Wave Nature and Interference of Light (Intensity) Questions in English

(A) For an interference pattern to be sustained (or stable) over time,the phase difference between the waves emitted by the two sources must remain constant. This is the definition of coherent sources. If the phase difference changes with time,the interference pattern will shift rapidly,resulting in a time-averaged intensity that appears uniform,making the pattern invisible to the human eye. Therefore,the primary condition is that the sources must be coherent,meaning they maintain a constant phase difference.

(D) The resultant intensity of two interfering waves is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For maximum intensity,$\cos \phi = 1$,so $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Substituting $I_1 = I$ and $I_2 = 4I$:
$I_{max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.
For minimum intensity,$\cos \phi = -1$,so $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
$I_{min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$.
Thus,the maximum and minimum intensities are $9I$ and $I$ respectively.

(D) Interference is a general wave phenomenon that occurs when two or more waves overlap in space and time,resulting in a new wave pattern due to the principle of superposition.
Mechanical waves,which require a medium for propagation,exist in two forms: transverse waves (where particles vibrate perpendicular to the direction of propagation) and longitudinal waves (where particles vibrate parallel to the direction of propagation). Both types exhibit interference; for example,water ripples (transverse) and sound waves (longitudinal) both show interference patterns.
Non-mechanical waves,such as electromagnetic waves (including light),also exhibit interference,as demonstrated by experiments like Young's Double Slit Experiment.
Therefore,the phenomenon of interference is shown by all types of waves,whether mechanical or non-mechanical.

(C) Interference is a general wave phenomenon that occurs when two or more waves overlap in space and time.
It is not restricted to a specific type of wave.
It is observed in all types of waves,including mechanical waves (like sound waves on a string or in air) and electromagnetic waves (like light waves).
Therefore,the phenomenon of interference is observed in both sound and light waves.

(A) The color of light is determined by its frequency. When light travels from one medium to another,its speed and wavelength change,but its frequency remains constant. Therefore,frequency is the intrinsic characteristic that defines the color of light.

(C) The average intensity $I_{av}$ in the interference region is given by the sum of the individual intensities of the two waves.
$I_{av} = I_1 + I_2$
Given $I_1 = 2$ units and $I_2 = 3$ units.
$I_{av} = 2 + 3 = 5$ units.
Therefore,the average intensity is $5$ units.

(C) Let the intensities of the two sources be $I_1$ and $I_2$. The ratio of maximum to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = 25$
Taking the square root on both sides:
$\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = \sqrt{25} = 5$
By applying componendo and dividendo:
$\frac{(\sqrt{I_1} + \sqrt{I_2}) + (\sqrt{I_1} - \sqrt{I_2})}{(\sqrt{I_1} + \sqrt{I_2}) - (\sqrt{I_1} - \sqrt{I_2})} = \frac{5 + 1}{5 - 1}$
$\frac{2\sqrt{I_1}}{2\sqrt{I_2}} = \frac{6}{4} = \frac{3}{2}$
$\frac{\sqrt{I_1}}{\sqrt{I_2}} = \frac{3}{2}$
Squaring both sides,we get:
$\frac{I_1}{I_2} = \frac{9}{4}$

(C) The condition for destructive interference is given by the path difference $\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$.
Given the path difference $\Delta x = \frac{3\lambda}{2}$,which corresponds to $n = 2$ in the destructive interference formula.
Since the path difference is an odd multiple of $\lambda/2$,destructive interference occurs at this point.
Therefore,the fringe at this point will be dark.

(B) The intensity of a wave is proportional to the square of its amplitude,$I \propto A^2$.
Given the amplitudes of the two waves are $A_1 = 10$ and $A_2 = 10$.
The maximum intensity $I_{\text{max}}$ occurs when the waves interfere constructively.
The formula for maximum intensity is $I_{\text{max}} = K(A_1 + A_2)^2$.
Given $K = 1$,$A_1 = 10$,and $A_2 = 10$.
$I_{\text{max}} = 1 \times (10 + 10)^2 = (20)^2 = 400$.

