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

The refractive index of the lens material is determined by the ratio of the speed of light in a vacuum to its speed within the medium. By dividing the speed of light in a vacuum, $3 \times 10^{8}\ \mathrm{ms}^{-1}$, by the given material speed of $2 \times 10^{8}\ \mathrm{ms}^{-1}$, we find the refractive index $\mu$ equals 1.5.

The physical geometry of a plano-convex lens allows us to determine its radius of curvature $R$ based on its central thickness $t$ and the radius of its circular aperture $r$. Because the thickness is small relative to the aperture of the lens, the relationship is expressed by the approximation $R = \frac{r^2}{2t}$. Converting the given thickness of $3\ \mathrm{mm}$ to $0.3\ \mathrm{cm}$ and using the aperture radius of $3\ \mathrm{cm}$, the calculation yields $R = \frac{3^2}{2 \times 0.3} = \frac{9}{0.6}$, which results in a radius of curvature of $15\ \mathrm{cm}$.

Applying the Lens Maker's formula, which relates the focal length $f$ to the refractive index and radii of curvature, we use $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$. Since one surface is flat, $R_2$ is infinite, reducing the equation to $\frac{1}{f} = \frac{\mu - 1}{R}$. Substituting the values of $\mu = 1.5$ and $R = 15\ \mathrm{cm}$ into this formula gives $\frac{1}{f} = \frac{1.5 - 1}{15}$, which simplifies to $\frac{1}{f} = \frac{0.5}{15}$, or $\frac{1}{f} = \frac{1}{30}$. This indicates that the focal length is $30\ \mathrm{cm}$.

The behavior of light at the prism interface is governed by the critical angle, where total internal reflection occurs if the angle of incidence exceeds $i_c = \arcsin(1/\mu)$. Given the prism geometry, light strikes the exit face at an angle of 45 degrees, which establishes a refractive index threshold of $\mu = 1 / \sin 45^\circ \approx 1.414$ for emergence. Light waves with a refractive index lower than this threshold will transmit through the surface, while those with a higher index will be trapped inside by total internal reflection. Because red light has a refractive index of 1.27, which is less than 1.414, it satisfies the condition to pass through the prism face. In contrast, both green and blue light have refractive indices of 1.42 and 1.49, respectively, which exceed this limit, causing them to reflect internally rather than emerge.

The nature of an image in a spherical mirror is determined by its orientation and magnification relative to the object. With an object height of 100 cm and an image height of 25 cm, the magnification $m = \frac{h_i}{h_o}$ evaluates to 0.25, indicating a diminished image. Because the image maintains the same orientation as the object, it is necessarily virtual. While concave mirrors only form erect images when the object is placed within the focal length of 40 cm, those images are always magnified. Conversely, a convex mirror consistently produces virtual, erect, and diminished images regardless of the object position, which means the image is always formed on the opposite side of the mirror.

The fundamental relationship governing electromagnetic waves is defined by the equation $c = \lambda v$, where $c$ represents the constant speed of light, $\lambda$ is the wavelength, and $v$ is the frequency. Because the speed of light in a vacuum remains identical for all colors, any difference in wavelength between two types of light must be accompanied by a corresponding difference in frequency to satisfy this equation. Consequently, since red light and blue light occupy different regions of the electromagnetic spectrum with distinct wavelengths, they inherently possess different frequencies as well.

When a point light source sits in front of a plane mirror, it forms a virtual image directly behind the mirror surface at the same distance, which in this case is 60 cm. The rays reflected from the mirror's edges define the field of view, determining the region where the image is visible to an observer. Since the man walks along a line 1.2 m in front of the mirror, the total distance from the virtual image behind the mirror to the observer's path is 0.6 m + 1.2 m = 1.8 m.

Applying the principle of similar triangles, the ratio of the total field of view width $W$ to the mirror width of 0.5 m is equal to the ratio of the distance from the virtual image to the observer's path to the distance from the virtual image to the mirror. Expressing this as an equation, we have:

$\frac{W}{0.5} = \frac{1.8}{0.6}$

Simplifying the right side yields a value of 3. Multiplying the mirror width by this factor results in a total field of view of 1.5 m. Converting this to the required units, the total distance between the extreme points where the image is visible corresponds to 150 cm.

The behavior of a lens is determined by the Lens Maker's formula, which describes how the focal length depends on the physical geometry of the lens surfaces and the relative refractive indices of the lens and the surrounding environment. When a lens with a refractive index of $1.4$ is placed in a medium that shares the same refractive index of $1.4$, the ratio of the refractive indices becomes equal to one, causing the optical power to vanish as shown in the equation:

$\frac{1}{f} = \left( \frac{\mu_{lens}}{\mu_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0$

Since $\frac{1}{f} = 0$, the focal length $f$ becomes infinite, effectively causing the lens to lose its ability to refract light and act like a simple plane sheet of glass.

