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

(D) Coherent sources are defined as sources that emit waves of the same frequency and maintain a constant phase difference over time.
Since the speed of a wave $v$ is related to its wavelength $\lambda$ and frequency $f$ by the equation $v = f \lambda$,if two waves have the same wavelength and frequency,they must have the same phase velocity.
In many contexts,the terms wavelength,frequency,and phase velocity are used to describe the characteristics of coherent sources.
Therefore,both options $(b)$ and $(c)$ describe the necessary conditions for coherence in standard wave physics.
Thus,the correct option is $(d)$.

(B) Given the ratio of intensities of two waves is $\frac{I_1}{I_2} = \frac{9}{4}$.
Let the amplitudes be $A_1$ and $A_2$. Since $I \propto A^2$,the ratio of amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
The ratio of maximum to minimum intensity is given by the formula $\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$.
Substituting the values,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{3 + 2}{3 - 2} \right)^2 = \left( \frac{5}{1} \right)^2 = \frac{25}{1}$.
Therefore,the ratio is $25:1$.

(A) The given equations for the two waves are $y_1 = a_1 \sin \omega t$ and $y_2 = a_2 \sin (\omega t + \phi)$.
Comparing these with the given equations,we have $a_1 = 4$,$a_2 = 3$,and the phase difference $\phi = \frac{\pi}{2}$.
The amplitude $A$ of the resultant wave is given by the formula $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos \phi}$.
Since $\cos(\frac{\pi}{2}) = 0$,the formula simplifies to $A = \sqrt{a_1^2 + a_2^2}$.
Substituting the values,$A = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

(C) The wavelength of light is given as $\lambda = 4000 \mathring A = 4000 \times 10^{-10} \ m = 4 \times 10^{-7} \ m$.
The total length is $L = 1 \ mm = 10^{-3} \ m$.
The number of waves $n$ is given by the ratio of the total length to the wavelength:
$n = \frac{L}{\lambda} = \frac{10^{-3} \ m}{4 \times 10^{-7} \ m}$.
$n = \frac{1}{4} \times 10^{-3 - (-7)} = 0.25 \times 10^4$.
Therefore,the number of waves is $0.25 \times 10^4$.

(D) $LASER$ is the acronym for Light Amplification by Stimulated Emission of Radiation.
$A$ laser beam is intense,monochromatic (meaning it consists of a single wavelength),collimated,and highly coherent.
Coherence implies that the waves are in phase and maintain a constant phase relationship,which is a characteristic of coordinated waves of a single wavelength.

(C) When light travels from a rarer medium (air) to a denser medium (glass) and reflects off the interface,it undergoes a phase change of $\pi$ radians. This is a standard result in wave optics known as Stokes' law of reflection,which states that reflection from a denser medium introduces a phase shift of $\pi$.

(C) When light travels from one medium to another,its speed and wavelength change due to the change in the refractive index of the medium.
However,the frequency of light depends only on the source of the light and remains constant regardless of the medium it travels through.
Mathematically,if $\nu$ is the frequency,$v$ is the speed,and $\lambda$ is the wavelength,then $v = \nu \lambda$.
In a medium with refractive index $\mu$,the new speed $v' = v/\mu$ and the new wavelength $\lambda' = \lambda/\mu$.
Thus,the new frequency $\nu' = v'/\lambda' = (v/\mu) / (\lambda/\mu) = v/\lambda = \nu$.
Therefore,the frequency remains unchanged.

(C) The relationship between wavelength $(\lambda)$ and frequency $(f)$ is given by $c = f\lambda$, where $c$ is the speed of light. Thus, $f = c/\lambda$. This implies that frequency is inversely proportional to wavelength $(f \propto 1/\lambda)$.
In the visible spectrum $(VIBGYOR)$, the wavelength increases from violet to red $(\lambda_V < \lambda_I < \lambda_B < \lambda_G < \lambda_Y < \lambda_O < \lambda_R)$.
Consequently, the frequency decreases from violet to red $(f_V > f_I > f_B > f_G > f_Y > f_O > f_R)$.
Option $A$ is correct because $\lambda_R > \lambda_G$.
Option $B$ is correct because $\lambda_B < \lambda_O$.
Option $C$ is incorrect because the frequency of green light is smaller than the frequency of blue light $(f_G < f_B)$.
Option $D$ is correct because $f_V > f_B$.
Therefore, the incorrect statement is $C$.

