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

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

  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 $b _{min} = \left (\dfrac {2\lambda^{2}}{L}\right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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:

  1. $10\ V/m$

  2. $24\ V/m$

  3. $30\ V/m$

  4. $6\ V/m$
Reveal answer Fill a bubble to check yourself
B Correct answer
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} $ $\times (30)^2$
$A _t\sqrt { \frac { 64 }{ 100 }\times(30)^2  } =24$

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

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: $\left( \lambda = 6000 \mathrm { A } ^ { 9 } \right)$ 

  1. 0.3$\mathrm { mm }$

  2. 0.1$\mathrm { mm }$

  3. 0.4$\mathrm { mm }$

  4. 0.2$\mathrm { mm }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

Conditions of diffraction is

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

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

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

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

Diffraction occurs significantly when the size of the aperture or obstacle is comparable to or smaller than the wavelength of the incident wave. Thus, a/lambda << 1 is the condition for significant diffraction.

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

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

  1. Bands will narrower and crowd
    full together

  2. Bands become broader and further apart

  3. No change will take place

  4. Bands disappear
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Fringe width beta = lambda * D / d. Since the wavelength of blue light is shorter than that of red light, the fringe width decreases, causing the bands to become narrower and more crowded.

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

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, 

  1. $3$ times

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

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

  4. $\displaystyle\frac{1}{3}$ times
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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 $4/3$).

  1. $0.3mm$

  2. $0.225mm$

  3. $0.4mm$

  4. $1.2mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $\omega _{water}=1.2mm$