Wave Optics – JEE Main PhysicsPractice Questions & PYQs

The intensity of light in an interference pattern is directly related to the amplitudes of the interfering waves, with the maximum intensity being proportional to the square of the sum of the individual amplitudes and the minimum intensity being proportional to the square of their difference. This relationship is mathematically described by the formula $\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2}$.

Given that the ratio of the amplitudes from the two slits is $2:1$, we can assign values of $A_1 = 2$ and $A_2 = 1$ for our calculation. Substituting these values into the ratio formula results in $\frac{(2 + 1)^2}{(2 - 1)^2}$, which simplifies to $\frac{3^2}{1^2}$. Consequently, the resulting ratio of maximum to minimum intensity is $9:1$.

The transmission of light through polarizers is governed by Malus's Law, which relates the intensity of transmitted light to the angle between the pass axis of the polarizer and the incident light's polarization direction. When unpolarized light with an intensity of 32 $\text{Wm}^{-2}$ hits the first polarizer, it becomes polarized and its intensity is reduced by half to 16 $\text{Wm}^{-2}$. As this polarized light passes through the subsequent polarizers, its final intensity follows the expression $I_{\text{final}} = 16 \cos^2 \theta \sin^2 \theta$, where $\theta$ is the angle between the pass axes of the first and second polarizers and the final polarizer is oriented perpendicularly to the first.

Since the first and third polarizers are perpendicular, the angle between the second and third polarizers is $90^\circ - \theta$, and because $\cos(90^\circ - \theta)$ is equivalent to $\sin \theta$, the intensity expression simplifies. Using the identity for the sine of a double angle, we can express the final intensity as follows:

$I_{\text{final}} = 4 \sin^2 (2\theta)$

Substituting the emerging intensity of 3 $\text{Wm}^{-2}$ gives $4 \sin^2 (2\theta) = 3$, which simplifies to $\sin^2 (2\theta) = 3/4$. Taking the square root, $\sin (2\theta) = \sqrt{3}/2$, so $2\theta$ can be 60 degrees or 120 degrees. This confirms that the possible angles between the pass axes of the first two polaroids are 30 degrees and 60 degrees.

In Young's double-slit experiment, the fringe width is defined by the formula $\beta = \frac{D\lambda}{d}$, where $D$ is the distance to the screen, $\lambda$ is the wavelength of light, and $d$ is the slit separation. Because the geometry of the experimental setup remains fixed, the fringe width is directly proportional to the wavelength, leading to the relationship $\beta \propto \lambda$. This allows us to compare the two scenarios using the ratio $\frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2}$, where we substitute the given values to obtain $\frac{2}{\beta_2} = \frac{400}{600}$. Solving this equation for the unknown width gives $\beta_2 = 2 \times \frac{600}{400}$, resulting in a new fringe width of 3 mm.

Malus's Law dictates that when polarized light passes through a polaroid, the transmitted intensity $I$ is given by $I = I_{\in} \cos^2 \theta$, where $\theta$ is the angle between the polarization of the incident light and the axis of the polaroid. When unpolarized light with intensity $I_0$ strikes the first polaroid, half of the intensity is transmitted, resulting in a polarized intensity of $I_A = \frac{I_0}{2}$. As this light passes through the intermediate polaroid oriented at 45 degrees, the intensity becomes $I_C = I_A \cos^2 45^\circ$, which simplifies to $I_C = \frac{I_0}{4}$. Because the final polaroid is perpendicular to the first, it sits at an angle of 45 degrees relative to the intermediate polaroid, causing the final transmitted intensity to be $I_B = I_C \cos^2 45^\circ$, resulting in a final value of $\frac{I_0}{8}$.

In the interference of two coherent light sources, the resultant intensity depends on the superposition of the waves from each source, where the maximum and minimum intensities are defined by the formulas $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$. Given that the intensity ratio is $\frac{I_1}{I_2} = \frac{1}{4}$, we can determine that $I_2 = 4I_1$.

