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

When light passes through a system of coaxial thin lenses, we analyze the refraction at each surface by treating the image formed by one lens as the object for the next, using the standard lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ at each step. For the first convex lens with a focal length of 24 cm, the object is placed 6 cm in front, meaning $u = -6$ cm. Substituting these into the formula yields $\frac{1}{v} - \frac{1}{-6} = \frac{1}{24}$, which simplifies to $\frac{1}{v} = \frac{1}{24} - \frac{1}{6} = \frac{1-4}{24} = -\frac{3}{24}$. This results in $v = -8$ cm, indicating that the first lens creates a virtual image 8 cm to the left of its optical center.

Because the lenses are separated by 10 cm, the virtual image formed by the first lens acts as an object for the second lens at a combined distance of 8 cm (from the first lens) plus 10 cm (the gap between lenses), placing it at $u = -18$ cm relative to the second lens. Applying the lens equation again for the second lens with a focal length of 9 cm, we have $\frac{1}{v} - \frac{1}{-18} = \frac{1}{9}$. Solving for the final image position gives $\frac{1}{v} = \frac{1}{9} - \frac{1}{18} = \frac{2-1}{18} = \frac{1}{18}$, so $v = 18$ cm, confirming that the final image is formed 18 cm to the right of the second lens.

To determine the final separation between the original object and this image, we sum the distinct segments along the optical axis. The object is 6 cm to the left of the first lens, the distance between the two lenses is 10 cm, and the final image is 18 cm to the right of the second lens. Adding these components together provides a total distance of 6 cm + 10 cm + 18 cm, which equals 34 cm.

When observing an object through stacked transparent media, the total apparent depth is the sum of the apparent depths calculated for each individual layer. Because the vessel has a total depth $d$ and is divided into two equal sections, each liquid layer possesses a real depth of $d/2$. The apparent depth of the first layer with refractive index $n_1$ is $\frac{d}{2n_1}$, and the apparent depth of the second layer with refractive index $n_2$ is $\frac{d}{2n_2}$. Adding these two components results in a total apparent depth of $\frac{d}{2n_1} + \frac{d}{2n_2}$, which simplifies to $\frac{d(n_1+n_2)}{2n_1n_2}$ when expressed with a common denominator.

The focal length of a lens is determined by the Lens Maker's Formula, which relates the physical curvature of the lens surfaces to the refractive indices of both the lens material and the surrounding medium. When the medium surrounding the lens changes from air to water, the effective refractive index of the lens relative to its environment shifts, which alters its converging power. This modification in the refractive index term causes the focal length to scale by a specific factor, provided the physical dimensions of the lens remain constant.

For a lens with a refractive index of 1.5 in air, the relationship follows $\frac{1}{f_{\text{air}}} = (1.5 - 1)K$ where $K$ represents the constant geometric factor of the lens, resulting \in $\frac{1}{f_{\text{air}}} = 0.5K$. When the lens is immersed \in water with a refractive index of 4/3, the new focal length $f_{\text{water}}$ satisfies the equation $\frac{1}{f_{\text{water}}} = \left( \frac{1.5}{4/3} - 1 \right) K$ which simplifies to $\frac{1}{f_{\text{water}}} = 0.125K$. Comparing these two expressions indicates that $f_{\text{water}} = 4f_{\text{air}}$, meaning the new focal length is 4 times 18 cm, or 72 cm. Subtracting the initial focal length from this new value gives a total change of 54 cm.

The resolving power of a compound microscope is determined by the formula $P = \frac{2 \mu \sin \theta}{1.22 \lambda}$, where $\mu$ represents the refractive index of the medium, $\theta$ is the angular aperture, and $\lambda$ is the wavelength of the light used. Because the resolving power $P$ is directly proportional to the refractive index of the medium, increasing $\mu$ between the specimen and the objective lens enhances the ability to distinguish fine details. Consequently, introducing a medium with a higher refractive index effectively boosts the resolving power, allowing for a clearer observation of the bacteria.

When sunlight strikes the surface of the pool, the light rays undergo refraction as they enter the water, with the angle of incidence relative to the normal being $60^{\circ}$ since the sunlight is incident at $30^{\circ}$ with the surface. By applying Snell's Law, $1 \cdot \sin 60^{\circ} = \frac{4}{3} \cdot \sin r$, we find that the sine of the refraction angle is $\sin r = \frac{3\sqrt{3}}{8}$. Calculating the tangent of this angle gives $\tan r = \sqrt{\frac{27}{37}} \approx 0.85$. This value represents the ratio of the shadow length underwater to the depth of the pool, meaning the underwater portion of the shadow in the 1.5 m deep pool is $1.5 \times 0.85 = 1.275$ m.

