Optics MCQs with Answers – Part 7 (NEET Physics)

Explanation: \(\textbf{Identify the two optical paths:}\) The direct ray family travels from the luminous mark through water and emerges into air. The reflected ray family first encounters the bottom plane mirror, then travels upwards through water and emerges into air. The two ray families therefore have different apparent source positions before reaching the water surface. \(\textbf{Calculate the direct apparent depth:}\) The actual mark depth below the surface is \[ d=10\,\text{cm} \] The refractive index of water is \[ n_w=\frac43 \] For near-normal viewing from air, \[ d_{\mathrm{app}}=\frac d{n_w} \] Therefore, \[ d_{\mathrm{direct}}=\frac{10}{4/3} \] \[ d_{\mathrm{direct}}=7.5\,\text{cm} \] \(\textbf{Locate the mirror image inside the water geometry:}\) The plane mirror is at depth \[ H=30\,\text{cm} \] The luminous mark is \(10\,\text{cm}\) below the water surface. Its distance in front of the mirror is \[ H-d=30-10 \] \[ H-d=20\,\text{cm} \] A plane mirror forms a virtual image at the same distance behind its reflecting plane. Therefore, the virtual mirror image lies at depth \[ d_m=H+(H-d) \] \[ d_m=2H-d \] \[ d_m=2(30)-10 \] \[ d_m=50\,\text{cm} \] \(\textbf{Refract the mirror-image rays into air:}\) The rays approaching the water surface after reflection behave as though they originate from the virtual mirror image at depth \(50\,\text{cm}\). The air observer sees this virtual source through the plane refracting surface. Using the near-normal relation, \[ d_{\mathrm{mirror,app}}=\frac{d_m}{n_w} \] \[ d_{\mathrm{mirror,app}}=\frac{50}{4/3} \] \[ d_{\mathrm{mirror,app}}=37.5\,\text{cm} \] \(\textbf{Determine the apparent image separation:}\) The two apparent depths are \[ 7.5\,\text{cm}\quad\text{and}\quad37.5\,\text{cm} \] Therefore, \[ S_{\mathrm{app}}=37.5-7.5 \] \[ S_{\mathrm{app}}=30\,\text{cm} \] \(\textbf{Check the physical mirror-image separation:}\) Inside the water geometry, the actual mark and its plane-mirror image are separated by \[ 50-10=40\,\text{cm} \] The apparent separation viewed from air is reduced by the water index: \[ \frac{40}{4/3}=30\,\text{cm} \] This agrees with the separate-depth calculation. \(\textbf{Physical interpretation:}\) The direct image depends only on refraction of light from the actual mark through the water surface. The second image requires plane-mirror image formation before applying the same apparent-depth relation. Treating the bottom reflection as though it merely doubled the water depth would ignore the actual position of the luminous mark. The correct optical sequence first locates the virtual mirror image relative to the mirror and then determines how air-water refraction changes its apparent depth. The observer therefore sees two images with different apparent depths, even though both originate from the same physical luminous mark. \(\textbf{Final answer:}\) \[ \boxed{d_{\mathrm{direct}}=7.5\,\text{cm},\quad d_{\mathrm{mirror,app}}=37.5\,\text{cm},\quad S_{\mathrm{app}}=30\,\text{cm}} \]

Explanation: \(\textbf{Establish the original refraction geometry:}\) The refractive indices are \[ n_1=1,\qquad n_2=1.50 \] The real object distance is \[ u=-60\,\text{cm} \] Initially, \[ R_1=+20\,\text{cm} \] The object height is \[ h_o=+2.0\,\text{cm} \] \(\textbf{Calculate the original image distance:}\) For refraction at a spherical boundary, \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Substituting, \[ \frac{1.50}{v_1}-\frac1{-60}=\frac{1.50-1}{20} \] \[ \frac{1.50}{v_1}+\frac1{60}=\frac1{40} \] \[ \frac{1.50}{v_1}=\frac1{40}-\frac1{60} \] \[ \frac{1.50}{v_1}=\frac1{120} \] Therefore, \[ v_1=+180\,\text{cm} \] \(\textbf{Calculate the original transverse magnification:}\) For paraxial refraction at a single spherical surface, the transverse magnification is \[ m=\frac{n_1v}{n_2u} \] Therefore, \[ m_1=\frac{(1)(180)}{(1.50)(-60)} \] \[ m_1=-2 \] The signed image height is \[ h_{i1}=m_1h_o \] \[ h_{i1}=-2(2.0) \] \[ h_{i1}=-4.0\,\text{cm} \] The initial image is real, inverted and enlarged. \(\textbf{Change the radius while keeping the media and object fixed:}\) The replacement surface has \[ R_2=+30\,\text{cm} \] Using the same spherical-refraction equation, \[ \frac{1.50}{v_2}-\frac1{-60}=\frac{0.50}{30} \] \[ \frac{1.50}{v_2}+\frac1{60}=\frac1{60} \] Therefore, \[ \frac{1.50}{v_2}=0 \] \[ \frac1{v_2}=0 \] The emergent rays are parallel in the ideal paraxial model. \(\textbf{Interpret the curvature change:}\) Increasing the positive radius from \(20\,\text{cm}\) to \(30\,\text{cm}\) makes the refracting surface less strongly curved. With the refractive-index contrast unchanged, its focusing strength decreases. At the larger radius, the surface provides exactly enough refraction to transform the incident divergent wavefront into a plane emergent wavefront. Its image is therefore at infinity rather than at a finite distance. \(\textbf{Check the role of the first image magnification:}\) The first image's height is determined from its finite real image distance. At the second condition, the ideal paraxial image is at infinity. The finite-image height calculated for the first condition must not be carried over unchanged to the collimated case. \(\textbf{Physical interpretation:}\) A curved refracting boundary can change from finite real imaging to collimation when its curvature changes. The object's distance and the refractive indices need not change. The original image height also depends on the refractive-index-weighted transverse magnification rather than on the mirror magnification formula. The calculation combines signed spherical refraction, image orientation and the limiting condition for parallel emergent rays. \(\textbf{Final answer:}\) \[ \boxed{v_1=+180\,\text{cm},\quad m_1=-2,\quad h_{i1}=-4.0\,\text{cm},\quad v_2\to\infty} \]

