Ray Optics And Optical Instruments – JEE Main PhysicsPractice Questions & PYQs

Snell's Law governs the refraction of light as it passes between different media. When the incoming ray strikes the air-medium interface at 45 degrees, the angle of incidence measured from the normal is also 45 degrees. Because the light ray deviates by 15 degrees from its original path upon entering the medium, the angle of refraction is 30 degrees. This interaction is modeled by the equation:

$\sin(45^\circ) = \mu \sin(30^\circ)$

Substituting the trigonometric values leads to the relationship $\frac{1}{\sqrt{2}} = \mu \times \frac{1}{2}$. Solving this expression for the refractive index yields $\mu = \sqrt{2}$, which is approximately 1.414.

The refractive index of a medium is defined as the ratio of the speed of light in a vacuum, $c$, to its speed in that medium, $v$, which is expressed by the relationship $v = c / n$. Given the difference in speeds between media $A$ and $B$ is $2.6 \times 10^7 \text{ m/s}$, we can establish the following equation:

$\frac{c}{n_A} - \frac{c}{n_B} = 2.6 \times 10^7$

To find the ratio of the refractive indices, we rearrange this equation to isolate the term for the ratio $n_B / n_A$ as shown below:

$\frac{n_B}{n_A} = 1 + \frac{2.6 \times 10^7 \times n_B}{c}$

Substituting the refractive index of medium $B$ as $1.47$ and the speed of light as $c = 3 \times 10^8 \text{ m/s}$ into this expression results in $1 + \frac{2.6 \times 10^7 \times 1.47}{3 \times 10^8}$. Calculating this value yields $1 + 0.1274$, which provides a final ratio of approximately $1.13$.

When the object is placed in front of the convex lens, we apply the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ to determine where the image would form. With an object distance of 60 cm and a focal length of 20 cm, the equation is $\frac{1}{v} - \frac{1}{-60} = \frac{1}{20}$. Rearranging for $v$ gives $\frac{1}{v} = \frac{1}{20} - \frac{1}{60}$, which simplifies to $\frac{1}{v} = \frac{2}{60}$, indicating that the light rays are converging toward a point 30 cm behind the lens.

Since the convex mirror is positioned 10 cm behind the lens, these converging rays are directed toward a point 20 cm behind the mirror surface. For the light rays to reflect back along their incident path and cause the final image to coincide with the original object, the rays must strike the mirror normally, which occurs when they are aimed directly at the mirror's center of curvature. Therefore, the radius of curvature of the mirror is 20 cm. Given that the focal length of a spherical mirror is half of its radius of curvature, the focal length is 10 cm.

In a refracting telescope set to normal adjustment, the final image is formed at infinity, meaning the separation between the objective and the eyepiece equals the sum of their focal lengths, while the magnification is defined by their ratio. Given a total length of 30 cm and an angular magnification of 2, the following conditions must be satisfied:
$f_o + f_e = 30$
$f_o = 2f_e$
Substituting the second relationship into the first results in $2f_e + f_e = 30$, which simplifies to $3f_e = 30$. This indicates that the focal length of the eyepiece is 10 cm, which leads to an objective focal length of $f_o = 2 \times 10 = 20$ cm.

The refractive index of a material is inversely proportional to the speed of light within that medium, which is represented by the relationship $\mu = c/v$, where $c$ is the speed of light in a vacuum. Since the time $t$ required for light to travel a specific thickness $d$ is defined by the ratio $t = d/v$, the difference in time for the two materials can be expressed as $\frac{d}{v_B} - \frac{d}{v_A} = 5 \times 10^{-10}$ s. Given that the ratio of the refractive indices is $\frac{\mu_A}{\mu_B} = \frac{1}{2}$, the corresponding ratio of velocities follows as $\frac{v_B}{v_A} = \frac{\mu_A}{\mu_B} = \frac{1}{2}$, which implies that $v_A = 2v_B$. Substituting $v_A = 2v_B$ into the time difference equation results in $\frac{d}{v_B} - \frac{d}{2v_B} = 5 \times 10^{-10}$. This simplifies to $\frac{d}{2v_B} = 5 \times 10^{-10}$, and solving for $d$ gives $d = 10 \times 10^{-10} \times v_B$. By finally substituting $v_B = v_A/2$ into this expression, we determine the thickness to be $d = 5 \times 10^{-10} \times v_A$ meters.

