Optics MCQs with Answers – Part 3 (NEET Physics)

Explanation: \(\textbf{Recognise the incident convergence:}\) The approaching rays are already convergent. Without the lenses, they would meet to the right of the optical centre. Their intended convergence point therefore acts as a virtual object. With positive distance measured along incident propagation, \[ u=+25\,\text{cm} \] \(\textbf{Initial contact combination:}\) The first two lens powers are \[ P_1=+4\,\text{D},\qquad P_2=-4\,\text{D} \] Their equivalent power is \[ P_{12}=P_1+P_2 \] \[ P_{12}=+4-4=0\,\text{D} \] The combination has zero net refracting power. The rays continue towards their original convergence point. \(\textbf{Initial image condition:}\) Zero power corresponds to \[ \frac1F=0 \] Using the thin-lens equation, \[ \frac1v-\frac1u=\frac1F \] gives \[ \frac1v=\frac1u \] Therefore, \[ v_{\text{initial}}=+25\,\text{cm} \] The image is real because the rays actually converge on the outgoing side. \(\textbf{Add the third lens:}\) The added power is \[ P_3=+5\,\text{D} \] The new equivalent power becomes \[ P_{123}=4-4+5 \] \[ P_{123}=+5\,\text{D} \] The corresponding focal length is \[ F=\frac1{5}\,\text{m} \] \[ F=+20\,\text{cm} \] \(\textbf{Determine the final image:}\) The virtual-object distance remains \(+25\,\text{cm}\). \[ \frac1v=\frac1F+\frac1u \] \[ \frac1v=\frac1{20}+\frac1{25} \] \[ \frac1v=\frac{5+4}{100} \] \[ \frac1v=\frac9{100} \] \[ v=+\frac{100}{9}\,\text{cm} \] \(\textbf{Determine transverse magnification:}\) For a thin lens, \[ m=\frac vu \] Substituting, \[ m=\frac{100/9}{25} \] \[ m=+\frac49 \] \(\textbf{Physical interpretation:}\) The original zero-power combination does not make an already convergent beam parallel. It leaves the beam's convergence unchanged. Adding positive power makes the rays converge sooner, bringing the real image closer to the combination. The magnification is positive because both the virtual-object distance and the real-image distance are positive. This does not contradict the negative magnification normally associated with a real image of a real object. Here, the image is formed from a virtual object represented by an incident converging bundle. \(\textbf{Final answer:}\) \[ \boxed{v=+\frac{100}{9}\,\text{cm},\quad m=+\frac49} \]

Explanation: \(\textbf{Image without the glass slab:}\) The lens focal length is \[ f=+20\,\text{cm} \] The object distance is \[ u=-60\,\text{cm} \] The thin-lens equation gives \[ \frac1v=\frac1f+\frac1u \] \[ \frac1v=\frac1{20}-\frac1{60} \] \[ v_0=+30\,\text{cm} \] The original magnification is \[ m_0=\frac{30}{-60}=-\frac12 \] \(\textbf{Optical effect of the slab:}\) For near-normal transmission through a plane-parallel slab in air, the effective axial shift is \[ \Delta=t\left(1-\frac1n\right) \] Substituting, \[ \Delta=30\left(1-\frac1{1.5}\right) \] \[ \Delta=30\left(1-\frac23\right) \] \[ \Delta=10\,\text{cm} \] \(\textbf{Arrangement P — slab before the lens:}\) The slab makes the object appear \(10\,\text{cm}\) closer to the lens. Its effective signed object distance is \[ u_P=-(60-10) \] \[ u_P=-50\,\text{cm} \] Applying the lens equation, \[ \frac1{v_P}=\frac1{20}-\frac1{50} \] \[ \frac1{v_P}=\frac{5-2}{100} \] \[ v_P=+\frac{100}{3}\,\text{cm} \] The transverse object height is unchanged by the plane interface in the paraxial construction. Therefore, \[ m_P=\frac{v_P}{u_P} \] \[ m_P=\frac{100/3}{-50} \] \[ m_P=-\frac23 \] \(\textbf{Arrangement Q — slab after the lens:}\) The lens initially produces a converging bundle directed towards \(v_0=+30\,\text{cm}\). The slab reduces the convergence angle while the rays are inside glass. After emergence into air, the rays regain their original angular directions but are laterally displaced. Their meeting point consequently shifts away from the lens by \(\Delta\). \[ v_Q=v_0+\Delta \] \[ v_Q=30+10 \] \[ v_Q=+40\,\text{cm} \] \(\textbf{Determine the height in arrangement Q:}\) A chief ray passing through the optical centre of the lens has its transverse propagation inside the slab reduced by the factor \(1/n\). The effective transverse propagation distance from the lens to the shifted image plane is \[ L_{\text{eff}}=(v_Q-t)+\frac tn \] \[ L_{\text{eff}}=(40-30)+\frac{30}{1.5} \] \[ L_{\text{eff}}=10+20 \] \[ L_{\text{eff}}=30\,\text{cm} \] Therefore, the transverse magnification remains that of the original lens image: \[ m_Q=-\frac{L_{\text{eff}}}{60} \] \[ m_Q=-\frac12 \] \(\textbf{Compare image positions:}\) \[ v_Q-v_P=40-\frac{100}{3} \] \[ v_Q-v_P=\frac{20}{3}\,\text{cm} \] Thus, Q produces the more distant image. \(\textbf{Compare image heights:}\) \[ \frac{|h_P|}{|h_Q|}=\frac{|m_P|}{|m_Q|} \] \[ \frac{|h_P|}{|h_Q|}=\frac{2/3}{1/2} \] \[ \frac{|h_P|}{|h_Q|}=\frac43 \] \(\textbf{Physical interpretation:}\) Placing a slab before the lens changes the effective object distance seen by that lens. Placing the same slab after the lens changes where an already converging bundle reaches its focus. Both arrangements shift the real image away from the lens, but not by the same amount. Their image magnifications also differ because the slab occupies different positions relative to the image-forming lens. The parallel faces ensure zero net angular deviation in air, not identical imaging effects for every placement. \(\textbf{Final answer:}\) \[ \boxed{v_Q-v_P=\frac{20}{3}\,\text{cm},\quad\frac{|h_P|}{|h_Q|}=\frac43} \]