Since the total shadow length is 2.15 m, the length of the shadow formed by the pole above the water surface must be $2.15 - 1.275 = 0.875$ m. In the region above the water, the sunlight travels in a straight line at an angle of $30^{\circ}$ to the horizontal, creating a right-angled triangle where the height of the pole $h$ is the opposite side to the $30^{\circ}$ angle. Using the relationship $\tan 30^{\circ} = \frac{h}{0.875}$, we find $h = 0.875 \times \tan 30^{\circ} \approx 0.875 \times 0.577 \approx 0.50$ m. Expressed in the requested units, the height of the pole above the water surface is 50 cm.

A fundamental principle of wave optics is that the frequency of a light wave is determined entirely by the source and remains unchanged during refraction, meaning $v_2 = v_1$. To find the new wavelength, we first determine the refractive index of the second medium using Snell's law: $1 \cdot \sin(45^\circ) = \mu \cdot \sin(30^\circ)$. Solving this equation gives $\frac{1}{\sqrt{2}} = \mu \cdot \frac{1}{2}$, which results in a refractive index of $\mu = \sqrt{2}$. Because the wavelength is inversely proportional to the refractive index when passing from air into a new medium, we calculate $\lambda_2 = \frac{\lambda_1}{\mu}$, leading to $\lambda_2 = \frac{1}{\sqrt{2}} \lambda_1$.

When light from a distant object strikes a convex lens, the parallel incoming rays converge to form an image precisely at the focal point. Since the first lens has a focal length of 20 cm, it creates a real image 20 cm behind it. This image acts as the object for the second lens, which is positioned 60 cm away from the first. Because the image from the first lens is formed 20 cm to its right, its distance relative to the second lens is 40 cm to the left, which gives an object distance of $u_2 = -40$ cm for the second lens.

Using the lens formula for the second lens:
$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$
Substituting the known values, the equation becomes:
$\frac{1}{v_2} - \frac{1}{-40} = \frac{1}{20}$
Solving for the image distance $v_2$, we calculate $\frac{1}{v_2} = \frac{1}{20} - \frac{1}{40}$, which simplifies to $\frac{1}{v_2} = \frac{1}{40}$, resulting in $v_2 = 40$ cm. This indicates the final image forms 40 cm to the right of the second lens. Since the second lens is placed 60 cm away from the first, the final distance from the first lens is 60 cm + 40 cm, totaling 100 cm.

The critical angle for a denser-to-rarer interface is defined by the relationship $\sin(i_c) = \frac{\mu_r}{\mu_d}$, where $\mu_r$ and $\mu_d$ are the refractive indices of the rarer and denser media, respectively. Because the refractive index of a medium is inversely proportional to the speed of light within it, this ratio is equivalent to the speed of light in the denser medium divided by the speed of light in the rarer medium, $\frac{v_d}{v_r}$. Consequently, the relationship can be expressed as $\sin(i_c) = \frac{v_d}{v_r}$.

Substituting the given values into this equation leads to $\sin(45^{\circ}) = \frac{v_d}{3 \times 10^8 \text{ m/s}}$. Given that $\sin(45^{\circ})$ is equal to $\frac{1}{\sqrt{2}}$, the equation becomes $\frac{v_d}{3 \times 10^8 \text{ m/s}} = \frac{1}{\sqrt{2}}$. Solving for $v_d$ gives $v_d = \frac{3 \times 10^8}{\sqrt{2}} \text{ m/s}$, which evaluates to approximately $2.12 \times 10^8 \text{ m/s}$.

The mirror formula establishes the relationship between object distance, image distance, and focal length for a concave mirror according to standard sign conventions. To begin, we calculate the focal length as half the radius of curvature; with a radius of 200 cm, the focal length is $f = -100$ cm. As the object travels towards the mirror at 2 cm/s for 10 s, it moves 20 cm closer, changing its position from 100 cm to 80 cm. Adhering to the sign convention for a concave mirror, the object distance is $u = -80$ cm.

Substituting these values into the mirror formula allows us to determine the position of the image:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{-80} = \frac{1}{-100}$
Solving for the image distance $v$ results in:
$\frac{1}{v} = \frac{1}{80} - \frac{1}{100} = \frac{5 - 4}{400} = \frac{1}{400}$
The final position of the image is 400 cm from the mirror.