The behavior of a lens is dictated by the Lens Maker's formula, expressed as:
$P = (\mu-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Cutting a lens along its principal axis splits the lens into two pieces, but since the radii of curvature and the refractive index remain identical to the original lens, the power of each resulting piece $P_1$ remains $P$. Slicing one of these halves perpendicular to the principal axis, however, converts one surface into a flat plane where the radius of curvature becomes infinite, which doubles the focal length and reduces the power of the resulting segments $L_2$ and $L_3$ to $\frac{P}{2}$ each. Consequently, option A is the incorrect statement because it mistakenly claims that $P_1 = \frac{P}{2}$ instead of its actual value of $P$.

The behavior of light as it passes through the globe is governed by the principle of refraction at spherical surfaces, which relates object and image distances to the surface radius and refractive indices through the formula $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$. For a parallel beam entering the first surface from air, we have $u = \infty$, $\mu_1 = 1$, $\mu_2 = 1.5$, and $R = 15$ cm. Substituting these values into the formula gives the equation $\frac{1.5}{v_1} - 0 = \frac{1.5 - 1}{15}$, which simplifies to $\frac{1.5}{v_1} = \frac{0.5}{15}$, resulting in $v_1 = 45$ cm from the first pole.

This image formed by the first surface serves as a virtual object for the second surface. Since the sphere diameter is 30 cm, this virtual object is located 15 cm beyond the second surface. Applying the refraction formula again at the exit surface where light travels from glass to air, we use $\mu_1 = 1.5$, $\mu_2 = 1$, and $R = -15$ cm, as the center of curvature for the second surface is positioned to the left. This yields the equation $\frac{1}{v_2} - \frac{1.5}{15} = \frac{1 - 1.5}{-15}$, which simplifies to $\frac{1}{v_2} = \frac{1}{10} + \frac{1}{30} = \frac{4}{30}$. Solving this gives $v_2 = 7.5$ cm, meaning the light converges 7.5 cm beyond the second pole. The distance from the center of the globe to the second pole is 15 cm, so the total distance from the center is $15 + 7.5 = 22.5$ cm, which is 225 mm.

The refractive index $\mu$ is related to the refracting angle $A$ and the angle of minimum deviation $\delta_m$ by the formula
$\mu = \frac{\sin\left(\frac{\delta_m+A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
Substituting the given value $\mu = \cot\left(\frac{A}{2}\right)$ into this expression yields
$\frac{\sin\left(\frac{\delta_m+A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
By eliminating the common $\sin\left(\frac{A}{2}\right)$ term from the denominators, we determine that $\sin\left(\frac{\delta_m+A}{2}\right) = \cos\left(\frac{A}{2}\right)$. Using the trigonometric identity $\cos\theta = \sin(90^\circ - \theta)$, the equation becomes $\sin\left(\frac{\delta_m+A}{2}\right) = \sin\left(90^\circ - \frac{A}{2}\right)$. Equating the arguments results in $\frac{\delta_m+A}{2} = 90^\circ - \frac{A}{2}$, which simplifies to $\delta_m + A = 180^\circ - A$, giving the final expression $\delta_m = 180 - 2A$.