(A) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{r^2}$,where $r$ is the distance from the source.
Given initial distance $r_1 = 60 \ cm$ and final distance $r_2 = 180 \ cm$.
The ratio of the intensities is given by $\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}$.
Substituting the values: $\frac{I_2}{I_1} = \left( \frac{60}{180} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9}$.
Therefore,the new intensity is $(1/9)$ times the original intensity.

(D) Coherent sources are defined as sources that emit light waves of the same frequency and maintain a constant phase difference over time.
Independent light sources,such as two different lamps,emit light due to independent atomic transitions.
These transitions occur randomly and independently in each source,leading to rapid and unpredictable changes in the phase of the emitted light.
Therefore,it is physically impossible to obtain coherent light from two independent light sources.
Thus,the correct option is $D$.

(D) Interference is a general property of waves. It occurs when two or more waves of the same frequency and constant phase difference superpose each other. This phenomenon is observed in all types of waves,including longitudinal mechanical waves (e.g.,sound waves),transverse mechanical waves (e.g.,waves on a string),and electromagnetic waves (e.g.,light waves). Therefore,the correct option is $D$.

(C) The intensity of the resultant wave in interference is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For maximum intensity, $\cos \phi = 1$, so $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Given $I_1 = I$ and $I_2 = 4I$, we have $I_{max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.
For minimum intensity, $\cos \phi = -1$, so $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Thus, $I_{min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$.
Therefore, the maximum and minimum intensities are $9I$ and $I$ respectively.

(C) The term $monochromatic$ is derived from the Greek words $mono$ (meaning single) and $chroma$ (meaning colour). However,in physics,a monochromatic wave is defined as a wave that consists of a single,specific wavelength or frequency. While monochromatic light often appears as a single colour to the human eye,the scientific definition strictly refers to the presence of a single wavelength. Therefore,option $C$ is the most accurate definition.

(A) For two light sources to produce a stable interference pattern,they must be coherent.
Coherent sources are defined as sources that emit light waves of the same frequency and wavelength,and maintain a constant phase difference over time.
If the phase difference between the two sources varies randomly with time,the interference pattern will shift rapidly,resulting in a uniform average intensity rather than a stable interference pattern.
Therefore,the condition for observing interference is that the phase difference between the light waves must remain constant.

(C) The wave nature of light is confirmed by phenomena such as interference,diffraction,and polarization,which cannot be explained by the particle theory of light.
Interference occurs when two or more light waves superpose to form a resultant wave of greater,lower,or the same amplitude.
Since interference is a characteristic property of waves,its observation in light confirms that light behaves as a wave.
Therefore,option $C$ is the correct answer.

(B) The coherence time $(\tau_c)$ is defined as the time interval over which the phase of a light wave remains predictable.
It is related to the coherence length $(L)$ and the velocity of light $(c)$ by the formula:
$\tau_c = \frac{L}{c}$
Therefore,the correct option is $B$.

(C) Given the ratio of amplitudes of two sources is $\frac{a_1}{a_2} = \frac{3}{5}$.
The intensity $I$ is proportional to the square of the amplitude,$I \propto a^2$.
The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2}$.
Substituting the given values:
$\frac{I_{\max}}{I_{\min}} = \frac{(3 + 5)^2}{(3 - 5)^2} = \frac{8^2}{(-2)^2} = \frac{64}{4} = \frac{16}{1}$.
Therefore,the ratio of intensities at maxima and minima is $16:1$.