Substituting $I_2 = 4I_1$ into these expressions gives $I_{\max} = (\sqrt{I_1} + \sqrt{4I_1})^2 = 9I_1$ and $I_{\min} = (\sqrt{I_1} - \sqrt{4I_1})^2 = I_1$. Using these values to evaluate the requested ratio $\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}}$, we calculate $\frac{9I_1 + I_1}{9I_1 - I_1} = \frac{10I_1}{8I_1}$, which simplifies to $\frac{5}{4}$. Equating this to the expression $\frac{2\alpha + 1}{\beta + 3}$ provides the relation $\frac{5}{4} = \frac{2\alpha + 1}{\beta + 3}$. Solving for the variables, we find that $2\alpha + 1 = 5$ yields $\alpha = 2$, and $\beta + 3 = 4$ yields $\beta = 1$. Therefore, the value of the ratio $\frac{\alpha}{\beta}$ is 2.

The resultant intensity of two interfering light beams is determined by the principle of superposition, where the total intensity depends on the individual intensities and the phase difference $\phi$ between the waves. The governing relationship is:

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

With the given intensities $I_1 = I$ and $I_2 = 4I$, we determine the intensity at point A where the phase difference is $\pi/2$. Inserting these values yields $I_A = I + 4I + 2\sqrt{I \cdot 4I}\cos(\pi/2)$, and since $\cos(\pi/2) = 0$, the resultant intensity simplifies to $I_A = 5I$.

At point B, the phase difference is $\pi/3$, so the intensity becomes $I_B = I + 4I + 2\sqrt{I \cdot 4I}\cos(\pi/3)$. Applying the value $\cos(\pi/3) = 1/2$, the expression becomes $I_B = 5I + 4I(1/2)$, which calculates to $I_B = 7I$. The difference between the resultant intensities at the two points is $7I - 5I = 2I$, meaning the value of $x$ is 2.

The fringe width in a Young's double slit experiment is defined by the equation $\beta = \frac{\lambda D}{d}$, where $\lambda$ represents the wavelength of light, $D$ is the distance from the slits to the screen, and $d$ is the separation between the slits. Because the wavelength and slit separation remain constant, a variation in the screen distance leads to a proportional change in fringe width, represented by the relationship $\Delta \beta = \frac{\lambda \Delta D}{d}$.

Converting all values to standard SI units results in a change in fringe width $\Delta \beta$ of 0.00003 meters, a slit separation $d$ of 0.001 meters, and a shift in screen distance $\Delta D$ of 0.05 meters. Substituting these values into the relationship yields the equation $0.00003 = \frac{\lambda \times 0.05}{0.001}$, which simplifies to $0.00003 = 50 \lambda$. Dividing by 50 to isolate the wavelength results in 0.0000006 meters, which corresponds to 600 nm.

When a light source moves relative to an observer, the observed wavelength shifts due to the Doppler effect. For relative velocities much smaller than the speed of light, this change is described by the formula $\frac{\Delta\lambda}{\lambda} = \frac{v}{c}$, where $\Delta\lambda$ is the shift, $\lambda$ is the original wavelength, $v$ is the velocity, and $c$ is the speed of light.

Since the wavelength increases from 670 nm to 670.7 nm, the shift $\Delta\lambda$ is 0.7 nm, which allows us to determine the velocity by rearranging the expression to $v = c \left( \frac{\Delta\lambda}{\lambda} \right)$. Using $c = 3 \times 10^8$ m/s and the given values, the calculation
$v = 3 \times 10^8 \times \left( \frac{0.7}{670} \right) \text{ m/s}$
results in a recession velocity of approximately $v \simeq 3.13 \times 10^5$ m/s.

The intensity of interference patterns arises from the superposition of light waves, where the resultant maximum and minimum intensities are defined by the sum and difference of the square roots of the individual beam intensities. Given an intensity ratio of 9:4, the corresponding amplitude ratio is 3:2. By substituting these values into the relationship $\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$ we find the ratio by squaring the sum of these amplitudes and dividing it by the square of their difference, resulting in 25:1.

