Wave Optics – JEE Main PhysicsPractice Questions & PYQs

The resolving power of a telescope, which dictates its ability to distinguish fine details, is governed by the diffraction limit known as Rayleigh's criterion. This relationship establishes that the minimum angular resolution $\theta$ is directly proportional to the wavelength $\lambda$ of the incident light and inversely proportional to the diameter $b$ of the objective lens, defined by the formula $\theta = \frac{1.22 \lambda}{b}$. Given an angular resolution of $3.0 \times 10^{-7}$ rad and a wavelength of 500 nm, or $5 \times 10^{-7}$ m, we rearrange the expression to solve for the objective diameter $b = \frac{1.22 \lambda}{\theta}$. Substituting these values into the equation leads to $b = \frac{1.22 \times 5 \times 10^{-7}}{3.0 \times 10^{-7}}$, which simplifies to 2.0333 m. Converting this result into centimeters gives 203.33 cm, which rounds to a nearest integer of 203.

Bright fringes for two different wavelengths $\lambda_1$ and $\lambda_2$ coincide at a distance $y$ from the central maximum when the fringe order conditions $y = \frac{n_1 \lambda_1 D}{d}$ and $y = \frac{n_2 \lambda_2 D}{d}$ result in $n_1 \lambda_1 = n_2 \lambda_2$. Given the wavelengths $\lambda_1 = 650 \times 10^{-9}\,\mathrm{m}$ and $\lambda_2 = 550 \times 10^{-9}\,\mathrm{m}$, the ratio of the fringe orders is $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$. Using the smallest integer values $n_1 = 11$ and $n_2 = 13$, the distance $y$ is determined by the formula:

$y = \frac{n_1 \lambda_1 D}{d} = \frac{11 \times 650 \times 10^{-9}\,\mathrm{m} \times 1.2\,\mathrm{m}}{2 \times 10^{-3}\,\mathrm{m}}$

Simplifying this calculation leads to $y = \frac{8580 \times 10^{-9}}{2 \times 10^{-3}}\,\mathrm{m} = 4290 \times 10^{-6}\,\mathrm{m} = 429 \times 10^{-5}\,\mathrm{m}$.

The shift in the central bright fringe in a Young's double slit experiment due to the introduction of a transparent sheet of thickness $t$ and refractive index $\mu$ is given by the formula $y = \frac{D}{d}(\mu - 1)t$, where $D$ is the distance to the screen and $d$ is the distance between the slits.

Substituting the given values $y = 20 \times 10^{-3}\ \text{m}$, $D = 1\ \text{m}$, $d = 0.4 \times 10^{-3}\ \text{m}$, $t = 20 \times 10^{-6}\ \text{m}$, and $\mu = \frac{\alpha}{10}$ results in the equation $20 \times 10^{-3} = \frac{1}{0.4 \times 10^{-3}} \left( \frac{\alpha}{10} - 1 \right) (20 \times 10^{-6})$. Simplifying the right side leads to $20 \times 10^{-3} = \frac{20 \times 10^{-3}}{0.4} \left( \frac{\alpha - 10}{10} \right)$, which reduces to $0.4 = \frac{\alpha - 10}{10}$. Solving this expression, $4 = \alpha - 10$, yielding $\alpha = 14$.

The interference pattern generated by two light beams depends on the superposition of their amplitudes, which are proportional to the square roots of the individual intensities. The ratio of maximum intensity to minimum intensity is governed by the following relationship:

$\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$

Given the intensity ratio of $1:9$, we determine that $\sqrt{I_1} = 1$ and $\sqrt{I_2} = 3$. Substituting these values into the equation leads directly to the solution:

$\frac{I_{max}}{I_{min}} = \frac{(1 + 3)^2}{(1 - 3)^2} = \frac{16}{4} = 4$

This yields a final intensity ratio of $4:1$.

The width of the central maximum in a single-slit diffraction pattern is given by the formula $w = \frac{2\lambda D}{a}$, where $a$ is the width of the slit, $\lambda$ is the wavelength, and $D$ is the distance to the screen. The spacing between adjacent maxima in a Young's double-slit interference pattern is defined by the fringe width $\beta = \frac{\lambda D}{d}$, where $d$ is the separation between the slits.