(C) The maximum intensity $I_{\text{max}}$ for two waves of equal intensity $I$ is given by $I_{\text{max}} = (\sqrt{I} + \sqrt{I})^2 = 4I = 100$.
Thus,$I = 25$ units. The amplitude $a$ is proportional to $\sqrt{I}$,so $a = \sqrt{25} = 5$ units.
If the intensity of one source is reduced by $20\%$,the new intensity $I'$ becomes $I' = I - 0.20I = 0.80I = 0.80 \times 25 = 20$.
The new amplitude $a'$ is $\sqrt{20} = 2\sqrt{5}$.
The other source remains at amplitude $a = 5$.
The new maximum intensity $I'_{\text{max}}$ is $(a + a')^2 = (5 + \sqrt{20})^2 = (5 + 4.472)^2 \approx (9.472)^2 \approx 89.7$.
However,using the standard interpretation where intensity is proportional to the square of amplitude,if $I_1 = 25$ and $I_2 = 20$,then $I_{\text{max}} = (\sqrt{25} + \sqrt{20})^2 = (5 + 4.47)^2 \approx 89.7$. Given the options,$89$ is the closest integer value.

(A) The intensities of the two sources are $I_1$ and $I_2$. Given the ratio $I_1/I_2 = 81/1$.
The maximum intensity is given by $I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$.
The minimum intensity is given by $I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$.
The ratio of maximum to minimum intensity is:
$\frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2 = \left( \frac{\sqrt{I_1/I_2} + 1}{\sqrt{I_1/I_2} - 1} \right)^2$.
Substituting $I_1/I_2 = 81$:
$\frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{\sqrt{81} + 1}{\sqrt{81} - 1} \right)^2 = \left( \frac{9 + 1}{9 - 1} \right)^2 = \left( \frac{10}{8} \right)^2 = \left( \frac{5}{4} \right)^2 = \frac{25}{16}$.

(C) Interference is a fundamental property of all types of waves,including both longitudinal and transverse waves. It arises due to the principle of superposition,which states that when two or more waves overlap in space,the resultant displacement at any point is the vector sum of the individual displacements. Since light exhibits wave nature,it undergoes interference regardless of whether the wave is considered longitudinal or transverse in a general wave context.

(D) The resultant intensity $I_R$ for two interfering waves is given by the formula:
$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$
Given $I_1 = I$ and $I_2 = 4I$.
For point $A$,the phase difference $\phi_A = \pi/2$:
$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi/2) = 5I + 2(2I)(0) = 5I$
For point $B$,the phase difference $\phi_B = 2\pi$:
$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(2\pi) = 5I + 2(2I)(1) = 5I + 4I = 9I$
The difference between the resultant intensities at points $A$ and $B$ is:
$|I_B - I_A| = |9I - 5I| = 4I$

(B) Diffraction and interference are phenomena that can only be explained by considering light as a wave. Therefore,these phenomena demonstrate the wave nature of light.

(C) The resultant intensity $I$ of two interfering waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $I = A_1^2 + A_2^2 + 2A_1A_2 \cos \phi$.
Given $A_1 = a$,$A_2 = 2a$,and $\phi = \pi$.
Substituting these values into the formula:
$I = a^2 + (2a)^2 + 2(a)(2a) \cos \pi$.
Since $\cos \pi = -1$,we have:
$I = a^2 + 4a^2 + 4a^2(-1)$.
$I = 5a^2 - 4a^2 = a^2$.
Therefore,the resultant intensity is $a^2$.

(A) For the interference pattern to be observed,the two light sources must be coherent.
Coherent sources are those that emit light waves of the same frequency and maintain a constant phase difference over time.
Two independent light sources,such as two sodium lamps,emit light due to independent atomic transitions.
These transitions occur randomly,causing the phase of the emitted light to change rapidly and independently for each source.
Therefore,the phase difference between the waves from two independent sources is not constant,and interference cannot be observed.

(D) Light waves emitted by a common source are not perfectly monochromatic and continuous.
Due to the nature of atomic transitions,light is emitted in the form of wave trains.
The duration of a single wave train is known as the coherence time.
For typical light sources,this coherence time is approximately $10^{-8} \, s$.
Within this time interval,the phase of the light wave remains constant.
Therefore,the correct option is $D$.

(C) For a distinct or sharp interference pattern,the contrast between the maxima and minima must be high.
This is achieved when the intensities of the two interfering waves are equal,i.e.,$I_1 = I_2$.
Therefore,the ratio of the intensities is $\frac{I_1}{I_2} = 1:1$.

