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

The total length of the light path is determined by summing the lengths of the two segments, $AO$ in the vacuum and $OB$ inside the glass slab. For the incident portion, the ray $AO$ forms the hypotenuse of a right triangle with an opposite side of length $a$, satisfying the equation $\sin 30^\circ = \frac{a}{AO}$, which gives $AO = 2a$. In the refracted section, the ray $OB$ acts as the hypotenuse of a triangle where the adjacent side is $b$, allowing us to use the relation $\cos 30^\circ = \frac{b}{OB}$ to find that $OB = \frac{2b}{\sqrt{3}}$. Combining these individual distances yields the total length $2a + \frac{2b}{\sqrt{3}}$.

The magnification $m$ of a lens is related to the object distance $u$ and focal length $f$ by the formula $m = \frac{f}{f+u}$. For a convex lens with $f = 20$ cm, the magnification can be either $-2$ for a real image or $+2$ for a virtual image. Substituting $m = -2$ into the formula gives $-2 = \frac{20}{u_1 + 20}$, which simplifies to $-2u_1 - 40 = 20$, yielding $u_1 = -30$ cm, meaning $x_1 = 30$ cm. Substituting $m = +2$ gives $2 = \frac{20}{u_2 + 20}$, which simplifies to $2u_2 + 40 = 20$, yielding $u_2 = -10$ cm, meaning $x_2 = 10$ cm. The ratio of the two distances is $\frac{x_1}{x_2} = \frac{30}{10} = 3:1$.

The combined focal length of these adjacent thin lenses is determined by calculating the total optical power, which is the algebraic sum of the powers of the two individual components. The plano-convex lens with refractive index $\mu_1$ acts as a converging element, contributing a power of $(\mu_1 - 1)/R$ since its curved surface has a radius of curvature $R$ and its flat surface has infinite radius. Conversely, the plano-concave lens with refractive index $\mu_2$ behaves as a diverging element, providing a negative power contribution of $-(\mu_2 - 1)/R$. Summing these two expressions simplifies the total power to $(\mu_1 - \mu_2)/R$, and taking the reciprocal of this equivalent power results in the combined focal length of $\frac{R}{\mu_1 - \mu_2}$.

The refraction of light at the spherical surface of the eye is governed by the formula $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$, where $n_1$ and $n_2$ are the refractive indices of the two media, $u$ is the object distance, $v$ is the image distance, and $R$ is the radius of curvature. Since the incident light consists of a parallel beam, the object distance $u$ is at infinity, which causes the term $\frac{n_1}{u}$ to become zero. Substituting the refractive indices $n_1 = 1$ and $n_2 = 1.34$ along with the radius of curvature $R = 7.8 \text{ mm}$, the equation simplifies to $\frac{1.34}{v} = \frac{1.34 - 1}{7.8}$. Solving this expression for $v$ yields $30.7 \text{ mm}$, which corresponds to 3.07 cm, placing the focal point at approximately 3.1 cm.

The focal length of an optical system formed by a lens and a silvered surface is determined by the effective power $P_{sys} = 2P_L + P_M$, where $P_L$ is the power of the lens and $P_M$ is the power of the mirror. For a plano-convex lens with refractive index $n$ and radius of curvature $R$, the power of the lens is $P_L = \frac{n-1}{R}$.

In the first case, where the plane surface is silvered, the mirror is flat with $P_M = 0$, so the effective power is $P_{sys1} = 2P_L = \frac{2(n-1)}{R} = \frac{1}{28}$, which gives $R = 56(n-1)$.

In the second case, where the curved surface is silvered, the mirror acts as a concave mirror with focal length $f_m = -\frac{R}{2}$, giving a power $P_M = -\frac{1}{f_m} = \frac{2}{R}$. The effective power is $P_{sys2} = 2P_L + P_M = \frac{2(n-1)}{R} + \frac{2}{R} = \frac{2n}{R} = \frac{1}{10}$. Substituting $R = 56(n-1)$ into this equation results in $\frac{2n}{56(n-1)} = \frac{1}{10}$, which simplifies to $20n = 56n - 56$. Solving for $n$ gives $36n = 56$, or $n = \frac{56}{36} \approx 1.55$.

The relationship between the object distance $u$ and image distance $v$ for a concave mirror is governed by the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, where $f = -5\,\text{cm}$. Since the mirror is at the origin and the object oscillates between $x = 8\,\text{cm}$ and $x = 12\,\text{cm}$, we set the object distance $u = -x$, leading to $\frac{1}{v} = \frac{1}{x} - \frac{1}{5}$, which implies $v = \frac{5x}{5-x}$. Because the particle's position $x(t)$ is a periodic function, the image distance $v(t)$ is also periodic, confirming the image executes periodic motion.