The resolving power of a telescope quantifies its ability to discern two closely spaced objects as separate entities. Based on the Rayleigh criterion, this metric is the inverse of the smallest angular separation between two distinct points that the optical system can distinguish, formulated as the diameter of the aperture divided by the product of the light's wavelength and the constant 1.22. The relationship is expressed as:

$\text{Resolving Power} = \frac{a}{1.22 \lambda}$

Converting the aperture of $24.4 \text{ cm}$ into meters results in $24.4 \times 10^{-2} \text{ m}$, and representing the wavelength of $2440 \text{ \AA}$ in meters gives $2440 \times 10^{-10} \text{ m}$. Substituting these values into the formula produces:

$\text{Resolving Power} = \frac{24.4 \times 10^{-2}}{1.22 \times 2440 \times 10^{-10}}$

Evaluating this expression involves simplifying the division of the leading constants, where $24.4$ divided by $1.22$ equals 20. This simplifies the equation to $20 \times 10^{-2} / (2440 \times 10^{-10})$, which effectively becomes $20 \times 10^8 / 2440$. Calculating this ratio yields approximately $0.0082$, which corresponds to $8.2 \times 10^5$.

Total internal reflection occurs when light travels from a denser medium into a rarer medium at an angle of incidence greater than the critical angle. To find the threshold for this effect, we first determine the refractive index $n$ of the medium using the formula $n = \sqrt{K \mu_r}$. With a dielectric constant $K$ of 4 and relative permeability $\mu_r$ of 1, the refractive index is $n = \sqrt{4 \times 1} = 2$. The critical angle $\theta_c$ is calculated by the relation $\sin(\theta_c) = \frac{1}{n}$, meaning $\sin(\theta_c) = \frac{1}{2}$, which corresponds to a critical angle of $30^\circ$. Because total reflection requires the incident angle to be greater than the critical angle, the incident ray must strike the boundary at an angle greater than $30^\circ$, identifying $60^\circ$ as the necessary value.

For total internal reflection to occur at the second surface, the light ray must reach the boundary at an angle of incidence $\theta''$ such that $\frac{4}{3} \sin \theta'' = 1 \cdot \sin 90^\circ$, meaning $\sin \theta'' = \frac{3}{4}$. From the geometry of the setup, the angle of refraction at the first surface $\theta'$ relates to this internal angle as $\theta' = 90^\circ - \theta''$. Applying Snell's law at the air-medium interface gives $1 \cdot \sin \theta = \frac{4}{3} \sin \theta'$, which simplifies to $\sin \theta = \frac{4}{3} \cos \theta''$ because $\sin(90^\circ - \theta'') = \cos \theta''$. By evaluating $\cos \theta'' = \sqrt{1 - \sin^2 \theta''} = \sqrt{1 - (\frac{3}{4})^2} = \frac{\sqrt{7}}{4}$, the final expression becomes $\sin \theta = \frac{4}{3} \cdot \frac{\sqrt{7}}{4} = \frac{\sqrt{7}}{3}$, leading to the maximum angle $\theta = \sin^{-1}\left(\frac{\sqrt{7}}{3}\right)$.

Understanding the longitudinal motion of an image requires applying the relationship between an object's velocity and the mirror's magnification. The velocity of an image $v_I$ in a spherical mirror is determined by the formula $v_I = -m^2 v_O$, where $v_O$ is the velocity of the object relative to the mirror and $m$ is the linear magnification factor.

The magnification $m$ is calculated using the mirror parameters and the object distance with the formula $m = \frac{f}{f - u}$. Given the mirror's focal length $f = 10 \text{ cm}$ and the object distance $u = -190 \text{ cm}$, the magnification becomes $m = \frac{10}{10 - (-190)} = \frac{10}{200} = \frac{1}{20}$.

Substituting this magnification value and the object's relative speed of $40 \text{ m/s}$ into the velocity relationship yields $v_I = -(\frac{1}{20})^2 \times 40$. Simplifying this expression leads to $v_I = -\frac{1}{400} \times 40$, which results in a speed magnitude of $0.1 \text{ m/s}$.

The standard mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ relates the object distance, image distance, and focal length under the accepted sign convention. For a concave mirror, when an object is placed at a distance $d_1$ beyond the center of curvature $C$, its position is defined as $u = -(R + d_1)$, where $R$ is the radius of curvature. Because the mirror forms a real image between $C$ and the focus in this configuration, the image distance is $v = -(R - d_2)$, and the focal length of the mirror is $f = -\frac{R}{2}$.

