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

The focal length of the lens is related to the object and image distances by the formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, where $u = 100 - 80 = 20 \text{ cm}$ and $v = 180 - 100 = 80 \text{ cm}$, yielding a focal length of $f = \frac{20 \times 80}{20 + 80} = 16 \text{ cm}$. Error propagation for this relationship is determined by differentiating the equation, resulting in $\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$. Because each distance is measured as the difference between two marks on a scale with a least count of $0.2 \text{ cm}$, the absolute uncertainty for both $u$ and $v$ is $\Delta u = \Delta v = 0.2 + 0.2 = 0.4 \text{ cm}$. Substituting these values into the rearranged expression $\frac{\Delta f}{f} = f \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \right)$ provides $\frac{\Delta f}{f} = 16 \left( \frac{0.4}{80^2} + \frac{0.4}{20^2} \right)$. Solving this arithmetic expression leads to a relative error of $0.017$, which corresponds to a percentage error of $1.7\%$.

When an observer looks from a denser medium like water into a rarer medium like air, objects appear to be at a different position and moving with a different velocity due to the effect of refraction. The apparent velocity of the bird as perceived by the rising fish is determined by the sum of the fish's own vertical velocity and the bird's actual velocity magnified by the refractive index of water. This relationship is expressed by the equation $v_{\text{app}} = v_f + \mu v$, where $v_{\text{app}}$ is the observed velocity, $v_f$ is the velocity of the fish, $\mu$ is the refractive index, and $v$ is the actual velocity of the bird.

Substituting the given values into the relationship, the equation becomes $12 = 8 + \frac{4}{3}v$. Subtracting 8 from both sides of the equation yields $4 = \frac{4}{3}v$. Solving this expression for $v$ results in 3 ms^-1 as the actual velocity of the diving bird.

When light refracts through a single spherical surface, the relationship between object distance $u$, image distance $v$, and the two refractive indices $n_1$ and $n_2$ is governed by the equation $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. Light passes from the rarer medium with an index of 1.0 to the denser medium with an index of 1.5, meaning $n_1 = 1.0$ and $n_2 = 1.5$. Following standard sign conventions, the object is placed at $u = -15$ cm, and because the center of curvature lies within the denser medium, the radius of curvature $R$ is taken as +30 cm. Substituting these values into the formula yields $\frac{1.5}{v} - \frac{1}{-15} = \frac{1.5 - 1}{30}$.

This simplifies to $\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$, which is equivalent to $\frac{1.5}{v} + \frac{1}{15} = \frac{1}{60}$. Rearranging to isolate the term with the image distance results in $\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15}$. Computing the difference on the right side gives $\frac{1.5}{v} = \frac{1 - 4}{60} = -\frac{3}{60}$, which further reduces to $\frac{1.5}{v} = -\frac{1}{20}$. Solving for the image distance $v$ gives a value of -30 cm, confirming that the image is formed 30 cm from the pole.

To determine the final image position, we begin by calculating the focal length of each lens using the lens maker's formula, $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. For the plano-concave lens, substituting a refractive index of 1.75 and a radius of curvature of 30 cm yields $\frac{1}{f_1} = (1.75 - 1)\left(-\frac{1}{30}\right)$, which simplifies to $f_1 = -40 \text{ cm}$. Because the object is at infinity, this diverging lens creates a virtual image 40 cm to its left.

Since the lenses are separated by 40 cm, this virtual image serves as the object for the plano-convex lens. The object distance for the second lens is the sum of the physical separation between the lenses and the distance from the concave lens to the virtual image, resulting in $u = -80 \text{ cm}$. The plano-convex lens, sharing the same material and curvature properties, has a focal length of $f_2 = 40 \text{ cm}$. Using the lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for the second lens gives $\frac{1}{v} - \frac{1}{-80} = \frac{1}{40}$, which leads to $v = 80 \text{ cm}$ to the right of the second lens. Consequently, the total distance from the concave lens to the final image is $40 + 80 = 120 \text{ cm}$.

In a system of two parallel mirrors, the distance from the mirror surface to an image is equal to the distance from the mirror surface to the object. The first image formed by mirror AAA is the reflection of the object, located 2 cm behind mirror AAA. To determine the location of the next image behind mirror AAA, we must consider the reflection of the image formed in mirror BBB; since the object is 8 cm from mirror BBB, that initial image sits 8 cm behind mirror BBB, which places it $10 + 8 = 18$ cm from mirror AAA. Consequently, mirror AAA creates this second image 18 cm behind its surface.

