining wave phenomena

Questions Related to explaining wave phenomena

A slit of width $d$ is illuminated by white light. For red light ($\lambda = 6500 $), the first minima is obtained at $\theta = 30$. Then, the value of $d$ will be

difference between interference and diffraction explaining wave phenomena diffraction

  1. $3250 \mathrm { A }$

  2. $6.5 \times 10 ^ { - 4 } \mathrm { mm }$

  3. $1.24$ microns

  4. $2.6 \times 10 ^ { - 4 } \mathrm { cm }$

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Correct Option: C

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

difference between interference and diffraction explaining wave phenomena diffraction

  1. $
    1.2 \mathrm{cm}
    $

  2. $
    1.2 \mathrm{mm}
    $

  3. $
    2.4 \mathrm{cm}
    $

  4. $
    2.4 \mathrm{mm}
    $

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Correct Option: D

A concave mirror having radius of curvature 40cm is placed in front of an illuminated source at a distance of 30cm. If find the location of the image.

difference between interference and diffraction explaining wave phenomena diffraction

  1. $4F$

  2. $6F$

  3. $3F$

  4. None of these

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Correct Option: C

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 

difference between interference and diffraction explaining wave phenomena diffraction

  1. $1.5\times { 10 }^{ -3 }m$

  2. $3\times { 10 }^{ -3 }m$

  3. $4.5\times { 10 }^{ -3 }m$

  4. $6\times { 10 }^{ -3 }m$

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Correct Option: A

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

difference between interference and diffraction explaining wave phenomena diffraction

  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 ${\lambda _2}$

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Correct Option: C

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 

difference between interference and diffraction explaining wave phenomena diffraction

  1. $\dfrac{\beta(\mu - 1)t}{2\lambda}$

  2. $\dfrac{\beta(\mu - 1)t}{\lambda}$

  3. $\dfrac{\beta(\mu - 1)t}{3\lambda}$

  4. $\dfrac{\beta(\mu - 1)t}{4\lambda}$

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Correct Option: B
Explanation:

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

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

difference between interference and diffraction explaining wave phenomena diffraction

  1. $ 2 \times 10^{-6} m $

  2. $ 2 \times 10^{-4} m $

  3. $ 4 \times 10^{-6} m $

  4. $ 4 \times 10^{-5} m $

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Correct Option: B

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

difference between interference and diffraction explaining wave phenomena diffraction

  1. $3000\overset {\circ}{A}$

  2. $4850\overset {\circ}{A}$

  3. $5892\overset {\circ}{A}$

  4. None of these

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Correct Option: A
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,
difference between interference and diffraction explaining wave phenomena diffraction

  1. ${ \beta } _{ G }$ > ${ \beta } _{ B }$ > ${ \beta } _{ R }$

  2. ${ \beta } _{ B }$ > ${ \beta } _{ G }$ >${ \beta } _{ R }$

  3. ${ \beta } _{ R }$ > ${ \beta } _{ B }$ > ${ \beta } _{ G }$

  4. ${ \beta } _{ R }$ > ${ \beta } _{ G }$ > ${ \beta } _{ B }$

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Correct Option: D
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 $(D)$

The Young's modulus of a wire is determined by the apparatus known as

difference between interference and diffraction explaining wave phenomena diffraction

  1. Kater's pendulum apparatus

  2. Poiseuille's apparatus

  3. Maxwell's needle

  4. Searle's apparatus

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Correct Option: D
Explanation:

It is the Searle's apparatus which is used to determine the Young's modulus of a wire.