Wave Optics – JEE Main PhysicsPractice Questions & PYQs

In a double-slit experiment, the diffraction pattern generated by the finite slit width modulates the interference pattern created by the slit separation. The diffraction minima occur at angles satisfying $a \sin \theta = n\lambda$, while bright interference fringes appear when the path difference satisfies $d \sin \theta = m\lambda$, where $m$ is an integer. By substituting the diffraction condition into the interference condition, the path difference at the angular position of the $n$-th diffraction minimum is given by the equation $d \sin \theta = \frac{d}{a} n\lambda$.

With green light of wavelength 5303 Å, a slit separation $d$ of 19.44 µm and a slit width $a$ of 4.05 µm, the ratio $\frac{d}{a} = 4.8$. Consequently, the first diffraction minimum occurs at a path difference of $4.8\lambda$, and the second diffraction minimum occurs at a path difference of $9.6\lambda$. Since bright fringes appear only at integer path differences, we count the integers 5, 6, 7, 8, and 9 that lie between these two values, which results in 5 bright fringes.

Interference patterns arise from the superposition of waves, where the resultant intensity is directly proportional to the square of the amplitude. When two coherent waves meet, the maximum intensity occurs where the amplitudes add together constructively, and the minimum intensity occurs where they subtract destructively, leading to the relationship $\frac{I_{max}}{I_{min}} = \left(\frac{a_1 + a_2}{a_1 - a_2}\right)^2$. Since we are given the amplitude ratio $\frac{a_1}{a_2} = \frac{1}{3}$, we can divide the terms inside the parentheses by $a_2$ to express the ratio as $\left(\frac{a_1/a_2 + 1}{a_1/a_2 - 1}\right)^2$. Substituting the provided value of $\frac{1}{3}$ yields $\left(\frac{1/3 + 1}{1/3 - 1}\right)^2$, which simplifies to $\left(\frac{4/3}{-2/3}\right)^2$, or $(-2)^2$, resulting in a final value of 4.

The ability of a telescope to distinguish between two close point sources is governed by the diffraction limit, which defines the smallest resolvable angular separation. This angular resolution, denoted as $\Delta\theta$, is directly proportional to the wavelength of incident light and inversely proportional to the diameter of the aperture. The fundamental relationship is expressed as:

$\Delta\theta = \frac{1.22 \lambda}{D}$

Converting the diameter of 200 cm to 2 meters and the wavelength of 500 nm to 500 x 10^-9 meters allows for direct substitution into the formula:

$\Delta\theta = \frac{1.22 \times 500 \times 10^{-9}}{2}$

The calculation results in 3.05 x 10^-7 radians, which is 305 x 10^-9 radians.

The relationship between the maximum and minimum intensity of interfering waves is determined by the amplitudes of the constituent waves, where intensity is proportional to the square of the amplitude, $I \propto A^2$. When two coherent sources interfere, the maximum intensity occurs when the amplitudes add together while the minimum intensity occurs when they subtract, leading to the expression $\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$. Given that this intensity ratio is $16$, taking the square root of both sides reveals that $\frac{A_1 + A_2}{A_1 - A_2} = 4$.

Solving for the ratio of the individual amplitudes involves expanding the equation to $A_1 + A_2 = 4(A_1 - A_2)$, which simplifies to $A_1 + A_2 = 4A_1 - 4A_2$. Rearranging the terms shows that $5A_2 = 3A_1$, establishing an amplitude ratio of $\frac{A_1}{A_2} = \frac{5}{3}$. Because the intensity ratio is equal to the square of the amplitude ratio, calculating $\left( \frac{5}{3} \right)^2$ results in an intensity ratio of $25 : 9$.