When light travels from an object at the bottom of a liquid-filled bucket up through the liquid and into the air, refraction causes the object to appear closer to the surface than its actual position. This phenomenon results in an apparent depth that is smaller than the real depth, creating an upward shift in the perceived location of the object. To maintain focus, the microscope must be raised by a distance equivalent to this shift, which is determined by the formula $S = d \left( 1 - \frac{1}{\mu} \right)$, where $S$ represents the shift, $d$ is the real depth of the liquid, and $\mu$ is the refractive index.

Given that the microscope must be raised by 30 cm and the refractive index of the liquid is $\frac{5}{3}$, we can set up the equation $30 = d \left( 1 - \frac{1}{5/3} \right)$. Simplifying the expression inside the parentheses leads to $30 = d \left( 1 - \frac{3}{5} \right)$, which further reduces to $30 = d \left( \frac{2}{5} \right)$. Multiplying both sides by the reciprocal to isolate the variable results in $d = 30 \times \frac{5}{2}$, which confirms that the height of the liquid in the bucket is 75 cm.

The refractive index of glass varies with the wavelength of incident light, a property fundamental to the lens maker's formula $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Because white light consists of multiple colors with distinct wavelengths, each color encounters a slightly different refractive index and therefore forms an image at a unique position along the principal axis. This effect, where a lens fails to bring all spectral components of light to a single focal point, is termed chromatic aberration.

A fundamental property of plane mirrors is that the image is always formed at the same distance behind the mirror as the object is placed in front of it. Initially, with the object at 12 cm, the image appears 12 cm behind the mirror. Moving the mirror 4 cm closer to the stationary object changes the object distance according to $12\ \text{cm} - 4\ \text{cm} = 8\ \text{cm}$, which forces the new image to form only 8 cm behind the mirror's new position. Since the image was originally 12 cm behind the starting mirror position and is now 8 cm behind the new position, the total displacement of the image is 8 cm towards the mirror.

When light travels from an object submerged in water into the air, refraction causes the object to appear at a shallower depth than its actual position. We calculate this apparent depth using the relationship $d' = \frac{d}{\mu}$, where the real depth $d = 8$ cm and the refractive index $\mu = \frac{4}{3}$, resulting in an apparent depth of 6 cm below the water surface. Because the mirror is fixed 50 cm from the bottom of the tank and the water occupies the first 8 cm, the distance from the water surface to the mirror is 42 cm.

Adding this air gap to the apparent depth of the bulb gives a total object distance of 48 cm from the mirror. Since a plane mirror forms an image at the same distance behind it as the object is in front, the image appears 48 cm above the mirror surface. Adding the mirror's height of 50 cm from the bottom, the image is located at a total distance of $48 + 50 = 98$ cm from the bottom of the tank.

To achieve dispersion without net deviation, the deviation produced by the first prism must equal the deviation produced by the second prism, expressed by the relationship $(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$. Inserting the known values gives the equation $(1.54 - 1) \times 6^{\circ} = (1.72 - 1) \times A_2$, which simplifies to $0.54 \times 6^{\circ} = 0.72 \times A_2$. Calculating the product on the left side yields $3.24^{\circ} = 0.72 \times A_2$, and solving for the unknown angle results in $A_2 = \frac{3.24^{\circ}}{0.72} = 4.5^{\circ}$.

The refractive power of a lens is determined by the Lens Maker's Formula, which establishes that power depends on both the physical geometry of the lens surfaces and the relative refractive index of the lens material with respect to the surrounding medium. Because the physical curvature remains constant for the biconvex lens, the power of the lens effectively scales with the factor $(\frac{\mu_L}{\mu_m} - 1)$, where $\mu_L$ is the refractive index of the lens and $\mu_m$ is the refractive index of the surrounding medium.

