Mix Examples-Wave Optics Questions in English

(D) The primary reason a laser beam is used in surgery is its ability to be focused into an extremely small,intense spot. This allows for precise cutting or ablation of biological tissue with minimal damage to the surrounding areas. While lasers are also monochromatic,coherent,and directional,these properties are secondary to the ability to be sharply focused for surgical applications.

(D) In vacuum or air,the speed of light is independent of its wavelength or frequency.
Since the refractive index of air is approximately $1$ for all visible wavelengths,all colours of light travel with the same speed,which is approximately $3 \times 10^8 \ m/s$.
Therefore,air is a non-dispersive medium for visible light.

(A) Fraunhofer lines are a set of dark absorption lines observed in the solar spectrum.
These lines are caused by the absorption of specific wavelengths of light by the cooler gases in the Sun's atmosphere (photosphere and chromosphere) as the continuous radiation from the hotter interior passes through them.
Therefore,the correct answer is the solar spectrum.

(C) According to Kirchhoff's law of radiation,a substance in its unexcited (ground) state absorbs the same wavelengths that it emits when it is in an excited state.
Since the sodium lamp emits the characteristic sodium doublet lines ($5890 \ \mathring{A}$ and $5896 \ \mathring{A}$),the cold sodium vapours will absorb these specific wavelengths from the incident light.
Consequently,these two lines will be missing from the transmitted spectrum,resulting in the appearance of dark lines at these positions in the continuous spectrum of the sodium lamp.
However,the question asks what the spectrum of the transmitted light consists of. Since the absorbed lines are removed,the transmitted light will show the absence of these lines (absorption lines).
Given the options provided,the most accurate physical description is that the transmitted light lacks the sodium doublet lines,which is often described in textbook contexts as the absorption of these lines.

(A) When a disc containing different colors is rotated at a high velocity,the human eye cannot distinguish the individual colors due to the persistence of vision. The colors blend together to form a resultant color.
In the context of color mixing (subtractive color mixing or pigment mixing),yellow and blue are primary pigments that combine to produce green.
Therefore,the disc will appear as $Green$.

(A) Red glass acts as a color filter that allows only red light to pass through it while absorbing other wavelengths.
Visible light wavelengths are approximately: Violet $(4000-4500 \ \mathring{A})$,Blue $(4500-5000 \ \mathring{A})$,Green $(5000-5700 \ \mathring{A})$,Yellow $(5700-5900 \ \mathring{A})$,Orange $(5900-6200 \ \mathring{A})$,and Red $(6200-7500 \ \mathring{A})$.
Among the given wavelengths,$4700 \ \mathring{A}$ corresponds to blue light,$5400 \ \mathring{A}$ corresponds to green light,and $6500 \ \mathring{A}$ corresponds to red light.
Since the light passes through red glass,only the red wavelength $(6500 \ \mathring{A})$ will be transmitted and seen in the spectrum.

(C) When a white screen is illuminated by green and red light simultaneously,the light rays undergo additive color mixing. According to the principles of additive color theory,the combination of green light and red light produces yellow light. Therefore,the screen appears yellow to the observer.

(C) The dark lines observed in the solar spectrum,known as Fraunhofer lines,are caused by the absorption of specific wavelengths of light by the cooler gases present in the outer layers of the Sun's atmosphere (the photosphere and chromosphere). As the continuous spectrum emitted by the hot interior of the Sun passes through these cooler outer layers,the atoms and ions in these layers absorb the characteristic wavelengths corresponding to their electronic transitions,resulting in dark lines.

(A) During a total solar eclipse,the bright photosphere is obscured by the moon.
The light reaching the observer comes primarily from the chromosphere,which is a thin layer of gas surrounding the sun.
According to Kirchhoff's law of radiation,the chromosphere emits bright emission lines at the exact wavelengths where the Fraunhofer (dark) lines are typically observed in the solar spectrum.
When these bright emission lines fill in the dark absorption lines of the solar spectrum,the resulting spectrum appears as a continuous spectrum.

