Wave Optics – JEE Main PhysicsPractice Questions & PYQs

When unpolarized light of intensity $I_0$ passes through any polarizer, it emerges with its intensity halved to $I_0/2$ while becoming linearly polarized along the transmission axis. Subsequent transmission through additional polarizers is governed by Malus's Law, which states that the emerging intensity is the product of the incident intensity and the square of the cosine of the angle between the incident polarization direction and the polarizer's transmission axis.

In the case without the intermediate polarizer, light polarized at $30^\circ$ encounters a polarizer set at $90^\circ$, resulting in an angle difference of $60^\circ$. The transmitted intensity is given by $\frac{I_0}{2} \cos^2(60^\circ) = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$.

When the third polarizer is placed at $60^\circ$, the light follows a sequence of transitions from $30^\circ$ to $60^\circ$ and then to $90^\circ$. The intensity after the $60^\circ$ polarizer is $\frac{I_0}{2} \cos^2(30^\circ) = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8}$. Passing this light through the final $90^\circ$ polarizer, which is at a $30^\circ$ angle relative to the previous axis, yields a final intensity of $\frac{3I_0}{8} \cos^2(30^\circ) = \frac{3I_0}{8} \times \frac{3}{4} = \frac{9I_0}{32}$. Dividing the intensity obtained with the third polarizer by the intensity without it, $\frac{9I_0/32}{I_0/8}$, results in a final ratio of $9/4$.

The resolution limit of a telescope's objective lens is governed by Rayleigh's criterion, which relates the angular resolution $\theta$ to the light wavelength $\lambda$ and the objective diameter $R$ through the equation:
$\theta = \frac{1.22 \lambda}{R}$
Rearranging this expression to solve for the diameter $R$ gives $R = \frac{1.22 \lambda}{\theta}$. Substituting the given values $\lambda = 500 \times 10^{-9} \text{ m}$ and $\theta = 5 \times 10^{-7} \text{ radian}$ results in $R = \frac{1.22 \times 500 \times 10^{-9}}{5 \times 10^{-7}}$. Simplifying the terms leads to $R = 1.22 \text{ m}$, which converts to 122 cm.

Single-slit diffraction creates an interference pattern where the locations of minima are defined by the equation $y_n = \frac{n \lambda D}{a}$, with $n$ representing the order of the minimum. To determine the linear distance between two specific points in the pattern, we find the difference between their individual positions relative to the central maximum. For the first minimum, we set $n = 1$ to get $y_1 = \frac{\lambda D}{a}$, and for the third minimum, we set $n = 3$ to obtain $y_3 = \frac{3 \lambda D}{a}$. Subtracting these two values yields the total separation:
$\Delta y = \frac{3 \lambda D}{a} - \frac{\lambda D}{a} = \frac{2 \lambda D}{a}$

The fringe width in Young's double slit experiment depends directly on the wavelength of light, characterized by the formula:

$W = \frac{\lambda D}{d}$

When the experiment is conducted in a medium with a refractive index $\mu$, the wavelength of light effectively decreases to $\lambda' = \frac{\lambda}{\mu}$, which causes the fringe width to decrease by the same factor. Consequently, the new fringe width $W'$ is given by the relation:

$W' = \frac{W}{\mu}$

Given an initial fringe width $W$ of 2.4 µm and a refractive index $\mu$ of 1.2, the new fringe width is $W' = \frac{2.4}{1.2} = 2$ µm.

A plane wavefront retains its planar form upon passing through a prism because the phase shift induced by varying material thickness preserves the parallelism of the rays, whereas a small pinhole acts as a point source, generating a spherical wavefront through the process of diffraction. According to the principles of diffraction, where the angular spread is proportional to $\frac{\lambda}{a}$, the curvature of an emerging wavefront is inversely related to the slit width $a$; as $a$ increases, the diffraction effect diminishes, causing the wavefront to approach the incident plane wave shape and thereby decreasing its curvature.

