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

The focal length of a spherical mirror is defined solely by its radius of curvature, making it entirely independent of the refractive index of the surrounding medium. Unlike lenses that rely on refraction to bend light, a mirror functions through the reflection of light, a process governed by the geometry of the surface rather than the optical properties of the environment. Therefore, whether the concave mirror is in air or submerged in a liquid with a refractive index of $\mu$, its ability to converge light remains unchanged, resulting in a focal length of $f$.

For a concave mirror, the magnification is defined as $m = -\frac{v}{u}$, where $v$ is the image distance and $u$ is the object distance. Given the magnification of $-3$, we can establish that $v = 3u$. Since the image is inverted, it is a real image formed on the same side as the object, meaning the distance between them is $|v - u| = 20\text{ cm}$. Substituting $v = 3u$ into this equation gives $|3u - u| = 20\text{ cm}$, which simplifies to $|2u| = 20\text{ cm}$, indicating that $|u| = 10\text{ cm}$ and $|v| = 30\text{ cm}$. Using the mirror formula in the form $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ with $u = -10\text{ cm}$ and $v = -30\text{ cm}$, we calculate the focal length:
$\frac{1}{f} = \frac{1}{-30} + \frac{1}{-10} = \frac{-1 - 3}{30} = -\frac{4}{30} = -\frac{1}{7.5}$
This calculation yields a focal length of $f = -7.5\text{ cm}$. Finally, because the radius of curvature $R$ relates to the focal length by $R = 2|f|$, the magnitude of the radius of curvature is $15\text{ cm}$.

The behavior of light at a spherical refracting surface is governed by the relation $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$, where $\mu_1$ is the refractive index of the surrounding air, $\mu_2$ is the refractive index of the glass, $u$ is the object distance, and $v$ is the image distance. Because both surfaces are concave with a radius of curvature $R$, we substitute the radius as $-R$ and set $\mu_1 = 1$ and $\mu_2 = 1.5$ in the formula. For surface $B$, with the object positioned at $u = -R/2$, the refraction equation becomes $\frac{1.5}{V_B} + \frac{2}{R} = -\frac{0.5}{R}$, which simplifies to $V_B = -0.6R$.

For surface $A$, applying the same refraction principle with the corresponding object geometry leads to the equation $\frac{1.5}{V_A} + \frac{2}{3R} = -\frac{0.5}{R}$. Solving this for $V_A$ gives $V_A = -\frac{9}{7}R$. Both resulting images are virtual and lie to the left of their respective vertices. To find the separation between these two images, we subtract the sum of their distances from the total distance separating the vertices, yielding a final value of $2R - (\frac{9}{7}R + 0.6R) = 0.114R$.

The focal length $f$ of a convex mirror is half its radius of curvature $R$. Given $R = 2\,\text{m}$, the focal length is $f = 1\,\text{m}$. We describe the object's position with the coordinate $u$ and the image's position with the coordinate $v$, connected by the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. With the object at a distance of $24\,\text{m}$ \in front of the mirror, we set $u = -24\,\text{m}$ and solve for the image distance: $\frac{1}{v} + \frac{1}{-24} = \frac{1}{1} \implies \frac{1}{v} = 1 + \frac{1}{24} = \frac{25}{24}$, resulting \in $v = \frac{24}{25}\,\text{m}$.

The car approaches the mirror at a uniform speed of $90\,\text{km/hr}$, which converts to a speed of $25\,\text{m/s}$. We differentiate the mirror equation with respect to time $t$ to relate the velocities: $-\frac{1}{v^2}\frac{dv}{dt} - \frac{1}{u^2}\frac{du}{dt} = 0$. Using the speed of the object $\frac{du}{dt} = 25\,\text{m/s}$ (as the distance $u$ decreases), we find the velocity of the image $\frac{dv}{dt}$: $\frac{dv}{dt} = -\frac{v^2}{u^2}\frac{du}{dt} = -\left(\frac{24/25}{-24}\right)^2(25) = -\frac{1}{625} \cdot 25 = -\frac{1}{25}\,\text{m/s}$.

