Wave Optics – JEE Main PhysicsPractice Questions & PYQs

Huygens' principle asserts that each point on a wavefront behaves as a point source of light, which defines how the wave front advances through a medium. Since the refractive index $\mu$ is smaller near the ground, the velocity $v = c/\mu$ is greater in the lower region of the beam than it is at higher altitudes where the refractive index is larger. Because the lower part of the horizontal wavefront advances faster than the upper part, the path of the light beam curves, causing it to bend upwards.

The limit of resolution for the human eye is determined by Rayleigh's criterion, which establishes that the minimum angular separation $\theta$ between two objects depends on the wavelength of light $\lambda$ and the pupil diameter $D$ according to the relation $\theta = \frac{1.22 \lambda}{D}$. Given that the pupil radius is 0.25 cm, the diameter is 0.5 cm, which converts to 0.005 m. Using the provided wavelength of 500 nm, or $5 \times 10^{-7}$ m, the angular resolution limit is found by substituting these values into the formula as $\theta = \frac{1.22 \times 5 \times 10^{-7}}{0.005}$, yielding an angular separation of $1.22 \times 10^{-4}$ radians. Finally, to find the minimum linear separation at a comfortable viewing distance of 25 cm, or 0.25 m, multiply the angular limit by the distance, resulting in $0.25 \times 1.22 \times 10^{-4}$ m. This calculation produces a final value of approximately 30 μm.

Malus's Law dictates that the intensity of polarized light passing through a polaroid is governed by the equation $I = I_0 \cos^2 \theta$, where $\theta$ represents the angle between the light's plane of polarization and the polaroid's transmission axis. Because the two beams are polarized in mutually perpendicular planes, a polaroid oriented at an angle $\theta$ relative to beam A will be oriented at $90^\circ - \theta$ relative to beam B. When the beams are observed to have equal brightness, their transmitted intensities must be identical, allowing the relationship $I_A \cos^2 \theta = I_B \cos^2 (90^\circ - \theta)$. Applying the specific condition where this equality occurs at $30^\circ$ for beam A leads to $I_A \cos^2 30^\circ = I_B \cos^2 60^\circ$. Substituting the trigonometric values $\cos^2 30^\circ = 3/4$ and $\cos^2 60^\circ = 1/4$ into the expression gives $I_A (3/4) = I_B (1/4)$, which simplifies to the ratio $\frac{I_A}{I_B} = \frac{1}{3}$.

The locus of points $P$ having a constant path difference $\Delta x = |S_2P - S_1P|$ from two point sources $S_1$ and $S_2$ is a hyperboloid of revolution centered on the axis connecting the sources. When a screen is positioned perpendicular to this axis, the intersection of these hyperboloid sheets with the flat surface results in concentric circles. These circular fringes are the loci of constant phase difference and reflect the radial symmetry of the wave emission from the two point sources.

The intensity of unpolarized light passed through a polarizer is reduced by half, expressed as $I_A = \frac{I_0}{2}$. When this polarized light subsequently encounters a second polarizer at an angle $\theta$ relative to the first, the Law of Malus determines the intensity of the emergent light using the formula $I = I_A \cos^2(\theta)$.

With $\theta = 45^\circ$, the final intensity is calculated as $I_B = \frac{I_0}{2} \cos^2(45^\circ)$. Since $\cos(45^\circ) = \frac{1}{\sqrt{2}}$, the square of the cosine is $\frac{1}{2}$, resulting in an emergent intensity of $I_B = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4}$.

The principle of superposition dictates that the resultant intensity $I_r$ of interfering light waves depends on individual intensities $I_1$ and $I_2$ and the phase difference $\phi$ via the expression $I_r = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. Given that the amplitude from one slit is $a$ and the other is $2a$, the corresponding intensities are $I_1 = a^2$ and $I_2 = 4a^2$, which transforms the formula into $I_r = a^2 + 4a^2 + 2\sqrt{4a^4} \cos \phi = 5a^2 + 4a^2 \cos \phi$.

Maximum intensity $I_m$ occurs during constructive interference and is calculated as $I_m = (a + 2a)^2 = 9a^2$, which implies $a^2 = I_m / 9$. Substituting this into the intensity equation yields $I_r = \frac{I_m}{9}(5 + 4 \cos \phi)$. Applying the trigonometric identity $\cos \phi = 2 \cos^2 (\frac{\phi}{2}) - 1$ leads to $I_r = \frac{I_m}{9}(5 + 4(2 \cos^2 \frac{\phi}{2} - 1))$, which simplifies to $I_r = \frac{I_m}{9}(1 + 8 \cos^2 \frac{\phi}{2})$.

Light reflecting from the air-glass interface undergoes a phase change of $\pi$ because the transition to a denser medium acts as a rigid boundary for the incident wave. In the setup of a plano-convex lens on a glass plate, the thickness of the air film at the point of contact is negligible, leading to an effective path difference of zero. Because of the phase shift introduced upon reflection, the waves interfere destructively at this point, which results in a dark center for the interference pattern. While both statements are true, the dark center is a specific physical consequence of the phase change and the experimental geometry, rather than the dark center providing the explanation for why the phase change occurs in the first place.