Inserting a transparent film in front of one slit alters the optical path length for light traveling through that slit, introducing an additional path difference defined as $(\mu - 1)t$. This extra path causes the interference pattern, including the central maximum, to shift by a distance equal to the product of $n$ and the fringe width. Given that the fringe width in a Young's double slit experiment is expressed as $\frac{D \lambda}{a}$, the shift of $n$ fringes corresponds to a total path difference of $n \frac{D \lambda}{a}$. Equating the introduced path difference $(\mu - 1)t$ to this value, the required thickness is determined by the equation $t = \frac{n D \lambda}{a (\mu - 1)}$

For light reflected from an interface to be completely polarized, the angle of incidence must specifically match the Brewster angle, $i_B$. In a scenario involving a liquid of refractive index $\mu$ and a glass plate of index $1.5$, light travels from the liquid to the glass. If the liquid is denser than the glass, it is possible for the light to undergo total internal reflection. If this total internal reflection occurs at an angle smaller than the Brewster angle, the light will never be incident at the Brewster angle, meaning the reflected light cannot become completely polarized.

The condition for this behavior is that the critical angle $C$ must be less than the Brewster angle $i_B$, or $C < i_B$, which mathematically implies $\sin C < \sin i_B$. Using the relations $\sin C = \frac{1}{\mu}$ and the Brewster angle defined by $\tan i_B = \frac{1.5}{\mu}$, we determine that $\sin i_B = \frac{1.5}{\sqrt{\mu^2 + (1.5)^2}}$. Substituting these into the inequality gives $\frac{1}{\mu} < \frac{1.5}{\sqrt{\mu^2 + (1.5)^2}}$. Squaring both sides of this inequality results in $\frac{1}{\mu^2} < \frac{2.25}{\mu^2 + 2.25}$. Algebraic rearrangement leads to $\mu^2 + 2.25 < 2.25 \mu^2$, which simplifies to $1.25 \mu^2 > 2.25$. Solving for $\mu$ gives $\mu^2 > \frac{2.25}{1.25}$, or $\mu^2 > \frac{9}{5}$, establishing the minimum value as $\mu > \frac{3}{\sqrt{5}}$.

The intensity distribution in a Young's double-slit experiment is determined by the phase difference between the interfering waves reaching a specific point on the screen. First, convert the provided path difference of $\frac{\lambda}{8}$ into a phase difference, denoted as $\Delta \phi$, by multiplying it by $\frac{2\pi}{\lambda}$, which results in $\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{8} = \frac{\pi}{4}$. The intensity at this point relative to the intensity at the central bright fringe follows the relationship:

$\frac{I}{I_0} = \cos^2\left(\frac{\Delta \phi}{2}\right)$

Substituting the calculated phase difference into this formula leads to $\frac{I}{I_0} = \cos^2\left(\frac{\pi/4}{2}\right) = \cos^2\left(\frac{\pi}{8}\right)$. Evaluating $\cos^2\left(\frac{\pi}{8}\right)$ results in approximately 0.85, which represents the ratio of the intensity at that location to the intensity at the central bright fringe.

In a Young's double slit experiment, bright fringes appear where the path difference between interfering light waves is an integer multiple of the wavelength, which is represented by the condition $d \sin\theta = n\lambda$. Because the given angle is quite small, we can simplify this using the approximation $\sin\theta \approx \theta$, meaning the path difference is determined by the expression $d\theta$. Using a slit separation of 0.1 mm and an angle of 1/40 radians, the path difference is 0.1 mm times 1/40, resulting in 0.0025 mm, which converts to 2500 nm.

Since bright fringes for two different wavelengths are observed at this same angular position, we equate 2500 nm to the products $n_1\lambda_1$ and $n_2\lambda_2$ for integer values $n_1$ and $n_2$. To find wavelengths within the visible range of 380 nm to 740 nm, we test integer values for the order of interference. Dividing 2500 nm by 4 gives a wavelength of 625 nm, and dividing 2500 nm by 5 gives a wavelength of 500 nm. Since both values fall within the required visible spectrum, the two wavelengths responsible for the observed fringes are 625 nm and 500 nm.