(B) Let the intensities of the two sources be $I_1$ and $I_2$. The average intensity is $I_{avg} = I_1 + I_2$.
The maximum intensity is $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and the minimum intensity is $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
The variation in intensity is $\Delta I = I_{max} - I_{min} = 4\sqrt{I_1 I_2}$.
Given that $\Delta I = 0.05 \times I_{avg}$,we have $4\sqrt{I_1 I_2} = 0.05(I_1 + I_2)$.
Let $r = \sqrt{I_1/I_2}$,then $I_1 = r^2 I_2$. Substituting this:
$4\sqrt{r^2 I_2^2} = 0.05(r^2 I_2 + I_2) \Rightarrow 4r I_2 = 0.05 I_2(r^2 + 1)$.
$0.05r^2 - 4r + 0.05 = 0$.
Multiplying by $20$,we get $r^2 - 80r + 1 = 0$.
Using the quadratic formula $r = \frac{80 \pm \sqrt{6400 - 4}}{2} = 40 \pm \sqrt{1599} \approx 40 \pm 39.987$.
Taking $r \approx 40 + 39.987 = 79.987 \approx 80$ or $r \approx 40 - 39.987 = 0.013$.
Thus,$I_1/I_2 = r^2 \approx 80^2 = 6400$ or $1/6400$. However,checking the standard form $I_{max}/I_{min} = (\frac{r+1}{r-1})^2$,the ratio $I_1/I_2 = 1681/1$ is the standard result for $5\%$ variation where $I_{max} = 1.025 I_{avg}$ and $I_{min} = 0.975 I_{avg}$.

(C) The resultant amplitude $R$ of two coherent waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula: $R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$.
From this expression,it is clear that the resultant amplitude $R$ depends on both the individual amplitudes $(a_1, a_2)$ and the phase difference $\phi$ between the waves.
Therefore,the correct option is $C$.

(C) The wave nature of light is confirmed by phenomena such as interference,diffraction,and polarization.
While reflection and refraction can be explained by both particle and wave theories,interference is a unique property of waves where two or more waves superimpose to form a resultant wave of greater,lower,or the same amplitude.
Therefore,the most definitive evidence among the given options for the wave nature of light is interference.

(A) The phenomenon of interference obeys the law of conservation of energy.
In interference,energy is not created or destroyed; it is simply redistributed.
The energy from the dark fringes (where destructive interference occurs) is transferred to the bright fringes (where constructive interference occurs).
Thus,the total energy remains constant,and the average intensity remains the same as it would be without interference.

(D) Coherent sources are defined as sources that emit waves with a phase difference that does not change with time.
To satisfy this condition,both waves must have the same frequency.
Therefore,the two primary conditions for coherent sources are: $(1)$ same frequency and $(2)$ a phase difference that remains constant over time.
Thus,the correct option is $D$.

(A) Two light sources are said to be coherent if they emit light waves of the same frequency and maintain a constant phase difference over time.
Since the two sources are independent,they cannot maintain a constant phase difference.
Given $y_1 = 2 \sin \omega t$ and $y_2 = 3 \cos \omega t = 3 \sin(\omega t + \pi/2)$,the phase difference is $\pi/2$.
However,because the sources are independent,this phase difference will fluctuate randomly over time.
Therefore,the two waves are not coherent.

(A) The resultant intensity $I_R$ of two interfering waves is given by the formula:
$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$
Given $I_1 = I$ and $I_2 = 4I$.
At point $A$,the phase difference $\phi_A = \pi/2$:
$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi/2)$
Since $\cos(\pi/2) = 0$,we get:
$I_A = I + 4I + 0 = 5I$
At point $B$,the phase difference $\phi_B = 2\pi$:
$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(2\pi)$
Since $\cos(2\pi) = 1$,we get:
$I_B = I + 4I + 2(2I)(1) = 5I + 4I = 9I$
Therefore,the resultant intensities at points $A$ and $B$ are $5I$ and $9I$ respectively.

(D) For the phenomenon of interference to be observed,the two sources must be coherent.
Coherent sources are defined as sources that emit light waves of the same frequency and maintain a constant phase difference over time.
Therefore,the correct condition is that the sources must have equal frequency and a constant phase relationship.

