Optics MCQs with Answers – Part 9 (NEET Physics)

Explanation: For a prism, the deviation is related to the external angles by \[ \delta=i+e-A \] In P, the specified angles give a deviation of \(40^\circ\). The ray enters at \(40^\circ\) and emerges at \(60^\circ\), so its path is not symmetric. The principle of reversibility allows the transmitted ray to travel back along the same optical path. When the directions are interchanged, incidence becomes \(60^\circ\) and emergence becomes \(40^\circ\). The total deviation magnitude remains unchanged, making P and Q mutually consistent. The third record also gives a deviation of \(40^\circ\) from the external-angle formula. However, equal incidence and emergence angles have a special significance for an ordinary prism surrounded by the same medium on both sides. They imply equal internal refraction angles and a symmetric ray path. Such a transmitted path corresponds to minimum deviation. The earlier unequal-angle path already has the same stated deviation, even though its incident and emergent directions differ. For a fixed ordinary prism and wavelength, an asymmetric transmitted path must have deviation greater than the minimum. It cannot have exactly the same deviation as the symmetric minimum-deviation path. The inconsistency cannot be detected by using only the expression \(i+e-A\). It requires combining the deviation relation with ray reversibility and the physical meaning of the minimum-deviation condition. Thus P and Q can describe the same prism, but the additional symmetric record is incompatible with them.

Explanation: \(\textbf{Recover the index of prism P:}\) At minimum deviation, \[ \mu=\frac{\sin[(A+\delta_m)/2]}{\sin(A/2)} \] For P, \[ A=60^\circ,\qquad\delta_{m,P}=30^\circ \] Therefore, \[ \mu_P=\frac{\sin45^\circ}{\sin30^\circ} \] \[ \mu_P=\sqrt2 \] \(\textbf{Recover the index of prism Q:}\) For Q, \[ \delta_{m,Q}=60^\circ \] Thus, \[ \mu_Q=\frac{\sin60^\circ}{\sin30^\circ} \] \[ \mu_Q=\sqrt3 \] \(\textbf{Analyse the common entry condition:}\) Both rays enter from air at \[ i_1=30^\circ \] Snell's law at the first face gives \[ \sin r_1=\frac{\sin30^\circ}{\mu} \] The internal second-face incidence is \[ r_2=A-r_1 \] \(\textbf{Determine the first internal angle for P:}\) \[ \sin r_{1,P}=\frac1{2\sqrt2} \] Therefore, \[ r_{1,P}\approx20.7^\circ \] \[ r_{2,P}=60^\circ-20.7^\circ \] \[ r_{2,P}\approx39.3^\circ \] \(\textbf{Check P's critical angle:}\) \[ \sin i_{c,P}=\frac1{\sqrt2} \] \[ i_{c,P}=45^\circ \] Since \[ 39.3^\circ\lt 45^\circ \] P permits ordinary emergence. \(\textbf{Determine Q's first internal angle:}\) \[ \sin r_{1,Q}=\frac1{2\sqrt3} \] Therefore, \[ r_{1,Q}\approx16.8^\circ \] \[ r_{2,Q}=60^\circ-16.8^\circ \] \[ r_{2,Q}\approx43.2^\circ \] \(\textbf{Check Q's critical angle:}\) \[ \sin i_{c,Q}=\frac1{\sqrt3} \] \[ i_{c,Q}\approx35.3^\circ \] Since \[ 43.2^\circ\gt 35.3^\circ \] Q undergoes TIR at its second face. \(\textbf{Interpret the comparison:}\) A larger refractive index bends the ray more strongly towards the normal at the first face. This produces a smaller \(r_1\) and, for fixed prism angle, a larger \(r_2\). The larger index also reduces the critical angle at the second glass-air boundary. Both changes make TIR more likely in Q. \(\textbf{Physical interpretation:}\) Minimum-deviation measurements determine refractive indices, but they do not directly specify how every other ray travels through the prisms. The changed incident condition must first be analysed with Snell's law. The internal prism relation then gives the angle of incidence at the second face. Only after that calculation can the critical-angle test be applied. Two prisms of equal geometrical angle can consequently have opposite emergence outcomes for the same external incidence. \(\textbf{Final answer:}\) \[ \boxed{\mu_P=\sqrt2,\quad r_{2,P}\approx39.3^\circ\lt i_{c,P}=45^\circ;\quad\mu_Q=\sqrt3,\quad r_{2,Q}\approx43.2^\circ\gt i_{c,Q}\approx35.3^\circ} \]

Explanation: \(\textbf{Identify the thin-prism relation in a medium:}\) For a thin prism, \[ \delta\approx(\mu_{\mathrm{rel}}-1)A \] where \[ \mu_{\mathrm{rel}}=\frac{n_g}{n_m} \] The prism refracting angle is \[ A=5.0^\circ \] The glass index is \[ n_g=1.60 \] \(\textbf{Calculate the deviation in air:}\) In air, \[ n_m=1 \] Therefore, \[ \delta_a\approx(1.60-1)(5.0^\circ) \] \[ \delta_a\approx+3.0^\circ \] The positive sign indicates bending towards the base. \(\textbf{Calculate the deviation in the first liquid:}\) For \[ n_m=\frac43 \] the relative index is \[ \mu_{\mathrm{rel}}=\frac{1.60}{4/3} \] \[ \mu_{\mathrm{rel}}=1.20 \] Hence, \[ \delta_1\approx(1.20-1)(5.0^\circ) \] \[ \delta_1\approx+1.0^\circ \] The prism still bends light towards its base, but less strongly than in air. \(\textbf{Calculate the deviation in the second liquid:}\) For \[ n_m=2.00 \] the relative index becomes \[ \mu_{\mathrm{rel}}=\frac{1.60}{2.00} \] \[ \mu_{\mathrm{rel}}=0.80 \] Thus, \[ \delta_2\approx(0.80-1)(5.0^\circ) \] \[ \delta_2\approx-1.0^\circ \] \(\textbf{Interpret the negative deviation:}\) The surrounding liquid now has a greater refractive index than the glass. The direction of the prism's net angular deviation reverses. With the stated sign convention, that produces deviation towards the apex. \(\textbf{Check the zero-deviation condition:}\) The thin-prism relation predicts \[ \delta=0 \] when \[ n_m=n_g \] This corresponds to refractive-index matching. The sign reversal between the two liquids is consistent with this limiting case. \(\textbf{Physical interpretation:}\) Thin-prism deviation depends on the refractive index of the prism relative to its surroundings. It is not determined by the absolute glass index alone. Increasing the surrounding index towards the glass index reduces the deviation. Once the surrounding index exceeds the glass index, the direction of the deviation reverses. The calculation also requires retaining the subtraction of unity in the thin-prism relation. \(\textbf{Final answer:}\) \[ \boxed{\delta_a=+3.0^\circ,\quad\delta_{4/3}=+1.0^\circ,\quad\delta_{2.00}=-1.0^\circ} \]

