Wave Optics – JEE Main PhysicsPractice Questions & PYQs

In a Young's double slit experiment, the angular width of a fringe is directly proportional to the wavelength of the light being used. When the entire experimental setup is immersed in a medium with a refractive index $\mu$, the wavelength of light decreases by this factor, changing to $\lambda' = \frac{\lambda}{\mu}$. Consequently, the angular width is reduced by the same factor, such that the new angular width is equal to the original width divided by the refractive index.

Substituting the provided values, the new angular width is $\theta' = \frac{0.35^{\circ}}{7/5}$. This calculation simplifies to $0.35^{\circ} \times \frac{5}{7}$, which yields $0.25^{\circ}$. Since $0.25^{\circ}$ is equivalent to $\frac{1}{4}^{\circ}$, comparing this to the given expression $\frac{1}{\alpha}$ confirms that $\alpha = 4$.

The fringe width in a Young's double slit experiment is directly proportional to the wavelength of light, which decreases when the experimental setup is immersed in a denser medium. Because the wavelength of light in a medium is reduced by a factor equal to the refractive index of that medium, the fringe width also decreases by the same factor.

This relationship is defined by the formula $\beta' = \frac{\beta}{\mu}$. Here, $\beta'$ represents the new fringe width, $\beta$ is the original fringe width in air, and $\mu$ is the refractive index of the water. Given an initial fringe width of 12 mm and a refractive index of 4/3, dividing 12 by 4/3 results in a new fringe width of 9 mm.

Malus's Law governs the intensity of polarized light passing through an analyzer, stating that the transmitted intensity is proportional to the square of the cosine of the angle between the analyzer's axis and the incident light's polarization direction. With an initial intensity of $I_0 = 100$ lumens, rotating the analyzer by $\theta = 30^{\circ}$ implies the transmitted intensity follows $I = I_0 \cos^2 \theta$. Substituting the known values leads to $I = 100 \cos^2 30^{\circ}$, which simplifies to $100 \times (\frac{\sqrt{3}}{2})^2$. This calculation results in $100 \times \frac{3}{4}$, providing a final output intensity of 75 lumens.

In Young's double-slit experiment, bright fringes appear on the screen at regular, uniform intervals known as the fringe width, which is determined by the relationship $\beta = \frac{\lambda D}{d}$. In this expression, $\lambda$ represents the wavelength of the monochromatic light, $D$ corresponds to the distance from the slits to the screen, and $d$ is the separation between the two slits. Because bright fringes occur at consistent positions, the spatial gap between the first and third bright fringe spans exactly two such intervals, requiring a calculation of $2\beta$.

Standard SI units for these parameters are $\lambda = 5890 \times 10^{-10}$ m, $D = 0.5$ m, and $d = 0.0005$ m. Substituting these values into the formula for $\beta$ yields $\beta = \frac{5890 \times 10^{-10} \times 0.5}{0.0005}$, which simplifies to $5890 \times 10^{-7}$ m. Multiplying this value by two to find the distance between the first and third bright fringes results in $2 \times 5890 \times 10^{-7}$ m, which equals $1178 \times 10^{-6}$ m.

The number of waves contained within a given thickness is defined by the ratio of that distance to the wavelength of the light in the medium. When light travels through air versus a vacuum, the change in wavelength due to the refractive index creates a discrepancy in the total number of waves over the same path. If we denote the thickness as $t$, the number of waves in a vacuum is $t / \lambda_{\text{vac}}$, while in air it is $t / \lambda_{\text{air}}$. Since the wavelength in air is $\lambda_{\text{air}} = \lambda_{\text{vac}} / \mu_{\text{air}}$, the number of waves in air is effectively $t \mu_{\text{air}} / \lambda_{\text{vac}}$. Given that the difference between these counts is exactly one, we can write:

$\frac{t \mu_{\text{air}}}{\lambda_{\text{vac}}} - \frac{t}{\lambda_{\text{vac}}} = 1$

Rearranging this equation to solve for $t$ yields the relationship:

$t = \frac{\lambda_{\text{vac}}}{\mu_{\text{air}} - 1}$

Using the provided values where $\lambda_{\text{vac}} = 6000$ Å and $\mu_{\text{air}} = 1.0003$, the denominator becomes $1.0003 - 1 = 0.0003$. Calculating this gives a thickness of $6000 / 0.0003$ Å, which simplifies to $20,000,000$ Å. Converting this value to millimeters by accounting for the conversion factor results in a final thickness of 2 mm.

