Tag: wave optics

Questions Related to wave optics

The box of a pin hole camera, of length $L$, has a hole of radius $a$. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $b _{min}$) when

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $a = \sqrt {\lambda L}$ and $b _{min} = \left (\dfrac {2\lambda^{2}}{L}\right )$

  2. $a = \sqrt {\lambda L}$ and $b _{min} = \sqrt {4\lambda L}$

  3. $a = \dfrac {\lambda^{2}}{L}$ and $b _{min} = \sqrt {4\lambda L}$

  4. $a = \dfrac {\lambda^{2}}{L}$ and $b _{min} = \left (\dfrac {2\lambda^{2}}{L}\right )$

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Correct Option: D

A light wave is incident normally on a glass slab of refractive index $1.5$. If $4\%$ of light gets reflected and the amplitude of the electric field of the incident light is $30\ V/m$,then the amplitude of the electric field for the wave propogating in the glass medium will be:

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $10\ V/m$

  2. $24\ V/m$

  3. $30\ V/m$

  4. $6\ V/m$

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Correct Option: B
Explanation:

$P _{refracted}=\frac {  96}{ 100 } P _1$
$\Rightarrow K _2A^2 _t=\frac {  96}{ 100 }K _1A^2 _i$
$\Rightarrow r _2A^2 _t=\frac {  96}{ 100 }r _1A^2 _i$
$\Rightarrow A^2 _t=$ $ \frac {  96}{ 100 } \times$ $\frac { 1 }{ 3 } {2} $ $\times (30)^2$
$A _t\sqrt { \frac { 64 }{ 100 }\times(30)^2  } =24$

Light of wavelength 6328 A IS incident normally on a slit having a width of 0.2 mm. The width of the central maximum measured from minimum to minimum of diffraction pattern on a screen 9.0 meters away will be about

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. 0.72 degrees

  2. 0.09 degrees

  3. 0.36 degrees

    1. 18 degrees
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Correct Option: A

In a biprism experiment, interference bands are observed at a distance of one meter fromthe slit. A convex lens is put between the slit and the eyepiece gives two images of slit 0.7$\mathrm { cm }$ apart, the lens being 70$\mathrm { cm }$ from the eyepiece. The fringe width will be: $\left( \lambda = 6000 \mathrm { A } ^ { 9 } \right)$ 

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. 0.3$\mathrm { mm }$

  2. 0.1$\mathrm { mm }$

  3. 0.4$\mathrm { mm }$

  4. 0.2$\mathrm { mm }$

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Correct Option: A

Conditions of diffraction is

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $
    \frac{a}{\lambda}=1
    $

  2. $
    \frac{a}{\lambda}>>1
    $

  3. $
    \frac{a}{\lambda}<<1
    $

  4. None of these

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Correct Option: C

A diffraction is obtained by using a beam of red light. What will happen if the red light is replaced by the blue light 

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. Bands will narrower and crowd
    full together

  2. Bands become broader and further apart

  3. No change will take place

  4. Bands disappear

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Correct Option: A

Direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by (a is the width of the slit)

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $
    \operatorname{asin} \theta=\dfrac{\lambda}{2}
    $

  2. $
    a \cos \theta=\dfrac{3 \lambda}{2}
    $

  3. $
    \operatorname{asin} \theta=\lambda
    $

  4. $
    a \sin \theta=\dfrac{3 \lambda}{2}
    $

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Correct Option: A

If the whole bi-prism experiment is immersed in water then the fringe width becomes, if the refractive indices of bi-prism material and water are $1.5$ and $1.33$ respectively, 

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $3$ times

  2. $\displaystyle\frac{3}{4}$ times

  3. $\displaystyle\frac{4}{3}$ times

  4. $\displaystyle\frac{1}{3}$ times

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Correct Option: A
Explanation:
Separation between the coherent sources when the entire setup is in air $d _{ga} = 2a( \mu _g -\mu _a)A$
Separation between the coherent sources when the entire setup is put inside water $d _{gw} = 2a( \mu _g -\mu _w)A$

where  $a=$distance  between the single slit and the biprism, A is the prism angle and $\mu _g$ is the redfractive index of biprism.

Fringe width $\beta _{ga} = \dfrac{ D\lambda}{d _{ga}}$
Fringe width $\beta _{gw} = \dfrac{ D\lambda}{d _{gw}}$
$\therefore \dfrac{\beta _{ga}}{\beta _{gw}}=\dfrac{ d _{gw}}{d _{gw}}=\dfrac{ \mu _g - \mu _w}{ \mu _g - \mu _a}=\dfrac{1.5 -1.33}{1.5 -1}=\dfrac{0.17}{0.5}=\dfrac{0.17}{0.5}=\dfrac{1}{3}$
Hence, the fringe width increases 3 times.

In the Fresnel bi-prism experiment, the refractive index for the bi-prism is $\mu=3/2$ and fringe width obtained is $0.4mm$. If the whole apparatus is immersed as such in water then the fringe width will become(refractive index of water is $4/3$).

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $0.3mm$

  2. $0.225mm$

  3. $0.4mm$

  4. $1.2mm$

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Correct Option: D
Explanation:

In a fresnel bi-prism, fringe width,$\omega _{medium}=\dfrac{D\lambda _{medium}}{d _{medium}}$

where $\lambda _{medium}=\dfrac{\lambda _{vacuum}}{\mu _{medium}}$
and $d _{medium}\propto A(\dfrac{\mu _{bi-prism}}{\mu _{medium}}-1)$
Here $\mu _{medium}=\dfrac{4}{3}$ and $\mu _{bi-prism}=\dfrac{3}{2}$
Hence $\dfrac{\omega _{water}}{\omega _{vacuum}}=3$
Hence, $\omega _{water}=1.2mm$

In bi-prism experiment, fringes are obtained by white light sources. The fringe nearest to the central fringe will be

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. yellow

  2. green

  3. violet

  4. red

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Correct Option: C
Explanation:

Position of first fringe is given by $y _{1}=\dfrac{D\lambda}{d}$

Hence the nearest fringe is one with least wavelength, which is violet.