Wave Optics – JEE Main PhysicsPractice Questions & PYQs

Single slit diffraction minima occur at angles governed by the relationship:

$\sin\theta = \frac{n\lambda}{\omega}$

Here, $\omega$ represents the slit width, $\lambda$ is the wavelength, and $n$ indicates the order of the minimum. For the second diffraction minimum, the angle is given as $60^\circ$, so substituting $n=2$ into this formula gives $\sin 60^\circ = \frac{2\lambda}{\omega}$. Given that $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we equate this to $\frac{2\lambda}{\omega}$ and solve for the ratio $\frac{\lambda}{\omega}$, which simplifies to $\frac{\sqrt{3}}{4}$. For the first minimum, where $n=1$, the equation becomes $\sin\theta_1 = \frac{\lambda}{\omega}$. Substituting the previously derived ratio results in $\sin\theta_1 = \frac{\sqrt{3}}{4}$. Since $\frac{\sqrt{3}}{4}$ is approximately $0.433$, the corresponding angle $\theta_1$ is close to $25^\circ$.

When two coherent sources emit sound waves in phase, the interference pattern is determined by the path difference $\Delta x$ between the waves reaching a specific point. Destructive interference occurs when the path difference satisfies the condition $\Delta x = (n + 0.5)\lambda$, resulting in an intensity minimum, while constructive interference occurs when $\Delta x = n\lambda$, resulting in an intensity maximum, with $n$ representing an integer.

Initially, with the two sources $1.5\text{ m}$ apart and the listener positioned $2\text{ m}$ directly in front of $S_2$, the distance from $S_1$ to the listener forms the hypotenuse of a right-angled triangle with legs of $1.5\text{ m}$ and $2\text{ m}$. Using the Pythagorean theorem, this distance is $S_1L = \sqrt{1.5^2 + 2^2} = 2.5\text{ m}$. Comparing this to the fixed distance of $S_2L = 2\text{ m}$, the path difference is $\Delta x = S_1L - S_2L = 0.5\text{ m}$. Since the wavelength is $\lambda = 1\text{ m}$, this path difference is exactly $0.5\lambda$, which confirms why the listener detects an intensity minimum at this location.

To find the next intensity maximum as the listener moves, the path difference must increase from $0.5\lambda$ to the next condition for a maximum, which is $\Delta x = 1\lambda = 1\text{ m}$. Keeping the distance from $S_2$ fixed at $2\text{ m}$, we define the new path difference as $S_1L - S_2L = 1\text{ m}$. Substituting the known distance $S_2L = 2\text{ m}$ into this expression gives $S_1L - 2\text{ m} = 1\text{ m}$, which determines that the distance $d$ from $S_1$ must be $3\text{ m}$.

In a Young's double-slit experiment, the spatial extent of a fixed screen segment is defined by the product of the number of fringes and the fringe width. Given that the fringe width is determined by $w = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the slit separation, keeping the screen segment length constant implies that the product of the number of fringes and the wavelength must remain invariant, expressed as $N_1 \lambda_1 = N_2 \lambda_2$. Substituting the initial values of 16 fringes at 700 nm and the final wavelength of 400 nm into the relationship $16 \times 700 = N_2 \times 400$ allows for the calculation of the new number of fringes. Solving this relationship yields $N_2 = 28$.

In Young's double-slit experiment, the path difference between light waves arriving at a point on the screen is determined by the geometry of the setup. For a point at a distance $y$ from the central fringe, the path difference is given by the formula:
$\Delta x = \frac{yd}{D}$
where $d$ is the slit separation and $D$ is the distance to the screen.

Given $y$ = 1.27 mm, $d$ = 1 mm, and $D$ = 1 m, we convert these into consistent units of meters, yielding $y$ = 1.27 x 10^-3 meters, $d$ = 1 x 10^-3 meters, and $D$ = 1 meter. Plugging these into the equation provides:
$\Delta x = \frac{1.27 \times 10^{-3} \times 1 \times 10^{-3}}{1}$
$\Delta x = 1.27 \times 10^{-6}$
This calculation shows the path difference is 1.27 micrometers. This result corresponds to the second order bright fringe, as dividing the total path difference by the light wavelength of 632.8 nm results in $n = 2$.