When light travels from a denser to a rarer medium, the reflected and refracted rays, along with the angle of reflection $r$ and refraction $r'$, satisfy the geometric condition $r + 90 + r' = 180$, which simplifies to $r' = 90 - r$. By applying Snell's law, $n_1 \sin i = n_2 \sin r'$, and utilizing the fact that $i = r$, we find the ratio of refractive indices to be $n_2 / n_1 = \sin r / \cos r$, which is equivalent to $\tan r$. Given the definition of the critical angle $C$ as $\sin C = n_2 / n_1$, it follows that $C = \sin^{-1}(\tan r)$.

When an image perfectly coincides with its object, the light rays are reflecting directly back along their incident path, which only occurs when they strike the mirror perpendicular to its surface. This configuration requires the lens to direct light rays toward the center of curvature of the mirror. Given a focal length of 15 cm, the radius of curvature is found using the relationship $R = 2f$, which results in 30 cm.

Because the rays are inherently focused on this point 30 cm behind the mirror, removing the mirror allows the lens to converge light at that same spatial position. Summing the segments of the path from the object to the lens, the lens to the mirror, and the mirror to its center of curvature, the total distance from the object to the final image is $12 + 8 + 30 = 50$ cm.

The magnification $m$ of an image produced by a convex lens relates the object distance $u$ and the focal length $f$ through the expression $m = \frac{f}{f+u}$. Because the image size remains equal to the object size in both cases, the magnitude of the magnification must be unity, implying that the two possible magnifications are $1$ and $-1$. Setting the ratio of these expressions for the given object distances equal to the ratio of the magnification values gives $\frac{f-10}{f-20} = -1$. Solving this equation, we find $20 - f = -10 + f$, which leads to $2f = 30$, confirming that the focal length of the convex lens is $15$ cm.

When an incident ray represented by the unit vector $\vec{a}$ strikes a surface with a unit normal $\vec{c}$, it can be decomposed into components parallel and perpendicular to the normal. The component of the incident ray along the normal is given by the projection $(\vec{a} \cdot \vec{c})\vec{c}$. Because reflection reverses only the component parallel to the normal while leaving the tangential component unchanged, we must subtract twice this projection from the incident vector to obtain the reflected unit vector $\vec{b}$. This geometric transformation is mathematically expressed as $\vec{b} = \vec{a} - 2(\vec{a} \cdot \vec{c})\vec{c}$.

Snell's Law dictates that light passing from one medium to another changes direction only if the refractive indices differ. When a light ray encounters an interface between two materials without bending, the path remains continuous, which occurs solely when the two media possess identical refractive indices at that specific wavelength. Consequently, we determine the required wavelength by equating the refractive index expressions for the two prisms.

Setting the refractive indices equal to each other gives:
$1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2} = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}$
Subtracting the terms related to the refractive index constant from one side and the variable terms from the other yields:
$\frac{10.8 \times 10^{-14}}{\lambda^2} - \frac{1.8 \times 10^{-14}}{\lambda^2} = 1.45 - 1.2$
This simplifies to:
$\frac{9 \times 10^{-14}}{\lambda^2} = 0.25$
Rearranging to solve for the square of the wavelength results in:
$\lambda^2 = \frac{9 \times 10^{-14}}{0.25} = 36 \times 10^{-14}$
Taking the square root leads to a wavelength of 600 nm.

The Lens Maker's Formula, $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, defines the focal length of a lens in terms of its refractive index and the radii of curvature of its surfaces. For the plano-convex lens with refractive index $\mu_1$ and curved surface of radius $R$, the focal length $f_1$ satisfies the equation $\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{R} - \frac{1}{\infty }\right) = \frac{\mu_1 - 1}{R}$.

For the plano-concave lens with refractive index $\mu_2$ and curved surface of radius $R$, the focal length $f_2$ satisfies the equation $\frac{1}{f_2} = (\mu_2 - 1)\left(-\frac{1}{R} - \frac{1}{\infty }\right) = \frac{1 - \mu_2}{R}$.

When these lenses are placed in combination, the effective focal length $f$ is given by the sum of the reciprocal focal lengths, $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$. Substituting the individual focal lengths results in $\frac{1}{f} = \frac{\mu_1 - 1}{R} + \frac{1 - \mu_2}{R} = \frac{\mu_1 - \mu_2}{R}$. Calculating the ratio of the radius of curvature $R$ to the combined focal length $f$ yields $\frac{R}{f} = \mu_1 - \mu_2$.

Minimum deviation in a prism is achieved when the light path maintains perfect symmetry, such that the angle of incidence is equal to the angle of emergence ($i = e$). When this symmetry is established, the light ray traveling inside the prism aligns parallel to the base, a characteristic outcome for equilateral or isosceles prisms. Because these observations all describe the same geometric configuration necessary for minimizing the deviation of light, statements (A), (B), and (C) are all true.

The relationship between the angle of deviation $\delta$ and the angle of incidence $i$ in a prism is governed by the equation $\delta = (i + e) - A$, where $e$ represents the angle of emergence and $A$ is the angle of the prism. As $i$ increases from a low value, the deviation $\delta$ first decreases until it reaches a specific minimum value known as the angle of minimum deviation $\delta_m$, which occurs precisely when the ray passes through the prism symmetrically with $i = e$. Past this minimum point, any further increase in the angle of incidence causes the deviation to rise again, creating a distinctive asymmetric curve that slopes downward and then upward.