Given that 20 maxima of the interference pattern are contained within the central maximum of the diffraction pattern, the total width $w$ must equal the span of 20 interference fringes, expressed as $w = 20\beta$. Equating these two expressions leads to $\frac{2\lambda D}{a} = 20 \frac{\lambda D}{d}$, which simplifies to $a = \frac{d}{10}$. Substituting the given value of $d = 1.5 \, \text{mm} = 0.15 \, \text{cm}$ into this result yields $a = \frac{0.15 \, \text{cm}}{10} = 0.015 \, \text{cm}$. Since $0.015 \, \text{cm}$ is equivalent to $15 \times 10^{-3} \, \text{cm}$, the value of $x$ is 15.

In Young's double slit experiment, bright fringes occur at positions determined by the relationship $y = n \lambda \frac{D}{d}$. For two different wavelengths to produce coinciding bright fringes at the same location, their respective path conditions must satisfy $n_1 \lambda_1 = n_2 \lambda_2$, where $n_1$ and $n_2$ are the orders of the bright fringes for wavelengths $\lambda_1 = 480$ nm and $\lambda_2 = 600$ nm, respectively.

Rearranging this equality to determine the ratio of the fringe numbers gives $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$. Substituting the given wavelengths into this expression results in $\frac{n_1}{n_2} = \frac{600}{480}$, which simplifies to the ratio $\frac{5}{4}$. Since the problem asks for the least number of fringes required for coincidence, we use the smallest integer values that satisfy this ratio, meaning the 5th bright fringe of the 480 nm light coincides with the 4th bright fringe of the 600 nm light.

The spacing between interference fringes in Young's double slit experiment depends on the wavelength of light, the separation between the slits, and the distance from the slits to the screen. When the apparatus is submerged in a liquid, the effective wavelength of the light decreases by the factor of the liquid's refractive index. This change in wavelength results in a corresponding reduction in the fringe width, described by the expression $\Delta W = \frac{\lambda D}{\mu d}$.

Substituting the provided values-where the wavelength is 690 nm, the screen distance is 0.72 m, the slit separation is 1.5 mm, and the refractive index is 1.44-leads to the following calculation:
$\Delta W = \frac{690 \times 10^{-9} \times 0.72}{1.5 \times 10^{-3} \times 1.44}$
Evaluating this expression produces a value of 0.00023 m, which is equivalent to 0.23 mm.

The fringe width in Young's double slit experiment is defined by $\beta = \frac{\lambda D}{d}$, where $\lambda$ represents the wavelength, $D$ the distance to the screen, and $d$ the slit separation. Because the fringe width is inversely proportional to the slit separation, increasing the separation from $d_1 = 0.2 \text{ mm}$ to $d_2 = 0.4 \text{ mm}$ implies that the new fringe width $\beta_2$ relates to the initial width $\beta_1$ as $\beta_2 = \frac{d_1}{d_2} \beta_1 = \frac{0.2}{0.4} \beta_1 = 0.5 \beta_1$. Consequently, the magnitude of the percentage change is calculated as $\frac{|\beta_2 - \beta_1|}{\beta_1} \times 100\% = \frac{|0.5\beta_1 - \beta_1|}{\beta_1} \times 100\% = 50\%$.

In a single slit diffraction experiment, the condition for obtaining intensity minima is given by the equation $a \sin \theta = n \lambda$, where $a$ is the width of the slit, $\lambda$ is the wavelength of the light, and $n$ is the integer order of the minimum. For small angles typical in these experiments, we use the approximation $\sin \theta \approx \theta$, which simplifies the expression for the angular position of the $n$-th minimum to $\theta = \frac{n \lambda}{a}$. Consequently, the second minimum on the left corresponds to an angular position of $\frac{2\lambda}{a}$, and the third minimum on the right corresponds to $\frac{3\lambda}{a}$.

The total angular separation between these two positions is the sum of their angles relative to the central maximum, resulting in $\frac{2\lambda}{a} + \frac{3\lambda}{a} = \frac{5\lambda}{a}$. Since the measured separation is $30^{\circ}$, we convert this value to radians to maintain consistency in our calculation, yielding $\frac{\pi}{6}$ radians. Equating these two expressions gives $\frac{5\lambda}{a} = \frac{\pi}{6}$, which allows us to solve for the slit width as $a = \frac{30\lambda}{\pi}$. Substituting the given wavelength of $628 \text{ nm}$, or $0.628 \mu m$, we calculate $a = \frac{30 \times 0.628 \mu m}{3.14}$, which simplifies to a final slit width of $6 \mu m$.