To determine the acceleration, we differentiate the expression for the image velocity with respect to time, keeping in mind that the object's acceleration $\frac{d^2u}{dt^2}$ is zero because the speed is uniform: $\frac{d^2v}{dt^2} = -\frac{d}{dt}\left(\frac{v^2}{u^2}\right)\frac{du}{dt} = -\left( \frac{2v\frac{dv}{dt}u^2 - 2u\frac{du}{dt}v^2}{u^4} \right) \frac{du}{dt}$. Substituting our known values $v = \frac{24}{25}\,\text{m}$, $u = -24\,\text{m}$, $\frac{dv}{dt} = -\frac{1}{25}\,\text{m/s}$, and $\frac{du}{dt} = 25\,\text{m/s}$ yields $\frac{d^2v}{dt^2} = -2\left( \frac{v\frac{dv}{dt}u - v^2\frac{du}{dt}}{u^3} \right) \frac{du}{dt} = -2\left( \frac{(\frac{24}{25})(-\frac{1}{25})(-24) - (\frac{24}{25})^2(25)}{(-24)^3} \right) (25) = -\frac{2}{25}\,\text{m/s}^2$. The magnitude of the acceleration is $a = 0.08\,\text{m/s}^2$, and calculating $100a$ gives $100 × 0.08 = 8$.

When a thin convex lens made of a material with refractive index $\mu_2$ is submerged in a liquid with refractive index $\mu_1$ and its rear surface is silvered, the device functions as a compound system equivalent to a concave mirror. A real and inverted image coincident with the object is produced only when the object is placed at the center of curvature of this equivalent mirror, which corresponds to a distance of twice the effective focal length from the system. The total optical power of this combination is determined by the sum of the power from two refractions through the lens and one reflection from the back silvered surface, expressed as $P_{eq} = 2 P_l + P_m$. Using the power contribution from the lens in the medium, $P_l = (\frac{\mu_2 - \mu_1}{\mu_1}) (\frac{1}{R_1} + \frac{1}{R_2})$, and adding the power of the spherical mirror surface $P_m = \frac{2}{R_2}$, the total equivalent power becomes $P_{eq} = 2 (\frac{\mu_2 - \mu_1}{\mu_1}) (\frac{1}{R_1} + \frac{1}{R_2}) + \frac{2}{R_2}$. By calculating the reciprocal of this equivalent power to find the focal length and then doubling that value to determine the placement, the required object distance is found to be $\frac{\mu_1 |R_1| \cdot |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_2|}$.

The focal length of a lens is fundamentally tied to the refractive index of the surrounding medium, as described by the generalized lens maker's formula, $\frac{n_m}{f} = (n_l - n_m) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$, where $n_m$ is the refractive index of the medium and $n_l$ is the index of the lens. When the lens is in air, with $n_m = 1$ and $f = 12$ cm, the relationship simplifies to $\frac{1}{12} = (1.6 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$, which allows us to determine that the geometric factor $\left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{12 \times 0.6} = \frac{1}{7.2}$ cm$^{-1}$.

When the lens is submerged in water, where $n_m = 1.28$, the expression modifies to $\frac{1.28}{f_w} = (1.6 - 1.28) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. By substituting the previously calculated geometric factor into this equation, we get $\frac{1.28}{f_w} = 0.32 \times \frac{1}{7.2}$. Solving for the new focal length gives $f_w = \frac{1.28 \times 7.2}{0.32} = 28.8$ cm, which converts to a final value of 288 mm.

To be visible at the rim of the vessel, the light rays from the coin must graze the surface, which occurs only if the angle of incidence at the liquid-air interface equals the critical angle $\theta_c$. In a hemispherical vessel of radius $R$, the path from the bottom point $O$ to the rim point $E$ forms a triangle with the center of the top face $C$, where the segments $OC$ and $CE$ are both equal to $R$. This geometry establishes an isosceles right triangle, meaning the angle of incidence between the ray $OE$ and the surface normal $CE$ is $45^{\circ}$. Substituting $\theta_c = 45^{\circ}$ into the refractive index formula $\mu = \frac{1}{\sin \theta_c}$ yields the minimum refractive index of $\sqrt{2}$.