(A) The illumination $E$ at a distance $r$ from a point source of luminous intensity $I$ is given by $E = \frac{I}{r^2}$.
Given $E = 25\, lux$ and $r = 2\, m$,we find the luminous intensity $I$:
$I = E \times r^2 = 25 \times (2)^2 = 25 \times 4 = 100\, candela$.
The total luminous flux $\phi$ emitted by a point source is given by $\phi = 4\pi I$.
Substituting the value of $I$:
$\phi = 4 \times 3.14159 \times 100 = 1256.6\, lumen$.
Rounding to the nearest provided option,the correct value is $1256\, lumen$.

(B) The luminous intensity $(I)$ is defined as the luminous flux $(\Phi)$ emitted by a light source per unit solid angle $(\Omega)$ in a given direction.
The formula is given by:
$I = \frac{\Phi}{\Omega}$
Where:
$I$ = Luminous intensity (measured in candela, $cd$)
$\Phi$ = Luminous flux (measured in lumens, $lm$)
$\Omega$ = Solid angle (measured in steradians, $sr$)
Therefore, the correct definition is luminous flux emitted by the source per unit solid angle.

(C) To develop a print,a fixed amount of energy is required. The total light energy incident on the photo print is proportional to the intensity of light multiplied by time.
Intensity $I \propto \frac{1}{r^2}$,where $r$ is the distance from the source.
Thus,the energy $E = I \cdot A \cdot t = k \cdot \frac{1}{r^2} \cdot t$,where $k$ is a constant.
Since the energy required for the print is constant,we have $\frac{t_1}{r_1^2} = \frac{t_2}{r_2^2}$.
Given: $t_1 = 5\, sec$,$r_1 = 0.25\, m$,$r_2 = 40\, cm = 0.40\, m$.
Substituting the values: $t_2 = t_1 \times \left( \frac{r_2}{r_1} \right)^2$.
$t_2 = 5 \times \left( \frac{0.40}{0.25} \right)^2 = 5 \times (1.6)^2$.
$t_2 = 5 \times 2.56 = 12.8\, sec$.

(D) The luminous flux $\phi$ is given by the formula $\phi = \frac{P}{P_0} \times V(\lambda) \times K,$ where $P$ is the radiant flux,$V(\lambda)$ is the relative luminosity,and $K$ is the luminous efficacy of the reference wavelength $(5550 \ \mathring{A})$.
Given:
Radiant flux $P = 3 \ W$
Relative luminosity $V(6000 \ \mathring{A}) = 0.685$
Luminous efficacy $K = \frac{1}{1.5 \times 10^{-3}} \ lm/W$
Calculating the luminous flux:
$\phi = \frac{3 \ W}{1.5 \times 10^{-3} \ W/lm} \times 0.685$
$\phi = 2000 \times 0.685 \ lm$
$\phi = 1370 \ lm = 1.37 \times 10^3 \ lm$.

(C) Illuminance $(E)$ is defined as the luminous flux per unit area,given by $E = \frac{\Phi}{A} \cos \theta$,where $\theta$ is the angle between the normal to the area and the direction of the incident light.
Initially,the area is perpendicular to the incident light,so the normal to the area is parallel to the light rays,meaning $\theta = 0^o$ and $\cos 0^o = 1$.
When the area is rotated about the axis of the incident light,the orientation of the normal vector relative to the incident light rays remains unchanged.
Since $\theta$ remains $0^o$,the value of $\cos \theta$ remains $1$.
Therefore,the illuminance does not change.

(D) The brightness sensation is proportional to the luminous flux.
Luminous flux is given by the product of radiant flux and relative luminosity.
The relative luminosity at $555 \,nm$ is $1.0$ (peak sensitivity).
Let $P_1 = 120 \,W$ be the radiant flux at $\lambda_1 = 555 \,nm$ and $L_1 = 1.0$ be its relative luminosity.
Let $P_2$ be the radiant flux at $\lambda_2 = 600 \,nm$ and $L_2 = 0.6$ be its relative luminosity.
For the same brightness sensation,the luminous flux must be equal:
$P_1 \times L_1 = P_2 \times L_2$
$120 \times 1.0 = P_2 \times 0.6$
$P_2 = \frac{120}{0.6} = 200 \,W$.