Explanation: \(\textbf{Lens geometry and surrounding medium:}\) The lens is symmetric and double-convex. \[ R_1=+20\,\text{cm},\qquad R_2=-20\,\text{cm} \] The liquid refractive index is \[ n_m=1.5 \] The real object has signed distance \[ u=-300\,\text{cm} \] \(\textbf{Curvature contribution:}\) The lens-maker formula in a surrounding medium is \[ \frac1f=\left(\frac{n_g}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \] The curvature factor is \[ \frac1{R_1}-\frac1{R_2}=\frac1{20}-\frac1{-20} \] \[ \frac1{R_1}-\frac1{R_2}=\frac1{10}\,\text{cm}^{-1} \] \(\textbf{Red component:}\) For red light, \[ n_R=1.4 \] Its relative refractive index is \[ \mu_R=\frac{1.4}{1.5}=\frac{14}{15} \] Therefore, \[ \frac1{f_R}=\left(\frac{14}{15}-1\right)\frac1{10} \] \[ \frac1{f_R}=-\frac1{150} \] \[ f_R=-150\,\text{cm} \] The geometrically convex lens behaves as a diverging lens for red light. \(\textbf{Red image position:}\) Using the thin-lens equation, \[ \frac1{v_R}=\frac1{f_R}+\frac1u \] \[ \frac1{v_R}=-\frac1{150}-\frac1{300} \] \[ \frac1{v_R}=-\frac1{100} \] \[ v_R=-100\,\text{cm} \] The red image is virtual. \(\textbf{Violet component:}\) For violet light, \[ n_V=1.6 \] Its relative refractive index is \[ \mu_V=\frac{1.6}{1.5}=\frac{16}{15} \] Hence, \[ \frac1{f_V}=\left(\frac{16}{15}-1\right)\frac1{10} \] \[ \frac1{f_V}=+\frac1{150} \] \[ f_V=+150\,\text{cm} \] The same lens behaves as a converging lens for violet light. \(\textbf{Violet image position:}\) \[ \frac1{v_V}=\frac1{150}-\frac1{300} \] \[ \frac1{v_V}=+\frac1{300} \] \[ v_V=+300\,\text{cm} \] The violet image is real. \(\textbf{Check image orientations:}\) For red light, \[ m_R=\frac{-100}{-300}=+\frac13 \] For violet light, \[ m_V=\frac{+300}{-300}=-1 \] The red image is erect and diminished, whereas the violet image is inverted and equal-sized. \(\textbf{Physical interpretation:}\) The surrounding liquid has a refractive index between the red and violet refractive indices of the lens material. Consequently, the relative refractive-index contrast has opposite signs for the two colours. This reverses the sign of the lens's refracting power between red and violet light. The result is more pronounced than ordinary longitudinal chromatic separation: the same geometrical lens acts as diverging for one colour and converging for the other. The colour frequencies remain individually unchanged at the interfaces. \(\textbf{Final answer:}\) \[ \boxed{v_R=-100\,\text{cm}\text{ (virtual)},\quad v_V=+300\,\text{cm}\text{ (real)}} \]

Explanation: \(\textbf{Initial lens and distant sources:}\) The original convex-lens power is \[ P_1=+5\,\text{D} \] The angular separation of the incoming stellar directions is \[ \Delta\theta=0.010\,\text{rad} \] Since the stars are very distant, each supplies an approximately parallel ray bundle. \(\textbf{Initial focal-plane position:}\) The focal length is \[ f_1=\frac1{P_1} \] \[ f_1=0.20\,\text{m} \] \[ f_1=20\,\text{cm} \] The sharp spots lie in the focal plane at this distance. \(\textbf{Initial spot separation:}\) For small angular directions, the transverse coordinate of a focal-plane spot is approximately proportional to \(f\theta\). Therefore, \[ s_1\approx f_1\Delta\theta \] \[ s_1=20(0.010) \] \[ s_1=0.20\,\text{cm} \] \(\textbf{Add the concave lens:}\) The added lens power is \[ P_2=-3\,\text{D} \] For lenses in contact, \[ P_{\mathrm{eq}}=P_1+P_2 \] \[ P_{\mathrm{eq}}=+5-3 \] \[ P_{\mathrm{eq}}=+2\,\text{D} \] The combination remains converging. \(\textbf{New focal length:}\) \[ F=\frac1{P_{\mathrm{eq}}} \] \[ F=0.50\,\text{m} \] \[ F=50\,\text{cm} \] \(\textbf{New focal-plane separation:}\) \[ s_2\approx F\Delta\theta \] \[ s_2=50(0.010) \] \[ s_2=0.50\,\text{cm} \] \(\textbf{Required screen displacement:}\) The original focal plane was \(20\,\text{cm}\) from the lens. The new focal plane is \(50\,\text{cm}\) away. Therefore, \[ \Delta x=50-20 \] \[ \Delta x=+30\,\text{cm} \] \(\textbf{Physical interpretation:}\) Each distant star produces a focused spot rather than the extended image of a nearby object. The angular separation of their incident directions remains fixed. Reducing the converging power increases the effective focal length. A longer focal length converts the same angular separation into a greater physical separation on the focal plane. The screen must simultaneously be moved farther away to remain at that plane. The ratio of final to initial spot separation is \[ \frac{s_2}{s_1}=\frac{50}{20}=\frac52 \] This increase arises from angular-to-transverse image mapping, not from a change in the stars' actual separation. \(\textbf{Final answer:}\) \[ \boxed{s_2=0.50\,\text{cm},\quad \Delta x=+30\,\text{cm}} \]

Explanation: \(\textbf{Classify the incident ray bundle:}\) The incident rays are already converging. Their intended meeting point lies to the right of the lens. Therefore, the lens receives a virtual object with signed distance \[ u=+30\,\text{cm} \] The positive object distance is essential; a negative distance would describe a different incident ray condition. \(\textbf{Experiment P — convex lens:}\) The focal length is \[ f_P=+20\,\text{cm} \] The thin-lens formula gives \[ \frac1v=\frac1f+\frac1u \] Substituting, \[ \frac1{v_P}=\frac1{20}+\frac1{30} \] \[ \frac1{v_P}=\frac{3+2}{60} \] \[ \frac1{v_P}=\frac1{12} \] \[ v_P=+12\,\text{cm} \] The rays actually converge in the transmitted region. The image is real. \(\textbf{Experiment Q — concave lens:}\) The replacement lens has \[ f_Q=-20\,\text{cm} \] Using the same virtual object, \[ \frac1{v_Q}=-\frac1{20}+\frac1{30} \] \[ \frac1{v_Q}=\frac{-3+2}{60} \] \[ \frac1{v_Q}=-\frac1{60} \] \[ v_Q=-60\,\text{cm} \] The rays now emerge divergent. Their backward extensions meet on the left, forming a virtual image. \(\textbf{Compare transverse magnifications:}\) For P, \[ m_P=\frac{v_P}{u} \] \[ m_P=\frac{12}{30}=+\frac25 \] For Q, \[ m_Q=\frac{v_Q}{u} \] \[ m_Q=\frac{-60}{30}=-2 \] \(\textbf{Physical interpretation:}\) The convex lens adds convergence to the incoming beam, moving its real focus closer. The concave lens removes more convergence than the incoming bundle possesses, making the outgoing rays divergent. Thus, the same incident virtual object can produce either a real or a virtual image depending on lens power. For a virtual object, the sign of transverse magnification cannot be interpreted using the usual real-object image rules without identifying the object's nature. The two lenses have equal focal-length magnitudes but opposite powers and produce qualitatively different outgoing beams. \(\textbf{Final answer:}\) P forms a real image \(12\,\text{cm}\) to the right; Q forms a virtual image \(60\,\text{cm}\) to the left.