In the first scenario, the lens with an index of 1.8 is placed in air, where the surrounding medium has an index of 1.0. This yields a power factor proportional to $(\frac{1.8}{1} - 1)$, which calculates to $0.8$. When the lens is submerged in a liquid with an index of 1.5, the surrounding medium factor changes, resulting in a power factor proportional to $(\frac{1.8}{1.5} - 1)$. Simplifying the fraction $\frac{1.8}{1.5}$ gives $1.2$, making this second factor $(1.2 - 1)$, or $0.2$. The ratio of the power in air to the power in the liquid is determined by the ratio of these two factors, which is $\frac{0.8}{0.2}$, resulting in a value of 4.

A plane mirror creates a virtual image at a distance behind its surface equal to the distance of the object in front of it. Since the plane mirror is located 20 cm from the convex lens and forms an image 5 cm behind the mirror, the light rays arriving at the mirror must be converging toward a point 5 cm in front of the mirror. This means the convex lens creates an image at a distance of 15 cm from the lens itself.

Applying the standard lens equation, where $v$ is the image distance, $u$ is the object distance, and $f$ is the focal length, we use:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

With the lens having a focal length of $f = 10 \text{ cm}$ and forming an image at $v = 15 \text{ cm}$, we substitute these values into the equation. Following the sign convention where the object is positioned at a negative coordinate $-u$, the equation becomes:

$\frac{1}{15} - \frac{1}{-u} = \frac{1}{10}$

Simplifying this expression yields:

$\frac{1}{15} + \frac{1}{u} = \frac{1}{10}$

Solving for $u$ gives:

$\frac{1}{u} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$

This confirms that the object distance is 30 cm.

The refractive index of a glass slab is determined by the ratio of its real thickness $h$ to its apparent thickness $h'$, expressed as $\mu = \frac{h}{h'}$. The precision of this result relies on the least count of the travelling microscope used for measurement. With $20$ divisions per cm on the main scale, the value of one main scale division is $0.05$ cm. Since $50$ vernier scale divisions are equal to $49$ main scale divisions, the least count is the difference between one main scale division and one vernier scale division, yielding $\frac{1}{50}$ of a main scale division, which computes to $\frac{1}{50} \times \frac{1}{20}$ cm, or $0.01$ mm.

Errors in measurement propagate additively when dealing with division. Specifically, the relative uncertainty in the refractive index is the sum of the relative uncertainties in the real and apparent thicknesses:
$\frac{\Delta \mu}{\mu} = \frac{\Delta h}{h} + \frac{\Delta h'}{h'}$
Given the real thickness $h = 5.25$ mm and apparent thickness $h' = 5.00$ mm, along with an uncertainty $\Delta h = \Delta h' = 0.01$ mm, the absolute uncertainty $\Delta \mu$ is found by multiplying the relative uncertainty by the calculated refractive index:
$\Delta \mu = \mu \left( \frac{\Delta h}{h} + \frac{\Delta h'}{h'} \right) = \left( \frac{5.25}{5.00} \right) \left( \frac{0.01}{5.25} + \frac{0.01}{5.00} \right)$
Performing this calculation results in $0.0041$, which corresponds to $\frac{41}{10} \times 10^{-3}$, establishing that $x$ is equal to $41$.

The power of a lens is defined as the reciprocal of its focal length in meters, expressed by the formula $P = \frac{1}{f}$. When a bi-convex lens is sliced into two identical pieces along a plane perpendicular to its principal axis, the total power is the sum of the powers of these two parts, meaning $P = P_1 + P_1 = 2P_1$. Given that the original focal length is 10 cm, which converts to 0.1 m, the total power of the original lens is $P = \frac{1}{0.1} = 10\text{ D}$. Equating this total power to the sum of the parts results in the equation $2P_1 = 10\text{ D}$, which confirms that the power of each individual lens is 5 D.

In a reflecting telescope, the primary concave mirror collects incoming light and reflects it toward a focal point, but viewing the image directly at this location would require the observer to stand in front of the telescope, obstructing the aperture. A secondary mirror is therefore positioned to intercept these converging rays and redirect them to a more accessible location. This redirection allows the eyepiece to be mounted outside the main telescope tube, providing a convenient and unobstructed viewing position for the astronomer.

Reading glasses function by taking an object placed at the standard human near point of 25 cm and projecting a virtual image at the user's actual near point, allowing them to see clearly. The relationship between the lens power and these distances is defined by the thin lens formula, $\frac{1}{v} - \frac{1}{u} = P$, where $v$ represents the person's actual near point and $u$ is the standard near point.

