Wave Optics - Class 12 Engineering Physics - Extra Questions

According to the question,
Shift of central maxima = 5 $$\times$$(fringe width)

$$\therefore \dfrac{(\mu -1)tD}{d}= \dfrac{5\lambda D}{d} $$

$$\therefore t=\dfrac{5\lambda}{\mu -1}=\dfrac{25000}{1.5 - 1}= 50,000\mathring{A}$$

In reflected light $$2\mu t=(2n+1)\displaystyle\frac{\lambda}{2}$$
$$\displaystyle\lambda=\frac{4\mu t}{2n+1}=\displaystyle\frac{2970}{2n+1} nm=\displaystyle 594nm,424nm$$
In transmitted light $$\displaystyle 2\mu t=n\lambda$$ or $$\lambda=\displaystyle\frac{2\mu t}{n}=\frac{1485}{n}$$
$$=495nm$$.

Diverging rays form a curved wavefront. Rays diverging from a point sources form a spherical wavefront.

If $$t$$ be the thickness of the bubble,
then  $$ 4 \mu t = \lambda $$
Substituting the values,
$$ t = \dfrac{6 \times {10}^{-9}}{4 \times 1.25} $$
This gives, $$t =  13 \times {10}^{-8} $$
Thus, $$x = 8$$

A-2,3

0

The shape of the wave front in case of a point source is spherical.  

The light has a dual nature. Sometimes it behaves like a particle called photons, which explains how light travels in straight lines. Sometimes it behaves like a wave, which explains how the light bends around an object,.

For constructive interference from a thin film, 

We know,

For constructive Interference:

The speed of light in vacuum is a universal constant independent of all the factors listed and anything else. In particular, note the surprising fact that it is independent of the relative motion between the source and the observer. This fact is a basic axiom of Einsteins special theory of relativity.

No.

The path travelled by light for both YDSE and diffraction is same for the same point on the screen. However, in diffraction the initial phase difference at the two ends of the slits is $$\pi$$. Due to this additional phase difference of $$\pi$$, positions of bright fringe in YDSE is the position of dark fringe in ydse.

Variation of intensity of light with distance from the central maxima O is shown above in Young's double slit experiment. Bright and dark fringes are formed alternatively where bright fringe has maximum intensity and intensity of dark fringe is zero.

In order to identify whether the given beam is polarised or unpolarised, first of all pass it through a polariser and then through analiser.

The violet filter will allow only violet light to pass through it. Now, if the double slit experiment is performed with the white light and violet light, the fringe pattern will not be the same as obtained by just using white light as the source. To have interference pattern, the light waves entering from the slits should be monochromatic. So, in this case, the violet light will View superimpose with only violet light (of wavelength $$400 nm$$) in such a way that the bright bands will be of violet colour and the minima will be completely dark.

If the width of the slits are A₁ and A₂, the amplitude will also be $$A_1$$ and$$ A_2$$.

The maximum amplitude, A' will happen when constructive interference happens-

$$A' = (A_1 + A_2)$$

The maximum amplitude, A will happen when destructive interference happens-

$$A = (A_1 - A_2)$$

We know that intensity is proportional to the square of amplitude.

$$\dfrac{I_1}{I_2} = \dfrac{A'^2}{A^2} = \dfrac{(A_1+A_2)^2}{(A_1-A_2)^2} \\ \\ \dfrac{9}{25} =(\dfrac{A_1+A_2}{A_1-A_2} )^2\\ \\ \sqrt{ \dfrac{9}{25} }=\dfrac{A_1+A_2}{A_1-A_2} \\ \\\dfrac{A_1+A_2}{A_1-A_2} = \dfrac{3}{5} \\ \\ \dfrac{A_1}{A_2}= \dfrac{5+3}{5-3} = \dfrac{8}{2} \\ \\ {A_1:A_2= 4:1}$$

In case of candles we can't say that the sources are coherrent.

i.e. in same phase, their angular frequencies $$w$$ can be different which will make a pattern dependent on time and our eyes can't see the pattern formed clearly . The avg intensity becomes $$2I$$ because $$\cos{\theta}=0$$.