Explanation: \(\textbf{Initial lens power in air:}\) The glass refractive index is \[ n_g=1.5 \] The surface radii are \[ R_1=+20\,\text{cm},\qquad R_2=-20\,\text{cm} \] The curvature factor is \[ K=\frac1{R_1}-\frac1{R_2} \] \[ K=\frac1{20}-\frac1{-20} \] \[ K=\frac1{10}\,\text{cm}^{-1} \] The lens-maker formula in air gives \[ \frac1{f_1}=(1.5-1)K \] \[ \frac1{f_1}=0.5\left(\frac1{10}\right) \] \[ f_1=+20\,\text{cm} \] \(\textbf{Check the original magnification:}\) The original object distance is \[ u_1=-30\,\text{cm} \] Using the thin-lens formula, \[ \frac1{v_1}=\frac1{20}-\frac1{30} \] \[ v_1=+60\,\text{cm} \] Therefore, \[ m_1=\frac{60}{-30}=-2 \] \(\textbf{New focal length after immersion:}\) The surrounding-liquid refractive index is \[ n_m=1.25 \] The relative refractive index becomes \[ \mu=\frac{1.5}{1.25}=1.2 \] Thus, \[ \frac1{f_2}=(1.2-1)K \] \[ \frac1{f_2}=0.2\left(\frac1{10}\right) \] \[ f_2=+50\,\text{cm} \] \(\textbf{Preserve the signed magnification:}\) The desired magnification remains \[ m_2=-2 \] For a thin lens, \[ m_2=\frac{v_2}{u_2} \] Therefore, \[ v_2=-2u_2 \] Substitute this into the lens equation. \[ \frac1{v_2}-\frac1{u_2}=\frac1{f_2} \] \[ \frac1{-2u_2}-\frac1{u_2}=\frac1{50} \] \[ -\frac3{2u_2}=\frac1{50} \] Solving, \[ u_2=-75\,\text{cm} \] \(\textbf{Calculate the required object displacement:}\) The original signed object distance was \(-30\,\text{cm}\). The new distance is \(-75\,\text{cm}\). \[ \Delta u=u_2-u_1 \] \[ \Delta u=-75-(-30) \] \[ \Delta u=-45\,\text{cm} \] The negative displacement means the object moves left, away from the lens. \(\textbf{New image distance:}\) \[ v_2=-2u_2 \] \[ v_2=-2(-75) \] \[ v_2=+150\,\text{cm} \] \(\textbf{Physical interpretation:}\) Immersion decreases the lens's converging power and increases its focal length from \(20\,\text{cm}\) to \(50\,\text{cm}\). To maintain the same negative magnification, the object distance must increase in proportion to the focal length. The image distance likewise increases. Both initial and final images remain real, inverted and twice the object's height. Preserving magnification does not mean preserving the image position. The required mechanical object displacement is determined by the focal-length change together with the signed magnification condition. \(\textbf{Final answer:}\) \[ \boxed{\Delta u=-45\,\text{cm},\quad v_2=+150\,\text{cm}} \]

Explanation: \(\textbf{Identify the three axial object positions:}\) The lens focal length is \[ f=+20\,\text{cm} \] The endpoint positions are \[ u_A=-30\,\text{cm} \] \[ u_B=-60\,\text{cm} \] Their geometrical midpoint has coordinate \[ u_M=\frac{u_A+u_B}{2} \] \[ u_M=\frac{-30-60}{2} \] \[ u_M=-45\,\text{cm} \] \(\textbf{Image of endpoint A:}\) Using \[ \frac1v=\frac1f+\frac1u \] gives \[ \frac1{v_A}=\frac1{20}-\frac1{30} \] \[ \frac1{v_A}=\frac1{60} \] \[ v_A=+60\,\text{cm} \] \(\textbf{Image of endpoint B:}\) \[ \frac1{v_B}=\frac1{20}-\frac1{60} \] \[ \frac1{v_B}=\frac1{30} \] \[ v_B=+30\,\text{cm} \] \(\textbf{Geometrical midpoint of the endpoint images:}\) The two endpoint images lie at \(+60\,\text{cm}\) and \(+30\,\text{cm}\). Their midpoint is \[ x_{\text{mid}}=\frac{60+30}{2} \] \[ x_{\text{mid}}=+45\,\text{cm} \] \(\textbf{Image of the luminous rod midpoint:}\) The actual midpoint is at \(u_M=-45\,\text{cm}\). Apply the lens equation independently. \[ \frac1{v_M}=\frac1{20}-\frac1{45} \] \[ \frac1{v_M}=\frac{9-4}{180} \] \[ \frac1{v_M}=\frac1{36} \] \[ v_M=+36\,\text{cm} \] \(\textbf{Compare the two midpoint positions:}\) \[ x_{\text{mid}}-v_M=45-36 \] \[ x_{\text{mid}}-v_M=9\,\text{cm} \] The image of M lies nearer the lens than the midpoint of the endpoint images. \(\textbf{Physical interpretation:}\) Each point along the luminous rod has a different object distance from the lens. The lens relation between object and image positions is reciprocal rather than linear. Therefore, equal axial intervals in object space need not map to equal axial intervals in image space. Here, both endpoint images are real, but the order of their axial positions is reversed. The luminous midpoint does not map to the geometrical midpoint of the image interval. The difference is a consequence of non-uniform longitudinal image scaling, even though the lens is ideal and paraxial. The midpoint must be imaged by the lens equation, not by simply averaging the endpoint-image positions. \(\textbf{Final answer:}\) \[ \boxed{v_M=+36\,\text{cm},\quad x_{\text{mid}}-v_M=9\,\text{cm}} \]

Explanation: \(\textbf{Lens geometry and object position:}\) The glass refractive index is \[ n_g=1.5 \] The signed curvature radii are \[ R_1=+20\,\text{cm},\qquad R_2=-20\,\text{cm} \] The real object distance is \[ u=-50\,\text{cm} \] \(\textbf{Determine the curvature factor:}\) \[ K=\frac1{R_1}-\frac1{R_2} \] \[ K=\frac1{20}-\frac1{-20} \] \[ K=\frac1{10}\,\text{cm}^{-1} \] \(\textbf{Condition for parallel emergent rays:}\) For a real object at the first principal focus of a converging lens, the emergent rays are parallel. Therefore, the required focal length magnitude is \[ f=+50\,\text{cm} \] \(\textbf{Apply the lens-maker equation in liquid:}\) If the surrounding-liquid refractive index is \(n_m\), \[ \frac1f=\left(\frac{n_g}{n_m}-1\right)K \] Substitute the threshold focal length. \[ \frac1{50}=\left(\frac{1.5}{n_m}-1\right)\frac1{10} \] Multiplying by \(10\), \[ \frac15=\frac{1.5}{n_m}-1 \] Therefore, \[ \frac{1.5}{n_m}=\frac65 \] \[ n_m=1.5\left(\frac56\right) \] \[ n_m=1.25 \] \(\textbf{Liquid with refractive index 1.20:}\) The relative refractive index is \[ \mu_1=\frac{1.5}{1.20}=1.25 \] Therefore, \[ \frac1{f_1}=(1.25-1)\frac1{10} \] \[ f_1=+40\,\text{cm} \] The fixed object distance magnitude is \(50\,\text{cm}\), which exceeds \(f_1\). The lens therefore forms a real image. \(\textbf{Liquid with refractive index 1.30:}\) The relative refractive index is \[ \mu_2=\frac{1.5}{1.30}=\frac{15}{13} \] Thus, \[ \frac1{f_2}=\left(\frac{15}{13}-1\right)\frac1{10} \] \[ \frac1{f_2}=\frac2{13}\times\frac1{10} \] \[ f_2=+65\,\text{cm} \] The object distance magnitude is now smaller than the focal length. The lens therefore forms a virtual, erect image. \(\textbf{Physical interpretation:}\) Both liquids have refractive indices below that of the glass. Therefore, the double-convex lens remains converging in both liquids. Nevertheless, increasing the surrounding-liquid index reduces the refractive-index contrast and lengthens the focal distance. The fixed object crosses from outside to inside the effective focal length without moving. At the intermediate liquid index of \(1.25\), the emergent rays are parallel. The change from real to virtual imaging does not require the optical power to become negative. It occurs because the object position crosses the changing focal boundary. \(\textbf{Final answer:}\) The emergent rays become parallel at \(n_m=1.25\); the images are real at \(1.20\) and virtual at \(1.30\).

