Tag: wave optics

Questions Related to wave optics

The average path difference between two waves coming from third and fifth Fresnel zones of a wave front at the centre of the screen is :

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

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

  2. $2\lambda $

  3. $\lambda$

  4. $4\lambda$

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Correct Option: C
Explanation:

The third and fifth fresnel zone differ in phase by $2\pi $. Thus as $2\pi $ corresponds to path difference of $\lambda $.

In Young's double slit experiment, angular width of fringes is $0.20^o$ for sodium light of wavelength $5890\overset{o}{A}$. If complete system is dipped in water, then angular width of fringes becomes.

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

  1. $0.15^o$

  2. $0.22^o$

  3. $0.30^o$

  4. None of these

Show answer
Correct Option: A
Explanation:

b'Angular fringe width$=\theta=\displaystyle\frac{\lambda}{d}$
$\Rightarrow \theta\propto \lambda$
$\lambda _w=\displaystyle\frac{\lambda _a}{\mu _w}$
So, $\theta _w=\displaystyle\frac{\theta _{air}}{\mu _w}=\frac{0.20}{\displaystyle\frac{4}{3}}=0.15^o$'

In single slit diffraction pattern

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

  1. Centre fringe has negligible width than others.

  2. All fringes are of same width

  3. Central fringe does not exist

  4. None of the above

Show answer
Correct Option: D
Explanation:

In single silt diffraction, the central fringe has maximum intensity and has the width double than other fringes.

A transverse wave propagating along x-axis isrepresented by
$y\left (x, t \right ) = 8.0 \sin \left (0.5 \pi x - 4 \pi rt - \frac{\pi} {4}  \right )$
where $x$ is in metres and t is in seconds. The speed of the wave is:-

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

  1. $4 \pi$ m/s

  2. $0.5 \pi$ m/s

  3. $\frac{\pi} {4}$ m/s

  4. 8 m/s

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Correct Option: D
Explanation:

$V$ = $\frac{\omega} {k}$ = $\frac{4\pi} {0.5 \pi}$ = $8ms^{-1}$

The distance between two virtual images of slits in a biprism experiment is measured using the convex-lens which is

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

  1. Foucault's method

  2. Huygen's method

  3. Young's method

  4. conjugate foci method

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Correct Option: A

A biprism arrangement in air is immersed completely in a liquid. The fringe width

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

  1. remains same

  2. increases

  3. decreases

  4. none of the above

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Correct Option: C
Explanation:

$\beta =\cfrac { \lambda D }{ d } $

when immersed in a liquid, ${ \lambda  } _{ 1 }=\cfrac { \lambda  }{ \mu  } \ \therefore { \beta  }^{ 1 }=\cfrac { { \lambda  }^{ 1 } }{ d } =\cfrac { \lambda D }{ \mu d } =\cfrac { \beta  }{ \mu  } \ \therefore \beta \quad decreases $

A beam of natural light falls on a system of $6$ polaroids, which are arranged in succession such that each polaroid is turned through ${30}^{o}$ with respect to the preceeding one. The percentage of incident intensity that passes through the system will be

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

  1. $100$%

  2. $50$%

  3. $30$%

  4. $12$%

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Correct Option: D
Explanation:

If $I$ is the final intensity and $I _0$ is the initial intensity then,

$I=\dfrac{I _0}{2}(\cos^2{30})^5$
$\dfrac{I _0}{I}=\dfrac{1}{2}\times \left(\dfrac{\sqrt3}{2}\right)^{10}=0.12=12 percent$

In the diffraction of light of wavelength $\lambda$ through single slit of width d, the angle between the principal maxima and first minima will be:

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

  1. $\lambda/d$

  2. $\lambda/2d$

  3. $\lambda/4d$

  4. $\pi/2$

Show answer
Correct Option: A
Explanation:
theory of diffraction we know that minima condition is given by 
$n\lambda=d\sin \theta$
for first minima, $n=1$
$\lambda =d\sin\theta$
As $\theta\approx$ small
$\sin\theta\approx \theta$
$\lambda=d\theta$
$\theta=\dfrac{\lambda}{d}\rightarrow $ Required answer
$y=\dfrac{\lambda D}{d}$
$y=D\theta$
$\dfrac{\lambda D}{d}=D\theta$
$\theta=\dfrac{\lambda}{d}$
Required answer

A parallel beam of wavelength $\lambda =4500\mathring { A }$passes through a long slit of width $2\times 10^{-4}m$. The angular divergence for which most of the light is diffracted is (in $\times 10^{-5}$) radian)

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

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

  2. $\dfrac {5\pi}{4}$

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

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

Show answer
Correct Option: B

A double slit arrangement produces interference fringes for sodium light ($\lambda=5890 \mathring {A}$) that are $0.40^0$ apart. What is the angular fringe separation if the entire arrangement is immersed in water?

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

  1. $0.10^o$

  2. $0.20^o$

  3. $0.30^o$

  4. $0.40^o$

Show answer
Correct Option: C
Explanation:

We know the distance between two slits $d=\dfrac{\lambda}{\sin\theta}$, where $\lambda$=wavelength and $\theta$=angle from the original beam.

Given wavelength of sodium light ${\lambda} _{1}=5890 \mathring { A } =5890\times{10}^{-10} m$ and ${\theta} _{1}={0.40}^{0}$
Putting the above values in the given equation we get,
$d=\dfrac{5890\times {10}^{-10}}{\sin0.40}=8.43\times {10}^{-5} m$
Also ${\lambda} _{2}=\dfrac{{\lambda} _{1}}{n}$--------(A) 
where n= refractive index of water =1.33 
Thus A becomes,
${\lambda} _{2}=\dfrac{5890\times {10}^{-10}}{1.33}=4.42\times {10}^{-7} m$
Again,
$d=\dfrac{{\lambda} _{2}}{\sin{\theta} _{2}}$
${\theta} _{2}={\sin}^{-1} (\dfrac{4.42\times {10}^{-7}}{1.33})={0.30}^{0}$