The turning points of the image are found at the extreme positions of the object oscillation. At $x = 8\,\text{cm}$, the image position is $v_1 = \frac{5(8)}{5-8} = -\frac{40}{3}\,\text{cm}$, and at $x = 12\,\text{cm}$, the image position is $v_2 = \frac{5(12)}{5-12} = -\frac{60}{7}\,\text{cm}$. The image of the central point $x = 10\,\text{cm}$ is $v_0 = \frac{5(10)}{5-10} = -10\,\text{cm}$. The distances from the central image are $|v_1 - v_0| = |-\frac{40}{3} + 10| = \frac{10}{3}\,\text{cm}$ and $|v_2 - v_0| = |-\frac{60}{7} + 10| = \frac{10}{7}\,\text{cm}$. Since these distances are unequal, the oscillation of the image is asymmetric about the image of the central point. The distance between the turning points of the image is $|v_2 - v_1| = |-\frac{60}{7} - (-\frac{40}{3})| = |\frac{-180 + 280}{21}| = \frac{100}{21}\,\text{cm}$.

The angle of deviation $\delta$ produced by a prism is related to the angle of incidence $i$, the angle of emergence $e$, and the prism angle $A$ by the equation:
$\delta = i + e - A$
Substituting the given values $i = 60^\circ$, $\delta = 30^\circ$, and $A = 30^\circ$ results in $30^\circ = 60^\circ + e - 30^\circ$, which simplifies to $e = 0^\circ$. Since $e$ represents the angle between the emergent ray and the normal to the second surface, an angle of $0^\circ$ indicates that the emergent ray travels along the normal, perpendicular to the second face. The angle between the emergent ray and the surface of the second face is therefore $90^\circ - e$, which evaluates to $90^\circ - 0^\circ = 90^\circ$.

The equivalent focal length $F$ of two thin lenses with focal lengths $f_1$ and $f_2$ separated by a distance $d$ is determined by the equation:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Substituting the given values $F = 10$ cm and $d = 2$ cm results in $\frac{1}{10} = \frac{f_1 + f_2 - 2}{f_1 f_2}$, which simplifies to the condition $f_1 f_2 = 10(f_1 + f_2 - 2)$. The pair $f_1 = 20$ cm and $f_2 = 18$ cm satisfies this relation, as $20 \times 18 = 360$ and $10(20 + 18 - 2) = 360$. These values correctly identify the component focal lengths that produce the specified system focal length.

For the image to coincide with the object, the light rays incident on the convex mirror must be directed towards its center of curvature so that they strike the surface normally and retrace their path. Applying the lens formula $\frac{1}{v_L} - \frac{1}{u_L} = \frac{1}{f_L}$ to the convex lens with $u_L = -20$ cm and $f_L = 15$ cm, the image distance is determined by $\frac{1}{v_L} = \frac{1}{15} - \frac{1}{20} = \frac{1}{60}$, which gives $v_L = 60$ cm. This image acts as a virtual object for the convex mirror. Since the mirror is placed $5$ cm from the lens, the distance of this image from the mirror is $60 - 5 = 55$ cm. Because the light rays must be directed towards the center of curvature for normal incidence, the radius of curvature of the mirror is $R = 55$ cm. The focal length is thus $f_M = \frac{R}{2} = \frac{55}{2} = 27.5$ cm.

When light travels from a denser medium to a rarer medium, Snell's Law establishes that the ratio of the refractive indices is determined by the sines of the incident angle $A$ and the refracted angle $r$, expressed as $\frac{\mu_R}{\mu_D} = \frac{\sin A}{\sin r}$. Because the reflected and refracted rays are perpendicular, the sum of the incident angle $A$, the angle between them ($90^\circ$), and the angle of refraction $r$ must form a straight line, which implies $r = 90^\circ - A$.

Substituting this geometric relation into Snell's law yields the ratio of the refractive indices as:
$\frac{\mu_R}{\mu_D} = \frac{\sin A}{\sin(90^\circ - A)} = \frac{\sin A}{\cos A} = \tan A$
By the definition of the critical angle $\theta_C$, the ratio of the refractive indices is also defined by $\sin \theta_C = \frac{\mu_R}{\mu_D}$. Equating these two expressions, $\tan A = \sin \theta_C$, leads to the final result of $A = \tan^{-1}(\sin \theta_C)$.

When multiple lenses are arranged in a series, the image generated by the first lens effectively acts as the object for the second. A parallel beam of light incident on a diverging lens with a focal length of 25 cm creates a virtual image at a distance of 25 cm from that lens on the side where the light enters.