Explanation: \(\textbf{Equivalent power after adding lenses:}\) The initial convex-lens power is \[ P_0=+7\,\text{D} \] Each added concave lens contributes \[ P_c=-2\,\text{D} \] After adding \(k\) concave lenses, the equivalent power is \[ P_k=P_0+kP_c \] \[ P_k=7-2k \] \(\textbf{Relate image nature to the fixed object distance:}\) The real object distance is \[ u=-50\,\text{cm} \] For a converging lens, the boundary between finite real-image and virtual-image formation occurs when the object is at the first principal focus. At that boundary, \[ f=+50\,\text{cm}=+0.50\,\text{m} \] The corresponding power is \[ P_{\text{boundary}}=\frac1{0.50} \] \[ P_{\text{boundary}}=+2\,\text{D} \] A converging combination with power greater than \(+2\,\text{D}\) forms a finite real image of the fixed object. At exactly \(+2\,\text{D}\), the emergent rays are parallel. For positive power below \(+2\,\text{D}\), the object lies inside the focal length and produces a virtual image. \(\textbf{After two concave lenses:}\) \[ P_2=7-2(2) \] \[ P_2=+3\,\text{D} \] The corresponding focal length is \[ f_2=\frac13\,\text{m} \] \[ f_2=\frac{100}{3}\,\text{cm} \] Since the object is \(50\,\text{cm}\) away, it lies beyond the first focus. The image is real. \(\textbf{After three concave lenses:}\) \[ P_3=7-2(3) \] \[ P_3=+1\,\text{D} \] The focal length becomes \[ f_3=+100\,\text{cm} \] The object now lies inside the focal length. The image becomes virtual. Thus, three is the smallest number of added lenses producing a virtual image. \(\textbf{Find the onset of negative equivalent power:}\) The combination becomes diverging only when \[ P_k\lt 0 \] Therefore, \[ 7-2k\lt 0 \] \[ 2k\gt 7 \] \[ k\gt 3.5 \] Since the number of lenses must be an integer, the smallest possible value is \[ k=4 \] \(\textbf{Verify the four-lens case:}\) \[ P_4=7-2(4) \] \[ P_4=-1\,\text{D} \] The combination is now diverging. \(\textbf{Physical interpretation:}\) A virtual image can be produced by a converging lens as well as by a diverging lens. The first transition occurs when the decreasing positive power makes the focal length exceed the fixed object distance. The second transition occurs when the total power becomes negative. These thresholds are not identical. After three concave lenses, the combination is still converging but forms a virtual image. After four, it is diverging and continues to form a virtual image. The lens count must therefore be evaluated using two different physical conditions rather than by testing the power sign alone. \(\textbf{Final answer:}\) Three concave lenses first produce a virtual image; four first make the equivalent combination diverging.

Explanation: \(\textbf{Initial object and image positions:}\) The initial real object is at \[ u_1=-20\,\text{cm} \] The virtual image is at \[ v_1=-10\,\text{cm} \] \(\textbf{Check the original focal length:}\) Using \[ \frac1f=\frac1v-\frac1u \] gives \[ \frac1{f_1}=-\frac1{10}-\frac1{-20} \] \[ \frac1{f_1}=-\frac1{10}+\frac1{20} \] \[ \frac1{f_1}=-\frac1{20} \] \[ f_1=-20\,\text{cm} \] \(\textbf{Initial magnification:}\) \[ m_1=\frac{v_1}{u_1} \] \[ m_1=\frac{-10}{-20} \] \[ m_1=+\frac12 \] \(\textbf{Changed object position:}\) The object is moved farther from the lens. \[ u_2=-40\,\text{cm} \] The image must remain fixed at \[ v_2=-10\,\text{cm} \] \(\textbf{Determine the replacement focal length:}\) The lens formula requires \[ \frac1{f_2}=\frac1{v_2}-\frac1{u_2} \] Substituting, \[ \frac1{f_2}=-\frac1{10}-\frac1{-40} \] \[ \frac1{f_2}=-\frac1{10}+\frac1{40} \] \[ \frac1{f_2}=-\frac3{40} \] Therefore, \[ f_2=-\frac{40}{3}\,\text{cm} \] \(\textbf{Convert focal length to power:}\) For \(f\) measured in centimetres, \[ P=\frac{100}{f} \] Thus, \[ P_2=\frac{100}{-40/3} \] \[ P_2=-7.5\,\text{D} \] \(\textbf{New magnification:}\) \[ m_2=\frac{v_2}{u_2} \] \[ m_2=\frac{-10}{-40} \] \[ m_2=+\frac14 \] \(\textbf{Compare image heights:}\) The object height is unchanged. \[ \frac{|h_{i2}|}{|h_{i1}|}=\frac{|m_2|}{|m_1|} \] \[ \frac{|h_{i2}|}{|h_{i1}|}=\frac{1/4}{1/2} \] \[ \frac{|h_{i2}|}{|h_{i1}|}=\frac12 \] \(\textbf{Physical interpretation:}\) Moving the object farther away would ordinarily alter the virtual-image position. To preserve that position, the concave lens must become more strongly diverging. The required focal-length magnitude decreases from \(20\,\text{cm}\) to \(40/3\,\text{cm}\). The optical power accordingly changes from \(-5\,\text{D}\) to \(-7.5\,\text{D}\). Despite the unchanged image position, the magnification decreases because the object distance increases. The image remains virtual and erect but becomes half as high as the original image. \(\textbf{Final answer:}\) \[ \boxed{P_2=-7.5\,\text{D},\quad \frac{|h_{i2}|}{|h_{i1}|}=\frac12} \]

