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

The focal length of a lens is defined by the Lens Maker's formula, $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. For a symmetric biconvex objective lens with both surfaces having radius $R$, the initial focal length $F_0$ is given by $\frac{1}{F_0} = (n-1) \left( \frac{2}{R} \right)$. Cutting this lens vertically results in a plano-convex lens where one surface is flat and the other retains the curvature, resulting in a new focal length $F_0'$ calculated as $\frac{1}{F_0'} = (n-1) \left( \frac{1}{R} \right)$, which simplifies to $F_0' = 2F_0$.

To maintain the same total magnification $M = \frac{L}{F_0} \cdot \frac{D}{F_e}$ without changing the object distance or the eyepiece properties, the ratio $\frac{L}{F_0}$ must remain constant. Setting the initial state equal to the final state, $\frac{L}{F_0} = \frac{L'}{F_0'}$, and substituting the known relationship $F_0' = 2F_0$, we arrive at $\frac{L}{F_0} = \frac{L'}{2F_0}$. Solving this equation shows that the new tube length $L'$ must equal $2L$, meaning the tube length has to be increased two times.

The refraction at a spherical surface is governed by the equation $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$, where $n_1$ and $n_2$ are the refractive indices of the first and second media, $u$ is the object distance, $v$ is the image distance, and $R$ is the radius of curvature. For a parallel incident beam, the object distance $u = \infty$, so the equation simplifies to $\frac{n_2}{v} = \frac{n_2 - n_1}{R}$. Substituting the given values $n_1 = 1.0$, $n_2 = 1.5$, and $R = 50$ cm results in $\frac{1.5}{v} = \frac{1.5 - 1.0}{50}$, which simplifies to $\frac{1.5}{v} = \frac{0.5}{50}$. Solving for the image distance gives $v = \frac{1.5 \times 50}{0.5} = 150$ cm from the pole of the surface. Since the image forms 150 cm from the pole and the center of curvature is 50 cm from the pole, the distance $x$ from the center of curvature is $x = v - R = 150 - 50 = 100$ cm.

The magnification $M$ of a two-lens beam expander is determined by the ratio of the final beam diameter $D_2$ to the initial beam diameter $D_1$, which is equal to the ratio of the focal lengths of the second lens $f_2$ to the first lens $f_1$ as expressed by $M = \frac{D_2}{D_1} = \frac{f_2}{f_1}$. Substituting the given diameters into the magnification ratio yields $M = \frac{14}{2} = 7$. Utilizing the focal length of the first lens $f_1 = 40$, the focal length of the second lens is obtained from the relation $f_2 = M \times f_1 = 7 \times 40 = 280$ mm.

When two thin lenses are placed in physical contact along the same axis, their combined power is the algebraic sum of their individual powers, leading to the equivalent focal length relation $\frac{1}{f} = \frac{1}{f_{\text{convex}}} - \frac{1}{f_{\text{concave}}}$. The system functions as a concave lens when the effective focal length is negative, which requires the subtractive term to outweigh the positive power of the convex lens. Consequently, the combination behaves as a concave lens whenever the magnitude of the convex lens's focal length exceeds that of the concave lens, expressed as $|f_{\text{convex}}| > |f_{\text{concave}}|$.

The mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance. Since the image is real and magnified 3 \times , the magnification is $m = -\frac{v}{u} = -3$, which implies the relationship $v = 3u$. The distance between the object and the real image is given by $|v - u| = 40$ cm, so substituting $v = 3u$ results \in $|3u - u| = |2u| = 40$ cm, meaning $u = -20$ cm and consequently $v = -60$ cm. Substituting these values into the mirror equation provides $\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = -\frac{4}{60} = -\frac{1}{15}$, which determines the focal length to be $-15$ cm.

The refraction of light through a prism follows the geometric relationship $A = r_1 + r_2$, where $A$ is the prism angle and $r_1$ and $r_2$ are the angles of refraction at the first and second surfaces, respectively. When the emergent ray grazes the second surface, the angle of emergence is $e = 90^{\circ}$, and Snell's law at the second interface $\mu \sin(r_2) = \sin(90^{\circ})$ simplifies to $\sin(r_2) = \frac{1}{\mu}$.

For an equilateral prism, $A = 60^{\circ}$, and using the refractive index $\mu = \sqrt{2}$, the refraction angle at the second surface is determined by $\sin(r_2) = \frac{1}{\sqrt{2}}$, which yields $r_2 = 45^{\circ}$. Substituting these values into the prism formula $60^{\circ} = r_1 + 45^{\circ}$ results in the angle of refraction at the incident surface $r_1 = 15^{\circ}$.