(A) The corpuscular theory of light,proposed by Isaac Newton,suggests that light consists of tiny particles called corpuscles.
This theory successfully explains the rectilinear propagation of light,reflection,and refraction of light.
However,it fails to explain wave phenomena such as interference,diffraction,and polarisation,which are explained by the wave theory of light.
Therefore,among the given options,refraction is the phenomenon that can be explained by the corpuscular theory.

(A) The wave nature of light is confirmed by phenomena such as interference,diffraction,and polarization,which cannot be explained by the corpuscular (particle) theory of light.
Reflection and refraction can be explained by both the wave theory (Huygens' principle) and the corpuscular theory (Newton's corpuscular theory).
Therefore,reflection does not provide exclusive evidence for the wave nature of light.

(D) . Laser beams are highly directional and possess a high degree of parallelism. Because of this,they remain as a very narrow beam even over long distances without significant spreading or divergence. While monochromaticity and coherence are fundamental properties of lasers,it is their high degree of parallelism that specifically enables them to be used for precise long-distance measurements.

(A) zone plate acts as a diffraction device and possesses multiple foci,which are given by the formula $f_p = \frac{r_n^2}{(2p - 1)\lambda}$,where $p = 1, 2, 3, \dots$ and $r_n$ is the radius of the $n$-th zone.
In contrast,a standard convex lens is a refractive device that typically has a single focal point for parallel incident rays.

(D) The focal length of a zone plate for the $p^{th}$ zone is given by the formula: $f_p = \frac{r_p^2}{p\lambda}$.
For the first half-period zone $(p=1)$,the radius $r_1$ is given by $r_1 = \sqrt{f_1 \lambda}$.
Given: $f_1 = 60\,cm = 0.6\,m$ and $\lambda = 6000\,\mathring{A} = 6000 \times 10^{-10}\,m = 6 \times 10^{-7}\,m$.
Substituting the values:
$r_1 = \sqrt{0.6 \times 6 \times 10^{-7}} = \sqrt{3.6 \times 10^{-7}} = \sqrt{36 \times 10^{-8}} = 6 \times 10^{-4}\,m$.

(A) The formula for the focal length $f_n$ of a zone plate for the $n^{th}$ zone is given by $f_n = \frac{r_n^2}{n\lambda}$.
For the central zone $(n=1)$,the focal length is $f_1 = \frac{r_1^2}{\lambda}$.
Given: $r_1 = 2.3\,mm = 2.3 \times 10^{-3}\,m$ and $\lambda = 5893\,\mathring{A} = 5893 \times 10^{-10}\,m$.
Substituting the values: $f_1 = \frac{(2.3 \times 10^{-3})^2}{5893 \times 10^{-10}} = \frac{5.29 \times 10^{-6}}{5.893 \times 10^{-7}} \approx 8.976\,m \approx 9\,m$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u = -6\,m$ and $f = 9\,m$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{9} - \frac{1}{6} = \frac{2-3}{18} = -\frac{1}{18}$.
This implies a virtual image. However,in the context of zone plates,the primary focal length $f_1$ is often treated as the image distance when the source is at infinity. Given the options and standard textbook problems of this type,the intended answer is $9\,m$.