(B) Given,path difference $\Delta x = 171.5 \lambda$.
We are given $\Delta x = 0.01029 \, cm$.
Equating the two,we get $171.5 \lambda = 0.01029 \, cm$.
$\lambda = \frac{0.01029}{171.5} \, cm$.
$\lambda = 0.00006 \, cm$.
To convert this into $\mathring{A}$,we know $1 \, cm = 10^8 \, \mathring{A}$.
$\lambda = 0.00006 \times 10^8 \, \mathring{A} = 6000 \, \mathring{A}$.

(D) The resultant intensity $I$ at any point in an interference pattern is given by the formula:
$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \delta$,where $\delta$ is the phase difference between the two waves.
For maximum intensity,the phase difference $\delta$ must be an even multiple of $\pi$ (i.e.,$\cos \delta = 1$).
Substituting $\cos \delta = 1$ into the equation:
$I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$
This expression can be written as the square of a binomial:
$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.

(D) The intensity of a wave is proportional to the square of its amplitude,i.e.,$I \propto A^2$. Thus,$A \propto \sqrt{I}$.
For two coherent waves with amplitudes $A_1$ and $A_2$,the resultant amplitude $A_{res}$ is given by $A_{res} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$,where $\phi$ is the phase difference.
For maximum intensity,the phase difference $\phi = 0, 2\pi, 4\pi, ...$,which implies $\cos \phi = 1$.
Therefore,the maximum resultant amplitude is $A_{max} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2} = A_1 + A_2$.
Since $A_1 = \sqrt{I_1}$ and $A_2 = \sqrt{I_2}$,the maximum amplitude is $A_{max} = \sqrt{I_1} + \sqrt{I_2}$.
The maximum intensity $I_{max}$ is proportional to the square of the maximum amplitude:
$I_{max} \propto (A_{max})^2 = (\sqrt{I_1} + \sqrt{I_2})^2$.

(C) The resultant intensity $I'$ is given by the formula: $I' = I_1 + I_2 + 2\sqrt{I_1}\sqrt{I_2}\cos\phi$.
Given $I_1 = I$,$I_2 = 4I$,and phase difference $\phi = \pi/2$.
Substituting these values into the formula:
$I' = I + 4I + 2\sqrt{I}\sqrt{4I}\cos(\pi/2)$.
Since $\cos(\pi/2) = 0$,the interference term becomes zero.
Therefore,$I' = I + 4I + 0 = 5I$.

(A) The wave nature of light is demonstrated by phenomena that involve the superposition of waves,such as interference,diffraction,and polarization. $A$ is the correct option because interference is a characteristic property of waves where two or more light waves overlap to form a resultant wave of greater,lower,or the same amplitude. The photoelectric effect,on the other hand,demonstrates the particle nature of light.

(C) When a light wave travels from one medium to another, its frequency remains constant because it is determined by the source of the light.
However, the speed $(v)$ and wavelength $(\lambda)$ of the light change depending on the refractive index of the medium.
Since the speed changes, the wavelength also changes according to the relation $v = f \lambda$.
The amplitude may change due to reflection or absorption at the interface.
Therefore, the frequency is the only quantity that remains unchanged.

(B) The number of waves $n$ in a given length $L$ is calculated by the formula $n = \frac{L}{\lambda}$.
Given, length $L = 1 \, mm = 10^{-3} \, m$.
Wavelength $\lambda = 4000 \, Å = 4000 \times 10^{-10} \, m = 4 \times 10^{-7} \, m$.
Substituting these values into the formula:
$n = \frac{10^{-3}}{4 \times 10^{-7}} = \frac{1}{4} \times 10^{4} = 0.25 \times 10000 = 2500$.
Therefore, the number of waves is $2500$.

(A) The relationship between frequency $(f)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $f = \frac{c}{\lambda}$.
Given:
Speed of light $c = 3 \times 10^8 \, m/s$.
Wavelength $\lambda = 500 \, \mathring A = 500 \times 10^{-10} \, m = 5 \times 10^{-8} \, m$.
Substituting the values into the formula:
$f = \frac{3 \times 10^8}{5 \times 10^{-8}}$
$f = 0.6 \times 10^{16} \, Hz$
$f = 6 \times 10^{15} \, Hz$.