The passage of plane polarised light through a rotating polarizer is governed by Malus's Law, which states that the intensity of transmitted light depends on the square of the cosine of the angle between the light's polarization vector and the polarizer's axis. As the polarizer spins, the angle varies continuously, meaning we are interested in the average intensity transmitted over one full rotation. The time-averaged value of the squared cosine function over a complete cycle is exactly $\frac{1}{2}$, which implies that the polarizer transmits half of the incident intensity on average. Given the incident intensity $I = 3.3 \ \mathrm{Wm}^{-2}$ and the polarizer's cross-sectional area $A = 3 \times 10^{-4} \ \mathrm{m}^{2}$, the average rate of energy delivery is determined by the equation:

$P = \frac{1}{2} I A$

Substituting these values gives $P = \frac{1}{2} (3.3)(3 \times 10^{-4})$, which equals $4.95 \times 10^{-4}$. This resulting value is approximately $5.0 \times 10^{-4} \ \mathrm{J}$, representing the energy per revolution under the described conditions.

The angular width of interference fringes in a Young's double-slit experiment is defined by the ratio of the light's wavelength to the distance between the two slits. Begin by converting the wavelength of 500 nm to 5 x 10^-7 meters and the slit separation of 0.05 mm to 5 x 10^-5 meters. Using the formula $\Delta\theta = \frac{\lambda}{d}$ calculates the angular width in radians, which results in 0.01 radians. Multiplying this value by $\frac{180}{\pi}$ to convert from radians to degrees results in an angular width of approximately 0.57 degrees.

In a Young's double slit experiment, the physical distance covered by a series of interference fringes on the screen is proportional to the product of the number of fringes and the fringe width. Because the length of the screen portion, the distance to the screen, and the slit separation remain constant, the product of the number of fringes and the wavelength must be the same for both scenarios. This allows us to establish the equality $n_1 \lambda_1 = n_2 \lambda_2$.

By substituting the provided values into this relationship, we get the equation $15 \times 500 = 10 \times \lambda$. Dividing both sides by ten reveals that the unknown wavelength is 750 nm.

The net intensity of interfering waves from two coherent sources is determined by the total phase difference at the observation point, which is the sum of the initial phase difference between the sources and the phase shift caused by the path difference. The phase shift due to the path difference $\Delta x$ is calculated as $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$. Given a wavelength of 20 m and a separation of 5 m, the path difference corresponds to a phase shift of:
$\Delta \phi_{\text{path}} = \frac{2\pi}{20} \times 5 = \frac{\pi}{2}$
Since source P starts with a phase lead of $\frac{\pi}{2}$ over source Q, we use this as our baseline for the calculations.

At point A, the path geometry causes the wave from source Q to arrive before the wave from source P, introducing a path-based phase shift of $\frac{\pi}{2}$ in favor of Q. When combined with the initial phase lead of source P, the net phase difference becomes zero, resulting in constructive interference:
$I_A = I + I + 2\sqrt{I}\sqrt{I} \cos(0) = 4I$
At point B, which is equidistant from both sources, the path difference is zero. Consequently, the only phase difference is the initial $\frac{\pi}{2}$ lead of source P over source Q. The intensity at this point is:
$I_B = I + I + 2\sqrt{I}\sqrt{I} \cos\left(\frac{\pi}{2}\right) = 2I$
At point C, the geometry dictates that the wave from source P arrives before the wave from source Q. This adds an additional phase shift of $\frac{\pi}{2}$ to the initial phase difference, creating a total phase difference of $\pi$ and leading to destructive interference:
$I_C = I + I + 2\sqrt{I}\sqrt{I} \cos(\pi) = 0$
Comparing these results, the ratio of the intensities at points A, B, and C is $4I : 2I : 0$, which simplifies to $2:1:0$.

In a Young's double slit experiment, the distance between any two consecutive bright fringes, known as the fringe width, is determined by the geometric relationship between the wavelength of the light used, the distance to the screen, and the separation between the slits. This distance, denoted by $\beta$, is defined by the formula $\beta = \frac{D\lambda}{d}$. By substituting the given values of 1.5 m for the screen distance, 589 nm for the wavelength, and 0.15 mm for the slit separation, the calculation becomes $\beta = \frac{1.5 \times 589 \times 10^{-9}}{0.15 \times 10^{-3}}$. Solving this equation yields a result of 5.89 mm, which rounds to 5.9 mm.

The intensity of light in a double-slit experiment depends on the interference of two coherent waves, determined by their path difference and the resulting phase difference. This intensity is given by the formula $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$ where $I_1$ and $I_2$ are the intensities of the individual waves and $\phi$ is the phase difference between them. Since the maximum intensity $K$ occurs at a path difference of $\lambda$, where waves interfere constructively \in phase, we can set $I_1 = I_2 = I_0$. Given $I_{\text{max}} = 4I_0 = K$, we deduce that each source has an intensity of $I_0 = \frac{K}{4}$.