Explanation: \(\textbf{Connect the fixed object-screen separation to lens distances:}\) For a real object and a real image formed on opposite sides of a thin lens, \[ u\lt 0,\qquad v\gt 0 \] Let the positive object-lens distance be \(p=-u\). The fixed object-screen separation is \[ D=p+v \] Therefore, \[ p+v=90\,\text{cm} \] \(\textbf{Use the two interchangeable imaging positions:}\) The thin-lens formula is \[ \frac1f=\frac1v-\frac1u \] Since \(u=-p\), \[ \frac1f=\frac1v+\frac1p \] This equation is unchanged when \(p\) and \(v\) are interchanged. Consequently, the two sharp-image positions have interchanged object and image distances. \(\textbf{Apply the measured displacement:}\) The separation of the two lens positions is \[ d=|v-p| \] Taking the position with \(v\gt p\), \[ v-p=30\,\text{cm} \] Together with \[ v+p=90\,\text{cm} \] this gives \[ 2v=120\,\text{cm} \] \[ v=60\,\text{cm} \] \[ p=30\,\text{cm} \] The second position therefore has \(p=60\,\text{cm}\) and \(v=30\,\text{cm}\). \(\textbf{Calculate the lens focal length:}\) Using the first position, \[ \frac1f=\frac1{60}+\frac1{30} \] \[ \frac1f=\frac3{60} \] \[ f=20\,\text{cm} \] \(\textbf{Determine the first signed magnification:}\) For a thin lens, \[ m=\frac vu \] At the first position, \[ u_1=-30\,\text{cm},\qquad v_1=+60\,\text{cm} \] Therefore, \[ m_1=\frac{60}{-30} \] \[ m_1=-2 \] \(\textbf{Determine the second signed magnification:}\) At the second position, \[ u_2=-60\,\text{cm},\qquad v_2=+30\,\text{cm} \] Thus, \[ m_2=\frac{30}{-60} \] \[ m_2=-\frac12 \] \(\textbf{Check the two-image condition:}\) For this arrangement, \[ D=90\,\text{cm} \] and \[ 4f=80\,\text{cm} \] Since \(D\gt 4f\), two distinct sharp-image positions are physically possible. Both magnifications are negative because the images are real and inverted. \(\textbf{Physical interpretation:}\) Moving a converging lens between a fixed object and screen can produce two sharp images when their separation is sufficiently large. The two positions interchange the object and image distances. One position gives an enlarged image; the other gives a diminished image. Their magnifications are reciprocal in magnitude, while the lens focal length remains unchanged. The measured displacement between lens positions therefore provides an indirect method of determining focal length without measuring either image distance separately. \(\textbf{Final answer:}\) \[ \boxed{f=20\,\text{cm},\quad (u_1,v_1,m_1)=(-30\,\text{cm},+60\,\text{cm},-2),\quad (u_2,v_2,m_2)=(-60\,\text{cm},+30\,\text{cm},-\tfrac12)} \]

Explanation: \(\textbf{Locate the initial image:}\) For the convex lens, \[ f_1=+20\,\text{cm} \] The real object distance is \[ u=-40\,\text{cm} \] The thin-lens formula gives \[ \frac1{v_1}=\frac1{f_1}+\frac1u \] Substituting, \[ \frac1{v_1}=\frac1{20}-\frac1{40} \] \[ \frac1{v_1}=\frac1{40} \] Therefore, \[ v_1=+40\,\text{cm} \] \(\textbf{Identify the original image orientation:}\) The transverse magnification is \[ m_1=\frac{v_1}{u} \] \[ m_1=\frac{40}{-40} \] \[ m_1=-1 \] The original image is real, inverted and equal in height to the object. \(\textbf{Combine the lenses in contact:}\) The added concave lens has \[ f_2=-30\,\text{cm} \] For two thin lenses in contact, \[ \frac1F=\frac1{f_1}+\frac1{f_2} \] Therefore, \[ \frac1F=\frac1{20}-\frac1{30} \] \[ \frac1F=\frac1{60} \] Thus, \[ F=+60\,\text{cm} \] The combination remains converging but is weaker than the original convex lens. \(\textbf{Find the changed image position:}\) The object remains at \[ u=-40\,\text{cm} \] Using the equivalent focal length, \[ \frac1{v_2}=\frac1F+\frac1u \] \[ \frac1{v_2}=\frac1{60}-\frac1{40} \] \[ \frac1{v_2}=-\frac1{120} \] Therefore, \[ v_2=-120\,\text{cm} \] \(\textbf{Determine the changed magnification:}\) \[ m_2=\frac{v_2}{u} \] \[ m_2=\frac{-120}{-40} \] \[ m_2=+3 \] \(\textbf{Interpret the new image regime:}\) The equivalent focal length is \(60\,\text{cm}\), but the object is only \(40\,\text{cm}\) away. The object therefore lies inside the focal length of the combination. The rays emerging from the combined lens diverge. Their backward extensions form a virtual erect enlarged image on the object side. \(\textbf{Physical interpretation:}\) Adding a diverging lens to a converging lens does not necessarily make the combination diverging. Here the net power remains positive, but the focal length increases. That changed focal length places the unchanged object inside the new focal region. A lens combination can therefore change an image from real and inverted to virtual and erect without changing the object's position. The sign of the equivalent power and the object's distance relative to the equivalent focus must both be considered. \(\textbf{Final answer:}\) \[ \boxed{(v_1,m_1)=(+40\,\text{cm},-1),\quad F=+60\,\text{cm},\quad (v_2,m_2)=(-120\,\text{cm},+3)} \]

Explanation: \(\textbf{Identify the curvature factor that remains unchanged:}\) For a thin lens in a homogeneous surrounding medium, \[ \frac1f=\left(\frac{n_g}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \] The glass index is \[ n_g=1.50 \] Let the fixed curvature factor be \[ K=\frac1{R_1}-\frac1{R_2} \] Immersion changes the relative refractive-index factor, not the lens geometry. \(\textbf{Determine the refractive-index factor in air:}\) With surrounding air, \[ n_{m1}=1 \] Therefore, \[ \frac{n_g}{n_{m1}}-1=1.50-1 \] \[ =+\frac12 \] The original focal length is \[ f_1=+20\,\text{cm} \] \(\textbf{Determine the factor in the surrounding liquid:}\) The liquid index is \[ n_{m2}=1.80 \] Thus, \[ \frac{n_g}{n_{m2}}-1=\frac{1.50}{1.80}-1 \] \[ =\frac56-1 \] \[ =-\frac16 \] The negative factor shows that the immersed biconvex lens acts as a diverging lens. \(\textbf{Compare the focal lengths:}\) Since the curvature factor is unchanged, \[ \frac{1/f_2}{1/f_1}=\frac{-1/6}{+1/2} \] \[ \frac{1/f_2}{1/f_1}=-\frac13 \] Therefore, \[ \frac1{f_2}=-\frac13\left(\frac1{20}\right) \] \[ f_2=-60\,\text{cm} \] \(\textbf{Determine the immersed lens power:}\) Converting focal length into metres, \[ f_2=-0.60\,\text{m} \] Therefore, \[ P_2=\frac1{f_2} \] \[ P_2=-\frac53\,\text{D} \] \(\textbf{Calculate the immersed-lens image distance:}\) The real object is at \[ u=-30\,\text{cm} \] The thin-lens formula gives \[ \frac1v=\frac1{f_2}+\frac1u \] \[ \frac1v=-\frac1{60}-\frac1{30} \] \[ \frac1v=-\frac1{20} \] \[ v=-20\,\text{cm} \] \(\textbf{Determine image orientation and size:}\) \[ m=\frac vu \] \[ m=\frac{-20}{-30} \] \[ m=+\frac23 \] The image is virtual, erect and diminished. \(\textbf{Physical interpretation:}\) A lens's converging or diverging action is determined by its refractive index relative to the surrounding medium. A geometrically biconvex lens can become diverging when immersed in a liquid whose refractive index exceeds that of the lens material. The refractive-index factor changes sign, reversing the sign of optical power. The unchanged real object then produces a virtual diminished image, consistent with the negative equivalent focal length. The result distinguishes the physical shape of a lens from its actual optical behaviour in a specified medium. \(\textbf{Final answer:}\) \[ \boxed{P_2=-\frac53\,\text{D},\quad f_2=-60\,\text{cm},\quad v=-20\,\text{cm},\quad m=+\frac23} \]