A plane wave described by the equation $x+y+z = \text{constant}$ travels in a direction perpendicular to its wave fronts. This propagation direction is represented by the vector $\vec{A} = \hat{i} + \hat{j} + \hat{k}$, which is determined by the coefficients of the coordinate variables. The angle $\theta$ between this propagation direction and the $x$-axis is found using the direction cosine formula, where the cosine of the angle is the ratio of the $x$-component of the vector to its total magnitude. Calculating the magnitude gives $\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$, so the ratio becomes $\frac{1}{\sqrt{3}}$. Taking the inverse cosine of this value, the resulting angle is $\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

The phenomenon of interference requires the two overlapping light waves to possess the same frequency, which is essential to maintain a constant phase difference across the observation screen. Because red light and green light operate at different frequencies, their relative phase relationship changes too rapidly to establish a stable interference pattern. Consequently, when these two distinct light sources are used in a Young's double slit experiment, no interference fringes can be observed.

The intensity of the interference fringes in Young's double slit experiment depends on the amplitudes of the light waves originating from the two slits, and since amplitude is directly proportional to slit width, we can define the amplitudes as $A_1 = A$ and $A_2 = xA$ for slit widths $d$ and $xd$, respectively. The ratio of maximum to minimum intensity is governed by the relation
$\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2}$
Given that this ratio is $\frac{9}{4}$, we substitute the amplitude expressions to obtain
$\frac{(xA + A)^2}{(xA - A)^2} = \frac{9}{4}$
Taking the square root of both sides yields $\frac{xA + A}{xA - A} = \frac{3}{2}$. Cross-multiplying the equation results in $2(xA + A) = 3(xA - A)$, which expands to $2xA + 2A = 3xA - 3A$. Rearranging these terms isolates $5A = xA$, confirming that $x = 5$.

Unpolarized light incident on the first two polarizers results in an intensity of $I_0/2$ at each slit. By passing this light through the third polarizer $P_3$ oriented at $45^\circ$, Malus's Law dictates that the intensity emerging from each slit is $I_1 = I_2 = (I_0/2) \cos^2(45^\circ) = I_0/4$, and the corresponding amplitude is $A = A_0/2$. The path difference of $\lambda/3$ generates a phase difference of $\phi = 2\pi/3$ between the two waves. The resultant intensity is given by the superposition formula:

$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$

Substituting $I_1 = I_2 = I_0/4$ and $\phi = 2\pi/3$ results in $I = I_0/4 + I_0/4 + 2(I_0/4) \cos(2\pi/3)$, which simplifies to $I_0/2 - I_0/4 = I_0/4$.

Malus's Law states that the intensity of light transmitted through a polarizer is $I = I_{inc} \cos^2\theta$, where $\theta$ is the angle between the polarization direction and the polarizer's axis. When unpolarized light of intensity $I_0$ passes through $P_1$, the intensity becomes $I_1 = I_0/2$, and introducing $P_3$ at an angle $\theta$ relative to $P_1$ followed by $P_2$ at an angle $(\pi/2 - \theta)$ relative to $P_3$ results in a final intensity of
$I_{final} = I_1 \cos^2\theta \cos^2\left(\frac{\pi}{2} - \theta\right) = \frac{I_0}{2} \cos^2\theta \sin^2\theta = \frac{I_0}{8} \sin^2(2\theta)$
This intensity is maximized when $\sin^2(2\theta) = 1$, which implies $2\theta = \pi/2$, yielding an angle of $\theta = \pi/4$ between $P_1$ and $P_3$ and consequently an angle of $\pi/4$ between $P_3$ and $P_2$.

In Young's double slit experiment, the fringe width is governed by the relation $\beta = \frac{\lambda D}{d}$, which shows that fringe spacing is directly proportional to the wavelength of light. When light moves from air into an optically denser medium, its speed decreases, but because the frequency of a light wave is determined by the source and remains constant, the wavelength must decrease according to the relation $\lambda' = \frac{\lambda}{\mu}$, where $\mu$ is the refractive index of the medium. Consequently, this reduction in wavelength directly causes the fringe width to decrease, bringing the fringes closer together, which makes the assertion a direct consequence of the physical changes described in the reason.