At a path difference of $\frac{\lambda}{6}$, the corresponding phase difference $\phi$ is calculated as $\frac{2\pi}{\lambda} \times \frac{\lambda}{6}$, which equals $\frac{\pi}{3}$ radians. Substituting the known values into the interference intensity equation yields $I = \frac{K}{4} + \frac{K}{4} + 2\sqrt{\frac{K}{4} \cdot \frac{K}{4}} \cos\left(\frac{\pi}{3}\right)$ Simplifying this expression, we find $I = \frac{K}{2} + 2(\frac{K}{4})(\frac{1}{2})$, which reduces further to $\frac{K}{2} + \frac{K}{4} = \frac{3K}{4}$. Converting this to a denominator of 12 by multiplying both the numerator and denominator by 3 results \in $\frac{9K}{12}$, which indicates that $n = 9$.

Interference in a double-slit experiment depends on the phase difference between waves, which determines the resultant intensity $I$ relative to the central maximum intensity $I_0$ through the relation $I = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)$. The phase difference $\Delta\phi$ is directly related to the path difference $\Delta x$ by the expression $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$. Given that the path difference is $\frac{\lambda}{8}$, we find the phase difference is $\Delta\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{8} = \frac{\pi}{4}$. Substituting this value into our intensity formula gives the ratio

$\frac{I}{I_0} = \cos^2\left(\frac{\pi/4}{2}\right) = \cos^2\left(\frac{\pi}{8}\right)$

which simplifies to the numerical value $0.853$.

When light travels through a medium, it slows down compared to its speed in a vacuum, which means a physical distance $L$ in that medium is equivalent to a longer distance in a vacuum. This effective distance, known as the optical path length, is calculated by multiplying the refractive index $n$ by the physical path length $L$. Because the two waves originate in phase, any phase difference between them arises solely from the discrepancy between their respective optical path lengths as they traverse their specific media.

For the first wave, the optical path is $n_1 L_1$, and for the second wave, it is $n_2 L_2$. The path difference, which represents the effective distance advantage of one wave over the other, is given by $\Delta p = n_1 L_1 - n_2 L_2$. Since a full wavelength $\lambda$ corresponds to a phase change of $2\pi$ radians, the phase difference is found by scaling this path difference relative to the wavelength. Applying this relationship yields a phase difference of $\Delta \phi = \frac{2\pi}{\lambda}(n_1 L_1 - n_2 L_2)$.

Diffraction minima in a single slit arrangement appear at specific angles governed by the condition $d \sin \theta = n \lambda$, where $d$ is the slit width, $\lambda$ is the wavelength, and $n$ represents the order of the minima. Because $\sin \theta$ is physically limited to a maximum value of 1, the order $n$ must satisfy the inequality $n < \frac{d}{\lambda}$. Substituting the provided values, where the slit width is $0.6 \times 10^{-4}$ m and the wavelength is $6 \times 10^{-7}$ m, the ratio $\frac{d}{\lambda}$ evaluates to 100. This implies that $n$ can take any integer value from 1 up to 99, meaning there are 99 distinct minima on one side of the central maximum. Since the diffraction pattern is symmetric, accounting for both sides results in a total of 198 minima.

Malus's Law describes the intensity of light transmitted through an analyser as $I = I_{max} \cos^2 \theta$, where $I_{max}$ is the intensity of the polarized light incident on the analyser and $\theta$ represents the angle between the transmission axes of the polarizer and the analyser. Given that the output intensity is $10\%$ of the incident intensity, the relationship becomes $0.1 I_{max} = I_{max} \cos^2 \theta$, which simplifies to $\cos^2 \theta = 0.1$. Solving this equation gives an angle $\theta$ of approximately 71.6 degrees. Because the output intensity reaches zero when the analyser is rotated to be perpendicular to the polarizer, at an angle of 90 degrees, the additional rotation required to achieve this is $90 - 71.6$, resulting in 18.4 degrees.

The intensity of light emerging from a slit in a Young's double-slit experiment is directly proportional to the width of that slit. Because the intensity of a wave is also proportional to the square of its amplitude, expressed as $I \propto A^2$, we can conclude that the amplitude is proportional to the square root of the slit width. Given the ratio of the slit widths is $4:1$, the ratio of the amplitudes $A_1$ and $A_2$ is determined by $\frac{A_1}{A_2} = \sqrt{\frac{4}{1}} = \frac{2}{1}$.