The intensity of light at any point on the screen is determined by the phase difference between the interfering waves according to the following equation:

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

Since the intensity at the given point is three-fourths of the maximum intensity, substituting this value results in $I_{max} \cos^2 \left( \frac{\Delta \phi}{2} \right) = \frac{3}{4} I_{max}$, which simplifies to $\cos^2 \left( \frac{\Delta \phi}{2} \right) = \frac{3}{4}$. Taking the square root of both sides gives $\cos \left( \frac{\Delta \phi}{2} \right) = \frac{\sqrt{3}}{2}$, indicating that the phase difference is $\Delta \phi = 60^\circ$ or $\frac{\pi}{3}$ radians. Using the fundamental relationship between phase difference and path difference, $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$, we substitute the obtained phase difference to find $\frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x$. Solving for the path difference yields $\Delta x = \frac{\lambda}{6}$, which confirms that the value of $x$ is 6.

When light is incident on a surface at the Brewster angle, the reflected and refracted rays are mutually perpendicular, which implies that the sum of the angle of incidence $i_p$ and the angle of refraction $r$ is $i_p + r = 90^{\circ}$. Given that the Brewster angle is $i_p = \tan^{-1}(1.52) = 57.7^{\circ}$, the angle of the refracted beam is calculated using the relation $r = 90^{\circ} - i_p$. Substituting the known value results in $r = 90^{\circ} - 57.7^{\circ} = 32.3^{\circ}$.

The angular width of the central maximum in a single-slit diffraction pattern represents the angular separation between the first minima on either side, given by the expression $\theta = \frac{2\lambda}{d}$. Substituting the values $\lambda = 628 \times 10^{-9} \text{ m}$ and $d = 0.2 \times 10^{-3} \text{ m}$ into the equation $\theta = \frac{2\lambda}{d}$ provides an angular width of $6.28 \times 10^{-3}$ radians. Converting this to degrees by multiplying with the factor $\frac{180}{\pi}$ and using the approximation $\pi \approx 3.14$ yields $360 \times 10^{-3}$ degrees. This simplifies to $36 \times 10^{-2}$ degrees, identifying $\alpha$ as 36.

The resultant intensity at any point in an interference pattern is determined by the phase difference between the superposing waves, given by the expression $I_P = 4I_0 \cos^2 (\frac{\Delta \phi}{2})$, where $I_0$ represents the intensity contribution from each individual slit. Since the resultant intensity at point $P$ is equal to the intensity of an individual slit, setting $I_P = I_0$ leads to the equation $I_0 = 4I_0 \cos^2 (\frac{\Delta \phi}{2})$. Dividing both sides by $4I_0$ results in $\cos^2 (\frac{\Delta \phi}{2}) = \frac{1}{4}$, which implies that $\cos (\frac{\Delta \phi}{2}) = \frac{1}{2}$. Consequently, the phase difference is determined by $\frac{\Delta \phi}{2} = \frac{\pi}{3}$, which simplifies to $\Delta \phi = \frac{2\pi}{3}$. Using the standard relation $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$ to connect phase difference to path difference, we calculate $\Delta x = \frac{\Delta \phi \lambda}{2\pi} = \frac{\lambda}{3}$. With an incident wavelength of 6000 Å, the path difference is $\frac{6000}{3} = 2000$ Å.

When unpolarized light strikes a dielectric interface at the Brewster angle, the reflected wave becomes linearly polarized, with its oscillations constrained to the direction perpendicular to the plane of incidence. Given the phase term $kx - kz - \omega t$, the reflected wave propagates such that the electric field vector must be transverse to the direction of travel while possessing no $z$-component, as dictated by its specific polarization orientation. This physical requirement necessitates an electric field expression involving only the $x$ and $y$ unit vectors, identifying the correct representation as $(E_x \hat{i} + E_y \hat{j}) \sin(kx - kz - \omega t)$.

The resultant intensity at any point on the screen in Young's double slit experiment is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$, where $I_1 = I_2 = I_0$ and $\phi$ is the phase difference between the waves reaching that point. The position on the screen directly in front of one of the slits corresponds to a transverse distance $y = d/2$ from the central maxima, where $d$ is the distance between the slits, resulting in a path difference $\Delta x = \frac{yd}{D} = \frac{d^2}{2D}$.