Explanation: \(\textbf{Assign signs to the individual prism deviations:}\) For a thin prism, \[ \delta\approx\left(\frac{n_g}{n_m}-1\right)A \] P's base defines the positive transverse direction. Q's base points oppositely, so Q's deviation enters the net result with a negative sign when both prisms are optically denser than their surroundings. \(\textbf{Calculate P's deviation in air:}\) \[ \delta_{P,a}=(1.50-1)(4.0^\circ) \] \[ \delta_{P,a}=+2.0^\circ \] \(\textbf{Calculate Q's deviation in air:}\) \[ \delta_{Q,a}=\left(\frac53-1\right)(3.0^\circ) \] \[ \delta_{Q,a}=2.0^\circ \] This deviation is towards Q's base and therefore contributes negatively. \(\textbf{Determine the initial net deviation:}\) \[ \delta_{\mathrm{net},a}=\delta_{P,a}-\delta_{Q,a} \] \[ \delta_{\mathrm{net},a}=2.0^\circ-2.0^\circ \] \[ \delta_{\mathrm{net},a}=0 \] The two prisms compensate each other's angular deviations in air. \(\textbf{Recalculate P's deviation in liquid:}\) The surrounding index becomes \[ n_m=1.25=\frac54 \] For P, \[ \frac{n_P}{n_m}=\frac{1.50}{1.25} \] \[ \frac{n_P}{n_m}=\frac65 \] Thus, \[ \delta_{P,l}=\left(\frac65-1\right)(4.0^\circ) \] \[ \delta_{P,l}=0.80^\circ \] \(\textbf{Recalculate Q's deviation in liquid:}\) \[ \frac{n_Q}{n_m}=\frac{5/3}{5/4} \] \[ \frac{n_Q}{n_m}=\frac43 \] Therefore, \[ \delta_{Q,l}=\left(\frac43-1\right)(3.0^\circ) \] \[ \delta_{Q,l}=1.0^\circ \] \(\textbf{Determine the changed net deviation:}\) \[ \delta_{\mathrm{net},l}=0.80^\circ-1.0^\circ \] \[ \delta_{\mathrm{net},l}=-0.20^\circ \] The negative sign indicates net deviation towards Q's base. \(\textbf{Explain why initial compensation is lost:}\) The two prisms have different refractive indices. Immersion changes their relative refractive-index contrasts by different factors. Although their air deviations were equal, their immersed deviations are no longer equal. \(\textbf{Physical interpretation:}\) Oppositely oriented thin prisms can be arranged to produce zero net deviation at one surrounding refractive index. That compensation is not necessarily preserved when the surrounding medium changes. Each prism responds according to its own material index relative to the new medium. The final direction must be determined by algebraically adding the signed deviations. \(\textbf{Final answer:}\) \[ \boxed{\delta_{\mathrm{net},a}=0,\quad\delta_{P,l}=+0.80^\circ,\quad\delta_{Q,l}=-1.0^\circ,\quad\delta_{\mathrm{net},l}=-0.20^\circ} \]

Explanation: \(\textbf{Calculate red-light deviation:}\) For a thin prism in air, \[ \delta\approx(\mu-1)A \] The prism angle is \[ A=6.0^\circ \] For red light, \[ \mu_R=1.50 \] Therefore, \[ \delta_R=(1.50-1)(6.0^\circ) \] \[ \delta_R=3.0^\circ \] \(\textbf{Calculate violet-light deviation:}\) For violet light, \[ \mu_V=1.54 \] Thus, \[ \delta_V=(1.54-1)(6.0^\circ) \] \[ \delta_V=3.24^\circ \] \(\textbf{Determine the angular dispersion:}\) The angular separation is \[ \Delta\delta=\delta_V-\delta_R \] \[ \Delta\delta=3.24^\circ-3.0^\circ \] \[ \Delta\delta=0.24^\circ \] \(\textbf{Convert the separation into radians:}\) Using \[ 1^\circ\approx\frac{\pi}{180}\,\text{rad} \] gives \[ \Delta\delta\approx0.24\frac{\pi}{180} \] \[ \Delta\delta\approx4.19\times10^{-3}\,\text{rad} \] \(\textbf{Find the separation on the screen:}\) The screen distance is \[ D=1.5\,\text{m} \] For small angles, the approximate transverse separation is \[ s\approx D\Delta\delta \] \[ s\approx1.5(4.19\times10^{-3}) \] \[ s\approx6.28\times10^{-3}\,\text{m} \] \[ s\approx6.28\,\text{mm} \] \(\textbf{Interpret the initial colour ordering:}\) The higher violet refractive index produces a larger deviation. Both colours bend towards the prism's base. Violet therefore lies farther towards the base than red. \(\textbf{Reverse the prism orientation:}\) Turning the prism so that its base points oppositely reverses the signs of both deviations. The magnitudes remain \[ |\delta_R|=3.0^\circ,\qquad|\delta_V|=3.24^\circ \] Their angular difference retains magnitude \(0.24^\circ\). However, the colour that lies farther along the original positive transverse direction changes. \(\textbf{Physical interpretation:}\) Dispersion arises because the refractive index of a material varies with wavelength. For an ordinary dispersive glass prism, violet light deviates more than red light. The difference between their deviations determines the angular separation of the emergent beams. A screen converts that angular separation into a transverse distance. Reversing the prism reverses the spatial direction of dispersion without changing the absolute angular separation. The calculation distinguishes individual deviations from the much smaller difference between them. \(\textbf{Final answer:}\) \[ \boxed{\delta_R=3.0^\circ,\quad\delta_V=3.24^\circ,\quad\Delta\delta=0.24^\circ,\quad s\approx6.28\,\text{mm}} \]