Substituting these expressions into the mirror formula gives $\frac{1}{-\frac{R}{2}} = \frac{1}{-(R - d_2)} + \frac{1}{-(R + d_1)}$, which simplifies to $\frac{2}{R} = \frac{1}{R - d_2} + \frac{1}{R + d_1}$. Combining the terms on the right side, we get $\frac{2}{R} = \frac{2R + d_1 - d_2}{(R - d_2)(R + d_1)}$. Cross-multiplying leads to $2(R^2 + d_1R - d_2R - d_1d_2) = R(2R + d_1 - d_2)$. Expanding both sides yields $2R^2 + 2d_1R - 2d_2R - 2d_1d_2 = 2R^2 + d_1R - d_2R$. After canceling the $2R^2$ terms and rearranging, we obtain $d_1R - d_2R = 2d_1d_2$, which simplifies to:

$R = \frac{2d_1d_2}{d_1 - d_2}$

The behavior of light passing through a prism is governed by the relationship between the refractive index, the angle of the prism, and the angle of minimum deviation. In the state of minimum deviation, the refractive index is given by the formula $\mu = \frac{\sin(\frac{A+\delta_{\min}}{2})}{\sin(\frac{A}{2})}$. Given that the angle of minimum deviation is equal to the angle of the prism, we can set $\delta_{\min} = A$, transforming the numerator to $\sin(\frac{A+A}{2}) = \sin(A)$. Using the double-angle identity $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$, the expression simplifies to $\mu = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})}$, which reduces further to $\mu = 2\cos(\frac{A}{2})$. Rearranging this equation to solve for the angle of the prism results in $A = 2\cos^{-1}(\frac{\mu}{2})$.

The behavior of light at a convex refracting surface is governed by the relation $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$, which links the refractive indices of the two media, the image position, the object position, and the radius of curvature. We are told the light enters the surface from air, so $n_1 = 1$, and since the wavelength inside the surface is $2/3$ of the wavelength in air, the refractive index of the medium must be $n_2 = 1.5$ or $3/2$, because refractive index is inversely proportional to wavelength. The real image is formed $10$ m behind the surface, giving us $v = +10$ m. Given that this image distance is $2/3$ of the object's distance, the magnitude of the object's distance is $15$ m, which we express as $u = -15$ m according to the standard sign convention for refracting surfaces.

Substituting these values into the refraction formula results in $\frac{1.5}{10} - \frac{1}{-15} = \frac{1.5 - 1}{R}$. This simplifies to $\frac{3}{20} + \frac{1}{15} = \frac{0.5}{R}$, and by finding a common denominator of $60$ for the left side, we get $\frac{9}{60} + \frac{4}{60} = \frac{13}{60}$. Setting this equal to $\frac{0.5}{R}$ leads to $\frac{13}{60} = \frac{1}{2R}$, which means $26R = 60$. Solving for $R$ yields $R = \frac{30}{13}$ m. Comparing this result directly to the given expression $\frac{x}{13}$ m, we can determine that the value of $x$ is $30$.

When light passes from one transparent medium to another through a spherical boundary, the image location is determined by the refraction formula $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$. This equation relates the image distance $v$ and object distance $u$ to the respective refractive indices $\mu_1$ and $\mu_2$, alongside the radius of curvature $R$ of the surface.

Standard sign conventions assign a negative value to measurements made against the direction of incident light, so the object distance $u$ is $-40 \text{ cm}$ and the radius of curvature $R$ is $-25 \text{ cm}$. Substituting these parameters into the formula with $\mu_1 = 1.25$ and $\mu_2 = 1.4$ results in the equation $\frac{1.4}{v} - \frac{1.25}{-40} = \frac{1.4 - 1.25}{-25}$. Isolating the term containing $v$ gives $\frac{1.4}{v} = -\frac{0.15}{25} - \frac{1.25}{40}$. Computing these values leads to $\frac{1.4}{v} \approx -0.03725$, which yields an image distance of $v \approx -37.58 \text{ cm}$. The negative sign signifies that a virtual image forms $37.58 \text{ cm}$ from the surface on the same side as the object.