The ratio of the intensity of maxima to minima is calculated using the interference relationship $\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2$. By substituting the amplitude values of $A_1 = 2$ and $A_2 = 1$ into this formula, the calculation becomes $\left(\frac{2 + 1}{2 - 1}\right)^2$, which simplifies to $3^2 : 1^2$, resulting in a final ratio of $9:1$.

The resolving power of a microscope, which determines the smallest distance between two points that can be seen as distinct, relies on the Rayleigh criterion. This limit of resolution is calculated using the formula $d = \frac{1.22 \lambda}{2 \cdot \text{NA}}$, where $\lambda$ represents the wavelength of the light and $\text{NA}$ is the numerical aperture of the objective lens.

Using the provided values, where $\lambda = 5000 \times 10^{-10} \text{ m}$ and $\text{NA} = 1.25$, we substitute them into the expression to find $d = \frac{1.22 \times 5000 \times 10^{-10}}{2 \times 1.25}$. Since the denominator simplifies to $2.5$, the calculation becomes $\frac{6100 \times 10^{-10}}{2.5}$, which equals $2440 \times 10^{-10} \text{ m}$. Converting this to micrometers results in $0.244 \text{ }\mu\text{m}$, which rounds to $0.24 \text{ }\mu\text{m}$.

In a Young's double slit experiment, destructive interference occurs at a point when the path difference between the two waves arriving from the sources is an odd multiple of half the wavelength, expressed as $\lambda/2$ for the first minimum. When a point is situated directly in front of one slit, the distance from that slit to the screen is $D$, while the distance from the second slit to the same point forms the hypotenuse of a right-angled triangle with sides $D$ and $d$. According to the Pythagorean theorem, the distance from the second slit is $\sqrt{D^2 + d^2}$. For the configuration where the distance to the screen is $D=2d$, the path difference is $\sqrt{(2d)^2 + d^2} - 2d$, which simplifies to $d\sqrt{5} - 2d$. Setting this path difference equal to the condition for the first minimum, $\lambda/2$, allows us to solve the equation $d(\sqrt{5}-2) = \lambda/2$ for the slit separation, resulting in $d = \frac{\lambda}{2(\sqrt{5}-2)}$.

In a Young's double-slit experiment, constructive interference resulting in bright fringes occurs when the path difference between light waves satisfies the condition $d \sin \theta = n\lambda$. In this equation, $d$ represents the separation between the slits, $\theta$ is the angle at which the fringe is observed, $\lambda$ is the wavelength of the incident light, and $n$ is an integer representing the order of the bright fringe.

To determine the number of bright fringes within the range $-30^\circ \leq \theta \leq 30^\circ$, we first calculate the maximum fringe order $n$ at the boundary angle of $30^\circ$ by computing $n = \frac{d \sin 30^\circ}{\lambda} = \frac{0.320 \times 10^{-3} \times 0.5}{500 \times 10^{-9}}$, which yields $n = 320$. This result confirms that there are 320 bright fringes on the positive side of the central axis and 320 bright fringes on the negative side. Summing these with the single central bright fringe located at $\theta = 0^\circ$ provides a total of $320 + 320 + 1 = 641$ bright fringes.

Placing a thin film of refractive index $\mu$ and thickness $t$ in front of one slit introduces an extra optical path length equal to $(\mu - 1)t$ relative to the light traveling through air. In an interference pattern, this added path difference causes the entire fringe system to shift, and a shift of exactly one fringe width corresponds to an optical path difference equal to the wavelength $\lambda$. Equating these two values gives the condition $(\mu - 1)t = \lambda$, which directly allows us to determine the required thickness by rearranging the terms.

$t = \frac{\lambda}{\mu - 1}$

The resolving power of a telescope is limited by the diffraction of light at the objective lens, defined by Rayleigh's criterion as $\theta = \frac{1.22 \lambda}{D}$, where $\lambda$ is the wavelength and $D$ is the lens diameter. Before calculating, ensure units are consistent by converting the wavelength to $6 \times 10^{-7} \text{ m}$ and the diameter to $2.5 \text{ m}$. Applying these values to the formula, the resolution is $\theta = \frac{1.22 \times 6 \times 10^{-7}}{2.5}$. This division results in a value of $2.928 \times 10^{-7}$ radians, which is approximately $3.0 \times 10^{-7}$ radians.