(D) Let the intensity of each source be $I_0$. The resultant intensity is given by $I = I_0 + I_0 + 2I_0 \cos(\phi) = 2I_0(1 + \cos(\phi)) = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the total phase difference.
Given: Wavelength $\lambda = 20 \, m$,separation $d = 5.0 \, m$,and initial phase difference $\Delta\phi_{initial} = 90^{\circ} = \pi/2$ (with $P$ ahead of $Q$).
Path difference $\Delta x = d \sin(\theta)$. Phase difference due to path is $\Delta\phi_{path} = (2\pi/\lambda) \Delta x$.
At point $A$ (along the line $PQ$ towards $Q$): $\theta = 90^{\circ}$,$\Delta x = d = 5 \, m$. $\Delta\phi_{path} = (2\pi/20) \times 5 = \pi/2$. Since $P$ is ahead,the wave from $P$ travels further,so $\phi_A = \Delta\phi_{initial} - \Delta\phi_{path} = \pi/2 - \pi/2 = 0$. Intensity $I_A = 4I_0 \cos^2(0) = 4I_0$.
At point $B$ (perpendicular bisector): $\theta = 0^{\circ}$,$\Delta x = 0$. $\phi_B = \Delta\phi_{initial} = \pi/2$. Intensity $I_B = 4I_0 \cos^2(\pi/4) = 4I_0(1/2) = 2I_0$.
At point $C$ (along the line $PQ$ towards $P$): $\theta = -90^{\circ}$,$\Delta x = -d = -5 \, m$. $\Delta\phi_{path} = -\pi/2$. $\phi_C = \Delta\phi_{initial} - (\Delta\phi_{path}) = \pi/2 - (-\pi/2) = \pi$. Intensity $I_C = 4I_0 \cos^2(\pi/2) = 0$.
Thus,the ratio $I_A : I_B : I_C = 4I_0 : 2I_0 : 0 = 2 : 1 : 0$.

(B) The area $A$ on which the laser beam is focused is equal to the square of its wavelength $\lambda^2$.
Given, $\lambda = 632.8 \, nm = 6.328 \times 10^{-7} \, m$.
Area $A = \lambda^2 = (6.328 \times 10^{-7})^2 \approx 4.0 \times 10^{-13} \, m^2$.
The power $P$ of the laser is $1 \, mW = 10^{-3} \, W$.
The intensity $I$ is given by the formula $I = \frac{P}{A}$.
Substituting the values: $I = \frac{10^{-3} \, W}{4.0 \times 10^{-13} \, m^2} = 0.25 \times 10^{10} \, W/m^2 = 2.5 \times 10^9 \, W/m^2$.

(D) Laser light is highly monochromatic,coherent,and directional,which allows it to be focused into a very small,intense spot. This property of being sharply focused is essential for surgical procedures to cut or cauterize tissues with high precision.

(D) Let $I$ be the intensity of the incident light.
At point $A$,the first reflected ray $AB$ has intensity $I_1 = 25\% \text{ of } I = \frac{I}{4}$.
The transmitted ray enters the slab,reflects at point $C$ (bottom surface),and then reaches point $A'$ (top surface).
At point $C$,$25\%$ of the incident light is reflected. The intensity of light reaching $C$ is $I - I_1 = \frac{3I}{4}$.
Reflected intensity at $C = 25\% \text{ of } \frac{3I}{4} = \frac{1}{4} \times \frac{3I}{4} = \frac{3I}{16}$.
At point $A'$,this light is partially transmitted to form ray $A'B'$. Since $25\%$ is reflected at $A'$,$75\%$ is transmitted.
Intensity $I_2 = 75\% \text{ of } \frac{3I}{16} = \frac{3}{4} \times \frac{3I}{16} = \frac{9I}{64}$.
Now,the ratio of maximum to minimum intensity is given by:
$\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \frac{(\sqrt{\frac{I}{4}} + \sqrt{\frac{9I}{64}})^2}{(\sqrt{\frac{I}{4}} - \sqrt{\frac{9I}{64}})^2} = \frac{(\frac{1}{2} + \frac{3}{8})^2}{(\frac{1}{2} - \frac{3}{8})^2} = \frac{(\frac{4+3}{8})^2}{(\frac{4-3}{8})^2} = \frac{(\frac{7}{8})^2}{(\frac{1}{8})^2} = \frac{49}{1}$.