Explanation: \(\textbf{Apply the lens-maker equation in air:}\) For a thin lens, \[ \frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right) \] The glass refractive index is \[ n=1.50 \] Therefore, \[ n-1=0.50 \] \(\textbf{Determine the plano-convex focal length:}\) For lens P, one surface is plane, so its reciprocal radius is zero. The curved surface contributes a converging curvature factor of magnitude \[ K_P=\frac1{20}\,\text{cm}^{-1} \] Therefore, \[ \frac1{f_P}=0.50\left(\frac1{20}\right) \] \[ \frac1{f_P}=\frac1{40} \] \[ f_P=+40\,\text{cm} \] \(\textbf{Determine the symmetric biconvex focal length:}\) For lens Q, the signed surface radii are \[ R_1=+20\,\text{cm},\qquad R_2=-20\,\text{cm} \] Thus, \[ \frac1{f_Q}=0.50\left(\frac1{20}-\frac1{-20}\right) \] \[ \frac1{f_Q}=0.50\left(\frac2{20}\right) \] \[ f_Q=+20\,\text{cm} \] \(\textbf{Apply the common object condition:}\) In both experiments, \[ u=-30\,\text{cm} \] The image relation is \[ \frac1v=\frac1f+\frac1u \] \(\textbf{Calculate P's image:}\) \[ \frac1{v_P}=\frac1{40}-\frac1{30} \] \[ \frac1{v_P}=-\frac1{120} \] \[ v_P=-120\,\text{cm} \] The magnification is \[ m_P=\frac{v_P}{u} \] \[ m_P=\frac{-120}{-30} \] \[ m_P=+4 \] P forms a virtual erect enlarged image. \(\textbf{Calculate Q's image:}\) \[ \frac1{v_Q}=\frac1{20}-\frac1{30} \] \[ \frac1{v_Q}=\frac1{60} \] \[ v_Q=+60\,\text{cm} \] Therefore, \[ m_Q=\frac{60}{-30} \] \[ m_Q=-2 \] Q forms a real inverted enlarged image. \(\textbf{Physical interpretation:}\) A plano-convex lens has one curved refracting surface, whereas a symmetric biconvex lens has two curved surfaces contributing to its converging power. With the same material and curvature-radius magnitudes, the symmetric biconvex lens has twice the power of the plano-convex lens. The same object at \(30\,\text{cm}\) consequently lies inside P's focal length but outside Q's focal length. Thus, changing lens geometry changes not only the image distance but also whether the image is real or virtual and whether it is erect or inverted. The comparison requires combining lens curvature, focal length, signed image position and magnification. \(\textbf{Final answer:}\) \[ \boxed{(f_P,v_P,m_P)=(+40\,\text{cm},-120\,\text{cm},+4),\quad(f_Q,v_Q,m_Q)=(+20\,\text{cm},+60\,\text{cm},-2)} \]

Explanation: \(\textbf{Convert the first two focal lengths into powers:}\) For a thin lens, \[ P=\frac1f \] where the focal length is measured in metres. For lens P, \[ f_P=+0.20\,\text{m} \] \[ P_P=+5\,\text{D} \] For lens Q, \[ f_Q=-0.20\,\text{m} \] \[ P_Q=-5\,\text{D} \] \(\textbf{Determine the equivalent power of P and Q:}\) For thin lenses in contact, \[ P_{\mathrm{eq}}=P_P+P_Q \] \[ P_{\mathrm{eq}}=5-5 \] \[ P_{\mathrm{eq}}=0 \] The equivalent focal length has infinite magnitude. \(\textbf{Determine the image when the combination has zero power:}\) The object distance is \[ u=-30\,\text{cm} \] The thin-lens equation for the equivalent system is \[ \frac1v-\frac1u=\frac1F \] Since \[ \frac1F=0 \] we obtain \[ \frac1v=\frac1u \] Therefore, \[ v=-30\,\text{cm} \] The image is virtual and lies at the object-side position. \(\textbf{Determine its magnification:}\) \[ m=\frac vu \] \[ m=\frac{-30}{-30} \] \[ m=+1 \] In the ideal thin-lens contact approximation, the arrangement has unit transverse magnification. \(\textbf{Add the third convex lens:}\) Lens R has \[ f_R=+0.10\,\text{m} \] Its power is \[ P_R=+10\,\text{D} \] The new total power is \[ P'_{\mathrm{eq}}=5-5+10 \] \[ P'_{\mathrm{eq}}=+10\,\text{D} \] Therefore, \[ F'=+10\,\text{cm} \] \(\textbf{Calculate the changed image distance:}\) \[ \frac1{v'}=\frac1{F'}+\frac1u \] \[ \frac1{v'}=\frac1{10}-\frac1{30} \] \[ \frac1{v'}=\frac1{15} \] \[ v'=+15\,\text{cm} \] \(\textbf{Calculate the changed magnification:}\) \[ m'=\frac{v'}u \] \[ m'=\frac{15}{-30} \] \[ m'=-\frac12 \] The new image is real, inverted and diminished. \(\textbf{Physical interpretation:}\) Equal and opposite lens powers cancel in the thin-lens contact approximation. A zero-power combination does not automatically convert the divergent light from a finite real object into a parallel beam. Instead, it leaves the incident vergence unchanged, so the virtual image remains at the original object-side position. Adding a third converging lens creates nonzero net power and changes the image into a real inverted one. The result distinguishes zero optical power from the separate condition required for collimating light from a finite object. \(\textbf{Final answer:}\) \[ \boxed{P_{PQ}=0,\quad(v_{PQ},m_{PQ})=(-30\,\text{cm},+1),\quad P_{PQR}=+10\,\text{D},\quad(v_{PQR},m_{PQR})=(+15\,\text{cm},-\tfrac12)} \]