A simple microscope functions by creating a virtual image where the angle subtended by the object through the lens is equivalent to the angle subtended by the image at the eye, denoted as $\theta'$, which confirms the assertion. This device achieves magnification because it allows the object to be placed at a distance $u_0$ that is much smaller than the standard least distance of distinct vision, $D = 25$ cm. Since the object is positioned closer to the eye than the naked eye can normally focus, it subtends a significantly larger angle, resulting in a magnification described by the relationship $m = \frac{D}{u_0}$.

When a light ray passes through a prism at the angle of minimum deviation, the internal geometry becomes perfectly symmetric, which forces the angle of refraction $r$ to be exactly half the prism angle $A$, or $r = A/2$. We are given that the angle of incidence $i$ is double the angle of refraction, so substituting the relationship for $r$ gives $i = 2(A/2)$, which simplifies to $i = A$. According to Snell's law, the relationship between these angles and the refractive index $\mu$ at the prism interface is expressed as $\sin i = \mu \sin r$.

By substituting the known values, this relationship becomes $\sin A = \sqrt{3} \sin(A/2)$. To solve for the prism angle, we apply the trigonometric identity $\sin A = 2 \sin(A/2) \cos(A/2)$, which transforms the equation into $2 \sin(A/2) \cos(A/2) = \sqrt{3} \sin(A/2)$. Dividing both sides by the non-zero term $\sin(A/2)$ results in $\cos(A/2) = \frac{\sqrt{3}}{2}$. Since the cosine of 30 degrees is $\frac{\sqrt{3}}{2}$, it follows that $A/2$ is 30 degrees, meaning the prism angle is 60 degrees.

Determining the position of an image formed by a series of lenses involves applying the thin lens equation sequentially, where the image formed by each lens serves as the object for the subsequent lens. For the first lens with a focal length of $+10$ cm and an object at $-30$ cm, we calculate the image position $v_1$ using $\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}$. Substituting the known values, $\frac{1}{10} = \frac{1}{v_1} - \frac{1}{-30}$ results in $\frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30}$, which places the first image 15 cm to the right of the first lens.

The second lens is located 5 cm beyond the first, meaning this first image acts as an object for the second lens at a distance of $15 - 5 = 10$ cm to its right. Using the lens equation for the second lens with a focal length of $-10$ cm, we find $\frac{1}{v_2} = \frac{1}{-10} + \frac{1}{10}$, which equals $0$, indicating that the light rays emerge parallel from the second lens and are directed to infinity.

Since the rays arrive at the third lens from infinity, the object distance for this lens is treated as $u_3 = \infty$. Applying the formula $\frac{1}{f_3} = \frac{1}{v_3} - \frac{1}{u_3}$ with a focal length of $+30$ cm gives $\frac{1}{30} = \frac{1}{v_3} - \frac{1}{\infty }$, resulting in a final image distance of 30 cm from the third lens. To find the total displacement from the original object, we add the distance from the object to the first lens (30 cm), the distance between the first and second lenses (5 cm), the distance between the second and third lenses (10 cm), and the distance from the third lens to the final image (30 cm), which totals 75 cm.

When light transitions from a denser medium like diamond into a rarer one like air, it can only refract if the incident angle is less than the critical angle $\theta_C$. Calculating this threshold using the provided refractive indices, we find:

$\sin \theta_C = \frac{1}{2.42} \approx 0.413$

Since our incident angle of $30^\circ$ results in $\sin 30^\circ = 0.5$, which exceeds $0.413$, the condition $\theta > \theta_C$ is met. Consequently, the ray undergoes total internal reflection, meaning no refraction occurs.