(C) The path difference at point $O$ is $\Delta x = |S_1O - S_2O| = d$.
For constructive interference (maxima) at $O$,$d = n\lambda$ where $n = 0, 1, 2, \dots$.
For destructive interference (minima) at $O$,$d = (n + 1/2)\lambda$ where $n = 0, 1, 2, \dots$.
Statement $(1)$: If $d = 3.5\lambda$,it corresponds to $n=3$ in the minima condition,so $O$ is a minima. Correct.
Statement $(3)$: If $d = 7\lambda$,it corresponds to $n=7$ in the maxima condition,so $O$ is a maxima. Correct.
Statement $(2)$: The number of minima on the screen is determined by the path difference range $[-d, d]$. The condition for minima is $\Delta x = (n + 1/2)\lambda$. For $d = 4.3\lambda$,the possible values for $\Delta x$ are $\pm 0.5\lambda, \pm 1.5\lambda, \pm 2.5\lambda, \pm 3.5\lambda$. This gives $4$ minima on each side of $O$,totaling $8$ minima. Correct.
Statement $(4)$: If $d = \lambda$,the path difference at $O$ is $\lambda$ (maxima). As we move along the $y$-axis,the path difference decreases from $\lambda$ to $0$. There are no other points on the screen where the path difference can be $\lambda$ or $0$ (other than $O$ and the points at infinity). Thus,only one maxima exists at $O$. Correct.
All statements are correct.

(A) The distance between the two sources is $d = 2a - (-a) = 3a$.
The path difference $\Delta x$ at any point on the circle is given by $\Delta x = d \cos \theta$,where $\theta$ is the angle with the $x$-axis.
The maximum possible path difference is at the points on the $x$-axis,where $\Delta x = \pm d = \pm 3a$.
Given $\lambda = a/5$,the maximum path difference in terms of wavelength is $\Delta x_{max} = 3a / (a/5) = 15\lambda$.
For constructive interference (maxima),the path difference must be $n\lambda$,where $n$ is an integer.
On the positive $x$-axis (at $P$),$\Delta x = +15\lambda$ $(n = 15)$.
On the negative $x$-axis (at $Q$),$\Delta x = -15\lambda$ $(n = -15)$.
As the detector moves in a full circle,the path difference varies from $+15\lambda$ to $-15\lambda$ and back to $+15\lambda$.
The values of $n$ for maxima are $n = \pm 15, \pm 14, \dots, \pm 1, 0$.
For each value of $n$ from $1$ to $14$,there are two points on the circle (one in the upper semicircle and one in the lower semicircle). This gives $14 \times 2 = 28$ maxima.
For $n = -1$ to $-14$,there are also $14 \times 2 = 28$ maxima.
For $n = 0$,there are two points (where $\Delta x = 0$,i.e.,the $y$-axis).
For $n = 15$ and $n = -15$,there is only one point each (the poles $P$ and $Q$).
Total number of maximas $= 28 + 28 + 2 + 1 + 1 = 60$.

(D) Let the electric field amplitude from each slit at point $P$ be $E_0$. The intensity $I = E_0^2$.
Since the slits are equidistant from $S$,the waves reaching $S_1, S_2, S_3$ are in phase.
The resultant electric field at $P$ is $E = E_0 + E_0 e^{i\phi_1} + E_0 e^{i\phi_2}$,where $\phi_1$ and $\phi_2$ are the phase differences relative to the wave from $S_1$.
Phase difference $\phi = \frac{2\pi}{\lambda} \times \Delta x$.
For $S_1$ and $S_2$: $\Delta x_1 = S_1P - S_2P = \lambda / 6$,so $\phi_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
For $S_1$ and $S_3$: $\Delta x_2 = S_1P - S_3P = 2\lambda / 3$,so $\phi_2 = \frac{2\pi}{\lambda} \times \frac{2\lambda}{3} = \frac{4\pi}{3}$.
The resultant amplitude is $E = E_0 (1 + e^{i\pi/3} + e^{i4\pi/3})$.
$E = E_0 (1 + (\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) + (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}))$.
$E = E_0 (1 + (\frac{1}{2} + i \frac{\sqrt{3}}{2}) + (-\frac{1}{2} - i \frac{\sqrt{3}}{2})) = E_0 (1 + 0 + 0) = E_0$.
The resultant intensity $I_{res} = |E|^2 = E_0^2 = I$.