And if candles are replaced by light sources, interference happen due to coherrent sources.

Given that,

Soap film thickness, $$d=0.4\times {{10}^{-6}}\,m$$

Wavelength, $$\lambda =580\times {{10}^{-9}}m$$

Soap film appear dark (minimum interference of light)

For, minimum illuminiation reflection of light, $$2\mu d=n\lambda $$

At $$n=2$$

 $$\Rightarrow \mu =\dfrac{n\lambda }{2d}=\dfrac{2\times 580\times {{10}^{-9}}}{2\times 0.4\times {{10}^{-6}}}=1.45$$

Refractive index film, $$\mu =1.45$$ 

$$ \large \underline{\textbf{Doppler Effect}}$$

Path diff. = $$x = {\dfrac{yd}{\Delta} }$$

                      $$ = \dfrac{ 0.5 \times 10^{ - 2} \times 2 \times 10^{ - 3}}{2}$$

                      $$ = 5 \times {10^{ - 6}}m.$$

   Phase difference ,

                $$\phi  = {\dfrac{2\pi x} {\lambda} }$$

                   $$ = \dfrac{2\pi  \times 5 \times 10^{ - 6}}{6 \times 10^{ - 7}}$$

                   $$ = {\dfrac{50\pi }{ 3}}$$

                   $$ = 10\pi  + {\dfrac{2\pi }{3}}$$

                   $$ = {\dfrac{2\pi }{3}}$$

    Amplitude ,

             $$A = \sqrt {{r^2} + {r^2} + 2{r^2}\cos \left( {{{2\pi } \over 3}} \right)} $$

                $$ = r$$

                                $${\dfrac{I} {{I_{\max }}}} = {\dfrac{{A^2}}  {{2^2}{r^2}}}$$

                                $${\dfrac{I} {0.2}} = {\dfrac{A}{4{r^2}}} =  {\dfrac{r^2}{4{r^2}}}$$

                                $$I = {\dfrac{0.2} {4}} = 0.05\,w/{m^2}$$

Plane of vibration : The plane in which the vibration of electric field and magnetic field vibrate it is always perpendicular plane with direction of propagation. 

In an unpolarised light libration are symmetric about the direction of propagation for an unpolarised wave the displacement will be randomly changing with time through it will always be perpendicular to the direction of propagation. 

In YDSE the expression for intensity is given bu:-

$$r = 30^o$$

Given that

Path difference due to the glass slab,
      $$\triangle x=(\mu - 1)t$$

Thirty fringes are displaced due to the slab, Hence,
       $$\triangle x= 30\lambda$$

     $$(\mu-1)t=30\lambda$$

                  $$ t=\dfrac{30\lambda}{\mu-1}= \dfrac {30 \times 6 \times 10^{-5}}{1.5-1} = 3.6 \times 10^{-3}cm $$   

a) Change in the optical path $$=\mu t-t=(\mu -1)t$$

The intensity of interference fringes in a double slit arrangement is modulated by the diffraction pattern of each slit. Alternatively, in double slit experiment the interference pattern on the screen is actually superposition of single slit diffraction for each slit. 

Graph $$(a)$$ represents the male voice. Usually the male voice has less pitch (or frequency) as compared to female.