To determine the refractive index of the lens, we apply the Lens Maker's Formula, which relates the focal length, the radii of curvature, and the refractive index of the material. For a thin biconvex lens, this relationship is defined by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$.

Substituting the known focal length of 12 cm and the radii of curvature 10 cm and 15 cm into the expression yields $\frac{1}{12} = (\mu - 1) \left( \frac{1}{10} + \frac{1}{15} \right)$. Combining the terms within the parentheses simplifies the equation to $\frac{1}{12} = (\mu - 1) \left( \frac{1}{6} \right)$. Isolating the refractive index term shows that $\mu - 1 = \frac{6}{12}$, which simplifies to $\mu - 1 = 0.5$, providing a final refractive index of 1.5.

When light passes through a prism at the angle of minimum deviation, the internal ray becomes parallel to the base, and the symmetry of the path ensures that the angle of incidence equals the angle of emergence. This geometric arrangement confirms that the internal refraction angle is strictly half the prism angle, expressed as
$r = \frac{A}{2}$
which directly invalidates the claim that the refraction angle becomes double the prism angle. Because the minimum deviation is defined by
$\delta = (\mu - 1)A$
a larger prism angle necessitates a larger deviation rather than a smaller one, while the general relationship
$\delta = i + e - A$
illustrates that any deviation greater than the minimum value can be produced by two distinct sets of incidence and emergence angles, as these values can be interchanged without affecting the total deflection.

Magnification in mirrors is defined by the ratio of image distance to object distance, expressed as $|v/u| = 1/4$, which implies that the object is placed at a distance $u = 4v$. Given that the separation between the object and the image is 40 cm, we can represent the total distance as $u + v = 40$; substituting the magnification relationship into this sum confirms that the object is located at 32 cm and the image at 8 cm. Applying these positions to the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ leads to a calculated focal length of 10.66 cm, which rounds to 10.7 cm.

Every point on a slanted object interacts with a convex lens independently according to the thin lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, where the resulting image position and transverse magnification depend on the specific object distance. By identifying the image coordinates for the endpoints of the slanted object, we can determine the angle formed by the image segment relative to the principal axis.

For point A located at an object distance of -30 cm with a focal length of 20 cm, the image distance is found via $\frac{1}{v_A} - \frac{1}{-30} = \frac{1}{20}$, yielding 60 cm. For point B at -29 cm, the image distance is calculated using $\frac{1}{v_B} - \frac{1}{-29} = \frac{1}{20}$, resulting in approximately 64.4 cm. Calculating the magnification $m = \frac{v}{u}$ for each point gives $m_A = \frac{60}{-30} = -2$ and $m_B = \frac{64.4}{-29} \approx -2.22$. With the vertical position of point B mapped to $y_B = -2.22 \times 2 = -4.44$ cm, the orientation of the image relative to the principal axis is determined by the ratio of the vertical change to the horizontal change. Specifically, the change in position is -4.44 cm vertically and 4.4 cm horizontally, resulting in a tangent value of $\frac{-4.44}{4.4} \approx -1$. This corresponds to an angle of -45 degrees.

The total focal length of a lens system formed by contact lenses is determined by the equation $\frac{1}{F} = \frac{1}{F_{lens}} + \frac{1}{F_{liquid}}$, where $F_{lens}$ is the focal length of the glass lens and $F_{liquid}$ is the focal length of the liquid layer.

The focal length of the glass lens is calculated using the lens maker's formula $\frac{1}{F_{lens}} = (n_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. With a refractive index of 1.5 and surface radii $R_1 = -30$ cm and $R_2 = -20$ cm, this becomes $\frac{1}{F_{lens}} = (1.5 - 1) \left( \frac{-1}{30} - \frac{1}{-20} \right) = 0.5 \left( \frac{-2 + 3}{60} \right) = \frac{1}{120}$ cm$^{-1}$. The liquid layer acts as a plano-concave lens with one curved surface radius of 30 cm and one plane surface radius of infinity, giving its focal length power as $\frac{1}{F_{liquid}} = (n_l - 1) \left( \frac{1}{30} \right)$. Using the refractive index 1.3, this yields $\frac{1}{F_{liquid}} = (1.3 - 1) \left( \frac{1}{30} \right) = \frac{0.3}{30} = \frac{1}{100}$ cm$^{-1}$. Combining these results, the total power is $\frac{1}{F} = \frac{1}{120} + \frac{1}{100} = \frac{5 + 6}{600} = \frac{11}{600}$ cm$^{-1}$, resulting in a total focal length $F = \frac{600}{11}$ cm.

