Wave Optics – JEE Main PhysicsPractice Questions & PYQs

When two plane-polarized light waves of the same frequency superimpose, the resultant amplitude is determined by vector addition, which accounts for both their individual amplitudes and the phase difference between them. The resultant amplitude $E_R$ is calculated using the following relationship:

$E_R = \sqrt{E_1^2 + E_2^2 + 2 E_1 E_2 \cos(\phi)}$

Given that both waves have an amplitude of $E_0$ and a phase difference $\phi = \pi/3$, we substitute these values into the formula to obtain $E_R = \sqrt{E_0^2 + E_0^2 + 2 E_0^2 \cos(\frac{\pi}{3})}$. Because the value of $\cos(\frac{\pi}{3})$ is $0.5$, the expression inside the square root simplifies to $2 E_0^2 + 2 E_0^2(0.5)$, which results in $\sqrt{3 E_0^2}$. This simplifies to $E_0\sqrt{3}$, which evaluates to approximately $1.7 E_0$.

The fundamental principle here is that the frequency of light remains constant as it transitions between media, while its speed and wavelength are reduced by the refractive index. Since the refractive index is 2, the light's velocity becomes half of its vacuum speed, specifically $v = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \, \text{m/s}$. We determine the new wavelength by taking the ratio of this reduced speed to the incident frequency of $5 \times 10^{14} \, \text{Hz}$. Applying the relationship $\lambda = \frac{v}{f}$, we perform the calculation $\lambda = \frac{1.5 \times 10^8}{5 \times 10^{14}}$, which results in $3 \times 10^{-7} \, \text{m}$, which is equivalent to 300 nm.

In a Young's double slit interference experiment, the amplitude of the light wave is proportional to the width of the slit, which implies that the intensity, being proportional to the square of the amplitude, relates to the square of the slit width. If the width of one slit is half that of the other, the amplitudes are related by the ratio $A_2 = 2A_1$.

Maximum intensity occurs during constructive interference at $(A_1 + A_2)^2$, while minimum intensity occurs during destructive interference at $(A_1 - A_2)^2$. The ratio of maximum to minimum intensity is found by substituting the amplitude relationship:

$\frac{I_{max}}{I_{min}} = \left( \frac{A_1 + 2A_1}{A_1 - 2A_1} \right)^2 = \left( \frac{3A_1}{-A_1} \right)^2 = (-3)^2 = 9$

The resulting intensity ratio is $9:1$.

Thin film interference dictates that transmission minima occur when the optical path difference $2\mu x$ corresponds to an integer multiple of the wavelength $\lambda$. As the fluid evaporates, the film thickness $x$ decreases, causing the interference condition to shift. Since a transmission minimum appears every 12 seconds, this interval represents the duration required for the thickness to change by an amount $\Delta x$ sufficient to satisfy $2\mu \Delta x = \lambda$.

The rate at which the film thins, $\frac{dx}{dt}$, is calculated by dividing this necessary change in thickness, $\Delta x = \frac{\lambda}{2\mu}$, by the time period of 12 seconds. Using the provided values of $\lambda = 560 \times 10^{-9}$ m and $\mu = 1.4$:
$\frac{dx}{dt} = \frac{560 \times 10^{-9}}{2 \times 1.4 \times 12} = \frac{5}{3} \times 10^{-8} \text{ m/s}$

Because the film occupies a cylindrical space with a constant radius $r = 1.8 \times 10^{-2}$ m, the volume $V$ is given by $\pi r^2 x$. The evaporation rate $\frac{dV}{dt}$ is therefore $\pi r^2 \frac{dx}{dt}$. Substituting the radius and the previously determined thinning rate yields:
$\frac{dV}{dt} = \pi (1.8 \times 10^{-2})^2 \times \frac{5}{3} \times 10^{-8} = 54 \pi \times 10^{-13} \text{ m}^3/\text{s}$

The phenomenon of dispersion arises because the refractive index of a material is not constant but varies depending on the specific wavelength of the light traveling through it. Given the relationship $\lambda_{\text{red}} > \lambda_{\text{yellow}} > \lambda_{\text{violet}}$, light with a longer wavelength experiences a smaller refractive index and therefore bends less upon entering a prism. Consequently, red light deviates the least, validating that both the wavelength-dependent nature of refractive indices and the specific bending characteristics described in the statements are correct.