From Table $$ 35.1, $$ the index of refraction of ice is $$ 1.309 . $$ The pulses are in step with each other until one enters the ice, then that pulse slows down. The difference in the times of arrival of the pulses is

If the incident intensity is $$I$$,

Here path difference at a point P on the circle is given by,

YDS

Let us consider young's double slit experimental setup. Two narrow coherent light sources are obtained by wavefront splitting as mono chromatic light of wavelength $$\lambda$$ emerges out of two closely spaced, parallel slits $$S_1$$ and $$S_2$$ of equal widths. The separation $$S_1S_2=d$$ is very small. The interference pattern is observed on a screen placed parallel to the plane of $$S_1S_2$$ and some considerable distance.
$$D(D >> d)$$ from the slits. OP is the perpendicular bisector of segment $$S_1S_2$$. Consider point Q on the screen at a distance x from $$P(x << D)$$ the two light waves from $$S_1, S_2$$ reach Q along paths $$S_1$$Q and $$S_2$$Q respectively.
From the figures
$$(S_2Q)^2=(S_2N)^2+(QN)^2=D^2+\left(x+\displaystyle\frac{d}{2}\right)^2$$    .....$$(1)$$
and $$(S_1Q)^2=(S_1M)^2+(QM)^2=D^2+\left(x-\displaystyle\frac{d}{2}\right)^2$$    .....$$(2)$$
$$\therefore (S_2Q)^2-(S_1Q)^2=D^2+\left(x+\displaystyle\frac{d}{2}\right)^2-D^2-\left(x-\displaystyle\frac{d}{2}\right)^2$$
$$\therefore (S_2Q+S_1Q)(S_2Q-S_1Q)=2xd$$
$$\therefore S_2Q-S_1Q=\displaystyle\frac{2xd}{S_2Q+S_1Q}$$
In partice, $$D >> x$$ and $$D >> d$$, so that
$$S_2Q-S_1Q\simeq 2D$$
$$\therefore$$ Path difference, $$(S_2Q-S_1Q)=\displaystyle\frac{2xd}{2D}=\frac{xd}{D}$$   .....$$(3)$$
Expression for the fringe width or band width: The distance between consecutive bright(or dark) fringes is called the fringe width or band width X.
Point Q will be bright(maximum intensity), if the path difference $$\displaystyle\frac{xd}{D}=n\lambda$$, where $$n=0,1, 2, 3,...$$
point Q will be dark(minimum intensity equal to zero) if $$\displaystyle\frac{xd}{D}=(2m-1)\frac{\lambda}{2}$$, where $$m=1, 2, 3,...$$
These conditions show that the bright and dark fringes occur alternately and are equally speed.
For point P, the path difference $$(S_2P-S_1P)=0$$. Hence, point P will be bright, the central bright fringe or band. On both sides of point P, the interference pattern consists of alternate bright and dark bands or fringes parallel to the slit.
Let $$x_n$$ and $$x_{n+1}$$ be the distances of the $$n^{th}$$ and $$(n+1)^{th}$$ bright fringes from the central bright fringe.
$$\therefore \displaystyle\frac{x_nd}{D}=n\lambda$$ or $$x_n=\displaystyle\frac{n\lambda D}{d}$$     ....$$(4)$$
and $$\displaystyle\frac{x_{n+1}d}{D}=(n+1)\lambda$$ or $$x_{n+1}=\displaystyle\frac{(n+1)\lambda D}{d}$$       ....$$(5)$$
$$\therefore$$ The distance between consecutive bright fringes
$$=x_{n+1}-x_n=\displaystyle\frac{\lambda D}{d}[(n+1)-n]=\displaystyle\frac{\lambda D}{d}$$     ....$$(6)$$

Fringe width is given by $$\beta = \dfrac {\lambda D}{d}$$
In the first case, $$\lambda_{1} = 6000\ A^{\circ} = 6000\times 10^{-10}m$$
$$\beta_{1} = 6\ mm = 0.006\ m$$
$$\therefore 0.006 = 6000\times 10^{-10} \dfrac {D}{d} ....(1)$$
In the second case, the distance between the slits and the screen is $$\dfrac {D}{2}, \beta_{2} = 4\ mm = 0.004\ m$$
$$\therefore 0.004 = \lambda_{2}\times \dfrac {\dfrac {D}{2}}{d} ....(2)$$
Directing equations $$(2)$$ by $$(1)$$ we get
$$\dfrac {0.004}{0.006} = \dfrac {\lambda_{2}}{2\times 6000\times 10^{-10}}$$
$$\lambda_{2} = \dfrac {0.004\times 2\times 6000\times 10^{-10}}{0.006}$$
$$\lambda_{2} = \dfrac {48\times 10^{-10}}{0.006}$$
$$\lambda_{2} = 8000\times 10^{-10}$$
$$\lambda_{2} = 8000\ A^{\circ}$$.