Explanation: \(\textbf{Lens-maker relation in a surrounding liquid:}\) For each thin lens, \[ P=\left(\frac{n_g}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \] The radii must be expressed in metres when calculating dioptres. \(\textbf{Curvature factor of lens P:}\) Lens P is symmetric double-convex. \[ R_{1P}=+0.20\,\text{m},\qquad R_{2P}=-0.20\,\text{m} \] Therefore, \[ K_P=\frac1{0.20}-\frac1{-0.20} \] \[ K_P=+10\,\text{m}^{-1} \] \(\textbf{Power of P in the unknown liquid:}\) The glass index is \[ n_P=1.6 \] Thus, \[ P_P=10\left(\frac{1.6}{n_m}-1\right) \] \(\textbf{Curvature factor of lens Q:}\) Lens Q is symmetric double-concave. \[ R_{1Q}=-0.40\,\text{m},\qquad R_{2Q}=+0.40\,\text{m} \] Therefore, \[ K_Q=\frac1{-0.40}-\frac1{+0.40} \] \[ K_Q=-5\,\text{m}^{-1} \] \(\textbf{Power of Q in the unknown liquid:}\) Its glass index is \[ n_Q=1.4 \] Hence, \[ P_Q=-5\left(\frac{1.4}{n_m}-1\right) \] \(\textbf{Impose zero equivalent power:}\) For contact lenses, \[ P_{\text{eq}}=P_P+P_Q \] The required condition is \[ 10\left(\frac{1.6}{n_m}-1\right)-5\left(\frac{1.4}{n_m}-1\right)=0 \] Expand the terms. \[ \frac{16}{n_m}-10-\frac7{n_m}+5=0 \] \[ \frac9{n_m}-5=0 \] Therefore, \[ \frac9{n_m}=5 \] \[ n_m=\frac95 \] \[ n_m=1.8 \] \(\textbf{Check cancellation of individual powers:}\) For lens P, \[ P_P=10\left(\frac{1.6}{1.8}-1\right) \] \[ P_P=10\left(\frac89-1\right) \] \[ P_P=-\frac{10}{9}\,\text{D} \] For lens Q, \[ P_Q=-5\left(\frac{1.4}{1.8}-1\right) \] \[ P_Q=-5\left(\frac79-1\right) \] \[ P_Q=+\frac{10}{9}\,\text{D} \] Their powers cancel exactly. \(\textbf{Determine the image magnification:}\) Zero equivalent power means \[ \frac1F=0 \] The thin-lens equation becomes \[ \frac1v-\frac1u=0 \] Therefore, \[ v=u \] The signed magnification is \[ m=\frac vu \] \[ m=+1 \] \(\textbf{Physical interpretation:}\) In this liquid, the geometrically double-convex lens P becomes diverging because its glass index is lower than the liquid index. The double-concave lens Q becomes converging for the same reason. Their powers have equal magnitudes despite the different glasses and surface radii. The contact combination consequently has zero net refracting power. For a real object, the outgoing rays continue to diverge as though from the original object position. The image is virtual, erect and equal-sized. This cancellation occurs at a particular surrounding-medium index, unlike cancellation between equal and opposite geometries made from identical materials. \(\textbf{Final answer:}\) \[ \boxed{n_m=1.8,\quad m=+1} \]

Explanation: \(\textbf{Lens curvature factor:}\) The lens is symmetric double-convex. \[ R_1=+20\,\text{cm},\qquad R_2=-20\,\text{cm} \] Therefore, \[ K=\frac1{R_1}-\frac1{R_2} \] \[ K=\frac1{10}\,\text{cm}^{-1} \] \(\textbf{Focal length for red light:}\) The red refractive index is \[ n_R=1.5 \] The lens-maker equation gives \[ \frac1{f_R}=(n_R-1)K \] \[ \frac1{f_R}=0.5\left(\frac1{10}\right) \] \[ f_R=+20\,\text{cm} \] \(\textbf{Determine the fixed screen position:}\) The original red object is at \[ u_R=-30\,\text{cm} \] Using the thin-lens formula, \[ \frac1{v_R}=\frac1{20}-\frac1{30} \] \[ \frac1{v_R}=\frac1{60} \] \[ v_R=+60\,\text{cm} \] The screen is therefore \(60\,\text{cm}\) to the right of the lens. \(\textbf{Focal length for violet light:}\) The violet refractive index is \[ n_V=1.6 \] Thus, \[ \frac1{f_V}=(1.6-1)\frac1{10} \] \[ \frac1{f_V}=\frac3{50} \] \[ f_V=+\frac{50}{3}\,\text{cm} \] \(\textbf{Find the required violet-object distance:}\) The image must remain on the same fixed screen. \[ v_V=+60\,\text{cm} \] Rearrange the thin-lens formula. \[ \frac1{u_V}=\frac1{v_V}-\frac1{f_V} \] Substituting, \[ \frac1{u_V}=\frac1{60}-\frac3{50} \] \[ \frac1{u_V}=\frac{5-18}{300} \] \[ \frac1{u_V}=-\frac{13}{300} \] Therefore, \[ u_V=-\frac{300}{13}\,\text{cm} \] \(\textbf{Determine the source displacement:}\) The source initially occupies \(u=-30\,\text{cm}\). The required displacement is \[ \Delta u=u_V-(-30) \] \[ \Delta u=-\frac{300}{13}+30 \] \[ \Delta u=\frac{-300+390}{13} \] \[ \Delta u=+\frac{90}{13}\,\text{cm} \] The positive displacement is towards the lens. \(\textbf{Calculate violet magnification:}\) \[ m_V=\frac{v_V}{u_V} \] \[ m_V=\frac{60}{-300/13} \] \[ m_V=-\frac{13}{5} \] \(\textbf{Physical interpretation:}\) The greater refractive index for violet light gives the lens greater converging power and a shorter focal length. If the source remained at its original position, the violet image would form closer to the lens than the red image. To bring the violet image back to the fixed screen, the source must move towards the violet first focus. The violet image remains real and inverted. Its magnification differs from that of the red image even though both are formed sharply on the same screen. This distinction follows because their object distances are different. \(\textbf{Final answer:}\) \[ \boxed{\Delta u=+\frac{90}{13}\,\text{cm},\quad m_V=-\frac{13}{5}} \]

