Mix Examples-Wave Optics Questions in English

(D) laser (Light Amplification by Stimulated Emission of Radiation) exhibits the following properties:
$1$. Highly coherent: The light waves are in phase with each other in both space and time.
$2$. Highly monochromatic: It emits light of a single wavelength or color.
$3$. Highly directional: The beam spreads very little over long distances.
Therefore,all the given statements are correct.

(A) The path difference at any point $P$ is given by $\Delta x = (S_1 P - S_2 P) - (\mu-1)t$. The phase difference is $\delta = (2\pi / \lambda) \Delta x$.
For $(A)$: $(\mu-1)t = 0$. $\delta(P_0) = (2\pi / \lambda)(0 - 0) = 0$ $(p)$. $\delta(P_1) = (2\pi / \lambda)(\lambda / 4) = \pi / 2$ (q is false).
For $(B)$: $(\mu-1)t = \lambda / 4$. $\delta(P_1) = (2\pi / \lambda)(\lambda / 4 - \lambda / 4) = 0$ $(q)$. $I(P_1) = I_{max} \cos^2(0) = I_{max}$.
For $(C)$: $(\mu-1)t = \lambda / 2$. $\delta(P_1) = (2\pi / \lambda)(\lambda / 4 - \lambda / 2) = -\pi / 2$. $I(P_1) = I_{max} \cos^2(-\pi / 4) = I_{max} / 2$. $\delta(P_2) = (2\pi / \lambda)(\lambda / 3 - \lambda / 2) = -\pi / 3$. $I(P_2) = I_{max} \cos^2(-\pi / 6) = 3I_{max} / 4$. Thus $I(P_2) > I(P_1)$ $(t)$.
For $(D)$: $(\mu-1)t = 3\lambda / 4$. $\delta(P_1) = (2\pi / \lambda)(\lambda / 4 - 3\lambda / 4) = -\pi$. $I(P_1) = I_{max} \cos^2(-\pi / 2) = 0$ $(r)$.

(D) The width of the central maximum of a single slit diffraction pattern is given by $w = \frac{2 \lambda D}{a}$,where $a$ is the slit width.
The distance between adjacent maxima in a Young's double slit experiment is the fringe width $\beta = \frac{\lambda D}{d}$.
The total width occupied by $20$ maxima (which corresponds to $20$ fringe widths) is $20 \times \beta = \frac{20 \lambda D}{d}$.
Given that these $20$ maxima are contained within the central maximum of the diffraction pattern:
$\frac{20 \lambda D}{d} = \frac{2 \lambda D}{a}$
$\frac{10}{d} = \frac{1}{a}$
$a = \frac{d}{10}$
Given $d = 1.5 \ mm = 0.15 \ cm = 1.5 \times 10^{-1} \ cm$.
$a = \frac{1.5 \times 10^{-1}}{10} \ cm = 15 \times 10^{-3} \ cm$.
Comparing this with $x \times 10^{-3} \ cm$,we get $x = 15$.

(D) In the Raman effect, when light is scattered by molecules, the scattered light contains frequencies different from the incident frequency.
Stokes' lines are the spectral lines observed in the scattered light that have a frequency lower than the incident (original) frequency.
Since frequency $(f)$ and wavelength $(\lambda)$ are inversely related by the equation $c = f\lambda$, a lower frequency corresponds to a longer (greater) wavelength.
Therefore, Stokes' lines have a wavelength greater than that of the original line.

(C) The wave theory of light,proposed by Huygens,successfully explains the phenomena of interference,diffraction,and the fact that the velocity of light in a denser medium is less than in a rarer medium.
However,the wave theory does not explain the scattering of light,which is better described by the particle nature of light or quantum theory.
Therefore,phenomena $(2)$,$(3)$,and $(4)$ support the wave theory of light.

(B) According to Newton's corpuscular theory,light consists of tiny particles called corpuscles. Newton proposed that these corpuscles travel faster in a denser medium because the attractive force exerted by the particles of the denser medium on the corpuscles increases their velocity. Therefore,according to this theory,the speed of light is lesser in a rarer medium and greater in a denser medium.