When a galaxy moves away from Earth, the light it emits undergoes a redshift, meaning its observed wavelength increases. This phenomenon is explained by the Doppler effect, where the relative shift in wavelength $\Delta\lambda$ is determined by the ratio of the source's velocity $v$ to the speed of light $c$, expressed by the equation $\Delta\lambda = \frac{v}{c} \times \lambda$. With the galaxy moving at a speed of $286 \times 10^3 \text{ m s}^{-1}$ and emitting light at an original wavelength of $630 \times 10^{-9} \text{ m}$, the shift is calculated as follows:

$\Delta\lambda = \frac{286 \times 10^3}{3 \times 10^8} \times 630 \times 10^{-9}$

Solving this expression results in a value of $6.006 \times 10^{-10} \text{ m}$. Because the problem defines the shift in the form $x \times 10^{-10} \text{ m}$, identifying the nearest integer for $x$ leads to a value of $6$.

The spacing between consecutive fringes in Young's double slit experiment is governed by the relationship $\beta = \frac{\lambda D}{d}$, where $\beta$ is the fringe width and $\lambda$ is the wavelength of the light source. Since blue light has a shorter wavelength compared to orange light, switching the source results in a decrease in the value of $\lambda$. Because the fringe width is directly proportional to the wavelength, this reduction in light wavelength causes the distance between consecutive fringes to decrease.

In a Young's double slit experiment, the intensity of light is directly proportional to the square of the amplitude, while the amplitude of the light emerging from a slit is directly proportional to the slit width. Given that one slit is three times as wide as the other, the amplitudes $A_1$ and $A_2$ are in a ratio of $3:1$. Consequently, the intensities $I_1$ and $I_2$ follow a ratio of $9:1$, which can be expressed as $I_1 = 9I_2$.

The ratio of minimum intensity to maximum intensity in an interference pattern is defined by the formula $\frac{I_{min}}{I_{max}} = \left( \frac{\sqrt{I_1} - \sqrt{I_2}}{\sqrt{I_1} + \sqrt{I_2}} \right)^2$ Substituting the relation $I_1 = 9I_2$ into this equation gives $\left( \frac{\sqrt{9I_2} - \sqrt{I_2}}{\sqrt{9I_2} + \sqrt{I_2}} \right)^2$ Simplifying the terms inside the parentheses results \in $\left( \frac{3\sqrt{I_2} - \sqrt{I_2}}{3\sqrt{I_2} + \sqrt{I_2}} \right)^2$ which reduces to $\left( \frac{2\sqrt{I_2}}{4\sqrt{I_2}} \right)^2$ or simply the square of $1/2$. This yields a final ratio of $1/4$. Comparing this result to the given expression $x:4$, the value of $x$ is 1.

In a Young's double slit experiment, the position of the $n$-th bright fringe on the screen relates to the wavelength of light through the formula $y_n = \frac{n \lambda D}{d}$, where $D$ is the distance to the screen and $d$ is the separation between the slits. For the first bright fringes of red and violet light, the order $n$ is equal to 1, allowing us to express their wavelengths as $\lambda = \frac{y_n d}{D}$. The difference in wavelength between these two colors is therefore directly proportional to the difference in their fringe positions on the screen.

Calculating the difference gives $\Delta \lambda = \frac{d}{D} (y_{\text{red}} - y_{\text{violet}})$. Substituting the provided values, where the difference in fringe positions is $3.5$ mm minus $2.0$ mm, or $1.5 \times 10^{-3}$ m, we compute $\Delta \lambda = \frac{0.3 \times 10^{-3} \text{ m}}{1.5 \text{ m}} \times (1.5 \times 10^{-3} \text{ m})$. This simplifies mathematically to $0.3 \times 10^{-6}$ m, which converts to $300$ nm.