The relationship between object distance $u$, image distance $v$, and focal length $f$ for a spherical mirror is given by the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, while the magnification is defined as $m = -\frac{v}{u}$. Given the initial object distance $u = -40$ cm and magnification $m = 1/2$, the image distance is $v = -m u = -(1/2)(-40) = 20$ cm. Substituting these values into the mirror formula gives $\frac{1}{20} + \frac{1}{-40} = \frac{1}{f}$, which yields $f = 40$ cm. For the target magnification $m = 1/3$, the new image distance becomes $v = -u/3$. Substituting this into the mirror formula results in $\frac{1}{-u/3} + \frac{1}{u} = \frac{1}{40}$, which simplifies to $-\frac{3}{u} + \frac{1}{u} = \frac{1}{40}$ and leads to $-\frac{2}{u} = \frac{1}{40}$. Solving for $u$ gives $u = -80$ cm, indicating that the object must be moved from its initial position of 40 cm to 80 cm, which is 40 cm away from the mirror.

When combining two thin prisms to achieve the condition of dispersion without deviation, the net deviation produced by the system must sum to zero. This implies that the deviation caused by the first prism must be perfectly balanced by the deviation of the second, which is mathematically described by the relationship $(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$, where $\mu$ and $A$ represent the refractive index and the prism angle, respectively.

For a first prism with an angle of $4^{\circ}$ and a refractive index of 1.54, combined with a second prism possessing a refractive index of 1.72, we substitute these parameters into the equation to determine the required angle $A_2$. This gives $(1.54 - 1) \times 4^{\circ} = (1.72 - 1) \times A_2$, which simplifies to $0.54 \times 4^{\circ} = 0.72 \times A_2$. Isolating the unknown variable results in $A_2 = \frac{0.54}{0.72} \times 4^{\circ}$, which calculates to $3^{\circ}$.

When light travels from an outside medium with refractive index $n_1$ into a block with refractive index $n_2$, its path is governed by Snell's Law, which relates the angles of incidence and refraction to the indices of the media. Since the angle of incidence relative to the surface normal is the complement of the given angle $\theta_1$, the relationship is expressed as $n_1 \sin(90^\circ - \theta_1) = n_2 \sin r$, where $r$ is the angle of refraction. Substituting the provided value $\cos \theta_1 = \frac{n_2}{2n_1}$ into this expression yields $n_1 \cos \theta_1 = n_2 \sin r$, which simplifies to $n_1 \left( \frac{n_2}{2n_1} \right) = n_2 \sin r$. This results in $\sin r = \frac{1}{2}$, confirming that $r = 30^\circ$.

The path of the two beams inside the block creates an isosceles triangle with a base equal to $d$ and a height $t$, which is the thickness of the block. By bisecting this triangle, we form a right-angled triangle where the angle of refraction $r$ is opposite the side of length $d/2$, and the adjacent side is the thickness $t$. The geometric relationship is defined by $\tan r = \frac{d/2}{t}$. Given $d = 4\sqrt{3}$ cm and $r = 30^\circ$, we substitute these values into the ratio $\frac{1}{\sqrt{3}} = \frac{2\sqrt{3}}{t}$. Solving for $t$ by multiplying both sides by $t\sqrt{3}$ gives $t = 2\sqrt{3} \times \sqrt{3}$, resulting in a thickness of 6 cm.

The mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ determines the image position for the convex mirror. Substituting the object distance $u = -30$ cm and focal length $f = +30$ cm gives $\frac{1}{v} + \frac{1}{-30} = \frac{1}{30}$, which simplifies to $\frac{1}{v} = \frac{2}{30}$, resulting in $v = 15$ cm. This virtual image forms 15 cm behind the convex mirror, placing it at a total distance of $30 + 15 = 45$ cm from the object. To form an image at the same location, the plane mirror must be placed such that its image distance $2x$ from the object is 45 cm, where $x$ is the distance from the object to the plane mirror. Solving $2x = 45$ cm gives $x = 22.5$ cm. The distance between the convex mirror, located 30 cm from the object, and the plane mirror, located 22.5 cm from the object, is $30 - 22.5 = 7.5$ cm.

The focal length of a lens is determined by the Lens Maker's formula, which depends on the refractive index of the lens material relative to the surrounding medium. When the refractive index of the surrounding medium differs from that of air, the focal power of the lens changes significantly. We can compare the focal length in the liquid, denoted as $f_m$, to the focal length in air, $f_a$, using the ratio derived from the Lens Maker's formula:

$\frac{f_m}{f_a} = \frac{(\mu_l - 1)\mu_m}{\mu_l - \mu_m}$

Here, $\mu_l$ is the refractive index of the lens and $\mu_m$ is the refractive index of the surrounding liquid. Substituting the given values, where $\mu_l = 1.5$, $\mu_m = 1.6$, and $f_a = 20$ cm, leads to the equation:

$\frac{f_m}{20} = \frac{(1.5 - 1) \times 1.6}{1.5 - 1.6}$

Calculating the terms on the right side involves finding the product of 0.5 and 1.6, which is 0.8, and dividing it by the difference of -0.1. This results in a value of -8. Multiplying this by 20 cm yields a final focal length of -160 cm, indicating that the lens changes its nature from converging to diverging when immersed in this particular medium.

A magnified virtual image formed by a curved mirror confirms the mirror is concave, as only concave mirrors can produce such images. The magnification formula $m = -\frac{v}{u}$ allows us to relate the image distance and object distance as $v = -2u$ when $m = 2$.

Since the image is virtual and forms behind the mirror, the total distance between the object and the image, $v - u$, is 15 cm. Substituting $v = -2u$ into this relationship gives $-2u - u = 15$, or $-3u = 15$, which solves to $u = -5$ cm. Consequently, the image distance $v$ is 10 cm.

Inserting these distances into the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ results in $\frac{1}{f} = \frac{1}{10} + \frac{1}{-5}$. This simplifies to $\frac{1}{f} = -\frac{1}{10}$, indicating that the focal length is -10 cm.

When a beam of light travels through a system of two convex lenses so that incident parallel rays also exit parallel, the focal points of the two lenses must coincide. Rays entering the first lens with a focal length of 10 cm converge at its focal point, which acts as the focal point for the second lens with a focal length of 15 cm. The separation distance between the lenses is $D = f_1 + f_2$. By adding the two focal lengths, the total distance between the lenses is 25 cm.

When a light ray passes through a prism at the angle of minimum deviation, the angle of incidence equals the angle of emergence, and the angle of refraction inside the prism is exactly half the prism angle, denoted as $r = \frac{A}{2}$. Since the ratio of the angle of minimum deviation to the prism angle is one, we know that $\delta_{\min} = A$. Substituting this relationship into the standard deviation formula $\delta_{\min} = 2i - A$ results in $A = 2i - A$, which simplifies to $i = A$.

By applying Snell's law, $\sin i = \mu \sin r$, and substituting the known values $\mu = \sqrt{3}$, $i = A$, and $r = \frac{A}{2}$, we obtain the equation $\sin A = \sqrt{3} \sin \left( \frac{A}{2} \right)$. Expanding $\sin A$ using the trigonometric identity $2 \sin \left( \frac{A}{2} \right) \cos \left( \frac{A}{2} \right)$ allows the equation to become $2 \sin \left( \frac{A}{2} \right) \cos \left( \frac{A}{2} \right) = \sqrt{3} \sin \left( \frac{A}{2} \right)$. Canceling the $\sin \left( \frac{A}{2} \right)$ terms from both sides leaves $\cos \left( \frac{A}{2} \right) = \frac{\sqrt{3}}{2}$, indicating that $\frac{A}{2} = 30^{\circ}$ and establishing the angle of the prism as $60^{\circ}$.