Explanation: \(\textbf{Derive the graph relation for a thin lens:}\) The thin-lens equation is \[ \frac1v-\frac1u=\frac1f \] The signed transverse magnification is \[ m=\frac vu \] Therefore, \[ v=mu \] Substituting, \[ \frac1{mu}-\frac1u=\frac1f \] \[ \frac{1/m-1}{u}=\frac1f \] Rearranging, \[ \frac1m=1+\frac uf \] \(\textbf{Identify the meaning of the graph slope:}\) Writing the equation as \[ Y=1+Su \] where \(Y=1/m\), its slope is \[ S=\frac1f \] The intercept is \(+1\), consistent with the described graph. \(\textbf{Calculate the experimental slope:}\) Using the two measured points, \[ S=\frac{0-(-2)}{-20-(-60)} \] \[ S=\frac2{40} \] \[ S=+0.05\,\text{cm}^{-1} \] \(\textbf{Recover the focal length:}\) \[ f=\frac1S \] \[ f=\frac1{0.05} \] \[ f=+20\,\text{cm} \] The positive focal length identifies a converging convex lens in air. \(\textbf{Predict the reciprocal magnification at the new object position:}\) The new signed object distance is \[ u=-30\,\text{cm} \] Thus, \[ \frac1m=1+\frac{-30}{20} \] \[ \frac1m=-\frac12 \] Therefore, \[ m=-2 \] \(\textbf{Recover the image distance:}\) Using \[ v=mu \] gives \[ v=(-2)(-30) \] \[ v=+60\,\text{cm} \] \(\textbf{Verify the lens equation:}\) \[ \frac1v-\frac1u=\frac1{60}-\frac1{-30} \] \[ =\frac1{60}+\frac2{60} \] \[ =\frac1{20} \] The result agrees with the recovered focal length. \(\textbf{Interpret the image:}\) A positive image distance corresponds to a real image on the opposite side of the lens. The negative magnification indicates inversion. Its magnitude of two indicates that the image height is twice the object height. The object is between the focal point and twice the focal distance, consistent with a real enlarged image. \(\textbf{Physical interpretation:}\) The reciprocal-magnification graph is a linear representation of thin-lens imaging. Its slope is the reciprocal focal length, while its intercept is fixed at unity by the lens equation and magnification definition. The sign of the slope identifies whether the lens is converging or diverging. Once the focal length is inferred from measured data, a new object's image position and orientation follow from the same graph. \(\textbf{Final answer:}\) \[ \boxed{f=+20\,\text{cm},\quad u=-30\,\text{cm},\quad v=+60\,\text{cm},\quad m=-2} \]

Explanation: \(\textbf{Translate the final observed magnification into an image distance:}\) The object distance is \[ u=-60\,\text{cm} \] The final image is erect and half the object's height. Therefore, its signed magnification is \[ m=+\frac12 \] For a thin lens, \[ m=\frac vu \] Thus, \[ v=mu \] \[ v=\frac12(-60) \] \[ v=-30\,\text{cm} \] The negative image distance is consistent with a virtual image on the object side. \(\textbf{Recover the equivalent focal length:}\) The thin-lens formula gives \[ \frac1F=\frac1v-\frac1u \] Substituting, \[ \frac1F=-\frac1{30}+\frac1{60} \] \[ \frac1F=-\frac1{60} \] Therefore, \[ F=-60\,\text{cm} \] The final combination is diverging. \(\textbf{Calculate the equivalent optical power:}\) Converting the equivalent focal length into metres, \[ F=-0.60\,\text{m} \] Therefore, \[ P_{\mathrm{eq}}=\frac1F \] \[ P_{\mathrm{eq}}=-\frac53\,\text{D} \] \(\textbf{Calculate the original convex-lens power:}\) The original focal length is \[ f_1=+30\,\text{cm}=+0.30\,\text{m} \] Therefore, \[ P_1=\frac1{0.30} \] \[ P_1=+\frac{10}{3}\,\text{D} \] \(\textbf{Recover the unknown added power:}\) For lenses in contact, \[ P_{\mathrm{eq}}=P_1+P_2 \] Thus, \[ P_2=P_{\mathrm{eq}}-P_1 \] \[ P_2=-\frac53-\frac{10}{3} \] \[ P_2=-5\,\text{D} \] \(\textbf{Determine the added lens focal length:}\) \[ f_2=\frac1{P_2} \] \[ f_2=-0.20\,\text{m} \] \[ f_2=-20\,\text{cm} \] The added lens is concave and diverging in air. \(\textbf{Check the original image condition:}\) Before adding the second lens, the object is at twice the original convex lens's focal length: \[ |u|=60\,\text{cm}=2f_1 \] Therefore, the original lens produces a real inverted equal-sized image at \(v=+60\,\text{cm}\). This is consistent with the stated observation. \(\textbf{Physical interpretation:}\) An unknown constituent lens can be identified from the changed image magnification of a contact combination. The image magnification first determines the signed image position. The image position then determines the equivalent focal length and power. Finally, subtracting the known lens power reveals the added lens. The final virtual diminished image establishes that the combined lens power has become negative despite the presence of the original convex lens. The inverse calculation must therefore preserve both image and power signs throughout. \(\textbf{Final answer:}\) \[ \boxed{f_2=-20\,\text{cm},\quad P_2=-5\,\text{D},\quad F=-60\,\text{cm}} \]

Explanation: \(\textbf{Convert the constituent focal lengths into powers:}\) For each thin lens, \[ P=\frac1f \] with \(f\) in metres. For P, \[ f_P=+0.25\,\text{m} \] \[ P_P=+4\,\text{D} \] For Q, \[ f_Q=-0.40\,\text{m} \] \[ P_Q=-2.5\,\text{D} \] For R, \[ f_R=+0.50\,\text{m} \] \[ P_R=+2\,\text{D} \] \(\textbf{Find the powers of the three possible pairs:}\) For P+Q, \[ P_{PQ}=4-2.5 \] \[ P_{PQ}=+1.5\,\text{D} \] For P+R, \[ P_{PR}=4+2 \] \[ P_{PR}=+6\,\text{D} \] For Q+R, \[ P_{QR}=-2.5+2 \] \[ P_{QR}=-0.5\,\text{D} \] \(\textbf{Translate the fixed object position into vergence:}\) The object distance is \[ u=-0.50\,\text{m} \] For a contact combination of power \(P_{\mathrm{pair}}\), \[ \frac1v=P_{\mathrm{pair}}+\frac1u \] Therefore, \[ \frac1v=P_{\mathrm{pair}}-2\,\text{D} \] A real image requires \(v\gt 0\), so the right-hand side must be positive. \(\textbf{Test P+Q:}\) \[ \frac1{v_{PQ}}=1.5-2 \] \[ \frac1{v_{PQ}}=-0.5\,\text{m}^{-1} \] Thus, \[ v_{PQ}=-2.0\,\text{m} \] This image is virtual. \(\textbf{Test P+R:}\) \[ \frac1{v_{PR}}=6-2 \] \[ \frac1{v_{PR}}=+4\,\text{m}^{-1} \] Therefore, \[ v_{PR}=+0.25\,\text{m} \] \[ v_{PR}=+25\,\text{cm} \] This image is real. \(\textbf{Determine its magnification:}\) \[ m_{PR}=\frac{v_{PR}}u \] \[ m_{PR}=\frac{+0.25}{-0.50} \] \[ m_{PR}=-\frac12 \] The image is inverted and diminished. \(\textbf{Test Q+R:}\) \[ \frac1{v_{QR}}=-0.5-2 \] \[ \frac1{v_{QR}}=-2.5\,\text{m}^{-1} \] \[ v_{QR}=-0.40\,\text{m} \] This image is virtual. \(\textbf{Physical interpretation:}\) A positive equivalent power does not guarantee a real image from every finite real-object position. The P+Q pair is converging, but its focal length exceeds the object's distance, so it produces a virtual image. The P+R pair has sufficient converging power to form a real image at the specified object distance. The Q+R pair is diverging and forms a virtual image as expected. Classifying the image therefore requires comparing each pair's power with the incident object's vergence, not merely checking whether the equivalent power is positive. \(\textbf{Final answer:}\) \[ \boxed{P_{PR}=+6\,\text{D},\quad v_{PR}=+25\,\text{cm},\quad m_{PR}=-\frac12} \]