In a Young's double-slit experiment, the intensity at any position on the screen is dictated by the interference of light waves, characterized by the relationship $I = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)$. For the intensity to be one-fourth of the maximum intensity, the condition becomes $\frac{I_0}{4} = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)$, which simplifies to $\cos^2\left(\frac{\Delta\phi}{2}\right) = \frac{1}{4}$. Taking the square root of both sides leads to $\cos\left(\frac{\Delta\phi}{2}\right) = \frac{1}{2}$, which corresponds to a phase difference of $\frac{\Delta\phi}{2} = \frac{\pi}{3}$, or $\Delta\phi = \frac{2\pi}{3}$.

The phase difference is linked to the path difference $\Delta x$ through the expression $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$, and since the path difference for a point at distance $y$ from the central maximum is $\Delta x = \frac{yd}{D}$, we can write $\frac{2\pi}{3} = \frac{2\pi}{\lambda} \left(\frac{yd}{D}\right)$. Rearranging this equation to solve for $y$ yields $y = \frac{\lambda D}{3d}$. Using the provided values of $\lambda = 600$ nm, $D = 1.0$ m, and $d = 1.0$ mm, the calculation is $y = \frac{600 \times 10^{-9} \text{ m} \times 1.0 \text{ m}}{3 \times 1.0 \times 10^{-3} \text{ m}}$. Solving this results in $y = 200 \times 10^{-6}$ m, which is 200 $\mu$m.

When unpolarized light of intensity $I_0$ passes through the first polaroid, it becomes linearly polarized and its intensity is reduced to exactly half, yielding $I_A = \frac{I_0}{2}$. This polarized light then enters the second polaroid, which is oriented at an angle of $45^{\circ}$ relative to the first. Applying Malus's Law, the final intensity of the emergent light is determined by the equation $I = I_A \cos^2(45^{\circ})$. Substituting the known values gives $I = \frac{I_0}{2} \times (\frac{1}{\sqrt{2}})^2$, which simplifies to $\frac{I_0}{2} \times \frac{1}{2}$, resulting in a final intensity of $\frac{I_0}{4}$.

In a single slit diffraction experiment, light spreads out and forms a pattern on a screen characterized by alternating bright and dark fringes. Dark fringes, or minima, occur when the path difference between light rays originating from the edges of the slit satisfies the condition $b \sin \theta = n \lambda$, where $b$ represents the width of the slit, $\theta$ is the angle of diffraction, $n$ is the integer order of the minimum, and $\lambda$ is the wavelength of the light. For the small angles typically encountered in this setup, the approximation $\sin \theta \approx \tan \theta = \frac{y}{D}$ applies, which allows us to define the position of the $n^{\text{th}}$ minimum as $y_n = \frac{n \lambda D}{b}$, where $D$ is the distance from the slit to the screen.

We can determine the slit width by evaluating the distance between the first and third minima. The position of the first minimum is $y_1 = \frac{\lambda D}{b}$, and the position of the third minimum is $y_3 = \frac{3 \lambda D}{b}$. Subtracting these positions gives a total separation of $\Delta y = \frac{2 \lambda D}{b}$. By substituting the given values-wavelength $\lambda = 6000 \times 10^{-10} \text{ m}$, screen distance $D = 0.5 \text{ m}$, and separation $\Delta y = 3 \times 10^{-3} \text{ m}$-into the formula, we solve for $b$. Rearranging the expression to $b = \frac{2 \lambda D}{\Delta y}$ yields $b = \frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}}$, which results in a final slit width of $2 \times 10^{-4} \text{ m}$.

In an interference pattern produced by two slits, the intensity at any given point on the screen is determined by the phase difference between the light waves, described by the relation $I = I_{max} \cos^2(\frac{\phi}{2})$. The condition for the intensity to drop to exactly half of the maximum value requires solving for $\cos^2(\frac{\phi}{2}) = \frac{1}{2}$, which simplifies to $\cos(\frac{\phi}{2}) = \frac{1}{\sqrt{2}}$, indicating that the phase difference must be $\phi = \frac{\pi}{2}$.

Because the phase difference is related to the position $y$ on the screen by $\phi = \frac{2\pi}{\lambda} \Delta x$ where the path difference is $\Delta x = \frac{yd}{D}$, we can equate the two expressions to find $\frac{2\pi}{\lambda} \cdot \frac{yd}{D} = \frac{\pi}{2}$. Rearranging this for the distance $y$ gives the formula $y = \frac{\lambda D}{4d}$. Using the provided wavelength 5000 Å, slit separation 1.0 mm, and screen distance 1.0 m, we substitute these values into the equation to get $y = \frac{5 \times 10^{-7} \text{ m} \times 1.0 \text{ m}}{4 \times 1 \times 10^{-3} \text{ m}}$. This calculation results in $y = 1.25 \times 10^{-4} \text{ m}$, which is equal to 125.

