Optics MCQs with Answers – Part 5 (NEET Physics)

Explanation: \(\textbf{Compare relative wave speeds at each colour:}\) The refraction angle is determined by the refractive-index ratio at the same wavelength. For propagation from P into Q, \[ n_P\sin i=n_Q\sin r \] Hence, \[ \sin r=\frac{n_P}{n_Q}\sin i \] \(\textbf{Red-light refraction in the first trial:}\) For red light, \[ \frac{n_{PR}}{n_{QR}}=\frac{1.5}{1.2} \] \[ \frac{n_{PR}}{n_{QR}}=\frac54 \] The incidence condition is \[ \sin i_1=\frac35 \] Therefore, \[ \sin r_R=\frac54\times\frac35 \] \[ \sin r_R=\frac34 \] \(\textbf{Violet-light refraction in the first trial:}\) For violet light, \[ \frac{n_{PV}}{n_{QV}}=\frac{1.6}{1.4} \] \[ \frac{n_{PV}}{n_{QV}}=\frac87 \] Thus, \[ \sin r_V=\frac87\times\frac35 \] \[ \sin r_V=\frac{24}{35} \] \(\textbf{Compare the transmitted directions:}\) \[ \frac34\gt \frac{24}{35} \] Both transmitted angles are acute, so \[ r_R\gt r_V \] Red bends farther away from the normal than violet in the first trial. This occurs because the relevant refractive-index ratio is greater for red. \(\textbf{Increase the common incidence angle:}\) In the second trial, \[ \sin i_2=\frac{17}{20} \] For red light, ordinary transmission would require \[ \sin r'_R=\frac54\times\frac{17}{20} \] \[ \sin r'_R=\frac{17}{16} \] This exceeds unity. Therefore, the red ray undergoes total internal reflection. \(\textbf{Test violet transmission:}\) For violet light, \[ \sin r'_V=\frac87\times\frac{17}{20} \] \[ \sin r'_V=\frac{34}{35} \] This is smaller than unity. Therefore, violet can still transmit ordinarily into Q. \(\textbf{Check the critical-angle interpretation:}\) For red, \[ \sin i_{cR}=\frac{1.2}{1.5} \] \[ \sin i_{cR}=\frac45 \] For violet, \[ \sin i_{cV}=\frac{1.4}{1.6} \] \[ \sin i_{cV}=\frac78 \] The second incidence lies between these critical-angle sines: \[ \frac45\lt \frac{17}{20}\lt \frac78 \] This independently confirms red total internal reflection and violet ordinary transmission. \(\textbf{Physical interpretation:}\) When both the incident and transmitted media are dispersive, the direction of colour separation depends on the ratio of their wavelength-dependent refractive indices. The colour with the greater absolute refractive index in P need not undergo the larger transmitted deviation. Here red has the greater ratio of incident to transmitted index. It therefore bends farther away from the normal and reaches its critical condition first. Increasing the common incidence angle produces a colour-selective transition from ordinary transmission to total internal reflection. The result follows from applying Huygens' wave-speed relation separately to the two wavelengths. \(\textbf{Final answer:}\) Initially \(r_R\gt r_V\); in the second trial, red undergoes total internal reflection while violet transmits ordinarily with \(\sin r'_V=\frac{34}{35}\).

Explanation: \(\textbf{Interpret spatial phase gradient:}\) For a monochromatic travelling plane wave, the phase changes by \(2\pi\) over one wavelength. The magnitude of the phase gradient along the propagation direction is the wave number: \[ k=\frac{2\pi}{\lambda} \] The graph slopes therefore determine the wavelengths. \(\textbf{Find the wavelength in P:}\) The measured phase gradient is \[ k_P=10\pi\,\text{rad m}^{-1} \] Hence, \[ \lambda_P=\frac{2\pi}{k_P} \] \[ \lambda_P=\frac{2\pi}{10\pi} \] \[ \lambda_P=0.20\,\text{m} \] \(\textbf{Find the wavelength in Q:}\) The measured phase gradient is \[ k_Q=15\pi\,\text{rad m}^{-1} \] Therefore, \[ \lambda_Q=\frac{2\pi}{15\pi} \] \[ \lambda_Q=\frac2{15}\,\text{m} \] \(\textbf{Convert spatial phase information into wave speeds:}\) The frequency is \[ f=6.0\times10^8\,\text{Hz} \] For P, \[ v_P=f\lambda_P \] \[ v_P=(6.0\times10^8)(0.20) \] \[ v_P=1.2\times10^8\,\text{m s}^{-1} \] For Q, \[ v_Q=f\lambda_Q \] \[ v_Q=(6.0\times10^8)\left(\frac2{15}\right) \] \[ v_Q=0.8\times10^8\,\text{m s}^{-1} \] \(\textbf{Determine the refractive indices:}\) Using \[ n=\frac cv \] gives \[ n_P=\frac{3.0\times10^8}{1.2\times10^8} \] \[ n_P=2.5 \] For Q, \[ n_Q=\frac{3.0\times10^8}{0.8\times10^8} \] \[ n_Q=3.75 \] \(\textbf{Apply the refraction relation in the oblique trial:}\) Huygens' construction gives \[ \sin r=\frac{v_Q}{v_P}\sin i \] Substituting, \[ \sin r=\frac{0.8}{1.2}\times\frac35 \] \[ \sin r=\frac23\times\frac35 \] \[ \sin r=\frac25 \] \(\textbf{Check the direction of bending:}\) The transmitted medium has the smaller wave speed and larger refractive index. Therefore, \[ r\lt i \] The result \[ \frac25\lt \frac35 \] is consistent with bending towards the normal. \(\textbf{Physical interpretation:}\) The measured phase gradient contains wavelength information, not frequency information by itself. Combining the spatial gradient with the independently specified temporal frequency determines the wave speed. The vacuum light speed then gives each medium's refractive index. Finally, the wave-speed ratio predicts the transmitted direction through Huygens' refraction relation. The question therefore connects a phase-versus-distance graph to material optical properties and an angular ray prediction. \(\textbf{Final answer:}\) \[ \boxed{n_P=2.5,\quad n_Q=3.75,\quad\sin r=\frac25} \]