Explanation: Statement I is correct because minimum deviation occurs for a symmetric ray path through an ordinary prism under the stated conditions. The internal refraction angles are equal. Together with the prism relation \(r_1+r_2=A\), this gives \(r_1=r_2=A/2\). The external incidence and emergence angles are also equal. Statement II is incorrect because the deviation relation is \(\delta=i+e-A\). A fixed sum \(i+e\) gives a fixed deviation for a fixed prism angle, but it does not establish that the deviation is the minimum possible value. Two asymmetric ray paths related by reversibility can have the same deviation while neither is the minimum-deviation path. Statement III is correct because the path of an ordinarily refracted ray is reversible. When its direction is reversed, the original emergence angle becomes the new incidence angle. The original incidence angle becomes the new emergence angle. Their sum remains unchanged, so the deviation magnitude remains the same. Statement IV is correct because the minimum-deviation relation is \[ \mu=\frac{\sin[(A+\delta_m)/2]}{\sin(A/2)} \] For a given refractive-index ratio, the minimum deviation is related to the prism angle. Changing \(A\) changes the geometry of the symmetric internal path. The minimum deviation therefore cannot be treated as a material property independent of prism shape. These statements distinguish three related but different ideas: deviation calculated from external angles, minimum deviation as a special symmetric condition, and reversibility of a transmitted ray. Only Statements I, III and IV are correct.

Explanation: \(\textbf{Apply the approximation only in its valid regime:}\) For a thin prism in air, \[ \delta\approx(\mu-1)A \] In P, \[ A_P=3^\circ,\qquad\mu=\sqrt2 \] Therefore, \[ \delta_P\approx(\sqrt2-1)(3^\circ) \] \[ \delta_P\approx1.24^\circ \] \(\textbf{Use the exact minimum-deviation relation for Q:}\) The second prism has \[ A_Q=60^\circ \] At minimum deviation, \[ \mu=\frac{\sin[(A_Q+\delta_m)/2]}{\sin(A_Q/2)} \] Substituting, \[ \sqrt2=\frac{\sin[(60^\circ+\delta_m)/2]}{\sin30^\circ} \] \(\textbf{Solve the exact angular condition:}\) Since \[ \sin30^\circ=\frac12 \] we obtain \[ \sin\left(\frac{60^\circ+\delta_m}{2}\right)=\frac1{\sqrt2} \] For the physical minimum-deviation branch, \[ \frac{60^\circ+\delta_m}{2}=45^\circ \] Therefore, \[ 60^\circ+\delta_m=90^\circ \] \[ \delta_m=30^\circ \] \(\textbf{Compare with the inappropriate extrapolation:}\) The thin-prism expression would give \[ \delta_{\mathrm{approx}}=60(\sqrt2-1)^\circ \] \[ \delta_{\mathrm{approx}}\approx24.85^\circ \] This differs from the exact minimum deviation of \(30^\circ\). \(\textbf{Identify why the results differ:}\) The thin-prism relation assumes small angles. The \(60^\circ\) prism has a substantially larger refracting angle. The trigonometric relations cannot be replaced by their small-angle forms without introducing appreciable error. \(\textbf{Physical interpretation:}\) A formula derived using a small-angle approximation is controlled by the geometry as well as the optical material. Keeping the refractive index unchanged does not make the approximation equally accurate for all prism angles. The narrow prism admits a useful nearly linear deviation relation. The wider prism requires the exact minimum-deviation equation. The question distinguishes approximation validity from the physical correctness of the underlying refraction laws. \(\textbf{Final answer:}\) \[ \boxed{\delta_P\approx1.24^\circ,\quad\delta_{m,Q}=30^\circ,\quad\delta_{\mathrm{thin\ estimate},Q}\approx24.85^\circ} \]

Explanation: The Assertion is true because a ray changes direction at an ordinary refracting surface only when the relevant refraction conditions produce an angular change. For red light, the prism and surrounding medium have equal refractive indices. There is no refractive-index contrast at either prism face for the red component. Its direction therefore remains unchanged through the ideal prism. The violet component experiences a different condition. The prism material has a greater refractive index than the surrounding medium for violet light. The violet ray consequently refracts at the prism faces and can emerge with a net angular deviation. The two colours no longer travel in the same direction. The Reason is true because Snell's law depends on the refractive indices of the media at each boundary. Refractive index may depend on wavelength in a dispersive material. Consequently, the relative index can equal unity for one colour while differing from unity for another. This difference directly explains the separation described in the Assertion. The experiment does not require the prism to act as an opaque red-light filter. Both colours may be transmitted, but their propagation directions are different. The key idea is that colour separation depends on wavelength-dependent refractive-index contrast between the prism and its surroundings. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Determine the internal ray geometry:}\) All rays enter their first prism faces normally. Therefore, \[ i_1=0^\circ \] By Snell's law, \[ r_1=0^\circ \] The internal prism relation is \[ r_1+r_2=A \] Hence, \[ r_2=A \] Thus, the internal incidence at the second face equals the prism refracting angle for these normally entering rays. \(\textbf{Calculate the critical angle:}\) The glass index is \[ \mu=\sqrt2 \] The surrounding medium is air. Therefore, \[ \sin i_c=\frac1{\sqrt2} \] \[ i_c=45^\circ \] \(\textbf{Classify prism P:}\) For P, \[ r_{2,P}=30^\circ \] Since \[ 30^\circ\lt 45^\circ \] ordinary refraction into air is possible. \(\textbf{Classify prism Q:}\) For Q, \[ r_{2,Q}=45^\circ \] Therefore, \[ r_{2,Q}=i_c \] The ray is exactly critically incident. Its limiting refracted direction lies along the second face. This condition must be distinguished from strict TIR. \(\textbf{Classify prism R:}\) For R, \[ r_{2,R}=60^\circ \] Since \[ 60^\circ\gt 45^\circ \] the ray undergoes TIR at the second face. \(\textbf{Check the physical distinction:}\) All three rays enter their first faces successfully. However, changing the prism angle changes the incidence at the second face. Their second-face behaviour can therefore range from ordinary refraction to critical incidence and TIR. \(\textbf{Physical interpretation:}\) Normal entry through one face of a prism does not imply normal emergence through the other. The normals to the two refracting faces have different directions. The angle between them is controlled by the refracting angle of the prism. Consequently, changing the prism angle changes the internal incidence at the second face even though the glass material and entry condition are unchanged. The three prisms illustrate the distinction between subcritical, critical and supercritical incidence. \(\textbf{Final answer:}\) \[ P:\ 30^\circ\lt 45^\circ\ \text{(refraction)};\quad Q:\ 45^\circ=i_c\ \text{(critical)};\quad R:\ 60^\circ\gt 45^\circ\ \text{(TIR)} \]