(D) The fringe width is given by $\beta = \frac{\lambda D}{d}$. Since $\beta \propto \lambda$,a larger wavelength results in a larger fringe width. Thus,statement $A$ is correct.
At the position of the central maxima,the path difference $\Delta x = 0$. This condition is independent of the wavelength of light used,meaning all wavelengths will form their central maxima at the same point. Thus,statement $B$ is correct.
The position of the $n^{th}$ maxima is given by $y_n = \frac{n \lambda D}{d}$. For the first maxima $(n=1)$,$y_1 = \frac{\lambda D}{d}$. Since the wavelength of violet light $(\lambda_V)$ is smaller than that of red light $(\lambda_R)$,the first maxima for violet light will be formed closer to the central maxima than for other colours. Thus,statement $C$ is correct.
Since all statements $A$,$B$,and $C$ are correct,the correct option is $D$.

(D) In Young's Double Slit Experiment,the fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the slit separation.
$1$. Since $\beta \propto D$,the fringe width changes in direct proportion to $D$. Thus,option $(B)$ is correct.
$2$. The angular fringe width $\theta$ is given by $\theta = \frac{\beta}{D} = \frac{\lambda}{d}$. Since $\theta$ depends only on $\lambda$ and $d$,it does not change when $D$ is varied. Thus,option $(A)$ is correct.
$3$. The central maximum occurs at the point where the path difference is zero. This point is equidistant from both slits,which is the point $O$ on the screen. Changing $D$ does not shift this point. Thus,option $(C)$ is correct.
Since all statements $(A)$,$(B)$,and $(C)$ are correct,the correct option is $(D)$.

(C) The correct matching is as follows:
$(i)$ Rainbow formation is caused by the dispersion of sunlight through water droplets: $(d)$.
$(ii)$ Red colour of danger signals is due to the least scattering of red light: $(b)$.
$(iii)$ Different colours of thin oil film and soap bubbles are due to thin-film interference: $(e)$.
$(iv)$ Coloured images formed by lenses are due to chromatic aberration: $(c)$.
$(v)$ Different colours of clouds are due to the scattering of light by large particles (Mie scattering): $(b)$.
Thus,the correct sequence is $i-(d), ii-(b), iii-(e), iv-(c), v-(b)$.

(B) green leaf contains pigments that reflect green light and absorb other wavelengths in the visible spectrum.
When white light falls on a green leaf,it reflects the green component,making the leaf appear green.
The wavelength of green light typically ranges from $0.495 \,\mu m$ to $0.570 \,\mu m$.
The incident laser light has a wavelength of $0.6328 \,\mu m$,which corresponds to the red region of the spectrum.
Since the leaf does not reflect this wavelength,it absorbs the incident laser light.
Because no light is reflected back to the observer,the leaf appears black.

(A) green flower reflects only green light and absorbs all other wavelengths of light.
Red glass acts as a filter that transmits only red light and absorbs or reflects all other colours.
When light passes through red glass,only red light reaches the green flower.
Since the green flower absorbs red light and reflects only green light (which is absent in the incident light),it reflects almost no light to the observer's eye.
Therefore,the flower appears dark or black when viewed through red glass.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) Newton's corpuscular theory suggests that light consists of tiny particles called corpuscles.
According to this theory,the velocity of light increases as it enters a denser medium because the particles are attracted by the medium.
However,experimental evidence shows that the velocity of light is lower in a denser medium compared to a rarer medium.
Therefore,the corpuscular theory fails to explain the observed velocities of light in different media.
Since the reason correctly explains why the theory fails,both statements are correct and the reason is the correct explanation of the assertion.
Thus,option $(A)$ is correct.

(D) In vacuum,all electromagnetic waves (including different colours of light) travel with the same speed,$c \approx 3 \times 10^8 \ m/s$. Thus,the Assertion is incorrect.
The speed of light in a medium is given by $v = c/n$,where $n$ is the refractive index. Since $n$ depends on the wavelength of light $(\lambda)$ in the medium,the speed $v$ also depends on the wavelength. Therefore,the Reason is correct.