(A) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit separation.
This implies that $\beta \propto \lambda$.
According to the photoelectric effect,the energy of a photon is $E = \frac{hc}{\lambda}$. For the photoelectric effect to occur,the energy of the incident photon must be greater than or equal to the work function of the metal $(\Phi)$,i.e.,$\frac{hc}{\lambda} \geq \Phi$.
This means that a smaller wavelength corresponds to higher energy.
Since $\lambda_{1}$ can produce the photoelectric effect and $\lambda_{2}$ cannot,it implies that $\lambda_{1}$ has higher energy than $\lambda_{2}$,which means $\lambda_{1} < \lambda_{2}$.
Since $\beta \propto \lambda$,it follows that $\beta_{1} < \beta_{2}$.

(A) In a Huygen's eye-piece,the cross-wires are placed between the field lens and the eye lens.
Because of this,the image of the object is formed by both lenses (field lens and eye lens),resulting in magnification by both.
However,the cross-wires are placed such that they are only magnified by the eye lens.
Since the magnification is not uniform for the object and the cross-wires,the measurements taken using this eye-piece are not accurate.
Therefore,both Statement $(S)$ and Reason $(R)$ are true,and $(R)$ correctly explains $(S)$.

(A) Huygen's eye-piece consists of two plano-convex lenses separated by a distance.
When used for measurements,the cross-wires are placed between the two lenses.
The final image of the object is formed by the combined effect of both lenses,resulting in a specific magnification.
However,the cross-wires are only magnified by the eye-lens.
Because the object and the cross-wires are magnified by different numbers of lenses,they are not magnified proportionately.
This leads to errors in measurement,making the statement $(S)$ correct and the reason $(R)$ the correct explanation for it.

(B) Let $L$ be the thickness of the air and vacuum columns.
The number of wavelengths in vacuum is $N_v = \frac{L}{\lambda_0}$.
The number of wavelengths in air is $N_a = \frac{L}{\lambda_a} = \frac{L}{\lambda_0 / \mu_a} = \frac{L \mu_a}{\lambda_0}$.
The difference in the number of wavelengths is given as $1$:
$N_a - N_v = 1$
$\frac{L \mu_a}{\lambda_0} - \frac{L}{\lambda_0} = 1$
$\frac{L}{\lambda_0} (\mu_a - 1) = 1$
$L = \frac{\lambda_0}{\mu_a - 1}$
Given $\lambda_0 = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}$ and $\mu_a = 1.0003$.
$L = \frac{6 \times 10^{-7}}{1.0003 - 1} = \frac{6 \times 10^{-7}}{0.0003} = \frac{6 \times 10^{-7}}{3 \times 10^{-4}} = 2 \times 10^{-3} \text{ m}$.
$L = 2 \text{ mm}$.

(A) Fraunhofer lines are a set of dark absorption lines observed in the solar spectrum.
These lines are produced when the continuous spectrum of light emitted by the hot,dense core of the sun (the photosphere) passes through the cooler,thinner gases of the solar atmosphere (the chromosphere).
The atoms in the chromosphere absorb specific wavelengths of light corresponding to their characteristic energy transitions,resulting in the dark lines observed in the spectrum.

(B) The Huygen eyepiece is designed such that the distance between the two plano-convex lenses is equal to half the sum of their focal lengths. This specific configuration satisfies the condition for achromatism (elimination of chromatic aberration) and also minimizes spherical aberration.

(C) In Young's double slit experiment,the fringe width is given by $\beta = \frac{\lambda D}{d} \Rightarrow \beta \propto \frac{1}{d}$. So,if the distance between the two slits $(d)$ increases,the fringe width reduces. This statement is correct.
$(b)$ In diffraction due to a single slit,the path difference between rays reaching the center of the screen is zero,so its intensity at the center is maximum,i.e.,a bright fringe is observed at the center. This statement is correct.
$(c)$ The resolving power of a microscope is defined as $\frac{1}{d_{\min }} = \frac{2 n \sin \theta}{1.22 \lambda}$. It is the reciprocal of the minimum separation $(d_{\min })$ of two points seen as distinct,not the maximum separation. Thus,this statement is incorrect.
$(d)$ Polarisation is possible only in those waves that have vibrations in different directions,which is the case for transverse waves. This statement is correct.
Therefore,the incorrect statement is $(c)$.

(B) Statement $A$ is correct because interference and diffraction involve the redistribution of energy in space,ensuring that the total energy remains constant,which is consistent with the law of conservation of energy.
Statement $B$ is false because interference and diffraction are wave phenomena that are not exclusive to light; they are exhibited by all types of waves,including sound waves,water waves,and matter waves.
Therefore,statement $A$ is true and statement $B$ is false. The correct option is $B$.