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

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 :

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

  2. $2\lambda $

  3. $\lambda$

  4. $4\lambda$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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.

  1. $0.15^o$

  2. $0.22^o$

  3. $0.30^o$

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

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

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

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:-

  1. $4 \pi$ m/s

  2. $0.5 \pi$ m/s

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

  4. 8 m/s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

  1. remains same

  2. increases

  3. decreases

  4. none of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

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

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

  1. $100$%

  2. $50$%

  3. $30$%

  4. $12$%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

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:

  1. $\lambda/d$

  2. $\lambda/2d$

  3. $\lambda/4d$

  4. $\pi/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
Multiple choice physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

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?

  1. $0.10^o$

  2. $0.20^o$

  3. $0.30^o$

  4. $0.40^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
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,
$d=\dfrac{{\lambda} _{2}}{\sin{\theta} _{2}}$
${\theta} _{2}={\sin}^{-1} (\dfrac{4.42\times {10}^{-7}}{1.33})={0.30}^{0}$