The fringe width in Young's double-slit experiment depends on the wavelength $\lambda$ of the light, the screen distance $D$, and the slit separation $d$ according to the relation $\beta = \frac{\lambda D}{d}$. Given that the screen distance remains constant throughout the experiment, the variation in fringe width is determined by the ratio of the wavelength change and the slit separation change, which can be expressed as:

$\beta_{new} = \beta_{initial} \times \left( \frac{\lambda_{new}}{\lambda_{initial}} \right) \times \left( \frac{d_{initial}}{d_{new}} \right)$

Since the wavelength increases from $5000 \text{ \AA}$ to $6000 \text{ \AA}$, the ratio of wavelengths is $1.2$. Simultaneously, the slit separation doubles, meaning the ratio of the initial separation to the new separation is $0.5$. Applying these factors to the initial fringe width of $0.5 \text{ mm}$ results in a calculation of $0.5 \text{ mm} \times 1.2 \times 0.5$, which yields a new fringe width of $0.3 \text{ mm}$.

When light is incident on the surface of a prism at Brewster's angle ($\theta_B$) with its electric field vector removed, the interaction between the wave and the interface changes such that the energy is not partitioned into a reflected component. This specific orientation allows the incident light to enter the prism entirely, resulting in complete transmission through the surface. Because the condition for reflection is not satisfied, the light continues into the medium, ensuring that only refraction takes place at the boundary.

Introducing a glass plate into the path of one of the interfering beams shifts the interference pattern but maintains the same intensity at the original central maximum if the new path difference is an integer multiple of the wavelength. This path difference introduced by the plate is calculated as $(\mu - 1)t$, where $\mu$ is the refractive index and $t$ is the thickness. For constructive interference to persist at that specific location, this path difference must be set equal to $n\lambda$, leading to the condition $(\mu - 1)t = n\lambda$.

Substituting the given values $\mu = 1.5$ and $t = x\lambda$ into this equation results in $(1.5 - 1) \times x\lambda = n\lambda$. This simplifies to $0.5x = n$, or $x = 2n$. Assigning the value $n = 1$ to obtain the corresponding integer solution confirms that $x = 2$.

When two light waves of intensities $I_1$ and $I_2$ superpose, the resultant intensity is determined by the phase difference between them. At point A, where the phase difference is zero, constructive interference occurs, leading to a resultant intensity of $I_A = (\sqrt{4I} + \sqrt{9I})^2 = (2\sqrt{I} + 3\sqrt{I})^2 = 25I$. At point B, where the phase difference is $\pi$, destructive interference takes place, resulting in an intensity of $I_B = (\sqrt{4I} - \sqrt{9I})^2 = (2\sqrt{I} - 3\sqrt{I})^2 = I$. The difference between these two resultant intensities is $25I - I = 24I$.

In Young's double slit experiment, the fringe width $\beta$, which is the separation between consecutive bright fringes, depends directly on the wavelength $\lambda$ of the light used. Assuming the geometry of the experiment remains constant, this relationship is described by the formula
$\beta = \frac{\lambda D}{d}$
Due to this direct proportionality, the ratio of the fringe widths is equal to the ratio of their respective wavelengths, expressed as $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$. With the first fringe width at 7.2 mm and the second at 8.1 mm, and an initial wavelength of 560 nm, the relationship becomes $\frac{8.1}{7.2} = \frac{\lambda_2}{560}$. Simplifying the ratio of the fringe widths results in the fraction $\frac{9}{8}$, and the new wavelength is determined by $\lambda_2 = \frac{9}{8} \times 560$, yielding 630 nm.

In an interference pattern, the maximum and minimum intensities generated by two coherent light sources are given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$. When we substitute these definitions into the ratio $\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}}$, the expression simplifies to $\frac{2(I_1 + I_2)}{4\sqrt{I_1 I_2}}$, which is equivalent to $\frac{\frac{I_1}{I_2} + 1}{2\sqrt{\frac{I_1}{I_2}}}$. Given that the ratio of the intensities $\frac{I_1}{I_2}$ is $4$, we substitute this value into the equation to get $\frac{4 + 1}{2\sqrt{4}}$. This yields $\frac{5}{2 \times 2}$, which equals $\frac{5}{4}$, identifying the value of $x$ as 4.

