Wave Optics – JEE Main PhysicsPractice Questions & PYQs

When a source of light is positioned exactly at the focal point of a convex lens, every ray emanating from that source is refracted to emerge traveling parallel to the principal axis. The wavefront is defined as a surface perpendicular to the direction of wave propagation, and because these refracted rays are strictly parallel, the phase front must propagate as a flat, two-dimensional surface. Consequently, the light emerging from the lens forms a plane wavefront.

When two coherent light beams superimpose, the intensity at any point in the interference pattern depends on the phase relationship between the waves, reaching extreme values at points of constructive and destructive interference. These maximum and minimum intensities are defined by the expressions $I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$. Substituting the provided intensities of $I$ and $4I$ leads to $I_{\text{max}} = (\sqrt{I} + \sqrt{4I})^2 = 9I$ and $I_{\text{min}} = (\sqrt{4I} - \sqrt{I})^2 = I$. The difference between these extreme intensities is $9I - I = 8I$, establishing that the value of xxx is 8.

In single slit diffraction, the central maximum occupies the angular region between the first two minima on either side of the center, defined by an angular width of $2\lambda / a$. When this pattern is focused onto a screen at the focal plane of a convex lens with focal length $f$, the linear width $w$ of the central maximum is determined by the formula $w = \frac{2\lambda f}{a}$

Converting the provided measurements to SI units yields a wavelength $\lambda = 6 \times 10^{-7}\ \text{m}$, a slit width $a = 1 \times 10^{-5}\ \text{m}$, and a focal length $f = 0.2\ \text{m}$. Substituting these values into the expression results in the calculation $w = \frac{2 \times (6 \times 10^{-7}) \times 0.2}{1 \times 10^{-5}}$ which simplifies to $w = 2.4 × 10^{-2}\ \text{m}$, resulting in a total linear width of 24 mm.

When unpolarized light of intensity $I_0$ encounters the first polarizer, it becomes linearly polarized with an intensity of $I_1 = \frac{I_0}{2}$. According to Malus's Law, when this polarized light subsequently passes through a second polarizer oriented at an angle $\theta$, the intensity is reduced by a factor of $\cos^2 \theta$. Placing a third polarizer such that it is crossed at $90^{\circ}$ relative to the first ensures that the angle between the second and third polarizers is $\phi = 90^{\circ} - \theta$.

The resulting intensity after the third polarizer is defined by the relation $I_3 = I_1 \cos^2 \theta \cos^2 \phi$. Substituting the expression for $\phi$ into this equation yields $I_3 = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta$. By applying the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$, this formula simplifies to $I_3 = \frac{I_0}{8} \sin^2 2\theta$. To obtain the maximum transmitted intensity, the term $\sin^2 2\theta$ must reach its peak value of $1$. This occurs when $2\theta = 90^{\circ}$, confirming that the intensity is maximized at a rotation of $45^{\circ}$.

In interference phenomena like Young's double slit experiment, the intensity at any point on the screen depends entirely on the phase difference between the two interfering light waves. The phase difference $\phi$ is calculated from the path difference $\Delta x$ using the relation:
$\phi = \frac{2\pi}{\lambda} \Delta x$
With a path difference of $\frac{7\lambda}{4}$, the phase difference becomes $\frac{2\pi}{\lambda} \times \frac{7\lambda}{4} = \frac{7\pi}{2}$ radians.

The relative intensity is determined by the interference formula:
$\frac{I}{I_{max}} = \cos^2\left(\frac{\phi}{2}\right)$
Substituting the phase difference yields $\frac{I}{I_{max}} = \cos^2\left(\frac{7\pi}{4}\right)$. Because the cosine function is periodic, $\cos(\frac{7\pi}{4})$ simplifies to $\cos(2\pi - \frac{\pi}{4})$, which equals $\cos(\frac{\pi}{4})$. Squaring $\cos(\frac{\pi}{4})$ results in $\left(\frac{1}{\sqrt{2}}\right)^2$, confirming the intensity ratio is $\frac{1}{2}$.

Destructive interference, which results in the formation of a dark fringe, occurs when the path difference between light waves originating from two coherent sources is an odd integer multiple of half the wavelength. For the specific geometry shown, the condition requires that the path difference between the rays traveling from the two slits to point $P$ satisfies the equation $2(\sqrt{D^{2+}d^2} - D) = \lambda/2$. Simplifying this expression leads to the relationship $\sqrt{D^{2+}d^2} - D = \lambda/4$, which can be rearranged as $\sqrt{D^{2+}d^2} = D + \lambda/4$.