In single-slit diffraction, the linear width $w$ of a secondary maximum is determined by its angular width $\frac{\lambda}{a}$, where $\lambda$ is the wavelength and $a$ is the slit width. When the diffraction pattern is projected onto the focal plane of a convex lens, the lens acts as a screen placed at a distance $D$ equal to its focal length $f$. The linear width of the first secondary maximum is defined by the formula $w = \frac{\lambda}{a} \cdot D$.

Substituting the given values $\lambda = 400 \times 10^{-9}$ m, $a = 0.2 \times 10^{-3}$ m, and $D = 100 \times 10^{-2}$ m into the equation yields $w = \frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} \times 100 \times 10^{-2}$. Solving this expression results in $w = 2 \times 10^{-3}$ m, which is equivalent to 2 mm.

Single-slit diffraction creates a characteristic pattern where light intensity drops to zero at specific angles, creating dark fringes known as minima. These minima occur when the light waves from different parts of the slit interfere destructively, which happens when the path difference between the waves originating from the edges of the slit is an integer multiple of the wavelength.

The condition for these minima is defined by the equation
$a \sin \theta = n \lambda$
where $a$ represents the slit width, $\theta$ is the diffraction angle, $n$ is the order of the minimum, and $\lambda$ is the wavelength of the light. For the first minimum, we set $n = 1$, simplifying the expression to $\sin \theta = \frac{\lambda}{a}$. Substituting the given values, the wavelength is 5000 angstroms, which is $5 \times 10^{-7} \text{ m}$, and the slit width is 0.001 millimeters, which is $1 \times 10^{-6} \text{ m}$. Placing these values into the equation gives $\sin \theta = \frac{5 \times 10^{-7}}{1 \times 10^{-6}}$, which simplifies to 0.5. Because the sine of 30 degrees is 0.5, the angle for the first minimum is 30 degrees.

Incoherent waves superpose by adding their intensities directly because their phase relationship varies randomly over time. Given the intensity ratio of 1:9, we can represent the intensities as $I_0$ and $9I_0$, which yields a combined intensity of $I_1 = I_0 + 9I_0 = 10I_0$ when the waves are incoherent.

When waves are coherent, they create an interference pattern that depends on their phase difference, as described by the formula $I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cos\phi$. With a phase difference of 60 degrees, we substitute the known values and use the fact that the cosine of 60 degrees is 0.5:

$I_2 = I_0 + 9I_0 + 2\sqrt{I_0 \cdot 9I_0} \cdot 0.5$

This simplifies to $I_2 = 10I_0 + 2(3I_0)(0.5)$, resulting in $I_2 = 13I_0$. Because the ratio $\frac{I_1}{I_2}$ is equal to $\frac{10}{13}$, it follows that $x = 13$.

The fringe width in Young's double slit experiment represents the constant separation between adjacent bright fringes, defined by the formula $\beta = \frac{\lambda D}{d}$. The position of the n-th bright fringe relative to the central maximum is given by $y_n = n\beta$. Converting the given parameters into standard SI units, the wavelength is 5 x 10^-7 m, the distance from the slits to the screen is 2 m, and the slit separation is 3 x 10^-4 m. Substituting these values into the expression for the fringe width provides the equation $\beta = \frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = \frac{10}{3} \times 10^{-3}$ m. Applying this to the third bright fringe, the calculation $y_3 = 3\beta$ yields a position of 10 mm.

In Young's double slit experiment, the intensity of light emanating from a slit is directly proportional to its physical width. If we designate the intensity of the narrower slit as $I$, the wider slit with four times the width corresponds to an intensity of $4I$. We calculate the resulting maximum and minimum intensities using the interference formulas $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$. Substituting the respective values gives $I_{\max} = (\sqrt{I} + 2\sqrt{I})^2 = 9I$ and $I_{\min} = (\sqrt{I} - 2\sqrt{I})^2 = I$, producing a ratio of $9:1$ between the maximum and minimum intensities.

In wave optics, the spatial extent of a double-slit interference pattern is defined by the fringe width $\beta = \frac{\lambda D}{d}$, while the diffraction pattern from a single slit is characterized by a central maximum with a width of $\frac{2 \lambda D}{a}$. Because the interference pattern is confined within this central diffraction peak, the total width spanned by 10 interference fringes, expressed as $10 \times \frac{\lambda D}{d}$, must equal the width of the diffraction maximum, $\frac{2 \lambda D}{a}$. Setting these expressions equal allows us to simplify the relationship to $\frac{10}{d} = \frac{2}{a}$, which isolates the slit width as $a = \frac{d}{5}$. Given the separation between the double slits is $d = 1 \text{ mm}$ or $10^{-3} \text{ m}$, substituting this value yields $a = \frac{10^{-3} \text{ m}}{5}$, which simplifies to $2 \times 10^{-4} \text{ m}$.