Considering the 15 cm separation between the lenses, this virtual image is located 40 cm to the left of the converging lens. By taking this position as the object distance, $u = -40$ cm, and using the focal length of the converging lens, $f = +20$ cm, the thin lens equation provides the image position:

$\frac{1}{v} - \frac{1}{-40} = \frac{1}{20}$

Rearranging this expression to solve for $v$ yields:

$\frac{1}{v} = \frac{1}{20} - \frac{1}{40}$

$\frac{1}{v} = \frac{1}{40}$

$v = 40 \text{ cm}$

The positive value for $v$ indicates that the final image is real and forms 40 cm away from the converging lens.

The setup requires analyzing light reflecting from the curved surface and then refracting through the flat boundary. The silvered curved surface acts as a concave mirror with a focal length of $f = -5\,\text{cm}$ because the radius of curvature is $10\,\text{cm}$. Since the bubble is $6\,\text{cm}$ below the flat surface, it sits $4\,\text{cm}$ from the concave mirror surface, placing the object distance at $u = -4\,\text{cm}$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, we solve $\frac{1}{v} + \frac{1}{-4} = \frac{1}{-5}$ to find $v = 20\,\text{cm}$, meaning an image is formed $20\,\text{cm}$ behind the silvered surface.

This virtual image serves as the object for the refraction occurring at the flat glass-air interface. Because the mirror surface is $10\,\text{cm}$ below the flat surface, the total distance of this virtual object from the flat surface is $20 + 10 = 30\,\text{cm}$. When observing this object through the refractive interface, the apparent depth is calculated using $d' = \frac{d}{\mu}$. Substituting the real depth of $30\,\text{cm}$ and the refractive index of $1.5$, we get $d' = \frac{30}{1.5} = 20\,\text{cm}$, which locates the final observed image $20\,\text{cm}$ below the flat surface.

The refractive index of a glass slab is calculated using the relationship between its real thickness and apparent thickness, defined by the formula $\mu = \frac{R_2 - R_1}{R_2 - R_3}$. This procedure necessitates recording three distinct positions with the travelling microscope: the reading of a mark on the table surface ($R_1$), the reading of the top surface of the glass slab ($R_2$), and the reading of the mark when viewed through the glass slab ($R_3$). Because these three independent measurements are essential to determine the real depth and the apparent depth of the mark, three is the minimum number of readings required.

Determining the final image location requires applying the thin lens formula sequentially to each optical component, where the image produced by one surface acts as the object for the next. The initial refraction through the convex lens, with an object distance of 60 cm and focal length of 30 cm, follows the equation:
$\frac{1}{v} - \frac{1}{-60} = \frac{1}{30}$
resulting in an image 60 cm beyond the lens.

As this light interacts with the concave lens, the previous image serves as an object positioned 40 cm away, leading to the calculation:
$\frac{1}{v} - \frac{1}{40} = \frac{1}{-120}$
which places the new image 60 cm behind the concave lens. Following the reflection by the plane mirror and a return pass through the concave lens where the object distance is 40 cm, the relation:
$\frac{1}{v} - \frac{1}{-40} = \frac{1}{-120}$
reveals a virtual image 30 cm in front of the concave lens.

Finally, this virtual image functions as an object 50 cm from the convex lens, and applying the lens formula:
$\frac{1}{v} - \frac{1}{-50} = \frac{1}{30}$
locates the ultimate real image at a distance of 75 cm from the convex lens. Since this result is absent from the provided options, the correct answer is not listed.

The magnifying power $M$ of a telescope is defined as the ratio of the angle subtended by the image at the eye ($\beta$) to the angle subtended by the object at the unaided eye ($\alpha$), represented by $M = \frac{\beta}{\alpha}$. Since angular magnification corresponds to reducing the perceived distance by a factor of $M$, a telescope with a magnifying power of 20 increases the visual angle of the object by a factor of 20. This makes the tree appear as if it is located at a distance reduced by 20, causing it to appear 20 times nearer to the observer.

The focal length of a convex lens is determined by the relationship between the object distance and the image distance. The object distance $u$ is the separation between the object pin at 22.2 cm and the convex lens at 32.2 cm, which equals 10 cm. The image distance $v$ is the distance from the lens at 32.2 cm to the image pin position at 71.2 cm, measuring 39 cm. Applying the standard lens formula $\frac{1}{f_1} = \frac{1}{v} - \frac{1}{u}$ with the proper sign convention where $u$ is negative and $v$ is positive, the focal length $f_1$ calculates to approximately 7.8 cm.

The convex mirror functions by reflecting light rays back along their path when they are directed toward its center of curvature, which occurs when the rays strike the mirror normally. The mirror is positioned at 45.8 cm, and the image point formed by the lens, which acts as the reference for the mirror's center of curvature, is at 71.2 cm. The distance between these two points represents the radius of curvature $R$ of the mirror, giving $R = 71.2 - 45.8 = 25.4$ cm. Given that the focal length $f_2$ of a spherical mirror is half its radius of curvature, we find $f_2 = 12.7$ cm.