The total deviation of a light ray after reflecting off a system of two inclined mirrors is simply the combined sum of the deviations occurring at each reflection point. This additive property holds true as the light changes direction sequentially through the optical setup.

In this specific configuration, the first mirror deflects the incident ray by $90^{\circ}$ from its original path. Upon striking the second mirror, the ray experiences an additional deflection of $120^{\circ}$. Combining these individual effects, the cumulative deviation of the light ray is calculated as $90^{\circ} + 120^{\circ} = 210^{\circ}$.

The deviation produced by a thin prism is determined by the formula $\delta = (\mu - 1)A$, where $\mu$ denotes the refractive index and $A$ is the prism angle. In an arrangement configured to produce no net dispersion, the two prisms must be oriented to oppose each other, which implies that their individual deviations also act in opposition. The first prism creates a deviation of $\delta_1 = (1.5 - 1) \times 6^{\circ} = 3^{\circ}$, while the second prism creates a deviation of $\delta_2 = (1.55 - 1) \times 5^{\circ} = 2.75^{\circ}$. Consequently, the net average deviation is the absolute difference between these two values, resulting in $\delta = |3^{\circ} - 2.75^{\circ}| = 0.25^{\circ}$. Because $0.25^{\circ}$ is equivalent to $\left(\frac{1}{4}\right)^{\circ}$, the value of $x$ is $4$.

The critical angle occurs when light passes from a denser medium to a rarer medium at an incident angle that results in a $90^\circ$ angle of refraction. Snell's law defines the sine of this critical angle as the ratio of the speed of light in the denser medium to its speed in the rarer medium, expressed as $\sin(i_c) = \frac{v_d}{v_r}$.

By inserting the given speeds of $1.5 \times 10^8$ m/s and $2.0 \times 10^8$ m/s, the calculation yields $\sin(i_c) = \frac{1.5 \times 10^8}{2.0 \times 10^8} = \frac{3}{4}$. Converting this to a tangent function requires visualizing a right-angled triangle where the opposite side is $3$ and the hypotenuse is $4$. Applying the Pythagorean theorem, the adjacent side is determined to be $\sqrt{4^2 - 3^2} = \sqrt{7}$. Since the tangent is the ratio of the opposite side to the adjacent side, the critical angle is $\tan^{-1}\left(\frac{3}{\sqrt{7}}\right)$.

The behavior of a biconvex lens immersed in a medium is determined by the lens maker's formula, which defines power $P$ as $P = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$, where $\mu_l$ is the refractive index of the lens, $\mu_m$ is the refractive index of the surrounding medium, and $R_1$ and $R_2$ are the radii of curvature for the lens surfaces. For a biconvex lens, we follow the standard sign convention where the first surface has a positive radius and the second has a negative radius, giving us $R_1 = 0.2 \, \text{m}$ and $R_2 = -0.4 \, \text{m}$.

Substituting the given values of $P = 1.25 \, \text{m}^{-1}$, $\mu_l = 1.5$, and the converted radii into the formula, we have $1.25 = (\frac{1.5}{\mu_m} - 1)(\frac{1}{0.2} - \frac{1}{-0.4})$. Simplifying the terms within the second parenthesis yields $5 + 2.5 = 7.5$. The equation then becomes $1.25 = (\frac{1.5}{\mu_m} - 1)(7.5)$. Dividing $1.25$ by $7.5$ results in $\frac{1}{6}$, such that $\frac{1.5}{\mu_m} - 1 = \frac{1}{6}$. Adding $1$ to both sides gives $\frac{1.5}{\mu_m} = \frac{7}{6}$, which allows us to solve for the refractive index of the medium as $\mu_m = 1.5 \times \frac{6}{7}$, resulting in a value of $\frac{9}{7}$.

Total internal reflection occurs when a light ray transitions from a denser medium to a less dense medium at an incident angle greater than the critical angle. The refractive index of a medium is determined by the ratio $\mu = \frac{c}{v}$, where $c$ is the speed of light in a vacuum and $v$ is the speed of light in the medium. Using $c = 3 \times 10^8$ m/s, the refractive index for medium A is $\mu_A = 1.5$ and for medium B is $\mu_B = 2$. Because the ray travels from medium B to medium A, it moves from a higher refractive index to a lower one, fulfilling the physical requirement for total internal reflection. The critical angle $i_c$ is established by the equation $\sin(i_c) = \frac{\mu_A}{\mu_B} = \frac{1.5}{2} = \frac{3}{4}$. Since total internal reflection happens when the angle of incidence $\theta$ exceeds this critical angle, the required condition is $\theta > \sin^{-1}\left(\frac{3}{4}\right)$.