Explanation: \(\textbf{Relate the graph to the thin-lens formula:}\) The real object remains at \[ u=-40\,\text{cm} \] The thin-lens equation is \[ \frac1v-\frac1u=\frac1f \] For power \(P\) in dioptres, \[ \frac1f=\frac{P}{100}\,\text{cm}^{-1} \] Therefore, \[ \frac1v=\frac{P}{100}+\frac1u \] Substituting the fixed object distance, \[ \frac1v=\frac{P}{100}-\frac1{40} \] \(\textbf{Interpret the graph intercept:}\) At zero power, \[ \frac1v=-\frac1{40}\,\text{cm}^{-1} \] This matches the given vertical intercept. The zero of \(1/v\) occurs when \[ \frac{P}{100}=\frac1{40} \] \[ P=+2.5\,\text{D} \] At this power, the emergent rays are parallel. \(\textbf{Lens of power }+2\,\text{D}\textbf{:}\) Its reciprocal image distance is \[ \frac1{v_1}=\frac2{100}-\frac1{40} \] \[ \frac1{v_1}=\frac1{50}-\frac1{40} \] \[ \frac1{v_1}=\frac{4-5}{200} \] \[ \frac1{v_1}=-\frac1{200} \] \[ v_1=-200\,\text{cm} \] The image is virtual. \(\textbf{Magnification at }+2\,\text{D}\textbf{:}\) \[ m_1=\frac{v_1}{u} \] \[ m_1=\frac{-200}{-40} \] \[ m_1=+5 \] The image is erect and enlarged five times. \(\textbf{Lens of power }+5\,\text{D}\textbf{:}\) \[ \frac1{v_2}=\frac5{100}-\frac1{40} \] \[ \frac1{v_2}=\frac1{20}-\frac1{40} \] \[ \frac1{v_2}=\frac1{40} \] \[ v_2=+40\,\text{cm} \] The image is real. \(\textbf{Magnification at }+5\,\text{D}\textbf{:}\) \[ m_2=\frac{40}{-40} \] \[ m_2=-1 \] The image is inverted and equal-sized. \(\textbf{Physical interpretation:}\) Both lenses have positive converging power. However, the \(+2\,\text{D}\) lens has focal length \(50\,\text{cm}\), placing the object inside its first focus. The \(+5\,\text{D}\) lens has focal length \(20\,\text{cm}\), placing the object at twice the focal length. The graph's horizontal intercept marks the boundary between virtual and real images for this fixed object. Increasing positive power across that threshold changes the image nature and orientation. A positive power alone is therefore insufficient to establish whether a particular real object produces a real image. \(\textbf{Final conclusion:}\) At \(+2\,\text{D}\), the image is virtual, erect and magnified five times; at \(+5\,\text{D}\), it is real, inverted and equal-sized.

Explanation: \(\textbf{Initial apparent object position:}\) The liquid refractive index is \[ n=1.5 \] The initial real depth is \[ d_1=30\,\text{cm} \] For near-normal observation from air, \[ d'_1=\frac{d_1}{n} \] \[ d'_1=\frac{30}{1.5} \] \[ d'_1=20\,\text{cm} \] \(\textbf{Effective distance seen by the lens:}\) The lens is located \(20\,\text{cm}\) above the stationary surface. The apparent object therefore lies \[ s_1=20+d'_1 \] \[ s_1=20+20 \] \[ s_1=40\,\text{cm} \] below the lens. Thus, \[ u_1=-40\,\text{cm} \] \(\textbf{Initial lens image:}\) The lens focal length is \[ f=+20\,\text{cm} \] Using \[ \frac1v=\frac1f+\frac1u \] gives \[ \frac1{v_1}=\frac1{20}-\frac1{40} \] \[ v_1=+40\,\text{cm} \] The initial magnification is \[ m_1=\frac{40}{-40} \] \[ m_1=-1 \] \(\textbf{Lower the object inside the liquid:}\) The object's real depth increases by \(15\,\text{cm}\). \[ d_2=30+15 \] \[ d_2=45\,\text{cm} \] Its new apparent depth from air is \[ d'_2=\frac{45}{1.5} \] \[ d'_2=30\,\text{cm} \] \(\textbf{New effective object distance:}\) The lens remains \(20\,\text{cm}\) above the surface. Therefore, \[ s_2=20+30 \] \[ s_2=50\,\text{cm} \] \[ u_2=-50\,\text{cm} \] \(\textbf{Final image position:}\) \[ \frac1{v_2}=\frac1{20}-\frac1{50} \] \[ \frac1{v_2}=\frac{5-2}{100} \] \[ v_2=+\frac{100}{3}\,\text{cm} \] \(\textbf{Calculate image displacement:}\) \[ \Delta v=v_2-v_1 \] \[ \Delta v=\frac{100}{3}-40 \] \[ \Delta v=-\frac{20}{3}\,\text{cm} \] The negative displacement indicates movement towards the lens. \(\textbf{Final magnification:}\) Plane-surface refraction changes the apparent axial position but not the paraxial transverse object height. The final lens magnification is \[ m_2=\frac{v_2}{u_2} \] \[ m_2=\frac{100/3}{-50} \] \[ m_2=-\frac23 \] \(\textbf{Compare image heights:}\) \[ \frac{|h_{i2}|}{|h_{i1}|}=\frac{|m_2|}{|m_1|} \] \[ \frac{|h_{i2}|}{|h_{i1}|}=\frac23 \] \(\textbf{Physical interpretation:}\) The object is physically lowered by \(15\,\text{cm}\), but its apparent position moves downward by only \(10\,\text{cm}\). Therefore, the effective object distance seen by the lens increases from \(40\,\text{cm}\) to \(50\,\text{cm}\). A real object moving farther from a convex lens beyond its first focus produces a real image closer to the lens. The magnification magnitude also decreases. The two stages of image formation must be handled sequentially: plane-surface refraction first determines the apparent object presented to the lens, and the lens then forms its real image. \(\textbf{Final answer:}\) \[ \boxed{\Delta v=-\frac{20}{3}\,\text{cm},\quad\frac{|h_{i2}|}{|h_{i1}|}=\frac23} \]

Explanation: \(\textbf{Initial object position:}\) Both real objects lie to the left of their respective optical elements. \[ u_1=-30\,\text{cm} \] \(\textbf{Concave mirror — initial image:}\) The mirror focal length is \[ f_m=-20\,\text{cm} \] The spherical-mirror equation is \[ \frac1{f_m}=\frac1u+\frac1v \] Therefore, \[ \frac1{v_{m1}}=-\frac1{20}-\frac1{-30} \] \[ \frac1{v_{m1}}=-\frac1{20}+\frac1{30} \] \[ v_{m1}=-60\,\text{cm} \] The mirror magnification is \[ m_{m1}=-\frac{v_{m1}}{u_1} \] \[ m_{m1}=-\frac{-60}{-30} \] \[ m_{m1}=-2 \] The mirror image is real, inverted and enlarged twice. \(\textbf{Convex lens — initial image:}\) The lens focal length is \[ f_l=+20\,\text{cm} \] The thin-lens equation is \[ \frac1v-\frac1u=\frac1{f_l} \] Thus, \[ \frac1{v_{l1}}=\frac1{20}-\frac1{30} \] \[ v_{l1}=+60\,\text{cm} \] The lens magnification is \[ m_{l1}=\frac{v_{l1}}{u_1} \] \[ m_{l1}=\frac{60}{-30} \] \[ m_{l1}=-2 \] Both initial images have the same orientation and height magnitude, but their signed image positions are opposite. \(\textbf{Move both objects inside their focal distances:}\) The new object distance is \[ u_2=-10\,\text{cm} \] \(\textbf{Concave mirror — final image:}\) \[ \frac1{v_{m2}}=-\frac1{20}-\frac1{-10} \] \[ \frac1{v_{m2}}=-\frac1{20}+\frac1{10} \] \[ \frac1{v_{m2}}=+\frac1{20} \] \[ v_{m2}=+20\,\text{cm} \] The positive mirror-image distance places the virtual image behind the mirror. Its magnification is \[ m_{m2}=-\frac{+20}{-10} \] \[ m_{m2}=+2 \] \(\textbf{Convex lens — final image:}\) \[ \frac1{v_{l2}}=\frac1{20}-\frac1{10} \] \[ \frac1{v_{l2}}=-\frac1{20} \] \[ v_{l2}=-20\,\text{cm} \] The negative lens-image distance places the virtual image on the object's side. Its magnification is \[ m_{l2}=\frac{-20}{-10} \] \[ m_{l2}=+2 \] \(\textbf{Physical comparison:}\) The mirror and lens have equal focal-length magnitudes but different image-distance sign conventions because one reflects light and the other transmits it. At the first object position, both form real, inverted images enlarged twice. At the second position, both form virtual, erect images enlarged twice. However, their virtual images lie on physically opposite sides of the optical element. For the mirror, the virtual image lies behind the reflecting surface. For the lens, it lies on the incident side. The equality of magnification does not imply equality of signed image positions. The distinction follows from the mirror magnification relation \(m=-v/u\) and the lens relation \(m=v/u\), together with their different imaging equations. \(\textbf{Final conclusion:}\) Both initially form real, inverted images magnified twice; both finally form virtual, erect images magnified twice, on opposite physical sides of their respective elements.