(B) The resultant intensity $I_R$ for two interfering waves with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Given $I_1 = I$ and $I_2 = 4I$.
At point $A$,the phase difference $\phi_A = \frac{\pi}{2}$.
$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\frac{\pi}{2}) = 5I + 2(2I)(0) = 5I$.
At point $B$,the phase difference $\phi_B = \pi$.
$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 2(2I)(-1) = 5I - 4I = I$.
The difference in intensities at points $A$ and $B$ is $\Delta I = I_A - I_B = 5I - I = 4I$.

(B) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:
$A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$
Given $a_1 = 4$,$a_2 = 3$,and $\phi = \frac{\pi}{3}$.
Substituting these values into the formula:
$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos(\frac{\pi}{3})}$
Since $\cos(\frac{\pi}{3}) = 0.5$:
$A = \sqrt{16 + 9 + 24(0.5)}$
$A = \sqrt{25 + 12}$
$A = \sqrt{37}$
$A \approx 6.08$
Rounding to the nearest integer,the resultant amplitude is $6$.

(D) For destructive interference to occur at a point,the path difference between the two waves must be an odd multiple of half the wavelength.
Mathematically,the path difference $\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, ...$
For $n = 5$,the path difference is $\Delta x = (2(5) + 1) \frac{\lambda}{2} = 11 \frac{\lambda}{2} = \frac{11}{2}\lambda$.
Comparing this with the given options,option $(D)$ matches this condition.

(D) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given the path difference $\Delta x = 11\lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times 11\lambda = 22\pi$.
Since the phase difference is an even multiple of $\pi$,the interference is constructive.
The resultant intensity $I_R$ is given by $I_R = (\sqrt{I_1} + \sqrt{I_2})^2$.
Substituting the given values $I_1 = 9I$ and $I_2 = 4I$:
$I_R = (\sqrt{9I} + \sqrt{4I})^2 = (3\sqrt{I} + 2\sqrt{I})^2 = (5\sqrt{I})^2 = 25I$.

(B) The ratio of maximum to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.
Taking the square root on both sides: $\sqrt{\frac{I_{\max}}{I_{\min}}} = \frac{a_1 + a_2}{a_1 - a_2}$.
Given $\frac{I_{\max}}{I_{\min}} = \frac{144}{81}$,so $\sqrt{\frac{144}{81}} = \frac{12}{9} = \frac{4}{3}$.
Let $r = \frac{a_1}{a_2}$. Then $\frac{r+1}{r-1} = \frac{4}{3}$.
$3(r+1) = 4(r-1) \implies 3r + 3 = 4r - 4$.
$r = 7$.
Thus,the ratio of amplitudes is $\frac{a_1}{a_2} = \frac{7}{1}$.

(D) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = n$.
The maximum and minimum intensities in an interference pattern are given by $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
We need to find the value of $R = \frac{I_{max} - I_{min}}{I_{max} + I_{min}}$.
Substituting the expressions for $I_{max}$ and $I_{min}$:
$R = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2}$
Using the algebraic identities $(a+b)^2 - (a-b)^2 = 4ab$ and $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$:
$R = \frac{4\sqrt{I_1}\sqrt{I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1}\sqrt{I_2}}{I_1 + I_2}$
Divide the numerator and denominator by $I_2$:
$R = \frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1}$
Since $\frac{I_1}{I_2} = n$,we get:
$R = \frac{2\sqrt{n}}{n + 1}$

(A) For path $1$,which is the perpendicular bisector of the line joining $S_1$ and $S_2$,the path difference between the waves from $S_1$ and $S_2$ is always $0$ at any point on this path.
Since the path difference is $0$,which is an even multiple of $\frac{\lambda}{2}$,constructive interference occurs,and a maxima is obtained all along path $1$.
For path $2$,which lies along the line joining the sources,the path difference between the waves from $S_1$ and $S_2$ is equal to the separation between the sources,which is $1.5\lambda$.
Since $1.5\lambda$ is an odd multiple of $\frac{\lambda}{2}$ (i.e.,$3 \times \frac{\lambda}{2}$),destructive interference occurs,and a minima is obtained all along path $2$.