Refraction causes light from the bottom of the water column to bend at the surface, creating an apparent depth that is less than the actual depth. With the refractive index of water $\mu = \frac{4}{3}$, an actual water height $h$ appears at a depth of $\frac{h}{\mu} = \frac{3h}{4}$. The student perceives the tumbler as half-filled when this apparent depth of the water column equals the height of the empty space above it, which is 17.5 - $h$. This equilibrium is defined by the equation $\frac{3h}{4} = 17.5 - h$. Rearranging this to solve for $h$ yields $\frac{7h}{4} = 17.5$, which simplifies to $h = \frac{17.5 \times 4}{7}$, confirming the actual water height is 10 cm.

To determine the time taken for light to traverse the path from the midpoint of the base to the apex of the prism, we must first account for how the material's refractive index alters the speed of light. Under the condition of minimum deviation, the refractive index $\mu$ is calculated using Snell's law, where the angle of incidence is $i = 60^{\circ}$ and the angle of refraction is $r = A/2 = 30^{\circ}$. This yields $\mu = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \sqrt{3}$.

Considering the geometry of the equilateral triangle with a side length of 10 cm, the segment $AP$ from the midpoint of base $BC$ to vertex $A$ acts as the triangle's altitude. By applying trigonometric ratios, the length of $AP$ is $10 \cos 30^{\circ}$, which simplifies to $5\sqrt{3}$ cm or $5\sqrt{3} \times 10^{-2}$ meters. Because the light travels through the medium of the prism, the effective optical distance $d$ is determined by the product of the refractive index and the geometric path, calculated as $d = \mu \times AP = \sqrt{3} \times 5\sqrt{3} \times 10^{-2} = 15 \times 10^{-2}$ meters. Dividing this optical distance by the speed of light $c = 3 \times 10^{8}$ m/s results in a time of $5 \times 10^{-10}$ seconds.

The focal point of any spherical mirror is located exactly halfway between the pole and the center of curvature, which establishes the magnitude of the focal length as half of the radius of curvature. Because a convex mirror reflects light from an outward-bulging surface, its focal point is positioned behind the mirror, necessitating a positive value according to standard Cartesian optics. These geometric properties combined result in the following relationship:

$f = +\frac{r}{2}$

A simple microscope provides magnification determined by the focal length of the lens relative to the near point of the human eye. When an object is positioned such that the image forms at a near point distance of 25 cm with a magnification of 6, the relationship $m = 1 + \frac{D}{f_0}$ allows us to determine the focal length of the objective lens. Substituting the known values $6 = 1 + \frac{25}{f_0}$ leads to $5 = \frac{25}{f_0}$, which identifies the focal length of the lens as 5 cm.

Transitioning to a compound microscope setup, the total magnification is calculated using the formula $m = \frac{L \cdot D}{f_0 \cdot f_e}$, where $L$ is the length of the microscope tube. Given the magnification is doubled to 12 and the tube length is 60 cm, substituting these values along with the previously determined focal length of the objective lens results in the equation $12 = \frac{60 \cdot 25}{5 \cdot f_e}$. This simplifies to $12 = \frac{300}{f_e}$, and dividing 300 by 12 reveals that the focal length of the eyepiece is 25 cm.

A plane mirror functions by producing a virtual, erect image located at a distance behind the reflective surface equal to the object's distance in front of it. When two such mirrors are positioned at a right angle, they essentially create a coordinate system where the intersection point of the mirrors serves as the origin. Given the source is a distance of $a$ from the first mirror and $2a$ from the second, the two primary images are formed at positions that correspond to the object's distance reflected across the mirror planes.

Considering the geometry of the setup, the images represent specific points in a 2D plane. If the mirrors are treated as the x and y axes, the first image is reflected $a$ behind the first mirror, and the second is reflected $2a$ behind the second mirror. Mapping these positions results in coordinates of $(-a, 2a)$ for the first image and $(a, -2a)$ for the second. The straight-line distance between these two images is the hypotenuse of a right-angled triangle, calculated as $\sqrt{(a - (-a))^2 + (-2a - 2a)^2}$. This simplifies to $\sqrt{(2a)^2 + (-4a)^2}$, which equals $\sqrt{4a^2 + 16a^2}$, or $\sqrt{20a^2}$. By factoring out the root, this becomes $2\sqrt{5}a$. Applying the provided value of $\sqrt{5} = 2.3$, the expression $2 \times 2.3 \times a$ leads to a final distance of $4.6a$.

The lens formula relates the focal length, object distance, and image distance, with a sign convention where both the focal length $f'$ and object distance $u$ are negative for a concave lens. Placing the object at the focus means $u = -f$ and $f' = -f$, which allows us to calculate the image distance $v$ as:
$\frac{1}{v} = \frac{1}{f'} + \frac{1}{u} = \frac{1}{-f} + \frac{1}{-f} = -\frac{2}{f}$
This yields an image distance of $v = -f/2$. Finally, the magnification $m$ is the ratio of image distance to object distance, calculated as:
$m = \frac{v}{u} = \frac{-f/2}{-f} = \frac{1}{2}$