(C) For constructive interference, the two light waves must arrive at a point in phase.
This occurs when the path difference $\Delta x$ between the waves is an integer multiple of the wavelength $\lambda$.
Mathematically, the condition for constructive interference is given by $\Delta x = n\lambda$, where $n = 0, 1, 2, 3, \dots$.
Therefore, the correct option is $C$.

(C) Two sources of waves are defined as coherent if they emit waves of the same frequency (or wavelength) and maintain a constant phase difference over time. If the phase difference changes with time,the interference pattern will not be stable,and the sources will be considered incoherent. Therefore,option $C$ is the correct definition.

(A) Given the ratio of intensities of two light waves is $\frac{I_1}{I_2} = \frac{4}{1}$.
Let the amplitudes be $A_1$ and $A_2$. Since $I \propto A^2$,the ratio of amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{4}{1}} = \frac{2}{1}$.
The ratio of maximum to minimum intensity is given by the formula $\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$.
Substituting the values,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{2 + 1}{2 - 1} \right)^2 = \left( \frac{3}{1} \right)^2 = \frac{9}{1}$.
Thus,the ratio is $9:1$.

(B) Two light sources are coherent if they emit light waves of the same frequency and maintain a constant phase difference over time.
When two sources are derived from a single original source (e.g.,using a slit or a mirror),the wavefront is divided into two parts.
These two resulting wavefronts act as if they emanated from two sources having a fixed phase relationship,thus satisfying the condition for coherence.

(C) Given the ratio of intensities of two waves is $\frac{I_1}{I_2} = \frac{25}{4}$.
Let the amplitudes of the waves be $A_1$ and $A_2$. Since intensity $I \propto A^2$,the ratio of amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
The ratio of maximum to minimum intensity is given by the formula $\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$.
Substituting the values,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{5 + 2}{5 - 2} \right)^2 = \left( \frac{7}{3} \right)^2 = \frac{49}{9}$.

(A) The essential condition for sustained interference is that the two sources of light 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.
If the phase difference between the two sources changes randomly with time,the interference pattern will shift rapidly,and the time-averaged intensity will be uniform,making the interference pattern invisible to the human eye.
Therefore,the correct condition is that the sources must have a constant phase difference.

(B) In the phenomenon of interference of light,the energy is redistributed in space. The energy that is missing from the points of destructive interference is transferred to the points of constructive interference. Therefore,the total energy of the system remains constant,meaning energy is conserved.

(C) The intensity $(I)$ of a light wave is directly proportional to the square of its amplitude $(a)$.
Mathematically,this is expressed as $I \propto a^2$.
Therefore,the intensity of light depends upon the amplitude of the wave.

(C) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
Given the ratio of amplitudes is $\frac{a_1}{a_2} = \frac{3}{4}$.
The ratio of their intensities is given by $\frac{I_1}{I_2} = \left( \frac{a_1}{a_2} \right)^2$.
Substituting the values: $\frac{I_1}{I_2} = \left( \frac{3}{4} \right)^2 = \frac{9}{16}$.
Therefore,the ratio of their intensities is $9:16$.

(D) The resultant intensity $I_R$ of two coherent sources is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference between the two waves.
For maximum intensity,the phase difference $\phi$ must be an even multiple of $\pi$,i.e.,$\phi = 0, 2\pi, 4\pi, \dots$,which makes $\cos \phi = 1$.
Substituting $\cos \phi = 1$ into the formula,we get: $I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$.
This expression can be written as a perfect square: $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.

(C) Coherent sources are defined as sources that emit light waves of the same frequency and wavelength,and maintain a constant phase difference over time.
If the phase difference between the waves emitted by two sources changes with time,the sources are called incoherent.
Therefore,for a source to be coherent,both the frequency must be constant and the phase difference must remain constant.

(A) The wave nature of light is primarily verified by phenomena such as interference,diffraction,and polarization.
$(a)$ Interference is a characteristic property of waves,confirming the wave nature of light.
$(b)$ The photoelectric effect verifies the particle nature (quantum nature) of light.
$(c)$ and $(d)$ Reflection and refraction can be explained by both the wave theory (Huygens' principle) and the particle theory (Fermat's principle/Newton's corpuscular theory),so they do not exclusively verify the wave nature.