Explanation: \(\textbf{Connect magnification with object position:}\) The thin-lens equation is \[ \frac1v-\frac1u=\frac1f \] The signed transverse magnification is \[ m=\frac vu \] Therefore, \[ v=mu \] Substituting into the lens equation, \[ \frac1{mu}-\frac1u=\frac1f \] Multiplying by \(u\), \[ \frac1m-1=\frac uf \] Hence, \[ u=f\left(\frac1m-1\right) \] \(\textbf{Experiment P — virtual enlarged image:}\) The first magnification is \[ m_P=+3 \] Therefore, \[ u_P=f\left(\frac13-1\right) \] \[ u_P=-\frac{2f}{3} \] The negative object distance is consistent with a real object on the incident side. \(\textbf{Experiment Q — real enlarged image:}\) The second magnification is \[ m_Q=-3 \] Thus, \[ u_Q=f\left(-\frac13-1\right) \] \[ u_Q=-\frac{4f}{3} \] The object lies farther from the lens in experiment Q. \(\textbf{Use the specified object displacement:}\) The object moves \(20\,\text{cm}\) farther to the left. Therefore, \[ u_Q-u_P=-20\,\text{cm} \] Substituting, \[ -\frac{4f}{3}+\frac{2f}{3}=-20 \] \[ -\frac{2f}{3}=-20 \] \[ f=+30\,\text{cm} \] \(\textbf{Recover both object distances:}\) \[ u_P=-\frac23(30) \] \[ u_P=-20\,\text{cm} \] \[ u_Q=-\frac43(30) \] \[ u_Q=-40\,\text{cm} \] \(\textbf{Determine both image positions:}\) From \(v=mu\), \[ v_P=(+3)(-20) \] \[ v_P=-60\,\text{cm} \] For experiment Q, \[ v_Q=(-3)(-40) \] \[ v_Q=+120\,\text{cm} \] \(\textbf{Calculate image separation:}\) The images lie on opposite sides of the lens. Therefore, \[ L=|v_Q-v_P| \] \[ L=|120-(-60)| \] \[ L=180\,\text{cm} \] \(\textbf{Physical interpretation:}\) The two images have equal height magnitudes but different orientations. The first object lies inside the convex lens's focal length and produces a virtual image. The second lies beyond the focal length and produces a real image. Equal absolute magnifications do not require equal object distances or equal image distances. The signed magnification conditions determine two different object positions relative to the same unknown focal length. The given mechanical displacement then determines that focal length uniquely. \(\textbf{Final answer:}\) \[ \boxed{f=30\,\text{cm},\quad L=180\,\text{cm}} \]

Explanation: \(\textbf{Initial beam produced by the lens:}\) The luminous point is at the first principal focus of the convex lens. Therefore, its rays emerge from the lens as a parallel beam. The beam travels along the principal-axis direction towards the plane mirror. \(\textbf{Return from the perpendicular mirror:}\) Initially, the mirror normal coincides with the principal axis. The incident parallel rays meet the mirror normally. After reflection, they retrace their paths towards the lens. The second passage through the convex lens brings these parallel returning rays to the focus on the incident side. The returned image therefore coincides with the original luminous point. \(\textbf{Experiment P — translate the mirror:}\) The mirror moves axially through \[ \Delta x_M=+10\,\text{cm} \] Its orientation remains perpendicular to the principal axis. The beam incident upon it is still parallel. Reflection therefore returns a parallel beam along the same axial direction. The lens focal length and returned beam direction are unchanged. Consequently, the returned focal image remains at the original object point. \(\textbf{Experiment Q — rotate the mirror:}\) The mirror is rotated through \[ \theta=0.010\,\text{rad} \] The incident beam direction remains fixed. By the law of reflection, rotation of a plane mirror through a small angle \(\theta\) changes the reflected-ray direction through \(2\theta\). Therefore, \[ \Delta\alpha=2\theta \] \[ \Delta\alpha=2(0.010) \] \[ \Delta\alpha=0.020\,\text{rad} \] \(\textbf{Focusing of the tilted returning beam:}\) A parallel paraxial beam incident at a small angle to the lens axis is brought to a point in the focal plane displaced transversely from the principal focus. For small angles, \[ y\approx f\Delta\alpha \] The focal length is \[ f=20\,\text{cm} \] Substituting, \[ y\approx20(0.020) \] \[ y\approx0.40\,\text{cm} \] \(\textbf{Physical interpretation:}\) Translating the perpendicular mirror changes the distance travelled by the parallel beam but not its direction. The final focal point is therefore unaffected by that translation. Rotating the mirror changes the returning beam direction. The lens converts this angular change into a transverse displacement in its focal plane. The return image remains in the focal plane under the paraxial approximation but no longer coincides with the original point. This arrangement connects the production of parallel rays by a convex lens, reflection from a plane mirror and angular-to-transverse focusing. \(\textbf{Final answer:}\) \[ \boxed{\Delta x_{\text{image,P}}=0,\quad |\Delta y_{\text{image,Q}}|\approx0.40\,\text{cm}} \]