When a thin glass plate is introduced in front of one of the slits in Young's double slit experiment, it increases the optical path length for that specific path, causing the entire interference pattern to shift. This shift occurs because the plate adds an extra optical path difference equivalent to $(\mu - 1)t$. Since the central maximum moves to the location previously occupied by the 4th bright fringe, this added path difference must equal the path difference associated with the 4th order, which is defined by the equation

$(\mu - 1)t = n\lambda$

Substituting the known values $\mu = 1.5$, $n = 4$, and $\lambda = 500 \times 10^{-9}$ into the expression yields $(1.5 - 1)t = 4 \times 500 \times 10^{-9}$ (where the resulting length is in meters). Simplifying this gives $0.5t = 2000 \times 10^{-9}$. Solving for $t$ results in $4000 \times 10^{-9}$, which corresponds to a final thickness of 4 $\mu$m.

When light strikes a transparent medium and the reflected ray is completely polarized, it occurs at the Brewster angle, which creates a specific geometric relationship between the rays. Under these conditions, the reflected ray and the refracted ray are perpendicular to one another, meaning their angles sum to $90^\circ$. This relationship is represented by the equation $i + r = 90^\circ$, where $i$ is the angle of incidence and $r$ is the angle of refraction. Given that the angle of incidence is $60^\circ$, the calculation becomes $60^\circ + r = 90^\circ$, which confirms the angle of refraction is $30^\circ$.

In a single slit diffraction experiment, the condition for the formation of minima is defined by $a \sin\theta = n\lambda$, where $a$ represents the slit width, $\lambda$ is the wavelength of light, $n$ is the order of the minimum, and $\theta$ is the angle of diffraction. Since the screen distance $D$ is typically much larger than the slit width, the small-angle approximation allows us to substitute $\sin\theta$ with $y/D$, leading to the formula $y = \frac{n\lambda D}{a}$ for the distance of the $n$-th minimum from the central maximum.

Substituting the given values into this expression with $n=1$, $\lambda = 550 \times 10^{-9} \text{ m}$, $D = 100 \times 10^{-2} \text{ m}$, and $a = 0.20 \times 10^{-3} \text{ m}$ provides the position $y$:

$y = \frac{550 \times 10^{-9} \times 100 \times 10^{-2}}{0.20 \times 10^{-3}}$

Calculating this yields $y = 2.75 \times 10^{-3} \text{ m}$. To match the target format of $x \times 10^{-5} \text{ m}$, we convert the result to $275 \times 10^{-5} \text{ m}$, identifying $x$ as 275.

The condition for obtaining minima in single slit diffraction is given by the relation $b \sin \theta = n \lambda$, where $b$ is the width of the slit, $\lambda$ is the wavelength, and $n$ denotes the order of the minimum. For small angles, $\sin \theta \approx \theta$, which simplifies the expression to the angular position formula $\theta = \frac{n \lambda}{b}$. Inserting the values $n = 2$, $\lambda = 600 \times 10^{-9} \text{ m}$, and $b = 4 \times 10^{-4} \text{ m}$ yields an angular position of $\theta = \frac{2 \times 600 \times 10^{-9}}{4 \times 10^{-4}} = 3 \times 10^{-3} \text{ rad}$ on one side of the central maximum. Since the total angular divergence spans the distance between minima on both sides of the pattern, doubling this single-sided angle results in a total divergence of $6 \times 10^{-3} \text{ rad}$.

When two wavelengths are used in a Young's double-slit experiment, a fringe from the first wavelength overlaps with one from the second wavelength when they arrive at the same position on the screen. Because the fringe position is proportional to the product of the order and the wavelength, we equate the conditions for both using the formula $n_1 \lambda_1 = n_2 \lambda_2$. Here, $n_1$ and $n_2$ are the fringe orders corresponding to the wavelengths of 450 nm and 650 nm, respectively.

Rearranging the relationship to obtain the ratio $\frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2}$, the expression becomes $\frac{n_2}{n_1} = \frac{450}{650}$. Reducing this fraction by dividing both components by 50 yields the simplified ratio of $\frac{n_2}{n_1} = \frac{9}{13}$. Since this represents the smallest integer ratio, the minimum order for the second wavelength is 9.