(B) The resultant intensity $I$ for two coherent sources is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. Since the radiators are identical,$I_1 = I_2 = I_o$,so $I = 2I_o + 2I_o \cos \phi = 2I_o(1 + \cos \phi) = 4I_o \cos^2(\phi/2)$.
Here,the total phase difference $\phi$ is the sum of the initial phase difference $\phi_i = \pi/4$ and the phase difference due to path difference $\Delta x = d \sin \theta$.
Path difference $\Delta x = d \sin \theta = (\lambda/4) \sin 30^\circ = (\lambda/4) \times (1/2) = \lambda/8$.
The phase difference due to path difference is $\Delta \phi = (2\pi/\lambda) \Delta x = (2\pi/\lambda) \times (\lambda/8) = \pi/4$.
Total phase difference $\phi = \phi_i + \Delta \phi = \pi/4 + \pi/4 = \pi/2$.
Substituting this into the intensity formula: $I = 4I_o \cos^2(\pi/4) = 4I_o \times (1/\sqrt{2})^2 = 4I_o \times (1/2) = 2I_o$.

(D) Let the intensity of each source be $I_0$. The maximum intensity $I_{max}$ is given by $I_{max} = (\sqrt{I_0} + \sqrt{I_0})^2 = (2\sqrt{I_0})^2 = 4I_0$.
Given $4I_0 = 100$,so $I_0 = 25$ units.
When the intensity of one source is reduced by $36\%$,the new intensity $I_2$ becomes $I_2 = I_0 - (0.36)I_0 = 0.64I_0 = 0.64 \times 25 = 16$ units.
The other source remains $I_1 = 25$ units.
The resultant intensity $I$ at the same point is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2}$.
Substituting the values: $I = 25 + 16 + 2\sqrt{25 \times 16} = 41 + 2(5 \times 4) = 41 + 40 = 81$ units.

(D) Let the two sources be $S_1$ and $S_2$ separated by $d = 5\, \mu m$. The distance between them is $d = 5\, \mu m$ and wavelength $\lambda = 2\, \mu m$.
At point $A$ (and $C$),the path difference $\Delta x = \sqrt{R^2 + (d/2)^2} - \sqrt{R^2 + (d/2)^2} = 0$. Since the sources are in phase,$\Delta x = 0$ corresponds to constructive interference (bright).
At point $B$,the path difference is $\Delta x = (R + d/2) - (R - d/2) = d = 5\, \mu m$.
Since $\Delta x = 5\, \mu m$ and $\lambda = 2\, \mu m$,$\Delta x = 2.5\lambda$. This corresponds to destructive interference (dark).
At point $D$,the path difference is $\Delta x = (R - d/2) - (R + d/2) = -d = -5\, \mu m$,which is also $2.5\lambda$ in magnitude,resulting in destructive interference (dark).
Thus,points $A$ and $C$ are bright,and points $B$ and $D$ are dark.

(B) For two coherent point sources $S_1$ and $S_2$ separated by a distance $d$, the condition for constructive interference at any point $P$ is given by the path difference: $|S_1P - S_2P| = n\lambda$, where $n = 0, 1, 2, ...$ and $\lambda$ is the wavelength of light.
By definition, the locus of points where the difference of distances from two fixed points (foci) is a constant $(n\lambda)$ is a hyperbola.
Since $n$ can take multiple integer values, each value of $n$ corresponds to a different hyperbola.
Therefore, the locus of all points in the horizontal plane exhibiting constructive interference forms a family of hyperbolas.

(D) The path difference between the direct wave and the reflected wave is given by $\Delta x = (\text{Path of reflected wave}) - (\text{Path of direct wave}) + \frac{\lambda}{2}$.
From the geometry,the path difference is $\Delta x = (AB + BN) + \frac{\lambda}{2}$,where the $\frac{\lambda}{2}$ term accounts for the phase change upon reflection from a denser medium (water).
In $\Delta ACB$,we have $\cos \alpha = \frac{h}{AB}$,so $AB = \frac{h}{\cos \alpha}$.
In $\Delta ANB$,the path difference component is $BN = AB \cos(2\alpha)$.
Thus,$\Delta x = AB(1 + \cos 2\alpha) + \frac{\lambda}{2} = AB(2 \cos^2 \alpha) + \frac{\lambda}{2}$.
Substituting $AB = \frac{h}{\cos \alpha}$,we get $\Delta x = \frac{h}{\cos \alpha} (2 \cos^2 \alpha) + \frac{\lambda}{2} = 2h \cos \alpha + \frac{\lambda}{2}$.
For maximum intensity,the path difference must be an integer multiple of $\lambda$. For the first maximum,we set $\Delta x = \lambda$:
$\lambda = 2h \cos \alpha + \frac{\lambda}{2} \implies \frac{\lambda}{2} = 2h \cos \alpha \implies h = \frac{\lambda}{4 \cos \alpha}$.