Explanation: Statement I is correct because each object point sends light through many different parts of the lens aperture. The lower unobstructed part can still refract rays from every visible object point towards its corresponding geometrical image point. Blocking the upper part of the lens does not remove the upper part of the object from the scene. Statement II is incorrect because different spatial regions of the lens are not assigned exclusively to correspondingly positioned parts of the object. The upper semicircle of the lens does not form only the upper half of the image. Image formation depends on the directions and convergence of rays from each object point. A sufficiently large remaining portion of the lens can continue to produce the complete geometrical image. Statement III is correct because covering part of the aperture does not change the refractive index, surface curvature or focal length of the remaining lens. The thin-lens equation remains \[ \frac1v-\frac1u=\frac1f \] The object distance also remains unchanged. Therefore, the image distance and transverse magnification remain unchanged in the stated ideal ray-optics model. Statement IV is correct because the opaque mask removes some rays that would otherwise reach the screen. With approximately uniform illumination over the original aperture, reducing the transmitting area reduces the light reaching corresponding image regions. The image is therefore generally dimmer. The mask changes aperture transmission, not the geometrical mapping between object points and image points. In a more detailed wave-optical description, aperture shape can influence diffraction, but that effect has explicitly been neglected here. Only Statements I, III and IV are correct.

Explanation: \(\textbf{Calculate the original meniscus-lens power:}\) For a thin lens in air, \[ \frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right) \] The refractive index is \[ n=1.50 \] The original signed radii are \[ R_1=+20\,\text{cm},\qquad R_2=+10\,\text{cm} \] Therefore, \[ \frac1{f_1}=0.50\left(\frac1{20}-\frac1{10}\right) \] \[ \frac1{f_1}=0.50\left(-\frac1{20}\right) \] \[ \frac1{f_1}=-\frac1{40} \] Thus, \[ f_1=-40\,\text{cm} \] The lens is diverging in air. \(\textbf{Determine how the signed surface radii change on turning:}\) Turning the lens around exchanges its physical surfaces. Because the positive propagation direction remains left to right, the signs of the corresponding radii also reverse. The new first and second radii are \[ R'_1=-R_2=-10\,\text{cm} \] \[ R'_2=-R_1=-20\,\text{cm} \] \(\textbf{Recalculate the focal length after turning:}\) \[ \frac1{f_2}=0.50\left(\frac1{-10}-\frac1{-20}\right) \] \[ \frac1{f_2}=0.50\left(-\frac1{10}+\frac1{20}\right) \] \[ \frac1{f_2}=-\frac1{40} \] Therefore, \[ f_2=-40\,\text{cm} \] \(\textbf{Apply the unchanged object distance:}\) The object is real and lies at \[ u=-20\,\text{cm} \] For either orientation, \[ \frac1v=\frac1f+\frac1u \] Substituting, \[ \frac1v=-\frac1{40}-\frac1{20} \] \[ \frac1v=-\frac3{40} \] Therefore, \[ v=-\frac{40}{3}\,\text{cm} \] \(\textbf{Determine the common magnification:}\) \[ m=\frac vu \] \[ m=\frac{-40/3}{-20} \] \[ m=+\frac23 \] The image is virtual, erect and diminished in either orientation. \(\textbf{Check the curvature invariance algebraically:}\) On reversal, \[ \frac1{R'_1}-\frac1{R'_2} = -\frac1{R_2}+\frac1{R_1} \] This equals the original curvature factor: \[ \frac1{R_1}-\frac1{R_2} \] Thus, the thin-lens focal length is unchanged. \(\textbf{Physical interpretation:}\) Turning an asymmetric thin lens around changes which physical surface the light encounters first. However, both signed curvature radii must be reassigned consistently. The resulting curvature factor in the lens-maker equation remains unchanged for the same material and surrounding medium. The negative focal length is not converted into a positive one merely because the lens is reversed. The image position and magnification consequently remain unchanged in the thin-lens approximation. \(\textbf{Final answer:}\) \[ \boxed{f_1=f_2=-40\,\text{cm},\quad v=-\frac{40}{3}\,\text{cm},\quad m=+\frac23} \]

Explanation: \(\textbf{Combine the original powers in air:}\) The individual air powers are \[ P_{A,\mathrm{air}}=+5.0\,\text{D} \] \[ P_{B,\mathrm{air}}=-3.0\,\text{D} \] For thin lenses in contact, \[ P_{\mathrm{eq,air}}=P_{A,\mathrm{air}}+P_{B,\mathrm{air}} \] \[ P_{\mathrm{eq,air}}=5.0-3.0 \] \[ P_{\mathrm{eq,air}}=+2.0\,\text{D} \] \(\textbf{Calculate the initial image position:}\) The real object distance is \[ u=-1.0\,\text{m} \] For a thin-lens combination, \[ \frac1v=P_{\mathrm{eq}}+\frac1u \] Thus, \[ \frac1{v_{\mathrm{air}}}=2.0-1.0 \] \[ v_{\mathrm{air}}=+1.0\,\text{m} \] The initial image is real and inverted. \(\textbf{Determine how A's power changes on immersion:}\) For fixed lens geometry, optical power is proportional to the relative-index factor \[ \left(\frac{n_g}{n_m}-1\right) \] For lens A, the factor in air is \[ 1.50-1=0.50 \] In liquid of index \(1.25\), it becomes \[ \frac{1.50}{1.25}-1=0.20 \] Therefore, \[ P_{A,\mathrm{liq}}=5.0\left(\frac{0.20}{0.50}\right) \] \[ P_{A,\mathrm{liq}}=+2.0\,\text{D} \] \(\textbf{Determine how B's power changes:}\) For B, the original relative-index factor is \[ 1.60-1=0.60 \] After immersion, \[ \frac{1.60}{1.25}-1=0.28 \] The biconcave geometry remains unchanged, so \[ P_{B,\mathrm{liq}}=-3.0\left(\frac{0.28}{0.60}\right) \] \[ P_{B,\mathrm{liq}}=-1.4\,\text{D} \] \(\textbf{Find the immersed equivalent power:}\) \[ P_{\mathrm{eq,liq}}=P_{A,\mathrm{liq}}+P_{B,\mathrm{liq}} \] \[ P_{\mathrm{eq,liq}}=2.0-1.4 \] \[ P_{\mathrm{eq,liq}}=+0.6\,\text{D} \] The immersed combination remains converging, but its equivalent focal length increases. \(\textbf{Determine the changed image position:}\) The object remains at \[ u=-1.0\,\text{m} \] Therefore, \[ \frac1{v_{\mathrm{liq}}}=0.6-1.0 \] \[ \frac1{v_{\mathrm{liq}}}=-0.4\,\text{m}^{-1} \] \[ v_{\mathrm{liq}}=-2.5\,\text{m} \] \(\textbf{Interpret the image changes:}\) In air, the combination has focal length \(0.50\,\text{m}\). The object at \(1.0\,\text{m}\) is outside that focal length and forms a real image. In liquid, the focal length is \[ F_{\mathrm{liq}}=\frac1{0.6}=\frac53\,\text{m} \] The same object lies inside the new focal distance. The immersed image is therefore virtual and erect, even though the net power remains positive. \(\textbf{Physical interpretation:}\) The powers of two thin lenses in contact add only after each constituent power has been evaluated in its actual surrounding medium. Because lenses A and B are made from glasses of different refractive indices, immersion changes their powers by different factors. The original positive and negative powers cannot simply be multiplied by one common factor. Their new algebraic sum remains positive but becomes small enough to change the image regime for the unchanged object. The result connects relative refractive index, lens curvature, equivalent power and real-to-virtual image transition. \(\textbf{Final answer:}\) \[ \boxed{(P_{\mathrm{eq}},v)_{\mathrm{air}}=(+2.0\,\text{D},+1.0\,\text{m}),\quad(P_{\mathrm{eq}},v)_{\mathrm{liq}}=(+0.6\,\text{D},-2.5\,\text{m})} \]