original path difference due to thin of oil=(u-1)t

white light is made up of seven colors VIBGYOR. All the seven colors will show interference, we can treat each color separately as a monochromatic light and study their interference separately. Each light will have their bright and dark fringes separately but at y=0 i.e for a center, each lights bright fringe will coincide. Two or more lights may have their bright and dark Fringe together or bright for one and dark for others may coincide depending upon the situation.

If the sources emit light waves of more than onewavelength (frequency) then constructive Interference or

Given :    $$D = 1 \ m$$      $$\lambda = 4500\times 10^{-10} \ m$$          $$d = 0.125\times 10^{-3} \ m$$

   $$x= dsin \theta=(2n-1)\dfrac{\lambda}{2}$$                  $$n=0,1,2,...$$        $$ \theta$$ is very small 

In this setup, we have $$n_2 > n_1$$ and $$n_2 > n_3$$, and the condition for constructive interference is 
$$2L =\left (m + \frac{1} {2} \right ) \frac{\lambda} {n_2} \Rightarrow \lambda = \frac{4Ln_2} {2m + 1} ,   m = 0, 1, 2, .............$$ 
Thus, we get 
$$\lambda$$ = $$\begin{cases} 4Ln_1 = 4\left (285 \space nm\right ) \left (1.60\right ) = 1824 \space nm  \space \left (m = 0  \right ) \\
4Ln_2 = 4\left (285 \space nm\right ) \left (1.60\right )/3 = 608 \space nm  \space \left (m = 1  \right ) \end{cases}$$
For the wavelength to be in the visible range, we choose $$m = 1$$ with $$\lambda = 608 \space nm$$.

Assume the wedge-shaped film is in air, so the wave reflected from one surface undergoes a phase change of $$\pi \space rad$$ while the wave reflected from the other surface does not. At a place where the film thickness is $$L$$, the condition for fully constructive interference is $$2nL = \left (m + \frac{1} {2}  \right )\lambda$$, where $$n$$ is the index of refraction of the film, $$\lambda$$ is the wavelength in vacuum, and $$m$$ is an integer. The ends of the film are bright. Suppose the end where the film is narrow has thickness $$L_1$$ and the bright fringe there corresponds to $$m= m_1$$. Suppose the end where the film is thick has thickness $$L_2$$ and the bright fringe there corresponds to $$m = m_2$$. Since there are ten bright fringes, $$m_2 = m_1 + 9$$. Subtract $$2nL_1 = \left (m_1 + \frac{1} {2}  \right ) \lambda$$ from $$2nL_1 = \left (m_1 + 9 + \frac{1} {2}  \right ) \lambda$$ to obtain $$2n\Delta L = 9\lambda$$, where $$\Delta L = l_2 - L_1$$ is the change in the film thickness over its length. Thus,
$$\Delta L = \frac{9\lambda} {2n} = \frac{9\left ( 630 \times 10^{-9} \space m \right )} {2 \left (1.50  \right )} = 1.89 \times 10^{-6} \space m$$.

Let us represent the natural light as a sum of two mutually perpendicular components, both with intensity $$I_0$$. Suppose that each polarizer transmits a fraction $$\alpha_1$$ of the light with oscillation plane parallel to be principle direction of the polarize and a fraction $$\alpha_2$$ with oscillation plane perpendicular to the principle direction of the polarizer . Then of light transmitted through the two polarizers is equal to