The wavefront of a light beam is defined as the locus of points possessing the same phase, and for a beam characterized as initially parallel, the rays must travel in exactly the same direction. Geometrically, this implies that the constant-phase surfaces are perpendicular planes, establishing that the wavefront begins as a planar surface. Although a medium with an intensity-dependent refractive index defined by $\mu(I) = \mu_0 + \mu_2 I$ eventually alters the wave propagation, the beam's starting condition dictates that the wavefront is planar at the outset.

In Young's double slit experiment, the position of bright fringes on the screen is determined by the relationship $y_n = \frac{n \lambda D}{d}$, where $n$ denotes the order of the fringe, $\lambda$ is the wavelength, and the ratio $\frac{D}{d}$ is a fixed constant for the experimental setup. When two distinct light sources create overlapping interference patterns, their bright fringes align at specific locations where the products of their respective fringe orders and wavelengths are equal because the geometric factors cancel out. By equating the third bright fringe of the 590 nm light with the fourth bright fringe of the unknown light, we establish the condition $3 \times 590 = 4 \times \lambda$. Dividing the product of 3 and 590 by 4 results in an unknown wavelength of 442.5 nm.

The intensity distribution in Young's double slit experiment is governed by the phase difference between the light waves meeting at a specific point on the screen, expressed as $I = I_0 \cos^2 \left(\frac{\phi}{2}\right)$, where $I_0$ is the maximum intensity. To find this ratio, we first convert the given path difference of $\frac{\lambda}{6}$ into a phase difference using $\phi = \frac{2\pi}{\lambda} \Delta x$, which results in $\phi = \frac{\pi}{3}$. Plugging this value into the intensity formula gives $\frac{I}{I_0} = \cos^2 \left(\frac{\pi}{6}\right)$, and since the cosine of $\frac{\pi}{6}$ is $\frac{\sqrt{3}}{2}$, squaring this value results in a final intensity ratio of $\frac{3}{4}$.

When unpolarized light with an initial intensity of $I_0$ interacts with a polarizing sheet, the filter functions as a selective barrier that permits only the component of the electric field oscillating parallel to the transmission axis to pass through. Because unpolarized light is composed of vibrations in all directions, the sheet naturally blocks the components perpendicular to its axis, reducing the transmitted intensity to exactly $\frac{I_0}{2}$. Since the energy not transmitted is the difference between the incident intensity and the transmitted portion, the final value is determined by the following calculation:

$I_{untransmitted} = I_0 - \frac{I_0}{2} = \frac{I_0}{2}$

In a standard Young's double slit experiment, the two slits act as coherent, parallel linear sources, and the resulting interference pattern on the screen is determined by the locus of points where the path difference follows the condition $\Delta x = d \sin \theta$. Because the slits are oriented parallel to each other, the set of points satisfying this condition forms a series of straight lines. In contrast, if the setup utilized two distinct point sources as in a double hole experiment, the geometric constraints would produce hyperbolic interference fringes instead.

The intensity of the principal maximum in a single-slit diffraction pattern is directly proportional to the square of the slit width, as the amplitude of the light reaching the central point depends linearly on the width of the aperture. Mathematically, this means $I \propto b^2$, where $b$ represents the slit width. If the slit width is doubled, the new intensity becomes proportional to $(2b)^2$, which equals $4b^2$. This transformation confirms that the intensity at the center of the screen increases by a factor of four, resulting in a final value of $4I_0$.

The resolution of two distinct points by the eye is governed by Rayleigh's criterion, which mandates that the angular separation between the objects must exceed the diffraction-limited resolution of the pupil. This limit is expressed by the relationship $\theta = 1.22 \frac{\lambda}{d}$, where $\lambda$ is the wavelength and $d$ is the pupil diameter. Since this angle also corresponds to the ratio of the dot separation to the viewing distance, the maximum distance is defined by the formula $D = \frac{yd}{1.22\lambda}$

Substituting the provided values of 1 mm for the dot separation, 3 mm for the pupil diameter, and 500 nm for the wavelength results in a distance of approximately 5 m, which is the threshold beyond which the eye can no longer distinguish the two points.

When unpolarized light strikes the surface of a transparent material like glass, it becomes partially polarized, but at a specific angle of incidence, the reflected light is completely plane-polarized. According to Brewster's law, the tangent of this polarizing angle is equal to the refractive index of the medium, which is expressed as $\tan(\theta) = n$. The angle of incidence that results in total polarization is therefore defined by the equation:

$\theta = \tan^{-1}(n)$

Constructive interference in Young's double-slit experiment occurs when the path difference between waves originating from the two slits is an integer multiple of the wavelength, defined by the condition $d \sin \theta = n \lambda$. Given that the separation between the slits is twice the wavelength, or $d = 2\lambda$, this relationship simplifies to $2\lambda \sin \theta = n \lambda$, or $\sin \theta = \frac{n}{2}$. Because the sine of any angle must physically fall within the range of -1 to 1, the possible integer values for $n$ are restricted to $-2, -1, 0, 1, 2$. This range produces a total of five points on the screen where the wave peaks align, representing the central maximum, the two first-order maxima, and the two second-order maxima.

To create a stable interference pattern, light waves must be coherent, meaning they need to maintain a constant phase difference over time. If the frequency of the two sources differs, the phase relationship would shift constantly, causing the interference pattern to blur into a uniform background intensity rather than distinct fringes. Therefore, the essential requirement is for both sources to operate at the exact same frequency while also maintaining a fixed, definite phase relationship to ensure the overlapping crests and troughs reinforce or cancel each other consistently.