The resolving power of a microscope is determined by the formula $RP = \frac{2 \mu \sin\theta}{1.22 \lambda}$, where $\mu$ is the refractive index of the medium, $\theta$ is the half-angle of the light cone, and $\lambda$ is the wavelength of the light used. Because the resolving power is directly proportional to the refractive index of the medium, replacing air, with a refractive index of 1, with oil, which has a refractive index of 2, increases the resolving power of the instrument. Consequently, the resolving power in oil becomes twice the resolving power it exhibited while in air.

When sunlight enters a spherical water droplet, it undergoes refraction, a single internal reflection, and refraction again upon exiting, which disperses the white light into its constituent colors. Because red light has a longer wavelength and is refracted less severely than violet light, it emerges at a higher angle relative to the incident sunlight. This geometric arrangement ensures that the observer perceives the red band along the upper edge of the primary rainbow, while the violet band appears at the bottom.

The relationship between the focal length of a lens, its material, and its physical shape is governed by the lens maker's formula, where the focal length $f$ is determined to be 10 cm from the graph. This principle connects $f$ to the refractive index $\mu$ of the lens material and the radii of curvature of its two surfaces. For a symmetric biconvex lens where both surfaces share an equal radius of curvature magnitude $R$, the sign convention requires the light to encounter one surface with a radius of $R$ and the other with a radius of $-R$. The governing equation is:

$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$

Substituting the known values of $f = 10$ cm and $\mu = 1.5$ into this expression yields:

$\frac{1}{10} = (1.5 - 1) \left( \frac{1}{R} + \frac{1}{R} \right)$

This simplifies to $\frac{1}{10} = 0.5 \times \frac{2}{R}$, which further reduces to $\frac{1}{10} = \frac{1}{R}$. Consequently, the radius of curvature $R$ is 10 cm.

When lenses are placed in contact, their powers add linearly, and the properties of each individual lens are defined by the lens maker's formula. For the biconvex lenses, applying the formula with a refractive index of 1.5 and a focal length of 15 cm reveals that the radius of curvature $R$ satisfies $\frac{1}{15} = (1.5 - 1)\left(\frac{2}{R}\right)$, which identifies $R$ as 15 cm. The liquid layer takes the shape of a biconcave lens with the same radius of curvature but a refractive index of 1.25, creating a negative focal power where $\frac{1}{f_{liquid}} = (1.25 - 1)\left(-\frac{2}{15}\right) = -\frac{1}{30}$. By combining the two biconvex lenses and the single liquid lens \in contact, the total power of the system sums to $\frac{1}{F} = \frac{1}{15} + \frac{1}{15} - \frac{1}{30} = \frac{3}{30}$, resulting in a final equivalent focal length of 10 cm.

The focal length of the lens is found using the lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, substituting the initial values of $v = 12 \text{ cm}$ and $u = -240 \text{ cm}$, which gives $\frac{1}{12} - \frac{1}{-240} = \frac{1}{f}$. A glass plate with thickness $t = 1 \text{ cm}$ and refractive index $\mu = 1.5$ creates an image shift towards the lens of $d = t(1 - \frac{1}{\mu})$, which is $1(1 - \frac{1}{1.5}) = \frac{1}{3} \text{ cm}$. To keep the image focused on the screen located at 12 cm, the lens must now produce an image at $v' = 12 - \frac{1}{3} = \frac{35}{3} \text{ cm}$. Using the lens formula again with this new image distance and the established focal length, $\frac{1}{v'} - \frac{1}{u'} = \frac{1}{f}$ becomes $\frac{3}{35} - \frac{1}{u'} = \frac{1}{12} + \frac{1}{240}$, meaning $\frac{1}{u'} = \frac{3}{35} - \frac{1}{12} - \frac{1}{240}$, which solves to $u' = -560 \text{ cm}$. The object must be shifted from its original position of 240 cm to 560 cm, resulting in a displacement of 320 cm, or 3.2 m.

Newton's lens formula provides a specialized way to relate image formation to distances measured from the focal points rather than the optical center. The principle dictates that the product of the object distance from the first focus and the image distance from the second focus equals the square of the focal length, denoted as $x_1 x_2 = f^2$. Given the experimental data $v'\mu' = 225$, we can directly equate this product to the squared focal length, $f^2 = 225$. Taking the square root of this value reveals that the focal length of the lens is 15 cm.