Explanation: \(\textbf{Separate the two aperture regions:}\) Rays passing through the lower region encounter only the original convex lens. Rays passing through the upper region encounter the convex lens and the additional concave lens in contact. These regions therefore have different equivalent focal lengths. However, either region can receive rays from every luminous point of the object. \(\textbf{Lower region — convex lens alone:}\) The convex lens focal length is \[ f_L=+20\,\text{cm} \] The real-object distance is \[ u=-60\,\text{cm} \] Using the thin-lens formula, \[ \frac1{v_L}=\frac1{f_L}+\frac1u \] \[ \frac1{v_L}=\frac1{20}-\frac1{60} \] \[ \frac1{v_L}=\frac1{30} \] \[ v_L=+30\,\text{cm} \] \(\textbf{Lower-region magnification:}\) \[ m_L=\frac{v_L}{u} \] \[ m_L=\frac{30}{-60} \] \[ m_L=-\frac12 \] The lower region forms a complete real, inverted and diminished image. \(\textbf{Upper region — contact-lens combination:}\) The additional concave lens has focal length \[ f_C=-60\,\text{cm} \] For thin lenses in contact, \[ \frac1F=\frac1{f_L}+\frac1{f_C} \] Substituting, \[ \frac1F=\frac1{20}-\frac1{60} \] \[ \frac1F=\frac1{30} \] \[ F=+30\,\text{cm} \] The upper region remains converging but has less optical power than the lower region. \(\textbf{Upper-region image position:}\) \[ \frac1{v_U}=\frac1F+\frac1u \] \[ \frac1{v_U}=\frac1{30}-\frac1{60} \] \[ v_U=+60\,\text{cm} \] \(\textbf{Upper-region magnification:}\) \[ m_U=\frac{v_U}{u} \] \[ m_U=\frac{60}{-60} \] \[ m_U=-1 \] \(\textbf{Compare image heights:}\) Both images are formed from the same object height. Therefore, \[ \frac{|h_U|}{|h_L|}=\frac{|m_U|}{|m_L|} \] \[ \frac{|h_U|}{|h_L|}=\frac1{1/2} \] \[ \frac{|h_U|}{|h_L|}=2 \] \(\textbf{Physical interpretation:}\) The two aperture regions have different optical powers and consequently form images in different axial planes. Neither region is restricted to imaging only the corresponding half of the object. Each forms a complete image, although the reduced collecting apertures affect brightness. The upper region produces the more distant and larger image because the added negative-power lens lengthens its effective focal length. A single screen cannot generally obtain both images sharply at the same axial position. This arrangement combines aperture reasoning with equivalent power and image magnification. \(\textbf{Final answer:}\) \[ \boxed{v_L=+30\,\text{cm},\quad v_U=+60\,\text{cm},\quad |h_U|:|h_L|=2:1} \]

Explanation: \(\textbf{Interpret the original ray intersection:}\) The original point object lies to the left of the lens. Its signed axial coordinate is \[ u_1=-30\,\text{cm} \] The two transmitted rays meet to the right at \[ v_1=+45\,\text{cm} \] Because the rays actually meet, the image is real. \(\textbf{Check the optical-centre ray:}\) The initial transverse object coordinate is \[ h_{o1}=+2\,\text{cm} \] For a thin lens, the ray through its optical centre continues approximately undeviated. The corresponding transverse magnification is \[ m_1=\frac{v_1}{u_1} \] Substituting, \[ m_1=\frac{45}{-30} \] \[ m_1=-\frac32 \] Therefore, \[ h_{i1}=m_1h_{o1} \] \[ h_{i1}=-\frac32(2) \] \[ h_{i1}=-3\,\text{cm} \] This agrees with the given intersection coordinate. \(\textbf{Determine the unknown focal length:}\) Apply the thin-lens formula. \[ \frac1f=\frac1v-\frac1u \] Substituting, \[ \frac1f=\frac1{45}-\frac1{-30} \] \[ \frac1f=\frac1{45}+\frac1{30} \] \[ \frac1f=\frac{2+3}{90} \] \[ \frac1f=\frac1{18} \] Therefore, \[ f=+18\,\text{cm} \] The lens is converging. \(\textbf{Changed object coordinates:}\) The new point lies at \[ u_2=-36\,\text{cm} \] Its transverse coordinate is \[ h_{o2}=+4\,\text{cm} \] \(\textbf{Find the new axial image position:}\) \[ \frac1{v_2}=\frac1f+\frac1{u_2} \] \[ \frac1{v_2}=\frac1{18}-\frac1{36} \] \[ \frac1{v_2}=\frac1{36} \] \[ v_2=+36\,\text{cm} \] \(\textbf{Find the new transverse coordinate:}\) \[ m_2=\frac{v_2}{u_2} \] \[ m_2=\frac{36}{-36} \] \[ m_2=-1 \] Hence, \[ h_{i2}=m_2h_{o2} \] \[ h_{i2}=(-1)(+4) \] \[ h_{i2}=-4\,\text{cm} \] \(\textbf{Physical interpretation:}\) The original image intersection provides both an axial conjugate pair and a check on transverse inversion. The two measurements are consistent with a converging lens of focal length \(18\,\text{cm}\). The new object lies at twice this focal length. Its real image consequently lies at twice the focal length on the opposite side and has equal height magnitude with inverted orientation. The optical-centre ray and lens equation give mutually consistent descriptions of both arrangements. \(\textbf{Final answer:}\) \[ \boxed{f=+18\,\text{cm},\quad (x_i,y_i)=(+36\,\text{cm},-4\,\text{cm})} \]

Explanation: \(\textbf{First passage through the convex lens:}\) The lens focal length is \[ f=+20\,\text{cm} \] The real-object distance is \[ u_1=-30\,\text{cm} \] Using the lens equation, \[ \frac1{v_1}=\frac1f+\frac1{u_1} \] \[ \frac1{v_1}=\frac1{20}-\frac1{30} \] \[ \frac1{v_1}=\frac1{60} \] \[ v_1=+60\,\text{cm} \] Without the mirror, the rays would converge \(60\,\text{cm}\) to the right of the lens. \(\textbf{Reflection before the first convergence point:}\) The plane mirror is effectively at the lens plane, before the rays have reached their intended convergence point. Reflection reverses their axial propagation while preserving their convergence. After reflection, the returning rays would meet \(60\,\text{cm}\) to the left of the lens if their second passage through the lens were omitted. \(\textbf{Classify the object for the return passage:}\) The returning rays approach the lens from the right and are already convergent. Their intended meeting point lies \(60\,\text{cm}\) beyond the lens along the reversed propagation direction. It therefore acts as a virtual object for the return passage. Using a new positive direction towards the left, \[ u_2=+60\,\text{cm} \] \(\textbf{Second passage through the lens:}\) The lens remains converging, with \[ f=+20\,\text{cm} \] The reverse-pass lens equation gives \[ \frac1{v_2}=\frac1f+\frac1{u_2} \] \[ \frac1{v_2}=\frac1{20}+\frac1{60} \] \[ \frac1{v_2}=\frac{3+1}{60} \] \[ \frac1{v_2}=\frac1{15} \] \[ v_2=+15\,\text{cm} \] The positive image distance is along the reversed direction, towards the left. Thus, the returned rays form a real image \(15\,\text{cm}\) to the left of the lens. \(\textbf{Determine the net transverse magnification:}\) The first lens passage gives \[ m_1=\frac{v_1}{u_1} \] \[ m_1=\frac{60}{-30}=-2 \] The plane mirror has unit transverse magnification. For the returning lens passage, \[ m_2=\frac{v_2}{u_2} \] \[ m_2=\frac{15}{60}=+\frac14 \] Therefore, \[ m_{\text{net}}=m_1(1)m_2 \] \[ m_{\text{net}}=(-2)\left(\frac14\right) \] \[ m_{\text{net}}=-\frac12 \] \(\textbf{Physical interpretation:}\) The plane mirror reverses the direction of propagation before the first lens image can form. The returning bundle is already convergent and must not be treated as though it came from a real object on the right. The lens strengthens this returning convergence, producing a real image on the original object's side. The two lens passages must be evaluated using their respective signed object conditions. The final image is inverted and diminished to half the original object height. \(\textbf{Final answer:}\) \[ \boxed{x_{\text{final}}=-15\,\text{cm},\quad m_{\text{net}}=-\frac12} \]