Explanation: \(\textbf{Identify the relaxed-eye condition:}\) The simple microscope has focal length \[ f=+5.0\,\text{cm} \] The least distance of distinct vision is \[ D=25\,\text{cm} \] For a final image at infinity, the object lies at the first focal point of the convex lens. Therefore, \[ u_P=-5.0\,\text{cm} \] \(\textbf{Calculate the relaxed-eye magnifying power:}\) For normal adjustment, \[ M_\infty=\frac Df \] Thus, \[ M_\infty=\frac{25}{5} \] \[ M_\infty=5 \] \(\textbf{Determine the near-point magnifying power:}\) For a simple microscope with its final virtual image at the near point, \[ M_D=1+\frac Df \] Therefore, \[ M_D=1+\frac{25}{5} \] \[ M_D=6 \] \(\textbf{Locate the near-point virtual image:}\) The image distance in trial Q is \[ v_Q=-25\,\text{cm} \] Its negative sign indicates an image on the object side of the lens. \(\textbf{Recover the required near-point object distance:}\) The thin-lens equation is \[ \frac1v-\frac1u=\frac1f \] Therefore, \[ \frac1u=\frac1v-\frac1f \] Substituting, \[ \frac1{u_Q}=-\frac1{25}-\frac15 \] \[ \frac1{u_Q}=-\frac6{25} \] \[ u_Q=-\frac{25}{6}\,\text{cm} \] \(\textbf{Calculate the object movement:}\) The original object distance magnitude is \(5.0\,\text{cm}\). The new magnitude is \[ \frac{25}{6}\,\text{cm} \] Therefore, the object moves towards the lens by \[ \Delta x=5-\frac{25}{6} \] \[ \Delta x=\frac56\,\text{cm} \] \(\textbf{Interpret the magnifying-power change:}\) The near-point arrangement gives a greater angular magnifying power than normal adjustment. However, normal adjustment allows a relaxed eye to observe the virtual image at infinity. The larger magnification at the near point is accompanied by a different accommodation requirement. \(\textbf{Physical interpretation:}\) A simple microscope is an angular magnifier rather than merely an object-enlarging lens. Its magnifying power depends on where the final virtual image is formed. The relaxed-eye condition requires the object at the focal point. To form a virtual image at the near point, the object must be placed slightly closer to the lens. The changed image distance and magnifying power are consistent with the thin-lens equation. \(\textbf{Final answer:}\) \[ \boxed{M_\infty=5,\quad M_D=6,\quad u_P=-5\,\text{cm},\quad u_Q=-\frac{25}{6}\,\text{cm},\quad\Delta x=\frac56\,\text{cm towards lens}} \]

Explanation: \(\textbf{Analyse image formation by the objective:}\) The objective focal length is \[ f_o=+1.0\,\text{cm} \] The real object distance is \[ u_o=-1.25\,\text{cm} \] The lens equation gives \[ \frac1{v_o}=\frac1{f_o}+\frac1{u_o} \] \[ \frac1{v_o}=1-\frac1{1.25} \] \[ \frac1{v_o}=1-0.8 \] \[ v_o=+5.0\,\text{cm} \] \(\textbf{Determine the objective's linear magnification:}\) \[ m_o=\frac{v_o}{u_o} \] \[ m_o=\frac{5.0}{-1.25} \] \[ m_o=-4 \] The objective produces a real, inverted and enlarged intermediate image. \(\textbf{Locate that image relative to the eyepiece:}\) The objective-eyepiece separation is \[ L=10\,\text{cm} \] The intermediate image lies \(5.0\,\text{cm}\) behind the objective. Its distance in front of the eyepiece is therefore \[ L-v_o=10-5 \] \[ L-v_o=5.0\,\text{cm} \] The signed eyepiece object distance is \[ u_e=-5.0\,\text{cm} \] \(\textbf{Determine the eyepiece's final-image condition:}\) The eyepiece focal length is \[ f_e=+5.0\,\text{cm} \] Since \[ |u_e|=f_e \] the intermediate image lies at the first focal point of the eyepiece. The eyepiece sends out parallel rays. The final image is therefore at infinity. \(\textbf{Calculate the eyepiece's angular magnification:}\) For final viewing at infinity, \[ M_e=\frac D{f_e} \] The near-point distance is \[ D=25\,\text{cm} \] Therefore, \[ M_e=\frac{25}{5} \] \[ M_e=5 \] \(\textbf{Determine the microscope's overall magnifying power:}\) The magnitude of the overall magnifying power is \[ |M|=|m_o|M_e \] \[ |M|=4(5) \] \[ |M|=20 \] \(\textbf{Identify the final orientation:}\) The objective first inverts the object. The eyepiece acts as a magnifier of that inverted intermediate image. The final microscope image remains inverted relative to the original object. \(\textbf{Physical interpretation:}\) A compound microscope forms an image in two distinct stages. The objective produces a real enlarged intermediate image. The eyepiece uses that intermediate image as its object and supplies additional angular magnification. The separation between the lenses determines whether the intermediate image lies at the eyepiece focus. In this experiment, it does, so the final image is at infinity. The overall magnifying power combines objective linear magnification with eyepiece angular magnification. \(\textbf{Final answer:}\) \[ \boxed{v_o=+5.0\,\text{cm},\quad m_o=-4,\quad u_e=-5.0\,\text{cm},\quad v_{\mathrm{final}}\to\infty,\quad |M|=20} \]