The effective focal length $f$ of the biconvex lens is calculated using the Lens Maker's Formula $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$, where for the given refractive index $\mu = 1.5$ and radius of curvature $R = 20 \text{ cm}$, the surface values are $R_1 = 20 \text{ cm}$ and $R_2 = -20 \text{ cm}$, resulting in $f = 20 \text{ cm}$. Applying the magnification formula $m = \frac{f}{f+u}$ for an object placed at $u = -30 \text{ cm}$ leads to $m = \frac{20}{20 - 30}$, which simplifies to $m = -2$.

The parallax method requires the formation of a real, inverted image that can be brought into coincidence with a search pin, a condition that cannot be met with the virtual, erect images produced when an object is placed between the pole and the focus. A concave mirror forms a real image only when the object is positioned at a distance from the mirror greater than its focal length ($u > f$), which allows for the parallax elimination technique to be performed successfully at any point beyond the focus.

When sunlight converges to a single point to ignite paper, the distance between the lens and the paper is equivalent to the focal length of the lens. Parallel rays originating from the sun are brought together at this focal point, which results in the concentration of heat necessary to start a fire.

For a biconvex lens with a focal length of 30 cm and radii of curvature of 60 cm, the lens maker's formula provides the relationship between the physical geometry of the lens and its refractive index. By adopting the standard sign convention where the first surface has a positive radius of 60 cm and the second surface has a negative radius of -60 cm, the governing equation is:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Substituting the specified values into this formula gives:
$\frac{1}{30} = (\mu - 1) \left( \frac{1}{60} - \frac{1}{-60} \right)$
Simplifying the expression within the parentheses results in:
$\frac{1}{30} = (\mu - 1) \left( \frac{1}{60} + \frac{1}{60} \right) = (\mu - 1) \left( \frac{2}{60} \right) = \frac{\mu - 1}{30}$
Equating both sides shows that $\mu - 1 = 1$, which means the refractive index is $\mu = 2$. Since the problem defines the refractive index as $\frac{\alpha}{10}$, setting $\frac{\alpha}{10} = 2$ leads to $\alpha = 20$.

The Lens Maker's formula states that the power $P$ of a lens is given by $P = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$, where $\mu$ is the refractive index and $R_1, R_2$ are the radii of curvature of the surfaces. For a biconvex lens with front radius $r_1$ and back radius $-r_2$, the power is $P_1 = (1.5 - 1)(\frac{1}{r_1} - \frac{1}{-r_2}) = 0.5(\frac{1}{r_1} + \frac{1}{r_2})$. For a plano-concave lens with refractive index $1.7$ and one surface radius $r_2$, the magnitude of power is $|P_2| = |(1.7 - 1)(\frac{1}{\infty } - \frac{1}{r_2})| = \frac{0.7}{r_2}$. Equating the magnitudes $|P_1| = |P_2|$ yields $0.5(\frac{1}{r_1} + \frac{1}{r_2}) = \frac{0.7}{r_2}$. Multiplying by 2 results in $\frac{1}{r_1} + \frac{1}{r_2} = \frac{1.4}{r_2}$, which simplifies to $\frac{1}{r_1} = \frac{0.4}{r_2}$. Therefore, the ratio of the radii of curvature is $\frac{r_1}{r_2} = \frac{1}{0.4} = \frac{5}{2}$.

The focal length of a lens is determined by the Lens Maker's formula, $\frac{1}{f} = \left(\frac{n_l}{n_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, where $n_l$ is the refractive index of the lens, $n_m$ is the refractive index of the surrounding medium, and $R_1, R_2$ are the radii of curvature. For the lens in air, where the medium is air with refractive index $1$, the expression $\frac{1}{18} = (1.5 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ simplifies to the geometric factor $\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{9}$.

When the lens is immersed in water with refractive index $\frac{4}{3}$, the focal length $f_w$ is given by $\frac{1}{f_w} = \left(\frac{1.5}{4/3} - 1\right) \times \frac{1}{9}$. Calculating this yields $\frac{1}{f_w} = \left(\frac{9}{8} - 1\right) \times \frac{1}{9} = \frac{1}{8} \times \frac{1}{9} = \frac{1}{72}$, which implies $f_w = 72 \text{ cm}$. The difference in focal lengths is $72 - 18 = 54 \text{ cm}$, and setting this equal to $\alpha \times 18$ gives $\alpha = \frac{54}{18} = 3$.