Squaring both sides of this equation results in $D^2 + d^2 = D^2 + D\lambda/2 + \lambda^2/16$. By subtracting $D^2$ from both sides, the equation becomes $d^2 = D\lambda/2 + \lambda^2/16$. Given that the wavelength of $400\text{ nm}$ is extremely small, the term $\lambda^2/16$ contributes negligibly to the total value, allowing us to approximate the slit separation as $d^2 \approx D\lambda/2$. Substituting the given values where $D = 0.2\text{ m}$ and $\lambda = 400 \times 10^{-9}\text{ m}$, the calculation becomes $d^2 \approx \frac{0.2 \times 400 \times 10^{-9}}{2}$, which simplifies to $d^2 \approx 400 \times 10^{-10}\text{ m}^2$. Taking the square root gives $d \approx 20 \times 10^{-5}\text{ m}$, which is equivalent to $0.20\text{ mm}$.

The angular spread of a central maximum in a single-slit diffraction pattern is determined by the positions of the first minima on either side of the central peak, which satisfy the condition $a \sin \theta = \lambda$. Given the slit width $a = 4.0$ cm and wavelength $\lambda = 2.0$ cm, we find the angular position $\theta$ of the first minimum by solving the equation $\sin \theta = \frac{2.0}{4.0} = 0.5$, which results \in $\theta = 30^\circ$. Since the central maximum spans the distance between these two symmetric minima, the total angular spread is given by $2\theta = 60^\circ$.

When unpolarized light passes through the first polarizing sheet, its intensity is reduced to half because the polarizer only transmits the component of light aligned with its transmission axis. For every subsequent sheet, Malus's Law dictates that the intensity of transmitted light is determined by multiplying the incident intensity by the square of the cosine of the angle between the polarization direction of the incoming light and the transmission axis of the new polarizer. Since each sheet is oriented at an angle of $45^{\circ}$ relative to the previous one, the intensity is multiplied by $\cos^2(45^{\circ}) = 1/2$ at each step.

Following this pattern, the first sheet results in an intensity of $I/2$, the second sheet results in $(I/2) \times (1/2) = I/4$, and this progression continues such that after $n$ sheets, the final intensity is expressed as $I_n = \frac{I}{2^n}$. Given that the final output intensity is observed to be $I/64$, we can set up the equation $\frac{I}{2^n} = \frac{I}{64}$. This leads to $2^n = 64$, which confirms that the number of polarizing sheets is $n = 6$.

When a light ray strikes the boundary between two dielectric media such that the reflected and refracted rays are mutually perpendicular, the angle of incidence satisfies Brewster's Law. At this specific orientation, the tangent of the incidence angle is equal to the ratio of the refractive indices of the two media. Because the refractive index is the square root of the dielectric constant for materials with a relative permeability of unity, the relationship is defined by the equation $\tan i = \sqrt{\frac{K_2}{K_1}}$.

Substituting the given dielectric constants $K_1 = 2.8$ and $K_2 = 6.8$ into this expression yields $\tan i = \sqrt{\frac{6.8}{2.8}}$. To determine the unknown value $\theta$ in the requested expression $\tan i = (1 + \frac{10}{\theta})^{1/2}$, the fraction inside the square root must be mathematically rearranged. Rewriting the numerator $6.8$ as the sum $2.8 + 4$ allows the expression to be split into $1 + \frac{4}{2.8}$. Converting the fraction $\frac{4}{2.8}$ to an equivalent form with a numerator of $10$ involves multiplying both the numerator and denominator by $2.5$, resulting in $\frac{10}{7}$. Comparing $1 + \frac{10}{7}$ to the given structure $1 + \frac{10}{\theta}$ demonstrates that $\theta = 7$.

In Young's double slit experiment, bright fringes occur due to constructive interference, where the position of the $n^{\text{th}}$ bright fringe relative to the central maximum is defined by the formula $y_n = \frac{n \lambda D}{d}$. In this expression, $y_n$ is the fringe position, $n$ is the fringe order, $\lambda$ is the wavelength of the light, $D$ is the distance between the slits and the screen, and $d$ is the separation between the slits.

To find the slit separation, we rearrange the formula to solve for $d$:
$d = \frac{n \lambda D}{y_n}$

Substituting the provided values where $n = 5$, $\lambda = 600 \times 10^{-9}$ m, $D = 1$ m, and $y_n = 5$ cm (which is $5 \times 10^{-2}$ m), the expression becomes:
$d = \frac{5 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-2}}$

Canceling the common factor of $5$ in the numerator and denominator simplifies the expression to $d = 600 \times 10^{-7}$ m. Adjusting the decimal places, this equals $60 \times 10^{-6}$ m, which corresponds to $60\ \mu\text{m}$.