The behavior of a ray passing through a prism is governed by the relation $\delta = i + e - A$, where $i$ and $e$ represent the angles of incidence and emergence, and $A$ denotes the prism angle. By inserting the given values of $35^\circ$, $79^\circ$, and $40^\circ$ into this equation, the angle of the prism is calculated to be $74^\circ$.

The refractive index $\mu$ is defined by the expression $\mu = \frac{\sin(\frac{A+\delta_m}{2})}{\sin(\frac{A}{2})}$, where $\delta_m$ is the angle of minimum deviation. Substituting $A = 74^\circ$, this becomes $\mu = \frac{\sin(37^\circ + \frac{\delta_m}{2})}{\sin(37^\circ)}$. Since the prism produces a deviation of $40^\circ$ for a specific incidence angle, the actual minimum deviation $\delta_m$ must be less than or equal to this observed value of $40^\circ$.

Using the approximation $\frac{1}{\sin(37^\circ)} \approx \frac{5}{3}$, the refractive index is bounded by $\frac{5}{3} \sin(37^\circ + \frac{\delta_m}{2})$. With $\delta_m$ being less than $40^\circ$, the value of $\mu$ is limited to being less than $\frac{5}{3} \sin(57^\circ)$. Because $\sin(57^\circ)$ is slightly less than $\sin(60^\circ) \approx 0.866$, the refractive index is constrained to be less than $\frac{5}{3} \times 0.866$, which is approximately $1.446$. Given these constraints, the most reasonable value among the provided options is 1.5.

For a ray of light entering a prism to successfully emerge through the second face, the angle of incidence at that internal surface must be strictly less than the critical angle defined by the refractive index. In a prism of angle $A$, the angle of refraction at the first surface $r$ and the angle of incidence at the second surface $r'$ are related by the geometry $r + r' = A$. Since the critical angle $C$ is determined by the relation $\sin C = 1/\mu$, the condition for transmission $r' < C$ can be expressed as $A - r < \sin^{-1}(1/\mu)$, which rearranges to $r > A - \sin^{-1}(1/\mu)$. By applying Snell's Law at the first interface, where $\sin \theta = \mu \sin r$, we substitute $\sin r = \frac{\sin \theta}{\mu}$ into the inequality to obtain $\frac{\sin \theta}{\mu} > \sin(A - \sin^{-1}(1/\mu))$. Isolating the incident angle $\theta$ leads to the required condition $\theta > \sin^{-1}[\mu \sin(A - \sin^{-1}(1/\mu))]$.

The behavior of a thin convex lens when placed in different media is governed by the Lens Maker's Formula, which demonstrates that the focal length is determined by the relative refractive index of the lens material $\mu$ compared to the surrounding medium $\mu_m$. By comparing the focal length in a medium $f_m$ to the focal length in air $f$, we use the relationship:

$\frac{f_m}{f} = \frac{\mu - 1}{\frac{\mu}{\mu_m} - 1}$

For the first liquid with a refractive index of $4/3$, the ratio becomes:

$\frac{f_1}{f} = \frac{\frac{3}{2} - 1}{\frac{3/2}{4/3} - 1} = \frac{\frac{1}{2}}{\frac{9}{8} - 1} = \frac{\frac{1}{2}}{\frac{1}{8}} = 4$

This result indicates that $f_1 = 4f$, meaning the focal length increases significantly while remaining positive. When the lens is submerged in the second liquid with a refractive index of $5/3$, the calculation changes:

$\frac{f_2}{f} = \frac{\frac{3}{2} - 1}{\frac{3/2}{5/3} - 1} = \frac{\frac{1}{2}}{\frac{9}{10} - 1} = \frac{\frac{1}{2}}{-\frac{1}{10}} = -5$

Because the refractive index of this second medium is greater than the refractive index of the glass, the denominator in the Lens Maker's Formula becomes negative. This causes the focal length $f_2$ to become negative, meaning the lens effectively behaves as a concave lens in that medium.

The refractive index $\mu$ of a material is not constant but increases as the frequency of the incident light rises. Because the critical angle $\theta_c$ is defined by $\sin(\theta_c) = \frac{1}{\mu}$, a higher refractive index directly results in a smaller critical angle. Since green light strikes the interface exactly at its specific critical angle, any spectral component with a higher frequency experiences a larger refractive index and thus a smaller critical angle, causing that light to undergo total internal reflection. Consequently, only the portions of the spectrum with a frequency lower than that of green light possess a critical angle larger than the angle of incidence, allowing them to successfully refract into the air.