The diffraction pattern generated by a single slit is governed by the interference of light, where the position of the first maximum is given by the expression $y = \frac{3 \lambda D}{2 a}$. When using two light sources with wavelengths 650 nm and 655 nm, the separation between their first maxima depends on the difference in wavelengths, expressed as $\Delta y = \frac{3 D}{2 a} (\lambda_2 - \lambda_1)$. Substituting the given values of 2.0 m for the distance, 0.5 mm for the aperture, and 5 nm for the difference in wavelength, the separation is calculated as $\Delta y = \frac{3 \times 2.0}{2 \times 0.5 \times 10^{-3}} \times 5 \times 10^{-9}$. This calculation simplifies to $\Delta y = \frac{3 \times 2.0 \times 5 \times 10^{-9}}{1.0 \times 10^{-3}}$, resulting in a separation of $3 \times 10^{-5}$ m.

When unpolarized light strikes a polarizing filter, the intensity of the light that emerges is reduced by exactly half. Starting with an initial unpolarized beam of intensity $2I_0$, passing it through the first polaroid P results in a linearly polarized beam with intensity $I_1 = \frac{1}{2} (2I_0) = I_0$.

The final intensity after passing through the second polaroid Q is found by applying Malus's Law, which states that the transmitted intensity equals the incident intensity multiplied by the square of the cosine of the angle between the two polarizers' axes. With the second polaroid oriented at $30^{\circ}$ relative to the first, the emergent intensity is calculated as $I_2 = I_1 \cos^2(30^{\circ})$. Substituting the values yields $I_2 = I_0 \times \frac{3}{4}$, which simplifies to $\frac{3I_0}{4}$.

Interference patterns arise because waves traveling different distances from two slits to a point on the screen interfere constructively or destructively. A dark fringe occurs specifically when the path difference equals an odd multiple of half the wavelength. For the first dark minimum, the path difference is exactly half the wavelength, or $\Delta x = \frac{\lambda}{2}$. Given that the fringe appears exactly opposite one of the slits, the distance from the center of the screen to this point corresponds to half the slit separation, represented as $y = \frac{d}{2}$. Using the geometric path difference formula $\Delta x = \frac{yd}{D}$ and substituting $y = \frac{d}{2}$, the relationship for the path difference at this specific location becomes $\Delta x = \frac{d^2}{2D}$.

Equating this path difference to the condition for destructive interference leads to the expression:
$\frac{d^2}{2D} = \frac{\lambda}{2}$
This simplifies to the relationship $\lambda = \frac{d^2}{D}$. Substituting the provided dimensions $d = 0.6 \times 10^{-3}$ m and $D = 0.8$ m into this equation allows for the calculation:
$\lambda = \frac{(0.6 \times 10^{-3})^2}{0.8} = 450 \times 10^{-9} \text{ m}$
The resulting wavelength of the light is 450 nm.

The superposition principle allows us to calculate the resultant intensity of two interfering light beams using the relation $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$, where $I_1$ and $I_2$ are the individual intensities and $\phi$ represents the phase difference. With individual intensities $I$ and $9I$, we calculate the intensity at point P by substituting the phase difference $\phi = \pi/2$ into this expression, which yields $I_P = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi/2) = 10I$ because the cosine term becomes zero. Similarly, at point Q where the phase difference is $\pi$, the expression becomes $I_Q = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi)$, which simplifies to $10I - 6I = 4I$ since the cosine of $\pi$ is negative one. Calculating the difference between these two results gives $10I - 4I = 6I$.

The intensity of light emerging from a slit in an interference experiment is directly proportional to its width, and because intensity is also proportional to the square of the light wave's amplitude, the ratio of the amplitudes can be determined by taking the square root of the ratio of the intensities. With the slit widths given as a ratio of $9:16$, the light intensities follow the same ratio, expressed as $I_1/I_2 = 9/16$. Taking the square root of this value yields the ratio of the amplitudes as $A_1/A_2 = \sqrt{9/16} = 3/4$.

The ratio of maximum intensity to minimum intensity in an interference pattern is determined by the amplitudes of the interfering waves according to the following formula:
$\frac{I_{max}}{I_{min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$
Substituting the proportional values of $3$ and $4$ for the amplitudes $A_1$ and $A_2$ into this expression results in:
$\frac{I_{max}}{I_{min}} = \left( \frac{3 + 4}{3 - 4} \right)^2 = \left( \frac{7}{-1} \right)^2 = \frac{49}{1}$
This yields a final ratio of $49:1$.