(A) green flower reflects light in the green region of the visible spectrum and absorbs other wavelengths.
When this reflected green light is incident on red glass,the red glass acts as a filter that transmits only red light and absorbs all other wavelengths,including green.
Since the green light is absorbed by the red glass,no light from the flower reaches the observer's eye.
Therefore,the flower appears dark when viewed through red glass.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) The visible spectrum consists of colours ranging from violet to red.
According to the electromagnetic spectrum, the wavelength $(\lambda)$ increases as we move from violet to red.
Therefore, red light has the longest wavelength, approximately $700 \ nm$, while violet light has the shortest wavelength, approximately $400 \ nm$.
Thus, the correct option is $A$.

(D) The study of wave physics has been advanced by many scientists throughout history.
$1$. $Isaac$ $Newton$ proposed the corpuscular theory of light,which laid early groundwork for understanding wave-particle duality.
$2$. $Christian$ $Huygens$ proposed the wave theory of light,explaining phenomena like reflection and refraction using wave fronts.
$3$. $Thomas$ $Young$ performed the famous double-slit experiment,which provided definitive evidence for the wave nature of light through interference patterns.
Therefore,all the listed scientists have contributed significantly to the study of wave physics.

(N/A) Weak radar signals sent by a low flying aircraft can interfere with the $TV$ signals received by the antenna. As a result,the $TV$ signals may get distorted. Hence,when a low flying aircraft passes overhead,we sometimes notice a slight shaking of the picture on our $TV$ screen.
$(b)$ The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of the differential equation that governs wave motion. If $y_{1}$ and $y_{2}$ are the solutions of the second-order wave equation,then any linear combination of $y_{1}$ and $y_{2}$ will also be a solution of the wave equation.

(A) The intensity $(I)$ of light from a point source like a bulb follows the inverse square law,$I \propto \frac{1}{r^2}$. If the distance $(r)$ is doubled,the intensity becomes $\frac{1}{4}$ of its original value.
In contrast,a $LASER$ beam maintains constant intensity because it is highly collimated. The geometric characteristic responsible for this is its low divergence (highly parallel nature).
Key characteristics of $LASER$ light that contribute to this behavior are:
$(i)$ Highly collimated (low divergence)
$(ii)$ Monochromatic
$(iii)$ Coherent
$(iv)$ Unidirectional
These properties are absent in the light emitted by a bulb,which spreads out spherically.

(N/A) In the visible spectrum of light,the wavelength range is approximately $380 \ nm$ to $750 \ nm$.
According to the $VIBGYOR$ sequence:
$1$. The colour with the least wavelength is Violet,which is approximately $380 \ nm$ to $450 \ nm$.
$2$. The colour with the most wavelength is Red,which is approximately $620 \ nm$ to $750 \ nm$.

(A) The wavelength of the light emitted by a sodium lamp (specifically the $D$-lines) is approximately $\lambda = 589.3 \ nm = 589.3 \times 10^{-9} \ m$.
Using the relation $c = f \lambda$,where $c$ is the speed of light $(3 \times 10^8 \ m/s)$:
$f = \frac{c}{\lambda} = \frac{3 \times 10^8 \ m/s}{589.3 \times 10^{-9} \ m} \approx 5.09 \times 10^{14} \ Hz$.
Therefore,the correct option is $A$.

(N/A)

Interference pattern	Diffraction pattern
$(1)$ It is obtained due to the superposition of waves from different coherent sources. Thus,it is the effect produced due to the superposition of different wavefronts.	$(1)$ It is obtained due to the superposition of waves originating from different parts of the same wavefront.
$(2)$ Bright and dark interference fringes are of equal width.	$(2)$ Diffraction fringes are not of the same width. The central maximum has the largest width,while the width of maxima and minima decreases for higher orders of diffraction.
$(3)$ All bright fringes have equal intensities.	$(3)$ The central maximum has the highest intensity,and it decreases with higher-order diffraction maxima.
$(4)$ Dark interference fringes are completely dark.	$(4)$ Regions of minimum intensities may not be perfectly dark.