Substituting the given values $d = 2 \times 10^{-3}$ m, $D = 10$ m, and $\lambda = 6 \times 10^{-7}$ m into the path difference expression yields:
$\Delta x = \frac{(2 \times 10^{-3})^2}{2 \times 10} = \frac{4 \times 10^{-6}}{20} = 2 \times 10^{-7} \text{ m}$
The phase difference is calculated as:
$\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{6 \times 10^{-7}} \times 2 \times 10^{-7} = \frac{2\pi}{3}$
Substituting this phase difference into the intensity equation with $\cos(2\pi/3) = -0.5$ results in:
$I = I_0 + I_0 + 2I_0 \cos\left(\frac{2\pi}{3}\right) = 2I_0 + 2I_0(-0.5) = 2I_0 - I_0 = I_0$

The angular separation of interference fringes in a Young's double slit experiment is determined by the relation $\theta = \frac{\lambda}{d}$, where $\lambda$ represents the wavelength of the light and $d$ is the distance between the slits. As this expression is independent of the distance $D$ between the slit plane and the screen, the angular separation remains constant when the screen is moved, making Statement I false. Because the angular separation $\theta$ is directly proportional to the wavelength $\lambda$, replacing the monochromatic source with one of a higher wavelength increases the angular separation, confirming that Statement II is true.

Inserting a transparent mica sheet of refractive index 1.56 and thickness $t$ into the path of one slit alters the optical path length, causing the central interference fringe to shift. The magnitude of this shift $\Delta y$ relates to the experimental setup parameters through the equation:
$\Delta y = (\mu - 1)t \frac{D}{d}$
Because the central fringe shifts to the location of the 7th bright fringe, which corresponds to the position $7 \frac{\lambda D}{d}$, equating these two expressions allows the geometric parameters $D$ and $d$ to cancel, resulting in the relationship:
$(\mu - 1)t = 7\lambda$
Substituting 1.56 for $\mu$ and 450 nm for $\lambda$ leads to the expression $(1.56 - 1)t = 7 \times 450 \times 10^{-9} \, \text{m}$. Solving for $t$ gives $0.56t = 3150 \times 10^{-9} \, \text{m}$, which simplifies to $t = 5625 \times 10^{-9} \, \text{m}$. Based on the definition $t = \alpha \times 10^{-9} \, \text{m}$, the value of $\alpha$ is 5625.

The intensity of light at any point on the screen in a Young's double slit experiment is determined by the interference of waves, calculated using the formula $I = I_{0} \cos^{2}\left(\frac{\Delta \phi}{2}\right)$, where $\Delta \phi$ is the phase difference between the waves and $I_{0}$ is the maximum intensity. A point located exactly opposite one of the slits corresponds to a vertical distance $y = \frac{d}{2}$ from the center, which results in a path difference defined by $\Delta x = \frac{yd}{D}$. The corresponding phase difference is given by $\Delta \phi = \frac{2\pi}{\lambda} \frac{yd}{D}$.

With the parameters $y = \frac{d}{2}$ and $D = 10d$, the phase difference evaluates to $\Delta \phi = \frac{2\pi}{\lambda} \frac{(d/2)d}{10d}$, which simplifies to $\frac{\pi d}{10 \lambda}$. Given the slit separation $d = 5\lambda$, the phase difference becomes $\frac{\pi (5\lambda)}{10 \lambda} = \frac{\pi}{2}$. Substituting this value into the intensity equation yields $I = I_{0} \cos^{2}\left(\frac{\pi/2}{2}\right) = I_{0} \cos^{2}\left(\frac{\pi}{4}\right)$, resulting in an intensity of $\frac{I_{0}}{2}$.

The fringe width in a Young's double-slit experiment is defined by the equation $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the slit separation. Given that the fringe widths of the two setups are equal, $\beta_1 = \beta_2$, we have the relationship $\frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2}$. Rearranging this to isolate the ratio of the distances gives $\frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2}$. Substituting the provided values $\frac{d_1}{d_2} = \frac{2}{1}$ and $\frac{\lambda_1}{\lambda_2} = \frac{1}{2}$ (which implies $\frac{\lambda_2}{\lambda_1} = \frac{2}{1}$), the calculation yields $\frac{D_1}{D_2} = \frac{2}{1} \times \frac{2}{1} = 4$.