Explanation: \(\textbf{Identify the fixed geometrical factor:}\) For a thin lens in air, \[ \frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right) \] The lens radii do not change with the colour of light. Therefore, the reciprocal focal length is proportional to \(n-1\). \(\textbf{Compare the two refractive-index factors:}\) For red light, \[ n_R-1=1.50-1 \] \[ n_R-1=0.50 \] For violet light, \[ n_V-1=1.60-1 \] \[ n_V-1=0.60 \] Thus, \[ \frac{1/f_V}{1/f_R}=\frac{0.60}{0.50} \] \[ \frac{f_R}{f_V}=\frac65 \] \(\textbf{Determine the violet focal length:}\) The red focal length is \[ f_R=+20\,\text{cm} \] Therefore, \[ f_V=20\left(\frac56\right) \] \[ f_V=\frac{50}{3}\,\text{cm} \] The violet focal length is shorter because the lens material has a greater refractive index for violet light. \(\textbf{Calculate the red image distance:}\) The real object is at \[ u=-60\,\text{cm} \] Using the thin-lens equation, \[ \frac1{v_R}=\frac1{f_R}+\frac1u \] \[ \frac1{v_R}=\frac1{20}-\frac1{60} \] \[ \frac1{v_R}=\frac1{30} \] \[ v_R=+30\,\text{cm} \] \(\textbf{Calculate the violet image distance:}\) \[ \frac1{v_V}=\frac1{f_V}+\frac1u \] Substituting, \[ \frac1{v_V}=\frac3{50}-\frac1{60} \] \[ \frac1{v_V}=\frac{18-5}{300} \] \[ \frac1{v_V}=\frac{13}{300} \] Therefore, \[ v_V=\frac{300}{13}\,\text{cm} \] \(\textbf{Determine the separation of the image planes:}\) The red image lies farther from the lens. Their separation is \[ S=v_R-v_V \] \[ S=30-\frac{300}{13} \] \[ S=\frac{390-300}{13} \] \[ S=\frac{90}{13}\,\text{cm} \] \(\textbf{Check the two image natures:}\) Both image distances are positive. Thus, both images are real and inverted. They are formed at different distances because the lens has different optical powers for the two wavelengths. \(\textbf{Physical interpretation:}\) A lens made from a dispersive material need not have the same focal length for all colours. A greater refractive index produces stronger refraction for the same lens geometry. The violet rays therefore converge more strongly than the red rays in this arrangement. With a fixed object position, the resulting image planes are separated. The effect illustrates how wavelength-dependent refractive index can prevent a simple lens from bringing all colours to the same image plane. \(\textbf{Final answer:}\) \[ \boxed{f_V=\frac{50}{3}\,\text{cm},\quad v_R=+30\,\text{cm},\quad v_V=+\frac{300}{13}\,\text{cm},\quad S=\frac{90}{13}\,\text{cm}} \]

Explanation: \(\textbf{Use the relative refractive-index factor:}\) For a thin lens immersed in a homogeneous medium, \[ \frac1f=\left(\frac{n_g}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \] The sign of the lens power depends on both the relative-index factor and the geometrical curvature factor. The surrounding liquid index is \[ n_m=1.60 \] Statement I is correct because P is biconvex, so its curvature factor is positive for the standard lens orientation. However, its material index is only \(1.50\). Therefore, \[ \frac{n_P}{n_m}-1=\frac{1.50}{1.60}-1 \] This factor is negative. The immersed biconvex lens therefore has negative power and acts as a diverging lens. Statement II is correct because Q is biconcave, so its geometrical curvature factor is negative. Its material index is \(1.40\), less than the liquid index. Thus, \[ \frac{n_Q}{n_m}-1=\frac{1.40}{1.60}-1 \] This factor is also negative. The product of the two negative factors is positive, so Q acts as a converging lens in the liquid. Statement III is correct because R has the same refractive index as its surroundings. Therefore, \[ \frac{n_R}{n_m}-1=\frac{1.60}{1.60}-1 \] \[ =0 \] Its lens power is zero irrespective of its plano-convex curvature. Statement IV is correct within the specified thin-lens model. For a real object at signed distance \(u\), the lens equation is \[ \frac1v-\frac1u=\frac1f \] When the refracting power is zero, \[ \frac1f=0 \] Thus, \[ \frac1v=\frac1u \] \[ v=u \] The signed image distance remains on the object side. The transverse magnification is \[ m=\frac vu \] \[ m=+1 \] The zero-power lens therefore does not create a new converging image from the divergent rays of a finite real object. The rays behave, in the ideal thin-lens description, as though the optically matched element were absent. The central lesson is that physical lens shape alone does not determine optical behaviour. Relative refractive index and curvature must be considered together. All four statements are correct.