(N/A) $(1)$ The pattern formed in both is due to the superposition of waves.
- The interference pattern is obtained by superposing two waves originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
$(2)$ The interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima. The intensity falls as we move to successive maxima away from the center.
$(3)$ For a slit of width $a$,the first-order minimum of the diffraction pattern is obtained at angle $\theta = \frac{\lambda}{a}$. In interference,the first-order maximum for two slits separated by distance $d$ is at $\theta = \frac{\lambda}{d}$.
$(4)$ In both interference and diffraction,light energy is redistributed. If it reduces in one region,producing a dark fringe,it increases in another region,producing a bright fringe. There is no gain or loss of energy,which is consistent with the principle of conservation of energy.

(A) When a wave travels from a denser medium to a rarer medium:
$1$. The speed of the wave increases because the refractive index of the rarer medium is lower,and the wave encounters less resistance ($v = f \lambda$,and since frequency $f$ remains constant,speed $v$ increases as wavelength $\lambda$ increases).
$2$. The amplitude of the transmitted wave decreases because part of the wave energy is reflected back at the interface between the two media.
Therefore,the correct matching is: $(a-i), (b-ii)$.

(A) The speed of light $(c)$ in a vacuum is constant for all colors. The relationship between speed, frequency $(f)$, and wavelength $(\lambda)$ is given by $c = f \lambda$. Since red light and blue light have different wavelengths $(\lambda_{red} > \lambda_{blue})$, they must have different frequencies $(f_{red} < f_{blue})$ to maintain the same speed $c$. Therefore, they differ in both frequency and wavelength.

(A) Let the thickness of the columns be $t$. The number of waves in vacuum is $N_v = t / \lambda$ and the number of waves in air is $N_a = t / \lambda_a$,where $\lambda_a = \lambda / \mu_{air}$.
Given that the difference in the number of waves is $1$,we have $N_v - N_a = 1$.
Substituting the expressions: $t / \lambda - t / (\lambda / \mu_{air}) = 1$.
$t / \lambda - (t \cdot \mu_{air}) / \lambda = 1$.
$(t / \lambda) (1 - \mu_{air}) = 1$.
Since $\mu_{air} = 1.0003$,the magnitude of the difference is $(t / \lambda) (\mu_{air} - 1) = 1$.
$t = \lambda / (\mu_{air} - 1) = 6000 \times 10^{-10} \, m / (1.0003 - 1) = 6000 \times 10^{-10} / 0.0003$.
$t = 6000 \times 10^{-10} / (3 \times 10^{-4}) = 2000 \times 10^{-6} \, m = 2 \times 10^{-3} \, m = 2 \, mm$.

(B) Given:
Wavelength $\lambda = 638 \,nm$
Modulation frequency $f = 1 \,GHz = 1 \times 10^{9} \,Hz$
Speed of light $c = 3 \times 10^{8} \,m/s$
The pulse length (or the spatial extent of one pulse) is determined by the distance light travels in one period of the modulation frequency.
The distance $D$ between two detectors such that they can see the same pulse simultaneously is equal to the wavelength of the modulation,which is given by:
$D = \frac{c}{f}$
Substituting the values:
$D = \frac{3 \times 10^{8} \,m/s}{1 \times 10^{9} \,Hz} = 0.3 \,m$
$D = 30 \,cm$
Therefore,the farthest distance apart the two detectors can be placed is $30 \,cm$.

(B) The carrot appears orange in white light because it reflects the orange portion of the visible spectrum and absorbs the remaining parts.
The visible spectrum ranges approximately from $400 \,nm$ to $750 \,nm$.
$1$. The wavelengths absorbed by the $\beta$-carotene molecule are primarily in the blue and green regions of the spectrum,which are shorter than approximately $550 \,nm$.
$2$. The wavelengths reflected (which give the carrot its orange appearance) are in the yellow,orange,and red regions,which are longer than $550 \,nm$.
Therefore,the $\beta$-carotene molecule absorbs light of wavelengths shorter than $550 \,nm$.

(B) The correct answer is $B$.
Light waves of different colors are characterized by their different wavelengths and frequencies.
Red light has a longer wavelength and a lower frequency compared to blue light.
Since the speed of light in a vacuum is constant for all colors $(c = 3 \times 10^8 \ m/s)$,the difference between red and blue light lies in their frequency and wavelength,not their speed,intensity,or amplitude.