Malus's Law governs the transmission of light through polarizers, stating that the intensity of light emerging from a filter depends on the square of the cosine of the angle between the polarization direction of the incident light and the transmission axis of the filter. When unpolarized light of intensity $I_0$ passes through the first polarizer, it becomes linearly polarized with an intensity of $\frac{I_0}{2}$. In the specified three-polarizer setup, the light then passes through subsequent polarizers at intermediate angles of $30^\circ$ and $60^\circ$, resulting in a final intensity of $I = \frac{I_0}{2} \cos^2 30^\circ \cos^2 60^\circ$. Substituting the trigonometric values gives $I = \frac{I_0}{2} \times \frac{3}{4} \times \frac{1}{4}$, which simplifies to $I = \frac{3I_0}{32}$, making the ratio $\frac{I_0}{I} = \frac{32}{3}$, or approximately 10.67.

Malus's Law states that when linearly polarized light of intensity $I_0$ passes through an ideal polarizer, the intensity of the transmitted light is $I = I_0 \cos^2 \theta$, where $\theta$ represents the angle between the polarization direction of the incident light and the transmission axis of the polarizer. For unpolarized light passing through the first polarizer in any setup, the emerging light becomes linearly polarized with an intensity of $\frac{I}{2}$, as the polarizer effectively filters out half of the intensity regardless of its orientation.

When multiple polarizers are placed in a series, the final intensity is calculated by successively applying Malus's Law, resulting in a transmitted intensity of $I = \frac{I}{2} \cos^2 \phi_1 \cos^2 \phi_2 \dots$ for each respective angle between consecutive polarizers. With polarizer C inserted between A and B, the final intensity is determined by the expression $\frac{I}{2} \cos^2 \phi_1 \cos^2 \phi_2$. Given the target intensity of $\frac{I}{8}$ and assuming a symmetric configuration where the angle between A and C is equal to the angle between C and B, denoted as $\phi$, the relationship becomes $\frac{I}{8} = \frac{I}{2} \cos^4 \phi$. Simplifying this equation gives $\cos^4 \phi = \frac{1}{4}$, which implies $\cos^2 \phi = \frac{1}{2}$. Taking the square root results in $\cos \phi = \frac{1}{\sqrt{2}}$, confirming that the angle between polarizer A and C is $45^\circ$.

According to Malus's law, the intensity of light transmitted through a polarizer is given by $I = I_0 \cos^2(\theta - \phi)$, where $\phi$ is the direction of the polarization and $\theta$ is the orientation of the polarizer's pass axis. Because the intensities are equal for the given values of $\theta$, the polarizer angles must be symmetric about the polarization direction $\phi$. Calculating the average of the two provided angles $\theta_1 = 8^\circ$ and $\theta_2 = 38^\circ$ gives the polarization direction as $\phi = \frac{8^\circ + 38^\circ}{2} = 23^\circ$. Given the $180^\circ$ periodicity of plane polarized light, the equivalent polarization angle is $23^\circ + 180^\circ = 203^\circ$.

The intensity of unpolarized light passing through a polarizer becomes $I/2$, and subsequent intensity through further polarizers follows Malus's Law, which states $I = I_{0} \cos^{2} \phi$, where $\phi$ is the angle between the transmission axis of the polarizer and the incident polarization. Since the intensity of light after passing through A and B is $I/2$, A and B are parallel, implying the angle between their transmission axes is $0^\circ$. Inserting polarizer C between them at an angle $\theta$ relative to A ensures that the angle between A and C is $\theta$, and the angle between C and B is also $\theta$ because B remains parallel to A.

The intensity after passing through C is $I_{C} = I_{A} \cos^{2} \theta = (I/2) \cos^{2} \theta$, and the final intensity after passing through B is $I_{B} = I_{C} \cos^{2} \theta = (I/2) \cos^{4} \theta$. Equating this expression to the given intensity $I/3$ yields $(I/2) \cos^{4} \theta = I/3$, which simplifies to $\cos^{4} \theta = 2/3$, resulting in $\cos \theta = (2/3)^{1/4}$.