In a Young's double slit experiment, the fringe width-the distance between consecutive bright or dark fringes-is determined by the relationship $\beta = \frac{\lambda D}{d}$. In this formula, $\lambda$ represents the wavelength of the light, $D$ is the distance from the screen to the slits, and $d$ is the separation between the two slits.

To determine the fringe separation for this specific setup, we must ensure all values are in consistent units, specifically meters. We are given a slit separation $d = 2 \times 10^{-3} \text{m}$, a screen distance $D = 1 \text{m}$, and a wavelength $\lambda = 500 \times 10^{-9} \text{m}$. Substituting these values into the formula, we have:
$\beta = \frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}}$
Performing the calculation results in $\beta = 250 \times 10^{-6} \text{m}$, which simplifies to $0.25 \text{mm}$.

The amplitude of light passing through a slit is directly proportional to its width, meaning if the first slit has amplitude $A$, the second slit with three times the width will have amplitude $3A$. In an interference pattern, the maximum intensity is determined by the square of the sum of these amplitudes, $I_{\max} \propto (A + 3A)^2$, while the minimum intensity is determined by the square of their difference, $I_{\min} \propto (3A - A)^2$. Calculating these values yields $I_{\max} \propto (4A)^2 = 16A^2$ and $I_{\min} \propto (2A)^2 = 4A^2$. Dividing the maximum intensity by the minimum intensity results in the ratio $16A^2 / 4A^2$, which simplifies to $4:1$.

When sunlight passes through a pinhole with a diameter such as $0.1\ \mu\mathrm{m}$, the resulting diffraction pattern is defined by the angular position of its minima, which follows the relationship $a \sin \phi = n \lambda$. Increasing the diameter $a$ requires the sine of the angle $\phi$ to decrease to maintain the equality with a constant wavelength $\lambda$, effectively tightening the spread of the light and causing the overall size of the diffraction pattern to decrease.

Simultaneously, because the total energy passing through the pinhole is now restricted to a smaller spatial region, the energy density at the center of the pattern becomes higher. This concentration of light means that the peak intensity is heightened as the central maximum shrinks. Thus, increasing the pinhole diameter results in a more compact pattern with a noticeably higher intensity.

The Doppler effect dictates that light waves from a receding source, such as a distant galaxy, are stretched, resulting in an observed wavelength longer than the emitted one. Because the recessional speed is small compared to the speed of light, the relative change in wavelength is directly proportional to the ratio of the object's velocity to the speed of light, expressed as $\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$.

The shift in wavelength is 5896 Å - 5890 Å = 6 Å. With the speed of light $c \approx 300,000 \text{ km/s}$, the velocity follows the relationship $v = c \times \frac{\Delta \lambda}{\lambda_0}$, leading to $v = 300,000 \text{ km/s} \times \frac{6}{5890} = 305.6 \text{ km/s}$ or approximately 306 km/s.

In a Young's double slit experiment, the bright fringes form a pattern where the position of the $n$-th fringe relative to the central maximum is determined by the formula $y_n = \frac{n \lambda D}{d}$. Here, $\lambda$ represents the wavelength of the light, $D$ is the distance from the slits to the screen, and $d$ is the separation between the two slits. By analyzing the spacing between the interference fringes on the screen, one can calculate the light's wavelength and, subsequently, its frequency by using the known speed of light.

The total distance between the fourth bright fringes on opposite sides of the central maximum is 2.4 cm, which means the position of the fourth fringe on a single side is 1.2 cm, or 0.012 m. Given the slit separation $d = 0.3 \times 10^{-3}$ m, the screen distance $D = 1.5$ m, and the fringe order $n = 4$, the wavelength is obtained by rearranging the position formula:
$\lambda = \frac{y_n d}{n D} = \frac{0.012 \times 0.3 \times 10^{-3}}{4 \times 1.5}$
This calculation results in a wavelength of $\lambda = 6 \times 10^{-7}$ m. Using the fundamental relationship $f = \frac{c}{\lambda}$ with the speed of light $c = 3 \times 10^8$ m/s, the frequency is determined to be $f = 5 \times 10^{14}$ Hz.