In Young's double slit experiment, bright fringes for different wavelengths coincide at positions that are common multiples of their respective fringe widths, which are calculated using the formula $\beta = \frac{\lambda D}{d}$. Evaluating the fringe width for the 800 nm wavelength yields $\beta_1 = \frac{800 \times 10^{-9} \times 7}{0.35 \times 10^{-3}}$, which equals 16 mm. Similarly, the fringe width for the 600 nm wavelength is calculated as $\beta_2 = \frac{600 \times 10^{-9} \times 7}{0.35 \times 10^{-3}}$, which results in 12 mm. Since the fringes overlap at a location equidistant from the central maximum that is an integer multiple of both calculated widths, the shortest distance is determined by the least common multiple of 16 mm and 12 mm, yielding 48 mm.

In Young's double slit experiment, bright fringes form at specific positions described by the formula $y = \frac{n \lambda D}{d}$, where $n$ is the fringe order, $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the slit separation. When bright fringes from two different wavelengths coincide, their positions must be equal, which leads to the condition $n_1 \lambda_1 = n_2 \lambda_2$, where $n_1$ and $n_2$ represent the orders of the fringes for wavelengths $\lambda_1$ and $\lambda_2$ respectively.

Given $\lambda_1 = 7000\ \mathring{\mathrm{A}}$ and $\lambda_2 = 5500\ \mathring{\mathrm{A}}$, the ratio of the fringe orders is given by $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5500}{7000} = \frac{11}{14}$. This indicates that the 11th bright fringe of the 7000 Å light coincides with the 14th bright fringe of the 5500 Å light. Using the fringe position formula with the screen distance $D = 150 \times 10^{-2}$ m and slit separation $d = 2.5 \times 10^{-3}$ m, the distance $y$ is determined as:
$y = \frac{n_1 \lambda_1 D}{d} = \frac{11 \times 7000 \times 10^{-10} \times 150 \times 10^{-2}}{2.5 \times 10^{-3}}$
Solving this expression yields $y = 462 \times 10^{-5}$ m, making the value of $n$ equal to 462.

When unpolarized light with an initial intensity of 256 W/m^2 encounters the first polaroid, its intensity is reduced by half to 128 W/m^2 because the polaroid only transmits the component of light parallel to its axis. According to Malus's Law, the transmitted intensity through subsequent polaroids follows the relationship $I = I_{incident} \cos^2(\theta)$, where $\theta$ is the angle between the polarization direction of the incident light and the axis of the next polaroid. The second polaroid is at 60° to the first, so the light emerging from it is $128 \times \cos^2(60^\circ)$, which results in 32 W/m^2. Since the third polaroid is at 90° to the first, the angle between the light polarized at 60° and the third axis is 30°, and applying the law again gives a final intensity of $32 \times \cos^2(30^\circ)$, which equals 24 W/m^2.

In a Young's double slit experiment, the intensity at any point on the screen is determined by the interference of waves from the two slits, with the phase difference linked to the path difference by the relation $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$. When each individual slit provides an intensity of $I_0$, the resulting intensity at any point is given by $I = 4 I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right)$.

For a path difference of $\frac{\lambda}{4}$, the phase difference is $\frac{\pi}{2}$, which yields an intensity of $I_1 = 4 I_0 \cos^2 \left( \frac{\pi}{4} \right) = 2 I_0$. For a path difference of $\frac{\lambda}{3}$, the phase difference is $\frac{2\pi}{3}$, resulting in an intensity of $I_2 = 4 I_0 \cos^2 \left( \frac{\pi}{3} \right) = I_0$. Adding these two intensities gives $I_1 + I_2 = 3 I_0$, so the ratio $\frac{I_1 + I_2}{I_0}$ equals $3$.

The insertion of transparent slabs into the light paths of a Young's double slit experiment alters the optical path length, leading to a shift in the central bright fringe. Each slab introduces an extra phase delay of $(\mu - 1)t$, and with two slabs present, the net path difference is given by $\Delta x = (\mu_2 - \mu_1)t$. Substituting the given values where $\mu_2 = 1.55$, $\mu_1 = 1.51$, and $t = 0.1 \text{ mm} = 10^{-4} \text{ m}$, we determine the path difference as:

$\Delta x = (1.55 - 1.51) \times 10^{-4} \text{ m} = 4 \times 10^{-6} \text{ m}$

The displacement of the central fringe on the screen is represented by $y = \frac{\Delta x D}{d}$. Because the fringe width is $\beta = \frac{\lambda D}{d}$, the total number of fringes shifted is the ratio of the displacement to the fringe width, which is expressed as $N = \frac{\Delta x}{\lambda}$. Using the wavelength $\lambda = 4000 \text{ \text{\AA}} = 4 \times 10^{-7} \text{ m}$, the final calculation is $\frac{4 \times 10^{-6} \text{ m}}{4 \times 10^{-7} \text{ m}}$, confirming that the pattern shifts by 10 fringes.