The condition for the $n$-th order minima in single slit diffraction is given by $a \sin \theta = n \lambda$ where $a$ is the slit width, $\theta$ is the angular position, $n = 2$ is the order of the minima, and $\lambda = 550 × 10^{-9} \text{ m}$ is the wavelength. Substituting the slit width $a = 22.0 × 10^{-5} \text{ cm} = 22.0 × 10^{-7} \text{ m}$ into the diffraction equation yields $\sin \theta = \frac{n \lambda}{a} = \frac{2 \times 550 \times 10^{-9}}{22.0 \times 10^{-7}} = 0.5$ Therefore, the angular position is $\theta = \arcsin(0.5) = \frac{\pi}{6}$ radians.

In a single-slit diffraction pattern, the angular spread of the central maximum is determined by the condition $a \sin\theta = \lambda$, where $a$ represents the slit width and $\theta$ is the half-angular width. Given that the total angular width is $60^\circ$, the half-angle is $30^\circ$, and with a slit width of $1\mu\text{m}$, the wavelength of the light is $\lambda = 1\mu\text{m} \times \sin(30^\circ) = 0.5\mu\text{m}$, which is equivalent to $0.5 \times 10^{-6}\text{m}$.

When a second slit is introduced for Young's double-slit experiment, the observed fringe width $\beta$ is governed by the relation $\beta = \frac{\lambda D}{d}$, where $D$ is the distance to the screen and $d$ is the separation between the centers of the slits. By substituting the known values-a fringe width of $1\text{ cm}$ and a screen distance of $50\text{ cm}$-along with the previously determined wavelength into the equation, we obtain $1\text{ cm} = \frac{0.5\mu\text{m} \times 50\text{ cm}}{d}$. Rearranging this expression to solve for $d$ results in $d = \frac{0.5\mu\text{m} \times 50}{1}$, which equals $25\mu\text{m}$.

The position of diffraction minima for a single slit is governed by the condition $b \sin \theta = n \lambda$, where $b$ is the slit width, $n$ is an integer representing the order of the minimum, and $\lambda$ is the wavelength. Under standard experimental conditions where the angle is very small, we can approximate $\sin \theta \approx \tan \theta = \frac{x_n}{D}$, which results in the linear relationship $x_n = \frac{n \lambda D}{b}$. This implies that the position of any minimum on the screen is directly proportional to its order $n$.

Given that the second minimum ($n=2$) occurs at 3 cm, we can determine that $2 \times (\frac{\lambda D}{b}) = 3$ cm, which establishes that the constant factor $\frac{\lambda D}{b}$ is 1.5 cm. The width of the central maximum corresponds to the distance between the first minima on either side of the center, spanning a total range of $2 \times x_1$. Since the first minimum ($n=1$) is at a position of $1 \times (\frac{\lambda D}{b})$, or 1.5 cm, doubling this distance to account for both sides yields a total width of 3.0 cm.

The position of the $n^{th}$ dark band in a single slit diffraction pattern is determined by the formula $a \sin\theta = n\lambda$, which, under the small-angle approximation $\sin\theta \approx \tan\theta = \frac{y}{D}$, transforms into $y = \frac{n \lambda D}{a}$, where $a$ is the slit width, $\lambda$ is the wavelength, $n$ is the band order, and $D$ is the distance to the screen.

Substituting the given values $n = 3$, $\lambda = 6 \times 10^{-7}$ m, $D = 0.5$ m, and $a = 10^{-4}$ m into the derived equation yields $y = \frac{3 \times (6 \times 10^{-7}) \times 0.5}{10^{-4}}$. Calculating this expression provides $y = \frac{9 \times 10^{-7}}{10^{-4}} = 9 \times 10^{-3}$ m, which corresponds to 9 mm.

For bright fringes of two different wavelengths to overlap, they must appear at the same distance from the central maximum. This condition occurs when the $n_1$-th bright fringe of the first wavelength coincides with the $n_2$-th bright fringe of the second, which is expressed by the equation $n_1 \lambda_1 = n_2 \lambda_2$.

Substituting the given wavelengths into this relationship, we find $n_1 \times 650 = n_2 \times 520$, which simplifies to the ratio $\frac{n_1}{n_2} = \frac{520}{650} = \frac{4}{5}$. The least distance corresponds to the smallest non-zero integers satisfying this ratio, identifying $n_1 = 4$ and $n_2 = 5$.