Explanation: \(\textbf{Calculate P's angular magnifying power:}\) For an astronomical telescope in normal adjustment, \[ |M|=\frac{f_o}{f_e} \] For P, \[ f_o=120\,\text{cm} \] \[ f_{e,P}=6.0\,\text{cm} \] Thus, \[ |M_P|=\frac{120}{6} \] \[ |M_P|=20 \] \(\textbf{Calculate Q's angular magnifying power:}\) For Q, \[ f_{e,Q}=12\,\text{cm} \] Therefore, \[ |M_Q|=\frac{120}{12} \] \[ |M_Q|=10 \] The angular magnifying power of P is twice that of Q. \(\textbf{Determine P's normal-adjustment length:}\) For a standard two-convex-lens refracting telescope, \[ L=f_o+f_e \] Therefore, \[ L_P=120+6 \] \[ L_P=126\,\text{cm} \] \(\textbf{Determine Q's normal-adjustment length:}\) \[ L_Q=120+12 \] \[ L_Q=132\,\text{cm} \] Thus, \[ L_Q-L_P=6.0\,\text{cm} \] Q is longer. \(\textbf{Compare objective light-gathering areas:}\) For a circular objective aperture, \[ A=\frac{\pi d^2}{4} \] The objective diameters are \[ d_P=4.0\,\text{cm} \] \[ d_Q=8.0\,\text{cm} \] Therefore, \[ \frac{A_Q}{A_P}=\left(\frac{d_Q}{d_P}\right)^2 \] \[ \frac{A_Q}{A_P}=\left(\frac84\right)^2 \] \[ \frac{A_Q}{A_P}=4 \] \(\textbf{Separate angular magnification from light gathering:}\) P has the smaller eyepiece focal length and therefore greater angular magnification. Q has the larger objective diameter and therefore greater light-gathering area. These are different optical design properties. A telescope can collect more light even while giving a smaller angular magnification. \(\textbf{Physical interpretation:}\) The eyepiece focal length controls angular magnifying power for a fixed objective focal length. The normal-adjustment tube length depends on the sum of the objective and eyepiece focal lengths. The objective aperture controls the amount of light collected from a distant source under comparable illumination conditions. Increasing objective diameter increases light collection according to aperture area. The comparison illustrates why telescope suitability for observing faint stars cannot be judged from angular magnifying power alone. \(\textbf{Final answer:}\) \[ \boxed{|M_P|=20,\quad |M_Q|=10,\quad L_P=126\,\text{cm},\quad L_Q=132\,\text{cm},\quad\frac{A_Q}{A_P}=4} \]

Explanation: Statement I is correct because the refractive index of ordinary optical glass generally depends on wavelength. A convex objective lens can therefore have different focal lengths for different colours. This produces longitudinal chromatic aberration unless suitable correction is provided. Statement II is correct because reflection does not depend on the same refractive-index dispersion mechanism that changes the focal length of an ordinary refracting lens. A primary reflecting objective therefore avoids chromatic aberration arising from passage through a dispersive objective lens. A reflecting telescope may still have other optical aberrations or alignment limitations, but those are distinct issues. Statement III is incorrect because light-gathering ability depends importantly on effective collecting area and transmission or reflection losses. The use of a mirror does not, by itself, require a lower light-gathering ability than a lens. A sufficiently large and efficiently reflecting objective can collect substantial light from faint celestial objects. Statement IV is correct because a larger effective aperture intercepts more light from the same distant source under comparable conditions. For a circular unobstructed aperture, collecting area increases with the square of diameter. An effective aperture comparison must also account for any blocked or otherwise unavailable region of the optical system. The two telescope designs therefore differ in their image-forming mechanisms but share the importance of effective aperture. The question distinguishes chromatic correction, optical architecture and light gathering as separate design considerations. Only Statements I, II and IV are correct.

Explanation: \(\textbf{Use the original normal-adjustment condition:}\) The objective-eyepiece separation is \[ L_1=12\,\text{cm} \] The original eyepiece focal length is \[ f_{e1}=5.0\,\text{cm} \] For the final image to be at infinity, the intermediate image must lie at the eyepiece's first focal plane. Therefore, its distance in front of the eyepiece is \(5.0\,\text{cm}\). \(\textbf{Locate the objective's intermediate image:}\) Measured from the objective, \[ v_o=L_1-f_{e1} \] \[ v_o=12-5 \] \[ v_o=+7.0\,\text{cm} \] The objective forms a real image \(7.0\,\text{cm}\) behind itself. \(\textbf{Determine why the objective image remains fixed:}\) The object and objective positions remain unchanged. The objective focal length and surrounding medium also remain unchanged. Therefore, the intermediate-image position produced by the objective does not move when the eyepiece is replaced. \(\textbf{Apply the new eyepiece focal length:}\) The replacement eyepiece has \[ f_{e2}=4.0\,\text{cm} \] To restore normal adjustment, the unchanged intermediate image must now lie \(4.0\,\text{cm}\) in front of the new eyepiece. \(\textbf{Calculate the required new separation:}\) \[ L_2=v_o+f_{e2} \] \[ L_2=7.0+4.0 \] \[ L_2=11.0\,\text{cm} \] The eyepiece must move \(1.0\,\text{cm}\) closer to the objective. \(\textbf{Compare eyepiece angular magnifications:}\) For final viewing at infinity, \[ M_e=\frac D{f_e} \] where \(D\) is the least distance of distinct vision. The ratio of the new to original eyepiece magnifying powers is \[ \frac{M_{e2}}{M_{e1}}=\frac{D/f_{e2}}{D/f_{e1}} \] \[ \frac{M_{e2}}{M_{e1}}=\frac{f_{e1}}{f_{e2}} \] \[ \frac{M_{e2}}{M_{e1}}=\frac54 \] \(\textbf{Identify the change in total microscope magnification:}\) The objective image and its linear magnification remain unchanged. The eyepiece supplies a greater angular magnification because its focal length is smaller. Consequently, the overall microscope magnifying-power magnitude also increases by the factor \(\frac54\) under the same viewing convention. \(\textbf{Check the final image condition:}\) The objective intermediate image remains at \[ v_o=+7.0\,\text{cm} \] The new eyepiece is at \[ L_2=11.0\,\text{cm} \] Thus, its signed object distance is \[ u_{e2}=7.0-11.0 \] \[ u_{e2}=-4.0\,\text{cm} \] This equals the negative of its focal length, confirming that the final rays emerge parallel. \(\textbf{Physical interpretation:}\) Replacing the eyepiece of a compound microscope does not necessarily require changing the objective's image. If the object and objective are undisturbed, that real intermediate image remains at its original location. The eyepiece must be repositioned so that the intermediate image lies at the new eyepiece's focal plane. A shorter-focal-length eyepiece then increases angular magnifying power while shortening the required objective-eyepiece separation in this arrangement. The problem combines the two-stage microscope image construction with a changed-condition focusing requirement. \(\textbf{Final answer:}\) \[ \boxed{v_o=+7.0\,\text{cm},\quad L_2=11.0\,\text{cm},\quad\frac{M_{e2}}{M_{e1}}=\frac54} \]