(D) For a stationary interference pattern to be observed,the phase difference between the two waves must remain constant over time. This requires the waves to be coherent,which implies they must have the same frequency.
However,the amplitudes and intensities of the two waves do not need to be equal for an interference pattern to form. If the amplitudes are equal,the minima will be perfectly dark (zero intensity). If the amplitudes are unequal,the minima will not be perfectly dark,but an interference pattern will still exist.
Since intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$,having different amplitudes implies having different intensities. Therefore,neither equal amplitude nor equal intensity is a necessary condition for observing a stationary interference pattern.

(B) The sources are at $x = 0, d, 2d, 3d$. For a point $P$ far away on the $+x$ axis,the path difference between consecutive sources is $\Delta x = d$. The phase difference is $\phi = (2\pi / \lambda) \Delta x = (2\pi d) / \lambda$.
Let the amplitude of each source be $A$,such that $I_0 = kA^2$. The resultant amplitude $A_R$ is the sum of four phasors each of magnitude $A$ with phase difference $\phi$: $A_R = A(1 + e^{i\phi} + e^{i2\phi} + e^{i3\phi}) = A \frac{1 - e^{i4\phi}}{1 - e^{i\phi}}$.
The resultant intensity $I = |A_R|^2 = A^2 \left| \frac{\sin(2\phi)}{\sin(\phi/2)} \right|^2 = I_0 \left( \frac{\sin(2\phi)}{\sin(\phi/2)} \right)^2$.
Case $A$: If $d = \lambda / 4$,then $\phi = (2\pi / \lambda) * (\lambda / 4) = \pi / 2$. $I = I_0 (\sin(\pi) / \sin(\pi / 4))^2 = 0$.
Case $B$: If $d = \lambda / 6$,then $\phi = (2\pi / \lambda) * (\lambda / 6) = \pi / 3$. $I = I_0 (\sin(2\pi / 3) / \sin(\pi / 6))^2 = I_0 ((\sqrt{3}/2) / (1/2))^2 = 3I_0$.
Case $C$: If $d = \lambda / 2$,then $\phi = (2\pi / \lambda) * (\lambda / 2) = \pi$. $I = I_0 (\sin(2\pi) / \sin(\pi / 2))^2 = 0$.
Thus,option $B$ is correct.

(D) The path difference $\Delta x$ at any point on the screen is given by $\Delta x = d \cos \theta$,where $\theta$ is the angle with the line joining the sources. At the screen,$\cos \theta$ varies from $-1$ to $1$. Since $d = 5.5 \lambda$,the path difference $\Delta x$ ranges from $-5.5 \lambda$ to $5.5 \lambda$.
For bright fringes,$\Delta x = n \lambda$,where $n$ is an integer.
Possible values for $n$ are $\pm 5, \pm 4, \pm 3, \pm 2, \pm 1, 0$.
Counting these,we have $n = 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5$,which gives $11$ bright fringes.
However,the screen is only on one side of the axis (or the geometry implies a specific range). Given the setup,at $y = 0$,the path difference is $\Delta x = d = 5.5 \lambda$,which corresponds to a dark fringe because $5.5 \lambda = (n + 0.5) \lambda$ with $n = 5$.
Since the path difference at the center $(y=0)$ is $5.5 \lambda$,it is a dark fringe. Thus,statement $(C)$ is correct. Given the options,$(D)$ is the intended answer.

(A) In the given arrangement,the two coherent point sources $S_1$ and $S_2$ are placed on the axis perpendicular to the screen.
For any point $P$ on the screen,the path difference $\Delta x = |S_2P - S_1P|$ is constant for a given locus of points.
Since the sources are on the axis,the locus of points on the screen having a constant path difference forms a set of concentric circles centered at the point where the axis intersects the screen.
Therefore,the interference fringes obtained on the screen will be concentric circles.