Explanation: \(\textbf{Use the known material factor:}\) The glass refractive index is \[ n=1.50 \] For a thin lens in air, \[ \frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right) \] Therefore, \[ n-1=0.50 \] \(\textbf{Extract the first-surface curvature from the plano-convex lens:}\) The second surface is plane after modification: \[ \frac1{R'_2}=0 \] The first surface remains unchanged. The modified focal length is \[ f'=+50\,\text{cm} \] Applying the lens-maker equation, \[ \frac1{50}=0.50\left(\frac1{R_1}-0\right) \] Thus, \[ \frac1{R_1}=\frac2{50} \] \[ \frac1{R_1}=\frac1{25} \] Therefore, \[ R_1=+25\,\text{cm} \] \(\textbf{Apply the original biconvex-lens focal length:}\) The original focal length is \[ f=+20\,\text{cm} \] Therefore, \[ \frac1{20}=0.50\left(\frac1{25}-\frac1{R_2}\right) \] Multiplying by two, \[ \frac1{10}=\frac1{25}-\frac1{R_2} \] \(\textbf{Recover the unknown second curvature:}\) Rearranging, \[ -\frac1{R_2}=\frac1{10}-\frac1{25} \] \[ -\frac1{R_2}=\frac{5-2}{50} \] \[ -\frac1{R_2}=\frac3{50} \] Therefore, \[ \frac1{R_2}=-\frac3{50} \] \[ R_2=-\frac{50}{3}\,\text{cm} \] \(\textbf{Check the original focal length:}\) Substituting both recovered radii, \[ \frac1f=0.50\left(\frac1{25}-\frac1{-50/3}\right) \] \[ \frac1f=0.50\left(\frac2{50}+\frac3{50}\right) \] \[ \frac1f=0.50\left(\frac1{10}\right) \] \[ f=20\,\text{cm} \] The original measurement is reproduced. \(\textbf{Interpret the surface contributions:}\) The first surface alone produces the weaker plano-convex power. The original second convex surface contributes additional positive converging power. Replacing it with a plane face therefore increases the focal length from \(20\,\text{cm}\) to \(50\,\text{cm}\). The recovered second radius is negative, as required by the stated biconvex geometry. \(\textbf{Physical interpretation:}\) Two focal-length measurements can reveal both unknown surface radii of a thin lens when one surface is deliberately made plane. The plano-convex measurement isolates the contribution of the unchanged first surface. The original biconvex measurement then determines the second curvature contribution. The signed radii are recovered from optical measurements rather than being supplied through a direct geometrical measurement. The process combines a controlled change in surface shape with an inverse application of the lens-maker formula. \(\textbf{Final answer:}\) \[ \boxed{R_1=+25\,\text{cm},\quad R_2=-\frac{50}{3}\,\text{cm}} \]

Explanation: \(\textbf{Verify the initial image condition:}\) The convex lens has \[ f_1=+20\,\text{cm} \] The real object and screen distances are \[ u_1=-30\,\text{cm} \] \[ v_1=+60\,\text{cm} \] The thin-lens formula gives \[ \frac1{v_1}-\frac1{u_1}=\frac1{60}+\frac1{30} \] \[ =\frac1{20} \] Thus, the initial screen position is consistent with the convex lens focal length. \(\textbf{Calculate the original signed magnification:}\) \[ m_1=\frac{v_1}{u_1} \] \[ m_1=\frac{60}{-30} \] \[ m_1=-2 \] The original image is real, inverted and twice the object's height. \(\textbf{Determine the contact combination's focal length:}\) The added concave lens has \[ f_2=-60\,\text{cm} \] For two thin lenses in contact, \[ \frac1F=\frac1{f_1}+\frac1{f_2} \] Substituting, \[ \frac1F=\frac1{20}-\frac1{60} \] \[ \frac1F=\frac2{60} \] Therefore, \[ F=+30\,\text{cm} \] The combination is converging but has a longer focal length than the original lens. \(\textbf{Keep the image at the original screen:}\) The screen cannot move. Therefore, the desired new signed image distance remains \[ v_2=+60\,\text{cm} \] The new object distance \(u_2\) must satisfy \[ \frac1{v_2}-\frac1{u_2}=\frac1F \] Rearranging, \[ \frac1{u_2}=\frac1{v_2}-\frac1F \] \(\textbf{Solve for the required object position:}\) \[ \frac1{u_2}=\frac1{60}-\frac1{30} \] \[ \frac1{u_2}=-\frac1{60} \] Therefore, \[ u_2=-60\,\text{cm} \] \(\textbf{Calculate the necessary object movement:}\) Initially, \[ u_1=-30\,\text{cm} \] Finally, \[ u_2=-60\,\text{cm} \] Thus, \[ \Delta u=u_2-u_1 \] \[ \Delta u=-60-(-30) \] \[ \Delta u=-30\,\text{cm} \] The negative displacement means the object must move \(30\,\text{cm}\) away from the lens. \(\textbf{Determine the restored image magnification:}\) \[ m_2=\frac{v_2}{u_2} \] \[ m_2=\frac{60}{-60} \] \[ m_2=-1 \] The restored image is real, inverted and equal in height to the object. \(\textbf{Check the changed focal geometry:}\) The new focal length is \(30\,\text{cm}\). The restored object distance is \(60\,\text{cm}\), exactly twice that focal length. Therefore, an equal-sized inverted image at twice the focal length on the opposite side is expected. This independently confirms the calculated magnification and screen position. \(\textbf{Physical interpretation:}\) Adding a negative-power lens weakens the original convex lens and increases the equivalent focal length. If the image screen must remain fixed, the object may need to be repositioned to compensate for the changed focusing power. The new object distance is determined from the specified final image plane rather than from a simple proportional scaling of the original distance. Here the object moves farther away, and the original enlarged real image becomes equal-sized. This is an inverse optical-adjustment problem involving lens power, fixed-screen imaging and transverse magnification. \(\textbf{Final answer:}\) \[ \boxed{F=+30\,\text{cm},\quad u_2=-60\,\text{cm},\quad\Delta u=-30\,\text{cm},\quad m_1=-2,\quad m_2=-1} \]

Explanation: \(\textbf{Locate the image along the principal axis:}\) The signed object distance and focal length are \[ u=-30\,\text{cm} \] \[ f=+20\,\text{cm} \] The thin-lens formula is \[ \frac1v-\frac1u=\frac1f \] Therefore, \[ \frac1v=\frac1{20}-\frac1{30} \] \[ \frac1v=\frac1{60} \] \[ v=+60\,\text{cm} \] The axial image distance remains unchanged when the lens is translated sideways without rotation. \(\textbf{Determine the transverse magnification:}\) \[ m=\frac vu \] \[ m=\frac{60}{-30} \] \[ m=-2 \] The negative sign means the image is inverted relative to the optical centre and principal axis of the lens. \(\textbf{Calculate the initial image height:}\) The initial lens-centre height is \[ y_{L1}=0 \] The object height relative to the fixed laboratory axis is \[ y_O=+2\,\text{mm} \] The image height relative to the lens centre is \[ y_{I1}-y_{L1}=m(y_O-y_{L1}) \] Substituting, \[ y_{I1}=(-2)(2) \] \[ y_{I1}=-4\,\text{mm} \] \(\textbf{Account for the translated optical centre:}\) The lens moves upwards by \[ \Delta y_L=+5\,\text{mm} \] Its new optical-centre height is \[ y_{L2}=+5\,\text{mm} \] The object remains at \(+2\,\text{mm}\). Therefore, its height relative to the new lens axis becomes \[ y_O-y_{L2}=2-5 \] \[ y_O-y_{L2}=-3\,\text{mm} \] \(\textbf{Calculate the final image height:}\) The image height relative to the translated lens axis is \[ y_{I2}-y_{L2}=m(y_O-y_{L2}) \] \[ y_{I2}-5=(-2)(-3) \] \[ y_{I2}-5=+6 \] \[ y_{I2}=+11\,\text{mm} \] \(\textbf{Determine the image displacement:}\) \[ \Delta y_I=y_{I2}-y_{I1} \] \[ \Delta y_I=11-(-4) \] \[ \Delta y_I=+15\,\text{mm} \] \(\textbf{Check the general displacement relationship:}\) For a stationary object and fixed axial magnification, \[ y_I=y_L+m(y_O-y_L) \] Therefore, \[ \Delta y_I=(1-m)\Delta y_L \] \[ \Delta y_I=[1-(-2)](5) \] \[ \Delta y_I=15\,\text{mm} \] \(\textbf{Physical interpretation:}\) Transverse magnification is defined relative to the lens's own principal axis, not an arbitrary fixed laboratory axis. When a lens is translated sideways, its optical centre and principal axis move together. The object consequently acquires a different signed height relative to that moving axis. The image's laboratory displacement includes both the direct displacement of the lens and the magnified change in relative object height. A lens translation of \(5\,\text{mm}\) can therefore move the image by \(15\,\text{mm}\) in this arrangement. The image remains real and has the same axial distance and transverse magnification throughout. \(\textbf{Final answer:}\) \[ \boxed{y_{I1}=-4\,\text{mm},\quad y_{I2}=+11\,\text{mm},\quad\Delta y_I=+15\,\text{mm}} \]