To determine the exact position on the screen, we apply the fringe position formula $y = \frac{n_1 \lambda_1 D}{d}$. Using the given experimental parameters, we convert the units to meters where $d = 0.5 \times 10^{-3}$ m, $D = 1.5$ m, and $\lambda_1 = 650 \times 10^{-9}$ m. Substituting these values into the calculation yields the following expression:

$y = \frac{4 \times 650 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}}$

Solving this provides a displacement of 0.0078 m, which is equivalent to 7.8 mm.

The ability of a telescope to distinguish two point sources as separate is determined by Rayleigh's criterion, which provides the minimum angular resolution $\theta$ using the formula $\theta = \frac{1.22 \lambda}{D}$. Here, the wavelength $\lambda$ is $600\text{ nm}$ (or $6 \times 10^{-7}\text{ m}$) and the objective diameter $D$ is $30\text{ cm}$ (or $0.3\text{ m}$). By calculating $\theta$ and relating it to the physical separation $x$ and the distance $d$ to the stars using the relation $x = \theta d$, we can find the minimum distance required for resolution.

Given that $d$ is $10$ light years, which is equivalent to $10 \times 9.46 \times 10^{15}\text{ m}$, we substitute the known values into the equation $x = \frac{1.22 \lambda d}{D}$. Carrying out this calculation leads to a minimum physical separation between the stars that is on the order of $10^6\text{ km}$.

The total width $b$ of the spot produced by a pinhole camera is the result of adding the geometrical spread, determined by the hole's radius $a$, to the diffraction spread, which is proportional to $\frac{\lambda L}{a}$. Combining these factors leads to the relationship $b = a + \frac{\lambda L}{a}$, where $L$ is the camera length and $\lambda$ is the wavelength of light.

The radius that minimizes the spot size is found by differentiating $b$ with respect to $a$ and setting the result to zero:
$\frac{db}{da} = 1 - \frac{\lambda L}{a^2} = 0$
This differentiation identifies the optimal hole radius as $a = \sqrt{\lambda L}$. Substituting this value back into the expression for the spot size yields $b_{min} = \sqrt{\lambda L} + \frac{\lambda L}{\sqrt{\lambda L}}$, which simplifies to $b_{min} = \sqrt{\lambda L} + \sqrt{\lambda L} = 2\sqrt{\lambda L}$, or written identically as $\sqrt{4\lambda L}$.

The visibility of an interference pattern in Young's double slit experiment depends on the angular width of the fringes relative to the angular resolution limit of the observer's eye. The fringe width $\beta$ at a distance $D$ from the slits is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength and $d$ is the separation between the slits. Consequently, the angular fringe width is defined by $\theta = \frac{\beta}{D} = \frac{\lambda}{d}$. When the angular fringe width becomes smaller than the angular resolution of the human eye, the interference fringes are no longer distinguishable and the pattern appears to vanish.

To find the distance $d_0$ where this occurs, we equate the angular fringe width to the eye's angular resolution of $\frac{1^\circ}{60}$. First, we convert the angular resolution from degrees to radians as follows:
$\frac{1^\circ}{60} = \frac{1}{60} \times \frac{\pi}{180} = \frac{\pi}{180 \times 60} \text{ rad}$
Setting the angular fringe width equal to this value, we have:
$\frac{\lambda}{d_0} = \frac{\pi}{180 \times 60}$
Substituting the wavelength $\lambda = 600 \text{ nm} = 6 \times 10^{-7} \text{ m}$ into the equation and solving for $d_0$:
$d_0 = \lambda \left( \frac{180 \times 60}{\pi} \right) = 6 \times 10^{-7} \text{ m} \times \frac{10800}{\pi} \approx 2.06 \times 10^{-3} \text{ m}$
This result is approximately $2 \text{ mm}$, representing the slit separation at which the interference pattern becomes unresolvable to the eye.