The phenomenon of interference occurs when coherent light waves superimpose, creating a pattern where the resulting intensity depends on the amplitudes of the combined beams. Since intensity is proportional to the square of the amplitude, we determine the resultant extremes by adding and subtracting the square roots of the individual intensities.

For beams with intensities $4I$ and $9I$, the constituent amplitudes are proportional to $\sqrt{4I} = 2\sqrt{I}$ and $\sqrt{9I} = 3\sqrt{I}$. The maximum intensity occurs during constructive interference, yielding $I_{max} = (2\sqrt{I} + 3\sqrt{I})^2 = 25I$, while the minimum intensity occurs during destructive interference, resulting in $I_{min} = (2\sqrt{I} - 3\sqrt{I})^2 = I$. The difference between the maximum and minimum intensities is $25I - I = 24I$. The numerical value of $x$ is 24.

The fringe width in Young's double slit experiment is defined by the formula $\beta = \frac{\lambda D}{d}$, which demonstrates that fringe width is directly proportional to the wavelength of light. Because this mathematical relationship is accurate, the reason statement provided is correct. Conversely, because red light possesses a longer wavelength than blue light, it generates fringes that are spaced further apart rather than closer, making the assertion statement false.

The critical angle $\theta_c$ for an optical interface between a denser medium with refractive index $n_d$ and a rarer medium with refractive index $n_r$ is determined by the relationship $\sin(\theta_c) = \frac{n_r}{n_d}$. Applying this to the first interface with materials of indices $n_1$ and $n_2$, we define the critical angle as $\sin(\theta_{1c}) = \frac{n_1}{n_2}$. For the second interface involving $n_1$ and $n_3$, the corresponding critical angle is expressed as $\sin(\theta_{2c}) = \frac{n_1}{n_3}$. Given the specific ratio $\frac{n_2}{n_3} = \frac{2}{5}$, we can establish that $n_3 = \frac{5}{2} n_2$, which allows us to express the second sine term in terms of the first:

$\sin(\theta_{2c}) = \frac{n_1}{ \frac{5}{2} n_2 } = \frac{2}{5} \left( \frac{n_1}{n_2} \right) = \frac{2}{5} \sin(\theta_{1c})$

Substituting this derived relationship into the provided equation $\sin(\theta_{2c}) - \sin(\theta_{1c}) = \frac{1}{2}$, we find that $\frac{2}{5} \sin(\theta_{1c}) - \sin(\theta_{1c}) = \frac{1}{2}$, which simplifies to $-\frac{3}{5} \sin(\theta_{1c}) = \frac{1}{2}$. Solving for the sine of the angle yields $\sin(\theta_{1c}) = -\frac{5}{6}$, a value that is physically impossible for a critical angle and does not match any of the provided numerical options.

To achieve maximum light transmission, a coating must minimize reflection through destructive interference between light waves bouncing off the film's top and bottom surfaces. The condition for this destructive interference in a thin film is $2\mu t = n\lambda$, where $n$ represents an integer. For the minimum thickness, we use the first-order interference with $n=1$, a refractive index $\mu = 2.0$, and a wavelength $\lambda = 550$ nm. Substituting these values into the expression $2 \times 2.0 \times t = 550$ nm leads to the calculation $t = \frac{550}{4}$ nm, which results in a minimum thickness of 137.5 nm.

In Young's Double Slit Experiment, the position of the $n$-th bright fringe from the central maximum is given by $y_n = \frac{n \lambda D}{d}$, which shows that the distance $y_n$ is directly proportional to the wavelength $\lambda$ of the light source. Because the fringe order and the geometric configuration of the apparatus remain unchanged, the relationship between the two scenarios simplifies to the ratio $\frac{y_2}{y_1} = \frac{\lambda_2}{\lambda_1}$. Using the given values, this proportionality yields the calculation $y_2 = \left( \frac{660 \text{ nm}}{600 \text{ nm}} \right) \times 10 \text{ mm}$, resulting in a new fringe position of 11 mm.