Explanation: \(\textbf{Assign magnification signs from image orientation:}\) The convex lens has \[ f=+15\,\text{cm} \] In experiment P, the real inverted image has magnification \[ m_P=-3 \] In experiment Q, the virtual erect image has magnification \[ m_Q=+3 \] The equal magnification magnitudes must not be substituted with the same sign. \(\textbf{Connect magnification to object distance:}\) For a thin lens, \[ \frac1v-\frac1u=\frac1f \] and \[ m=\frac vu \] Therefore, \[ v=mu \] Substituting, \[ \frac1{mu}-\frac1u=\frac1f \] \[ \frac{1-m}{mu}=\frac1f \] Rearranging, \[ u=f\left(\frac1m-1\right) \] \(\textbf{Determine the object position in P:}\) Using \[ m_P=-3 \] gives \[ u_P=15\left(-\frac13-1\right) \] \[ u_P=15\left(-\frac43\right) \] \[ u_P=-20\,\text{cm} \] \(\textbf{Determine P's real image distance:}\) \[ v_P=m_Pu_P \] \[ v_P=(-3)(-20) \] \[ v_P=+60\,\text{cm} \] The positive sign locates a real image on the opposite side of the lens. \(\textbf{Determine the object position in Q:}\) For \[ m_Q=+3 \] \[ u_Q=15\left(\frac13-1\right) \] \[ u_Q=15\left(-\frac23\right) \] \[ u_Q=-10\,\text{cm} \] This position lies inside the lens's focal distance. \(\textbf{Determine Q's virtual image distance:}\) \[ v_Q=m_Qu_Q \] \[ v_Q=(3)(-10) \] \[ v_Q=-30\,\text{cm} \] The negative sign identifies a virtual image on the object side. \(\textbf{Calculate the object displacement:}\) \[ \Delta u=u_Q-u_P \] \[ \Delta u=-10-(-20) \] \[ \Delta u=+10\,\text{cm} \] Thus, the object moves towards the lens. \(\textbf{Calculate the separation of the image positions:}\) The images lie on opposite sides of the fixed lens. Their axial separation is \[ S=|v_P-v_Q| \] \[ S=|60-(-30)| \] \[ S=90\,\text{cm} \] \(\textbf{Physical interpretation:}\) A convex lens can produce equally enlarged real and virtual images at different object distances. The image orientation changes as the object crosses the principal focus. The sign of magnification is therefore essential in identifying the correct object position. Although both images are three times the object's height, their signed positions differ substantially. One image can be captured on a screen, while the other is visible only through the backward extensions of divergent rays. The inverse calculation separates magnification magnitude, image orientation, object displacement and image-position separation. \(\textbf{Final answer:}\) \[ \boxed{u_P=-20\,\text{cm},\quad u_Q=-10\,\text{cm},\quad\Delta u=+10\,\text{cm},\quad S=90\,\text{cm}} \]

Explanation: \(\textbf{Establish the original lens imaging condition:}\) The convex lens has \[ f=+20\,\text{cm} \] Initially, \[ u_1=-30\,\text{cm} \] The thin-lens formula gives \[ \frac1{v_1}=\frac1f+\frac1{u_1} \] \[ \frac1{v_1}=\frac1{20}-\frac1{30} \] \[ \frac1{v_1}=\frac1{60} \] Thus, \[ v_1=+60\,\text{cm} \] \(\textbf{Determine the initial magnification:}\) \[ m_1=\frac{v_1}{u_1} \] \[ m_1=\frac{60}{-30} \] \[ m_1=-2 \] The original image is real, inverted and enlarged. \(\textbf{Move the object farther from the lens:}\) The object's distance increases by \(5\,\text{cm}\). Therefore, \[ u_2=-35\,\text{cm} \] The lens focal length remains fixed. \(\textbf{Calculate the new image distance:}\) \[ \frac1{v_2}=\frac1{20}-\frac1{35} \] \[ \frac1{v_2}=\frac{35-20}{700} \] \[ \frac1{v_2}=\frac3{140} \] Thus, \[ v_2=\frac{140}{3}\,\text{cm} \] This is approximately \(46.7\,\text{cm}\), not \(40\,\text{cm}\). \(\textbf{Determine the correct screen movement:}\) \[ \Delta v=v_2-v_1 \] \[ \Delta v=\frac{140}{3}-60 \] \[ \Delta v=-\frac{40}{3}\,\text{cm} \] The screen must move \(\frac{40}{3}\,\text{cm}\) towards the lens. \(\textbf{Determine the changed magnification:}\) \[ m_2=\frac{v_2}{u_2} \] \[ m_2=\frac{140/3}{-35} \] \[ m_2=-\frac43 \] \(\textbf{Interpret the motions:}\) The object moves away from the lens towards negative \(x\). The positive image distance decreases, so the image also moves towards negative \(x\) in the laboratory. The object and image therefore move in the same signed axial direction, although they lie on opposite sides of the lens. \(\textbf{Physical interpretation:}\) For a convex lens forming a real image, moving the object farther away generally moves its real image towards the principal focus. The image also becomes less enlarged. However, describing their motions as towards or away from the lens is not equivalent to describing their signed laboratory directions. Here the object moves away from the lens and the image moves towards the lens, but both move towards negative \(x\). The exact screen displacement must be obtained from the lens equation rather than inferred by multiplying the object movement by the original magnification. \(\textbf{Final answer:}\) \[ \boxed{v_1=+60\,\text{cm},\quad v_2=+\frac{140}{3}\,\text{cm},\quad\Delta v=-\frac{40}{3}\,\text{cm},\quad m_1=-2,\quad m_2=-\frac43} \]