To perform the calculation, we set the object distance to $u = -25$ cm and the corrective power to $P = 2$ diopters. Converting the power to cm$^{-1}$ units for consistency gives $P = \frac{2}{100}$ cm$^{-1}$. Substituting these values into the equation results in $\frac{1}{v} - \left(-\frac{1}{25}\right) = \frac{2}{100}$. Simplifying the equation leads to $\frac{1}{v} = \frac{2}{100} - \frac{1}{25}$, which becomes $\frac{1}{v} = \frac{2}{100} - \frac{4}{100} = -\frac{2}{100}$. Solving for $v$ yields $v = -50$ cm, indicating that the person's least distance of distinct vision is 50 cm.

When a thin rod is placed along the principal axis of a mirror, the length of its image is determined by calculating the specific image positions for each endpoint. Given that the 10 cm rod has its midpoint situated 40 cm from the pole, the two endpoints are located at distances of 35 cm and 45 cm from the mirror. Applying the sign convention for a concave mirror with a focal length of -20 cm, we define the object distances as $u_A = -45$ cm and $u_B = -35$ cm.

Using the mirror formula $v = \frac{uf}{u - f}$, we determine the image position for the endpoint at 45 cm as $v_A = \frac{(-45)(-20)}{-45 - (-20)} = \frac{900}{-25} = -36$ cm. Following the same procedure for the endpoint at 35 cm, the image position is $v_B = \frac{(-35)(-20)}{-35 - (-20)} = \frac{700}{-15} = -\frac{140}{3}$ cm.

The length of the image corresponds to the absolute difference between these two image distances. Computing this difference yields $|-36 - (-\frac{140}{3})| = |\frac{-108 + 140}{3}| = \frac{32}{3}$ cm. Since the problem states the image length is $\frac{x}{3}$ cm, we conclude that $x = 32$.

A plane mirror functions by reflecting light to form a virtual image that is identical in size to the object. Because the image maintains the same vertical orientation, it is considered erect, while the reversal of the horizontal axis creates the characteristic effect of lateral inversion. Since these mirrors never produce real or magnified images, the observation is accurately described only by the properties of being erect and laterally inverted.

The position of images created by a concave mirror is determined by the mirror equation, which relates the object distance, image distance, and focal length. For a concave mirror with a radius of curvature of 40 cm, the focal length is half the radius, set to $f = -20\ \text{cm}$ according to standard sign conventions.

For the first object placed at $u = -15\ \text{cm}$, applying the formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ gives $\frac{1}{v_1} + \frac{1}{-15} = \frac{1}{-20}$. Rearranging this expression results in $\frac{1}{v_1} = \frac{1}{15} - \frac{1}{20}$, which simplifies to $\frac{1}{v_1} = \frac{4-3}{60}$. Therefore, the first image forms at $v_1 = 60\ \text{cm}$.

Performing a similar calculation for the second object placed at $u = -25\ \text{cm}$ involves the equation $\frac{1}{v_2} + \frac{1}{-25} = \frac{1}{-20}$. Isolating the image distance yields $\frac{1}{v_2} = \frac{1}{25} - \frac{1}{20}$. By finding a common denominator, we get $\frac{1}{v_2} = \frac{4-5}{100}$, placing the second image at $v_2 = -100\ \text{cm}$.

Because one image is situated $60\ \text{cm}$ behind the mirror and the other is $100\ \text{cm}$ in front of the mirror, the total distance separating these two images is the sum of their individual distances from the pole, which is $60 + 100 = 160\ \text{cm}$.

When an object is viewed through a medium, the apparent distance is determined by the formula $d_{apparent} = \frac{d_{real}}{\mu}$, where $\mu$ is the refractive index of the material. By treating the bubble's position within the 24 cm ice cube as two distinct real distances, $d_1$ and $d_2$ from opposing faces, we can establish the total length constraint $d_1 + d_2 = 24$.

Viewing the bubble from one side gives an apparent distance of 12 cm, meaning the real distance is $d_1 = 12\mu$. Similarly, viewing from the opposite side results in an apparent distance of 4 cm, so the real distance is $d_2 = 4\mu$. Substituting these real distances into the sum equation provides $12\mu + 4\mu = 24$, which simplifies to $16\mu = 24$. Solving this yields a refractive index of $\mu = \frac{24}{16} = \frac{3}{2}$.