Explanation: \(\textbf{Establish the normal-adjustment condition:}\) The focal lengths are \[ f_o=100\,\text{cm},\qquad f_e=5.0\,\text{cm} \] For a distant object, the objective forms a real intermediate image at its focal plane. In normal adjustment, this image lies at the first focal plane of the eyepiece. Therefore, \[ L_\infty=f_o+f_e \] \[ L_\infty=100+5 \] \[ L_\infty=105\,\text{cm} \] \(\textbf{Determine the original angular magnification:}\) For an astronomical telescope in normal adjustment, \[ |M_\infty|=\frac{f_o}{f_e} \] \[ |M_\infty|=\frac{100}{5} \] \[ |M_\infty|=20 \] \(\textbf{Apply the near-point viewing condition:}\) The desired final image is virtual and lies at \[ v_e=-25\,\text{cm} \] relative to the eyepiece. The eyepiece lens equation is \[ \frac1{v_e}-\frac1{u_e}=\frac1{f_e} \] Therefore, \[ \frac1{u_e}=\frac1{v_e}-\frac1{f_e} \] \[ \frac1{u_e}=-\frac1{25}-\frac15 \] \[ \frac1{u_e}=-\frac6{25} \] \[ u_e=-\frac{25}{6}\,\text{cm} \] \(\textbf{Recover the new telescope length:}\) The objective image remains \(100\,\text{cm}\) behind the objective. It must now be \(\frac{25}{6}\,\text{cm}\) in front of the eyepiece. Hence, \[ L_D=100+\frac{25}{6} \] \[ L_D=\frac{625}{6}\,\text{cm} \] \(\textbf{Calculate the eyepiece displacement:}\) \[ L_\infty-L_D=105-\frac{625}{6} \] \[ L_\infty-L_D=\frac56\,\text{cm} \] The eyepiece moves towards the objective. \(\textbf{Determine the near-point magnifying power:}\) For the stated near-point viewing arrangement, \[ |M_D|=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right) \] Substituting, \[ |M_D|=\frac{100}{5}\left(1+\frac5{25}\right) \] \[ |M_D|=20\left(\frac65\right) \] \[ |M_D|=24 \] \(\textbf{Physical interpretation:}\) The objective continues to form the same real intermediate image of the distant star. Moving the eyepiece changes only the second imaging stage. For normal adjustment, the intermediate image lies at the eyepiece focus and the eye receives parallel rays. For near-point viewing, the intermediate image lies slightly inside that focus. The final image becomes virtual at the near point, and the angular magnifying power increases. The increased magnification is obtained with a shorter lens separation, although viewing at infinity is generally more comfortable for a relaxed eye. \(\textbf{Final answer:}\) \[ \boxed{L_\infty=105\,\text{cm},\quad L_D=\frac{625}{6}\,\text{cm},\quad\Delta L=-\frac56\,\text{cm},\quad |M_\infty|=20,\quad |M_D|=24} \]

Explanation: \(\textbf{Separate the two magnification stages:}\) A compound microscope combines the linear magnification of its objective with the angular magnification of its eyepiece. For normal adjustment, \[ |M|=|m_o|\frac D{f_e} \] The known quantities are \[ |M|=30,\qquad D=25\,\text{cm},\qquad f_e=5\,\text{cm} \] \(\textbf{Recover the objective magnification:}\) \[ 30=|m_o|\frac{25}{5} \] \[ 30=5|m_o| \] \[ |m_o|=6 \] The intermediate image is inverted, so \[ m_o=-6 \] \(\textbf{Relate the objective conjugate distances:}\) For the objective, \[ m_o=\frac{v_o}{u_o} \] Therefore, \[ v_o=-6u_o \] The objective focal length is \[ f_o=+1.0\,\text{cm} \] \(\textbf{Apply the objective lens equation:}\) \[ \frac1{v_o}-\frac1{u_o}=\frac1{f_o} \] Substituting \(v_o=-6u_o\), \[ -\frac1{6u_o}-\frac1{u_o}=1 \] \[ -\frac7{6u_o}=1 \] \[ u_o=-\frac76\,\text{cm} \] \(\textbf{Calculate the intermediate image distance:}\) \[ v_o=-6\left(-\frac76\right) \] \[ v_o=+7.0\,\text{cm} \] The objective forms a real enlarged intermediate image \(7.0\,\text{cm}\) behind itself. \(\textbf{Locate the eyepiece for normal adjustment:}\) To produce a final image at infinity, the intermediate image must be at the eyepiece's first focal plane. Thus, its distance from the eyepiece is \[ f_e=5.0\,\text{cm} \] The objective-eyepiece separation is \[ L=v_o+f_e \] \[ L=7.0+5.0 \] \[ L=12\,\text{cm} \] \(\textbf{Check the measured magnifying power:}\) Using the reconstructed geometry, \[ |M|=\left|\frac{7}{-7/6}\right|\frac{25}{5} \] \[ |M|=6(5) \] \[ |M|=30 \] The result agrees with the measurement. \(\textbf{Physical interpretation:}\) The observed overall magnifying power provides information about the objective's otherwise unknown conjugate distances. The eyepiece magnification can be calculated independently because its focal length and viewing condition are known. Dividing the overall magnification by that eyepiece factor recovers the objective magnification. The objective lens equation then determines its object and image distances. Finally, the normal-adjustment condition fixes the eyepiece position. This inverse problem reconstructs the microscope geometry from its observed performance rather than calculating performance from a known geometry. \(\textbf{Final answer:}\) \[ \boxed{u_o=-\frac76\,\text{cm},\quad v_o=+7.0\,\text{cm},\quad L=12\,\text{cm},\quad |M|=30} \]