The product of the refractive index ($n$) and the wavelength ($\lambda$) in a medium remains constant because the frequency of light does not change when passing through different media, allowing for the relation $n_1 \lambda_1 = n_2 \lambda_2$. Substituting the given values $n_1 = 4/3$, $\lambda_1 = 540 \text{ nm}$, and $n_2 = 3/2$ into the equation results in $\frac{4}{3} \times 540 = \frac{3}{2} \times \lambda_2$. Simplifying the expression leads to $720 = \frac{3}{2} \lambda_2$, which further solves to $\lambda_2 = \frac{720 \times 2}{3} = 480 \text{ nm}$.

The linear width of the central maxima in single-slit diffraction is defined by the relationship $W = \frac{2\lambda D}{a}$, where $\lambda$ represents the wavelength, $D$ is the distance from the slit to the screen, and $a$ is the slit width. Because $W$ is directly proportional to $\lambda$ and inversely proportional to $a$, the central maxima widens when the wavelength increases or the slit width decreases, confirming the validity of statements A and C. Furthermore, a decrease in wavelength results in a narrower central maxima where the light intensity is more concentrated, which leads to increased brightness for the central peak as stated in E.

The intensity at any point in an interference pattern is determined by the phase difference between the two interfering waves, which is directly proportional to the path difference according to the relationship $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$. When two waves of equal intensity $I_0$ superpose, the resultant intensity at a given point is governed by the formula $I = 2I_0 + 2I_0 \cos(\Delta \phi)$.

For point A, the path difference of $\lambda/3$ results in a phase difference of $\Delta \phi_A = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$. Substituting this phase into the intensity expression gives $I_A = 2I_0 + 2I_0 \cos\left(\frac{2\pi}{3}\right)$. Since $\cos(120^{\circ}) = -0.5$, this simplifies to $I_A = 2I_0 - I_0 = I_0$.

For point B, the path difference of $\lambda/6$ corresponds to a phase difference of $\Delta \phi_B = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3}$. Calculating the intensity at this point yields $I_B = 2I_0 + 2I_0 \cos\left(\frac{\pi}{3}\right)$. With $\cos(60^{\circ}) = 0.5$, the intensity becomes $I_B = 2I_0 + I_0 = 3I_0$. Dividing the intensity at A by the intensity at B provides the final ratio $\frac{I_A}{I_B} = \frac{I_0}{3I_0} = \frac{1}{3}$.

The shift in the central fringe $\Delta y$ in a Young's double-slit experiment when a transparent sheet of thickness $t$ and refractive index $n$ is placed in the path of one of the beams is given by $\Delta y = \frac{(n-1)tD}{d}$, where $D$ is the distance from the slits to the screen and $d$ is the distance between the slits. Rearranging this equation to solve for thickness yields $t = \frac{\Delta y \cdot d}{(n-1)D}$. Substituting the provided values $\Delta y = 0.2 \text{ cm}$, $d = 0.1 \text{ cm}$, $n = 1.5$, and $D = 50 \text{ cm}$ into the formula results in $t = \frac{0.2 \cdot 0.1}{(1.5 - 1) \cdot 50} = \frac{0.02}{0.5 \cdot 50} = \frac{0.02}{25} = 8 \times 10^{-4} \text{ cm}$.

When unpolarized light of intensity $I_0$ passes through a polarizer, it becomes linearly polarized with an intensity of $\frac{I_0}{2}$. As this light traverses the optically active solution, its plane of polarization rotates by an angle $\theta$, and since the analyzer is set at 0 degrees relative to the polarizer, the light hits the analyzer at an effective angle of $\theta$. Applying Malus's law to the system, we use the equation $\frac{3}{8} I_0 = \frac{I_0}{2} \cos^2 \theta$ to relate the final intensity to the transmitted light. Solving this expression results in $\cos^2 \theta = \frac{3}{4}$, which implies that $\cos \theta = \frac{\sqrt{3}}{2}$, corresponding to an angle of rotation of 30 degrees.