(C) Let the intensities of the two sources be $I_1$ and $I_2$. The maximum and minimum intensities in an interference pattern are given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Given $\frac{I_{\max}}{I_{\min}} = 25$,we have $\left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2 = 25$.
Taking the square root on both sides,$\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = 5$.
Let $x = \sqrt{\frac{I_1}{I_2}}$. Then $\frac{x+1}{x-1} = 5$.
$x + 1 = 5x - 5 \Rightarrow 4x = 6 \Rightarrow x = \frac{3}{2}$.
Therefore,$\frac{I_1}{I_2} = x^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.

(B) The relationship between frequency $(\nu)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by the formula: $\nu = \frac{c}{\lambda}$.
Given:
Speed of light,$c = 3 \times 10^8 \ \text{m/s}$.
Wavelength,$\lambda = 3000 \ \mathring A = 3000 \times 10^{-10} \ \text{m} = 3 \times 10^{-7} \ \text{m}$.
Substituting the values into the formula:
$\nu = \frac{3 \times 10^8}{3 \times 10^{-7}} = 10^{15} \ \text{Hz}$ (or $\text{cycles/sec}$).
Therefore,the correct option is $B$.

(D) For destructive interference,the waves must arrive at the point out of phase by an odd multiple of $\pi$ radians.
This corresponds to a path difference that is an odd multiple of half the wavelength $(\frac{\lambda}{2})$.
Mathematically,the path difference $\Delta x$ is given by $\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, 3, \dots$.
Therefore,the correct option is $(d)$.

(D) The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.
Given $\frac{I_{\max}}{I_{\min}} = \frac{36}{1}$.
Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = \sqrt{\frac{36}{1}} = 6$.
By applying componendo and dividendo,$\frac{(a_1 + a_2) + (a_1 - a_2)}{(a_1 + a_2) - (a_1 - a_2)} = \frac{6 + 1}{6 - 1}$.
This simplifies to $\frac{2a_1}{2a_2} = \frac{7}{5}$.
Therefore,the ratio of amplitudes is $\frac{a_1}{a_2} = 7:5$.

(B) The resultant intensity $I_{res}$ of two interfering waves is given by the formula: $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For maximum intensity,the phase difference $\phi$ must be $0$,so $\cos \phi = 1$.
Thus,the maximum intensity is given by: $I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Given $I_1 = I$ and $I_2 = 4I$.
Substituting these values: $I_{max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.

(B) In the phenomenon of interference,the total energy of the system is conserved.
When two coherent sources of light interfere,the light energy is not created or destroyed; it is simply redistributed in space.
This results in regions of constructive interference (maxima) and destructive interference (minima).
Since the sources are coherent,the phase difference between the waves at any given point remains constant over time.
Therefore,the interference pattern (the spatial distribution of energy) does not change with time.

(A) For the phenomenon of interference to be observed,the two sources of light must be coherent.
Coherent sources are defined as sources that emit radiation of the same frequency and maintain a constant phase difference over time.
If the phase difference changes randomly,the interference pattern will not be stable and will not be observable.
Therefore,the correct requirement is that the sources must have the same frequency and a definite (constant) phase relationship.

(D) The intensity $I$ of light from a point source follows the inverse square law with respect to the distance $r$ from the source.
Mathematically,$I \propto \frac{1}{r^2}$.
If the initial distance is $r_1 = r$ and the final distance is $r_2 = 2r$,the new intensity $I'$ is given by:
$I' \propto \frac{1}{(2r)^2} = \frac{1}{4r^2} = \frac{1}{4} I$.
Therefore,the intensity becomes one-fourth of its original value.