The condition for a combination of two thin prisms to produce dispersion without deviation is that their net angular deviation must be zero, which is expressed by the equation $(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$, where $\mu_1$ and $\mu_2$ are the refractive indices and $A_1$ and $A_2$ are the refracting angles of the respective prisms. Substituting the given values of $A_1 = 5^\circ$, $\mu_1 = 1.72$, and $\mu_2 = 1.9$ into this equation gives $(1.72 - 1) \times 5^\circ = (1.9 - 1) \times A_2$. This simplifies to $0.72 \times 5^\circ = 0.9 \times A_2$, which results in $3.6^\circ = 0.9 \times A_2$, and solving for $A_2$ yields $A_2 = 4^\circ$.

The refractive index of a prism is fundamentally related to the minimum angle of deviation and the apex angle of the prism. We first calculate the refractive index of the material by finding the ratio between the speed of light in vacuum, 3 × 10^8 m/s, and the velocity within the prism, 2.12 × 10^8 m/s, which gives $\mu = \sqrt{2}$. Given that the prism is equilateral, the apex angle is 60 degrees, so we apply the standard prism formula $\mu = \frac{\sin\left(\frac{\delta_{min} + A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Substituting the known values leads to $\sqrt{2} = \frac{\sin\left(\frac{\delta_{min} + 60^\circ}{2}\right)}{\sin 30^\circ}$. Since the sine of 30 degrees is 0.5, we simplify the expression to $\sin\left(\frac{\delta_{min} + 60^\circ}{2}\right) = \frac{1}{\sqrt{2}}$. Because the sine of 45 degrees equals $1/\sqrt{2}$, we set the argument to $\frac{\delta_{min} + 60^\circ}{2} = 45^\circ$, which results in a minimum angle of deviation of 30 degrees.

The refractive index $n$ of a prism is related to its refracting angle $A$ and the angle of minimum deviation $\delta_m$ by the formula $n = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$. Substituting the condition $\delta_m = A$ into this equation yields $n = \frac{\sin((A + A)/2)}{\sin(A/2)}$, which simplifies to $n = \frac{\sin(A)}{\sin(A/2)}$.

Applying the double-angle trigonometric identity $\sin(A) = 2\sin(A/2)\cos(A/2)$, the expression becomes $n = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)}$, reducing to $n = 2\cos(A/2)$. Since the refracting angle $A$ of a prism must satisfy $0 < A < \pi$, the term $\cos(A/2)$ lies strictly between $0$ and $1$. Consequently, the refractive index must be less than $2$. Combined with the physical requirement that $n > 1$ for a light-transmitting medium, the refractive index is bounded by the inequality $1 < n < 2$.

To find the magnification $m_1$ of the lens combination without any gap, we first determine the equivalent focal length $F$ of the two lenses in contact using the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. Given $f_1 = +5$ cm (convex) and $f_2 = -4$ cm (concave), we have $\frac{1}{F} = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20}$, so $F = -20$ cm. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$ with an object distance $u = -10$ cm, we get $\frac{1}{v} = -\frac{1}{20} + \frac{1}{-10} = -\frac{3}{20}$, resulting in $v = -\frac{20}{3}$ cm. The magnification is $m_1 = \frac{v}{u} = \frac{-20/3}{-10} = \frac{2}{3}$.

For the system with a gap of 1 cm, we calculate the magnification $m_2$ by finding the magnification of each lens separately. For the first (convex) lens, the image position is $\frac{1}{v_1} - \frac{1}{-10} = \frac{1}{5}$, which gives $v_1 = 10$ cm. This image acts as a virtual object for the second (concave) lens at a distance $u_2 = v_1 - d = 10 - 1 = 9$ cm. The magnification of the second lens is $m_2' = \frac{v_2}{u_2}$, where $\frac{1}{v_2} - \frac{1}{9} = \frac{1}{-4}$, giving $\frac{1}{v_2} = -\frac{5}{36}$ and $v_2 = -\frac{36}{5}$ cm. Thus, $m_2' = \frac{-36/5}{9} = -\frac{4}{5}$. The magnification of the convex lens is $m_1' = \frac{v_1}{u_1} = \frac{10}{-10} = -1$. The total magnification is $m_2 = m_1' \times m_2' = (-1) \times (-\frac{4}{5}) = \frac{4}{5}$.