In Young's double-slit experiment, the fringe width is determined by the interference pattern formed on the screen, expressed by the equation $\beta = \frac{D\lambda}{d}$.

To find the wavelength, we rearrange this formula to $\lambda = \frac{\beta d}{D}$. Using the provided values, we convert the fringe width to $\beta = 6 \times 10^{-3}$ m and the slit separation to $d = 10^{-3}$ m, with the screen distance as $D = 10$ m. Substituting these into the expression, $\lambda = \frac{(6 \times 10^{-3})(10^{-3})}{10}$, results in $\lambda = 6 \times 10^{-7}$ m. Expressed in nanometers, this value is 600 nm.

In an interference pattern, the resultant maximum and minimum intensities are governed by the relationship between the individual source intensities $I_1$ and $I_2$ as follows:
$I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2, \quad I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$
Subtracting these intensities yields a numerator of $4\sqrt{I_1 I_2}$, while adding them results in a denominator of $2(I_1 + I_2)$, which simplifies the total ratio to $\frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$. Dividing both the numerator and denominator by $I_2$ transforms the expression into $\frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1}$. Substituting the given intensity ratio $I_1/I_2 = 2x$ directly into this form results in the final value of $\frac{2\sqrt{2x}}{2x+1}$.

In Young's double slit experiment, the fringe width $\beta$ is defined by the formula $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the separation between the slits. Since the slit separation fluctuates over time according to $d(t) = d_0 + a_0 \sin \omega t$, it reaches a maximum value of $d_0 + a_0$ and a minimum value of $d_0 - a_0$. Because fringe width is inversely proportional to the slit separation, the largest fringe width occurs when the separation is at its minimum, while the smallest fringe width occurs when the separation is at its maximum.

Applying these limits to the fringe width formula yields the maximum width $\beta_{\text{max}} = \frac{\lambda D}{d_0 - a_0}$ and the minimum width $\beta_{\text{min}} = \frac{\lambda D}{d_0 + a_0}$. To determine the difference between these extreme values, we subtract the smallest fringe width from the largest: $\beta_{\text{max}} - \beta_{\text{min}} = \frac{\lambda D}{d_0 - a_0} - \frac{\lambda D}{d_0 + a_0}$. Simplifying this expression by finding a common denominator, which results in $d_0^2 - a_0^2$, the numerator becomes $\lambda D (d_0 + a_0 - (d_0 - a_0))$, ultimately reducing to $\frac{2 \lambda D a_0}{d_0^2 - a_0^2}$.

Initially, the light intensity on the screen is $I/2$ when the two polarizers are aligned. According to Malus's Law, when the second polarizer is rotated by an angle $\theta$ relative to the first, the transmitted intensity is described by the equation $I_{final} = I_{incident} \cos^2 \theta$, where $I_{incident} = I/2$.

To achieve a target intensity of $3I/8$, we solve the equation $3I/8 = (I/2) \cos^2 \theta$. Dividing both sides by $I/2$ results in $\cos^2 \theta = 3/4$, which gives $\cos \theta = \sqrt{3}/2$, meaning the second polarizer must be rotated by an angle of $30^\circ$.

The intensity of an interference pattern formed by two coherent light waves is governed by their constructive and destructive overlap, where the maximum intensity occurs at points of constructive interference. The mathematical expression for this maximum intensity is $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$. Since both coherent sources are provided with an identical intensity of $I_1 = I_2 = I_0$, substituting these values into the formula leads to $I_{max} = (\sqrt{I_0} + \sqrt{I_0})^2$. Squaring the sum of $2\sqrt{I_0}$ results in a final maximum intensity of $4I_0$.

The position of any bright fringe on the screen in a Young's double-slit experiment is determined by the formula $y_n = \frac{n \lambda D}{d}$, which demonstrates that the distance of each fringe from the central point is directly proportional to the wavelength of the light used. Because the wavelength of violet light is shorter than that of red light, substituting violet light into the experiment reduces the distance of each fringe from the central maximum. As a result, every fringe shifts toward the center, causing the distance between consecutive fringes to decrease and the entire interference pattern to appear more compressed.