(D) The resultant intensity $I_R$ of two interfering waves with intensities $I_1$ and $I_2$ is given by the formula:
$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$
where $\phi$ is the phase difference between the waves.
For maximum intensity,the phase difference $\phi$ must be $0^\circ$ (or an even multiple of $\pi$),which makes $\cos \phi = 1$.
Given $I_1 = I$ and $I_2 = I$,we substitute these values into the formula:
$I_{max} = I + I + 2\sqrt{I \cdot I} \cos 0^\circ$
$I_{max} = 2I + 2\sqrt{I^2} (1)$
$I_{max} = 2I + 2I = 4I$
Therefore,the maximum intensity obtained is $4I$.

(C) In the interference of light,the total energy is neither created nor destroyed.
Instead,energy is transferred from the regions of destructive interference (minima) to the regions of constructive interference (maxima).
The average energy over the entire pattern remains equal to the sum of the energies of the individual interfering waves.
Therefore,the phenomenon of interference is in complete agreement with the law of conservation of energy.

(B) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = \frac{1}{4}$.
Let $I_1 = k$ and $I_2 = 4k$.
The formula for fringe visibility $V$ is given by $V = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$.
Substituting the values,we get $V = \frac{2\sqrt{k \times 4k}}{k + 4k}$.
$V = \frac{2 \times 2k}{5k} = \frac{4k}{5k} = 0.8$.
Thus,the fringe visibility is $0.8$.

(B) Interference is a phenomenon resulting from the superposition of waves originating from two coherent sources.
Diffraction is a phenomenon resulting from the superposition of secondary wavelets originating from different parts of the same wavefront.
Both of these phenomena cannot be explained by the corpuscular theory of light.
Therefore,diffraction and interference provide strong evidence for the wave nature of light.

(D) The visible spectrum of light ranges approximately from $400 \, nm$ to $700 \, nm$.
Converting these values to meters,we get $400 \times 10^{-9} \, m$ to $700 \times 10^{-9} \, m$,which is $4 \times 10^{-7} \, m$ to $7 \times 10^{-7} \, m$.
Among the given options,$6 \times 10^{-7} \, m$ falls within this range.
Therefore,the correct option is $D$.

(B) From the geometry,the distance from $O$ to the line containing $P$ is $d$. Thus,$PO = d \sec \theta$.
Since $CP$ is a wavefront,the optical path length from $C$ to $P$ is equal to the path length from $O$ to $P$ along the ray $AO$. The path difference between the ray $BP$ and the reflected ray $OP$ is $\Delta = CO + OP$.
In $\triangle COP$,$CO = PO \cos 2\theta = d \sec \theta \cos 2\theta$.
Thus,$\Delta = d \sec \theta + d \sec \theta \cos 2\theta = d \sec \theta (1 + \cos 2\theta) = d \sec \theta (2 \cos^2 \theta) = 2d \cos \theta$.
Since the ray $OP$ undergoes reflection at the surface $QR$,there is an additional phase shift of $\pi$,which corresponds to a path difference of $\lambda / 2$.
For constructive interference,the total path difference must be an odd multiple of $\lambda / 2$ (due to the $\pi$ phase shift): $\Delta = \lambda / 2$.
$2d \cos \theta = \lambda / 2 \implies \cos \theta = \lambda / 4d$.

(D) For interference fringes to be observed,the superposing waves must have the same frequency and a constant phase difference.
$1$. Waves $(i)$ and $(ii)$ have the same angular frequency $\omega$. Thus,they can produce interference.
$2$. Waves $(iii)$ and $(iv)$ have the same angular frequency $2\omega$. Thus,they can also produce interference.
Therefore,interference fringes can be observed due to the superposition of $(i)$ and $(ii)$ or $(iii)$ and $(iv)$.

(B) For constructive interference (maxima),the net phase difference $\Delta \phi$ must be an integer multiple of $2\pi$.
Let $\phi_X$ and $\phi_Y$ be the phases of waves from $X$ and $Y$ at a point. The phase difference due to path difference $\Delta x = (YO - XO)$ is $\frac{2\pi}{\lambda} \Delta x$.
Given that $Y$ lags behind $X$ by $2\pi l$,the total phase difference is $\Delta \phi = \frac{2\pi}{\lambda}(YO - XO) - 2\pi l$.
For maxima,$\Delta \phi = 2\pi n$,where $n$ is an integer.
$2\pi n = \frac{2\pi}{\lambda}(YO - XO) - 2\pi l$.
Assuming the path difference $(YO - XO)$ is such that the condition simplifies to the distance $XO$ relative to the phase lag,the standard condition for maxima is $\frac{2\pi}{\lambda}(XO) = 2\pi(n + l)$.
Thus,$XO = \lambda(n + l)$.