Finally, the ratio of the magnifications is $\left| \frac{m_1}{m_2} \right| = \left| \frac{2/3}{4/5} \right| = \frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}$. Given that the calculated value $\frac{5}{6}$ is not among the options and considering the likelihood of a typographical error in the provided options where $\frac{5}{14}$ is indicated as correct, the standard derivation of the system's magnification results in $\frac{5}{6}$.

The relationship between the refractive index of a prism, the angle of the prism, and the angle of minimum deviation is defined by the formula $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$. When the angle of minimum deviation $\delta_m$ is equal to the angle of the prism $A$, the expression simplifies because the numerator becomes $\sin(\frac{A + A}{2})$, which is $\sin A$.

Applying the trigonometric identity $\sin A = 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})$, the formula for the refractive index becomes $\mu = \frac{2 \sin(\frac{A}{2}) \cos(\frac{A}{2})}{\sin(\frac{A}{2})}$. This reduces to the simplified expression $\mu = 2 \cos(\frac{A}{2})$. Given that the refractive index $\mu$ is $\sqrt{3}$, we set up the equation $2 \cos(\frac{A}{2}) = \sqrt{3}$, which means $\cos(\frac{A}{2}) = \frac{\sqrt{3}}{2}$. Since the angle whose cosine is $\frac{\sqrt{3}}{2}$ is $30^{\circ}$, we have $\frac{A}{2} = 30^{\circ}$, resulting in an angle of the prism $A = 60^{\circ}$.

The power of a thin lens is determined by the lens maker's formula, $P = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$. For a symmetric biconvex lens, the two surfaces have radii $R$ and $-R$, resulting in a total power of $P = (\mu - 1)(\frac{2}{R})$, which equals 4 D.

When the lens is cut into four equal segments, each piece acts as a plano-convex lens featuring one curved surface with radius $R$ and one flat surface. Consequently, the power for one of these parts is given by $P_{part} = (\mu - 1)(\frac{1}{R})$. Since this expression is exactly half of the original total power, the power of a single part is 2 D.

The optical relationship between an object and its image formed by a lens is governed by the standard thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. By rearranging this expression into the form $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$, we can see that when the object position satisfies $|u| > f$, the right-hand side is strictly positive, forcing the image distance $v$ to remain positive. This dependency creates a distinctive hyperbolic curve on the graph that illustrates how the image position responds to changes in the object position beyond the focal length.

Refraction at a spherical surface follows the relation $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$, where light travels from a medium with refractive index $n_1$ into one with index $n_2$. In this scenario, light passes from air, which has a refractive index of 1, into glass, which has a refractive index of 1.5. Because the center of curvature is situated within the glass, the radius $R$ is considered positive. We are given that the distance from the point $P$ to the object $O$ is equal to the distance from $P$ to the image $I$, which can be represented by the variable $x$. Following the standard Cartesian sign convention, the object distance is $u = -x$ because the object is in front of the surface, while the real image distance is $v = x$ because it forms behind the surface.

Substituting these values into the refraction equation yields $\frac{1.5}{x} - \frac{1}{-x} = \frac{1.5 - 1}{R}$. Simplifying the left side of this equation results in $\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R}$, which combines to $\frac{2.5}{x} = \frac{0.5}{R}$. Solving for the distance $x$ gives $x = \frac{2.5}{0.5} R$, which results in a value of $5 R$.

When two thin lenses are placed in contact, they function as a single lens system with an effective focal length determined by summing the reciprocals of the individual focal lengths. Given the convex lens with a focal length of $30 \text{ cm}$ and the concave lens with a focal length of $-20 \text{ cm}$, the equivalent focal length $F$ is calculated via:

$\frac{1}{F} = \frac{1}{30} + \frac{1}{-20} = -\frac{1}{60}$

This shows the combined system acts as a concave lens with a focal length of $-60 \text{ cm}$.

With this combined focal length established, we use the lens formula to determine the image distance $v$ for an object placed at $u = -20 \text{ cm}$. By utilizing the lens equation $\frac{1}{v} = \frac{1}{F} + \frac{1}{u}$, we substitute the known values:

$\frac{1}{v} = -\frac{1}{60} + \frac{1}{-20} = \frac{-1 - 3}{60} = -\frac{4}{60} = -\frac{1}{15}$

This calculation reveals that $v = -15 \text{ cm}$, confirming that the image is formed at a distance of $15 \text{ cm}$ from the lens.