Wave Optics

Given, wave eqn, $y=y_0 \cos (\omega t-k x) \mathrm{cm}$
$y_0, w, k$ are constants

we know, wave speed, $v=\sqrt{\frac{T}{\mu}}$, where $T=$ tension in string

$ \begin{aligned} &\text { So speeds : } v_1=\sqrt{\frac{T}{\mu}}, v_2=\sqrt{\frac{T}{4 \mu}}=\frac{1}{2} \sqrt{\frac{T}{\mu}}\\ &\begin{aligned}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & \Rightarrow v_2=\frac{v_1}{2}, v_3=\sqrt{\frac{\tau}{16 \mu}}=\frac{1}{4} \sqrt{\frac{\tau}{\mu}} \\ & \Rightarrow v_3=\frac{v_1}{4} \end{aligned} \end{aligned} $

wave numbers: $k=\frac{\omega}{v}$

so $k_1=k, k_2=\frac{\omega}{v_2}=2 k$

and $k_3=\frac{\omega}{v_3}=4 k$

Reflection of wave in string from fixed end, reflected wave phase change by $\pi_0$.
And reflected wave from free (rarer) end no phase change.

$\therefore$ For medium $P Q\left(s_2\right)$

$ k_{S_2}=k_{P Q}=2k $

For medium, $S_3, k_{s_3}=4 k$

Option ( $A$ ) reflection from $P$, phase change by ' $\pi$, No Change in ' $K$ '. Hence, ( $A$ ) is the correct option.

In option (B), $k_{s_2}=2 k$ but in option given $k_s=k$. Hence, wrong option.

In option (C), the direction of the reflected wave is in the negative direction. Hence wrong,

option (D), no phase change in transmitted wave and given $k_{s_3}=4 k$

Here, correct option.

So (A), (D) are correct.

$ \tan \theta =\frac{\mathrm{AB}}{d}$

And, $ \mathrm{AB} =d \tan \theta $ ........ (i)

$ \mathrm{BC} =\mathrm{AB} \sin \alpha $

$ =d \tan \theta \sin \alpha $

$ \mathrm{AD} =\mathrm{AB} \sin \theta $

Path difference in vacuum

$ \begin{aligned} \Delta x & =n_1 \mathrm{BC}-n_2 \mathrm{AD} \\\\ \Delta x & =n_1(\mathrm{AB}) \sin \alpha-n_2(\mathrm{AB} \sin \theta) \end{aligned} $

Hence, $\mathrm{AB}\left(n_1 \sin \alpha-n_2 \sin \theta\right)=0$

At any point P on circumference, the path difference, $\Delta$x = d sin$\theta$

At P₁, $\theta$ = 0$^\circ$ $\Rightarrow$ $\Delta$x = 0

At P₂, $\theta$ = 90$^\circ$ $\Rightarrow$ $\Delta$x = d

For constructive interference, $\Delta$x = n$\lambda$

$\therefore$ $n = {d \over \lambda } = {{1.8\,mm} \over {600\,nm}} = {{1.8 \times {{10}^{ - 3}}m} \over {600 \times {{10}^{ - 9}}m}} = 3000$

Since, n is integer, hence P₂ corresponds to bright spot and also it corresponds to maximum order of fringe.

Now, $\Delta$x = d sin$\theta$

d$\Delta$x = d cos$\theta$ d$\theta$ or R$\lambda$ = d cos$\theta$ d$\theta$

$\therefore$ $d\theta \propto {1 \over {\cos \theta }}$

Hence, as $\theta$ increases, cos $\theta$ decreases and consequently d$\theta$ increases.

At point O, the path difference is

${{2\pi } \over \lambda }({S_1}O - {S_2}O) = {{2\pi } \over \lambda }\left[ {\left( {D + {d \over 2}} \right) - \left( {D - {d \over 2}} \right)} \right]$

$ = {{2\pi d} \over \lambda } = {{2\pi (0.6003 \times {{10}^{ - 3}})} \over {600 \times {{10}^{ - 9}}}} = 2001\pi = (2n + 1)\pi $

Therefore, the dark fringe forms at O.

Hence, option (A) is correct.

The shape of wavefronts from both sources for D >> d is semi-circular, which are concentric with each other with the difference just in their phase resulting in concentric semicircular interference fringes. That is, the region nearby point O is dark.

Hence, option (B) is correct.

Fringe width $\beta = {{\lambda D} \over d}$; $\lambda$₁ = 400 nm; $\lambda$₂ = 400 nm

Since $\lambda$₂ > $\lambda$₁, $\beta$₂ > $\beta$₁

Number of fringes within a distance y is given by

${m_1} = {y \over {{\beta _1}}}$

${m_2} = {y \over {{\beta _2}}}$

Since, $\beta$₂ > $\beta$₁, we get m₂ < m₁.

Position of 3^rd maxima of $\lambda$₂ :

$y' = {D \over d}(3{\lambda _2}) = 1800{D \over d}$

Position of 5^th minima of $\lambda$₁ :

$y'' = {D \over d}\left( {{{9{\lambda _1}} \over 2}} \right) = 1800{D \over d}$

Hence, y' = y''.

Angular fringe width is $\theta = {\beta \over D} = {\lambda \over d}$

Since, $\beta$₂ > $\beta$₁ $\Rightarrow$ $\theta$₂ > $\theta$₁

If $d=\lambda$, the maximum path difference (path difference is given by $d \sin\theta$) will be less than $\lambda$. So there will be only central maximum on the screen, because in the equation $d\sin\theta=n\lambda,n$ can take only one value.

If $\lambda < d < 2\lambda$, then the maximum path difference will be less than 2$\lambda$. So there will be two more maximum on the screen in addition to the central maximum.

Intensity of dark fringes becomes zero when intensities at the two slits are equal. Initial intensity at both the slits is unequal so there will some brightness at dark fringes. Hence, when intensity of both slits is made equal, the intensity at dark fringes on screen will reduces to zero.