Explanation: \(\textbf{Identify the usable aperture of the reflector:}\) The outer diameter of P's primary mirror is \[ d_P=20\,\text{cm} \] Its central obstruction has diameter \[ d_b=8.0\,\text{cm} \] Only the unobstructed portion of the primary aperture contributes to the ideal effective collecting area. \(\textbf{Calculate P's available area:}\) \[ A_P=\frac{\pi}{4}(d_P^2-d_b^2) \] \[ A_P=\frac{\pi}{4}(20^2-8^2) \] \[ A_P=\frac{\pi}{4}(400-64) \] \[ A_P=84\pi\,\text{cm}^2 \] \(\textbf{Calculate Q's clear objective area:}\) For the unobstructed refracting objective, \[ d_Q=18\,\text{cm} \] Therefore, \[ A_Q=\frac{\pi d_Q^2}{4} \] \[ A_Q=\frac{\pi(18)^2}{4} \] \[ A_Q=81\pi\,\text{cm}^2 \] \(\textbf{Compare the collected light:}\) For equal incident irradiance and the specified ideal conditions, collected light is proportional to available aperture area. Thus, \[ \frac{A_P}{A_Q}=\frac{84\pi}{81\pi} \] \[ \frac{A_P}{A_Q}=\frac{28}{27} \] The reflector has a slightly greater effective collecting area. \(\textbf{Distinguish the objective imaging mechanisms:}\) P uses reflection at its primary mirror. Q uses refraction through a glass objective lens. The refractive index of ordinary optical glass generally varies with wavelength. This can cause different colours to have different focal positions in Q. \(\textbf{Identify the reflector's chromatic advantage:}\) The law of reflection does not produce the same wavelength-dependent focal-length variation associated with refraction through a dispersive objective. The reflecting primary therefore avoids that particular form of chromatic aberration. This does not mean that every optical imperfection of P disappears. \(\textbf{Check the role of magnifying power:}\) The telescopes are specified to have equal angular magnifying powers. That equality does not determine their light-gathering abilities. Collecting area must be evaluated separately from the objective and eyepiece focal-length ratio. \(\textbf{Physical interpretation:}\) A reflecting telescope may lose some effective aperture because a secondary assembly obstructs part of the incoming beam. Its collecting ability depends on the unobstructed area, not simply on the outer diameter. Even after allowing for obstruction, a larger reflecting objective can collect more light than a smaller refracting objective. Reflection also avoids chromatic aberration of the primary focusing element caused by material dispersion. The comparison distinguishes collecting area, angular magnification and chromatic performance as three separate properties. \(\textbf{Final answer:}\) \[ \boxed{A_P=84\pi\,\text{cm}^2,\quad A_Q=81\pi\,\text{cm}^2,\quad\frac{A_P}{A_Q}=\frac{28}{27}} \]

Explanation: \(\textbf{Locate the objective's intermediate image:}\) For the objective, \[ f_o=+2.0\,\text{cm} \] \[ u_o=-3.0\,\text{cm} \] The lens formula gives \[ \frac1{v_o}=\frac1{2}-\frac1{3} \] \[ \frac1{v_o}=\frac16 \] \[ v_o=+6.0\,\text{cm} \] \(\textbf{Determine the objective magnification:}\) \[ m_o=\frac{v_o}{u_o} \] \[ m_o=\frac6{-3} \] \[ m_o=-2 \] The objective forms a real inverted image enlarged twofold. \(\textbf{Verify the original normal adjustment:}\) The original lens separation is \[ L_1=10\,\text{cm} \] The intermediate image lies \[ 10-6=4.0\,\text{cm} \] in front of the eyepiece. Since \[ f_e=4.0\,\text{cm} \] the intermediate image is at the eyepiece focus. The final rays emerge parallel. \(\textbf{Calculate the original magnifying power:}\) For final viewing at infinity, \[ |M_\infty|=|m_o|\frac D{f_e} \] \[ |M_\infty|=2\left(\frac{20}{4}\right) \] \[ |M_\infty|=10 \] \(\textbf{Find the eyepiece object distance for near-point viewing:}\) The desired final image distance is \[ v_e=-20\,\text{cm} \] The eyepiece equation gives \[ \frac1{u_e}=\frac1{v_e}-\frac1{f_e} \] \[ \frac1{u_e}=-\frac1{20}-\frac14 \] \[ \frac1{u_e}=-\frac3{10} \] \[ u_e=-\frac{10}{3}\,\text{cm} \] \(\textbf{Determine the new microscope length:}\) The intermediate image remains \(6.0\,\text{cm}\) behind the objective. The eyepiece must be \(\frac{10}{3}\,\text{cm}\) beyond that image. Therefore, \[ L_2=6+\frac{10}{3} \] \[ L_2=\frac{28}{3}\,\text{cm} \] \(\textbf{Calculate the adjustment:}\) \[ L_1-L_2=10-\frac{28}{3} \] \[ L_1-L_2=\frac23\,\text{cm} \] The eyepiece moves towards the objective. \(\textbf{Calculate the near-point magnifying power:}\) \[ |M_D|=|m_o|\left(1+\frac D{f_e}\right) \] \[ |M_D|=2\left(1+\frac{20}{4}\right) \] \[ |M_D|=12 \] \(\textbf{Physical interpretation:}\) The objective continues to form the same intermediate image because neither the original object nor the objective is moved. Moving the eyepiece changes the final image location without changing objective magnification. A virtual final image at the near point requires the intermediate image to be slightly inside the eyepiece focal distance. The eyepiece consequently moves closer to the intermediate image. Its angular magnifying power increases from the relaxed-eye value, so the overall microscope magnification also increases. The calculation separates the objective's fixed linear magnification from the adjustable angular magnification of the eyepiece. \(\textbf{Final answer:}\) \[ \boxed{v_o=6.0\,\text{cm},\quad L_2=\frac{28}{3}\,\text{cm},\quad\Delta L=-\frac23\,\text{cm},\quad|M_\infty|=10,\quad|M_D|=12} \]