(C) The path difference between the beams reflected from two successive planes is given by $\Delta x = 2d \sin \theta$,where $\theta$ is the glancing angle (the angle between the incident beam and the plane).
For constructive interference,the path difference must be an integer multiple of the wavelength:
$\Delta x = n\lambda$
Substituting the expression for path difference:
$2d \sin \theta = n\lambda$
$\sin \theta = \frac{n\lambda}{2d}$
$\theta = \sin^{-1}\left(\frac{n\lambda}{2d}\right)$
Thus,the correct option is $C$.

(B) The path difference $\Delta x$ at a point $P$ on the circle is given by the projection of the separation $d$ onto the line connecting the point $P$ to the sources. From the geometry,$\Delta x = d \cos \theta$,where $\theta$ is the angle with the line joining the sources.
For constructive interference (maxima) at point $P$,the condition is $\Delta x = n\lambda$.
Equating the two expressions for path difference:
$d \cos \theta = n\lambda$
$\cos \theta = \frac{n\lambda}{d}$
$\theta = \cos^{-1} \left( \frac{n\lambda}{d} \right)$
Given $n = 4$,the angular position is $\theta = \cos^{-1} \left( \frac{4\lambda}{d} \right)$.

(B) Let the distance between the sources be $d = 4\lambda$. The detector is at a point $D$ on the $x$-axis at a distance $x$ from the origin (which is $S_2$).
The path difference $\Delta x$ between the waves reaching $D$ from $S_1$ and $S_2$ is given by $\Delta x = S_1D - S_2D$.
In the right-angled triangle $\Delta S_1S_2D$,$S_1D = \sqrt{d^2 + x^2} = \sqrt{(4\lambda)^2 + x^2}$ and $S_2D = x$.
For maxima,the path difference must be an integer multiple of the wavelength: $\Delta x = n\lambda$,where $n$ is an integer.
So,$\sqrt{16\lambda^2 + x^2} - x = n\lambda$.
Rearranging gives $\sqrt{16\lambda^2 + x^2} = n\lambda + x$.
Squaring both sides: $16\lambda^2 + x^2 = n^2\lambda^2 + 2nx\lambda + x^2$.
$16\lambda^2 - n^2\lambda^2 = 2nx\lambda$.
$x = \frac{(16 - n^2)\lambda}{2n}$.
For $x > 0$,we must have $16 - n^2 > 0$,which means $n^2 < 16$,so $n$ can be $1, 2, 3$.
If $n=1$,$x = \frac{15\lambda}{2} = 7.5\lambda$.
If $n=2$,$x = \frac{12\lambda}{4} = 3\lambda$.
If $n=3$,$x = \frac{7\lambda}{6} \approx 1.17\lambda$.
Thus,there are $3$ points where maxima are observed.

(A) The total phase difference at point $P$ is the sum of the initial phase difference and the phase difference caused by the path difference.
Initial phase difference $\phi_i = 66^\circ$ (where $A$ is ahead of $B$).
Path difference $\Delta x = PB - PA = \lambda / 4$.
The phase difference due to path difference is given by $\phi_p = \frac{2\pi}{\lambda} \times \Delta x = \frac{360^\circ}{\lambda} \times \frac{\lambda}{4} = 90^\circ$.
Since $A$ is ahead of $B$ by $66^\circ$,and the path $PB$ is longer than $PA$ by $\lambda/4$,the wave from $B$ lags behind the wave from $A$ due to the path difference.
The total phase difference $\Delta \phi = 66^\circ + 90^\circ = 156^\circ$.