Destructive interference occurs in a double slit experiment when the path difference between the waves originating from the two slits is an odd multiple of half the wavelength. For the first minimum, the condition is $\Delta x = \lambda / 2$.

In this specific geometric setup, the rays travel to point $P$ along paths of length $D$ and $\sqrt{D^2 + a^2}$. The path difference is $\sqrt{D^2 + a^2} - D$, and setting this equal to $\lambda / 2$ allows us to solve for the slit separation $a$.

Using the binomial approximation $(1 + x)^{1/2} \approx 1 + x/2$ for $a \ll D$, the path difference simplifies to $\frac{a^2}{2D} = \frac{\lambda}{2}$, which results in the formula $a = \sqrt{\lambda D}$.

By converting the given values to consistent units, we have $\lambda = 800 \times 10^{-6}$ mm and $D = 50$ mm. Substituting these into the formula $a = \sqrt{\lambda D}$ leads to $a = \sqrt{800 \times 10^{-6} \times 50}$, which simplifies to $0.2$ mm.

In a Young's double-slit experiment, the resultant intensity at any point on the screen depends on the phase difference between the two interfering waves, defined by the formula $I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$, where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference. For point P, where the phase difference is $\frac{\pi}{3}$, the intensity is calculated as $I_P = 4I_0 \cos^2\left(\frac{\pi}{6}\right)$. Since the value of $\cos\left(\frac{\pi}{6}\right)$ is $\frac{\sqrt{3}}{2}$, squaring this yields $\frac{3}{4}$, resulting in $I_P = 4I_0 \cdot \frac{3}{4} = 3I_0$.

For point Q, where the phase difference is $\frac{\pi}{2}$, the intensity is determined by $I_Q = 4I_0 \cos^2\left(\frac{\pi}{4}\right)$. Given that $\cos\left(\frac{\pi}{4}\right)$ is $\frac{1}{\sqrt{2}}$, squaring this term provides $\frac{1}{2}$, leading to an intensity of $I_Q = 4I_0 \cdot \frac{1}{2} = 2I_0$. Comparing these two results, the ratio of the intensity at point P to the intensity at point Q is $\frac{3I_0}{2I_0}$, which simplifies to $3:2$.

Inserting a thin plate of refractive index $\mu$ and thickness $t$ into the path of one of the slits increases the optical path length for that beam, causing the entire interference pattern to shift towards the side of the inserted plate. The displacement of the central maximum, denoted as $\Delta x$, is related to the fringe-width $\beta_0$ by the expression $\Delta x = \frac{t(\mu - 1)}{\lambda} \beta_0$. Since the problem defines this shift as $x \beta_0$, the value we need to determine is $x = \frac{t(\mu - 1)}{\lambda}$. Using the provided values, we convert the thickness $t$ of 10 micrometers to $10 \times 10^{-6}$ meters and the wavelength $\lambda$ of 500 nanometers to $5 \times 10^{-7}$ meters. Calculating the value of $x$ gives:

$x = \frac{10 \times 10^{-6} (1.2 - 1)}{5 \times 10^{-7}} = \frac{10^{-5} \times 0.2}{5 \times 10^{-7}} = 4$

The central maximum shifts by four times the fringe-width.

In single-slit diffraction, destructive interference occurs at specific angles where light waves originating from different points across the slit width cancel one another out. This physical principle is described by the equation $a \sin \theta = n \lambda$, where $a$ represents the slit width, $\theta$ is the diffraction angle, $n$ is an integer for the order of the minimum, and $\lambda$ is the wavelength of the incident light. Since the problem asks for the first minimum, we set $n = 1$.

By substituting the given wavelength of 600 nm and the angle of 30 degrees into the expression, we obtain $a \sin 30^{\circ} = 600$ nm. Because the value of $\sin 30^{\circ}$ is 0.5, the calculation $0.5a = 600$ nm results in a slit width of 1200 nm, which is equivalent to 1.2 $\mu$m.

Brewster's law establishes that the tangent of the angle of incidence for perfectly polarized reflected light equals the ratio of the refractive index of the second medium to that of the first. When light travels from air to glass, the condition is $\tan \theta_B = \mu_g / \mu_a$, but reversing the path to travel from glass to air necessitates inverting the ratio to $\tan i_2 = \mu_a / \mu_g$, which simplifies to $1 / \tan \theta_B$ or $\cot \theta_B$. Because $\cot \theta_B$ is mathematically equivalent to $\tan(\pi/2 - \theta_B)$, Statement I correctly identifies the complementary relationship between the two angles, whereas Statement II is inaccurate because the correct formula for the reverse path requires the reciprocal of the glass refractive index rather than $\mu_g$ itself.