Explanation: \(\textbf{Identify the initial telescope adjustment:}\) The objective focal length is \[ f_o=80\,\text{cm} \] The eyepiece focal length is \[ f_e=5.0\,\text{cm} \] For a very distant source, the objective forms its intermediate image at the objective focus. The initial normal-adjustment separation is \[ L_1=f_o+f_e \] \[ L_1=80+5 \] \[ L_1=85\,\text{cm} \] \(\textbf{Apply the finite laboratory object distance:}\) The target is now \[ 400\,\text{cm} \] in front of the objective. Using the Cartesian convention, \[ u_o=-400\,\text{cm} \] The objective remains the same convex lens. \(\textbf{Locate its new intermediate image:}\) \[ \frac1{v_o}=\frac1{f_o}+\frac1{u_o} \] \[ \frac1{v_o}=\frac1{80}-\frac1{400} \] \[ \frac1{v_o}=\frac{5-1}{400} \] \[ \frac1{v_o}=\frac1{100} \] \[ v_o=+100\,\text{cm} \] \(\textbf{Determine the new eyepiece position:}\) For the final image to remain at infinity, the objective's intermediate image must lie at the first focal plane of the eyepiece. Therefore, \[ L_2=v_o+f_e \] \[ L_2=100+5 \] \[ L_2=105\,\text{cm} \] \(\textbf{Calculate the eyepiece displacement:}\) \[ \Delta L=L_2-L_1 \] \[ \Delta L=105-85 \] \[ \Delta L=+20\,\text{cm} \] The eyepiece must move away from the objective. \(\textbf{Check the second-stage image condition:}\) With the new separation, \[ u_e=v_o-L_2 \] \[ u_e=100-105 \] \[ u_e=-5.0\,\text{cm} \] Since \[ u_e=-f_e \] the eyepiece produces a final image at infinity. \(\textbf{Explain why the ordinary telescope-length formula changes:}\) The relation \[ L=f_o+f_e \] assumes an object at infinity. The laboratory target is at a finite distance. Its objective image lies beyond the objective's focal plane, so the correct eyepiece position must be calculated from the actual intermediate image. \(\textbf{Physical interpretation:}\) A telescope adjusted for distant stars does not necessarily remain correctly focused on a nearby laboratory target. The objective image moves as the object distance changes. For the specified target, that intermediate image forms farther behind the objective. To retain relaxed-eye viewing, the eyepiece must follow the displaced intermediate image while preserving the eyepiece focal separation. The experiment illustrates the limits of the standard normal-adjustment tube-length formula and the practical need for telescope focusing. \(\textbf{Final answer:}\) \[ \boxed{v_o=+100\,\text{cm},\quad L_1=85\,\text{cm},\quad L_2=105\,\text{cm},\quad\Delta L=+20\,\text{cm}} \]

Explanation: \(\textbf{Locate the red objective image:}\) The red objective focal length is \[ f_{o,R}=100\,\text{cm} \] A distant source sends approximately parallel rays to the objective. Therefore, the red intermediate image lies \[ v_{o,R}=+100\,\text{cm} \] behind the objective. \(\textbf{Determine the telescope's fixed separation:}\) The eyepiece focal length is \[ f_e=+4.0\,\text{cm} \] For red normal adjustment, \[ L=f_{o,R}+f_e \] \[ L=100+4 \] \[ L=104\,\text{cm} \] \(\textbf{Confirm the red final-image condition:}\) The red intermediate image is \(4.0\,\text{cm}\) in front of the eyepiece. Thus, \[ u_{e,R}=-4.0\,\text{cm} \] This equals the negative of the eyepiece focal length. The final red rays are parallel, corresponding to an image at infinity. \(\textbf{Locate the violet objective image:}\) The violet objective focal length is \[ f_{o,V}=96\,\text{cm} \] For the same distant-object condition, \[ v_{o,V}=+96\,\text{cm} \] The violet intermediate image lies closer to the objective than the red intermediate image. \(\textbf{Determine the violet object distance for the eyepiece:}\) The lens separation remains \[ L=104\,\text{cm} \] Therefore, \[ u_{e,V}=96-104 \] \[ u_{e,V}=-8.0\,\text{cm} \] The violet intermediate image is now beyond twice the eyepiece's focal distance from the eyepiece. \(\textbf{Calculate the violet final image:}\) Using the eyepiece lens equation, \[ \frac1{v_{e,V}}=\frac1{f_e}+\frac1{u_{e,V}} \] \[ \frac1{v_{e,V}}=\frac14-\frac18 \] \[ \frac1{v_{e,V}}=\frac18 \] \[ v_{e,V}=+8.0\,\text{cm} \] \(\textbf{Interpret the positive image distance:}\) The violet rays actually converge after passing through the eyepiece. They form a real image \(8.0\,\text{cm}\) beyond it rather than emerging parallel. Therefore, the red-adjusted telescope is not simultaneously in normal adjustment for violet light. \(\textbf{Physical interpretation:}\) A dispersive objective can produce different intermediate focal planes for different wavelengths. An eyepiece fixed relative to one colour's intermediate image does not necessarily give the same final-image condition for another colour. Here the red intermediate image lies at the eyepiece focal plane, while the violet one lies farther from the eyepiece. The eyepiece consequently produces parallel red rays but converging violet rays. This changed-condition analysis demonstrates longitudinal chromatic aberration in the complete two-lens telescope, not merely in an isolated objective. \(\textbf{Final answer:}\) \[ \boxed{L=104\,\text{cm},\quad v_{\mathrm{final},R}\to\infty,\quad u_{e,V}=-8.0\,\text{cm},\quad v_{\mathrm{final},V}=+8.0\,\text{cm}} \]