Explanation: \(\textbf{Locate the two selected boundary points:}\) The sphere radius is \[ R=12\,\text{cm} \] The source displacement from the centre is \[ OS=3\,\text{cm} \] The distance from S to the nearer boundary point is \[ d_{\mathrm{near}}=R-OS \] \[ d_{\mathrm{near}}=12-3 \] \[ d_{\mathrm{near}}=9\,\text{cm} \] The distance to the farther boundary point is \[ d_{\mathrm{far}}=R+OS \] \[ d_{\mathrm{far}}=12+3 \] \[ d_{\mathrm{far}}=15\,\text{cm} \] \(\textbf{Find the speed inside the sphere:}\) The sphere's refractive index is \[ n_1=1.5 \] Therefore, \[ v_1=\frac c{n_1} \] \[ v_1=\frac{3.0\times10^8}{1.5} \] \[ v_1=2.0\times10^8\,\text{m s}^{-1} \] \(\textbf{Determine the two boundary-arrival times:}\) For the nearer point, \[ t_{\mathrm{near}}=\frac{0.09}{2.0\times10^8} \] \[ t_{\mathrm{near}}=0.45\,\text{ns} \] For the farther point, \[ t_{\mathrm{far}}=\frac{0.15}{2.0\times10^8} \] \[ t_{\mathrm{far}}=0.75\,\text{ns} \] \(\textbf{Compare with the observation time:}\) The observation is made at \[ t=0.60\,\text{ns} \] Therefore, \[ t_{\mathrm{near}}\lt t\lt t_{\mathrm{far}} \] The wavefront has crossed the boundary at the nearer point but has not yet reached the farther point. \(\textbf{Calculate the transmitted advance at the nearer point:}\) The surrounding-liquid index is \[ n_2=1.2 \] The transmitted-wave speed is \[ v_2=\frac{3.0\times10^8}{1.2} \] \[ v_2=2.5\times10^8\,\text{m s}^{-1} \] The elapsed time since the nearer boundary contact is \[ \Delta t=0.60-0.45 \] \[ \Delta t=0.15\,\text{ns} \] The transmitted distance beyond that boundary point is \[ s=v_2\Delta t \] \[ s=(2.5\times10^8)(0.15\times10^{-9}) \] \[ s=0.0375\,\text{m} \] \[ s=3.75\,\text{cm} \] \(\textbf{Determine the position of the opposite wavefront portion:}\) Along the opposite direction, the wavefront has travelled within the sphere for \(0.60\,\text{ns}\). Its distance from S is \[ s_{\mathrm{far}}=v_1t \] \[ s_{\mathrm{far}}=(2.0\times10^8)(0.60\times10^{-9}) \] \[ s_{\mathrm{far}}=12\,\text{cm} \] It remains \[ 15-12=3\,\text{cm} \] short of the farther boundary point. \(\textbf{Interpret the wavefront geometry:}\) The source is not at the sphere's centre. Different parts of its initial spherical wavefront therefore encounter the curved interface at different times and generally at different incidence angles. The transmitted portions travel with a different speed and undergo direction changes according to their local incidence. Consequently, the emerging wavefront is not generally a sphere centred on S. \(\textbf{Physical interpretation:}\) A concentric source and spherical interface would produce simultaneous normal incidence around the entire boundary. Displacing the source removes that symmetry. At the specified observation time, part of the disturbance has already entered the surrounding liquid while another part remains inside the sphere. Huygens' construction must account for the different propagation times, local interface normals and medium-dependent secondary-wavelet speeds. The unequal boundary-arrival times provide direct evidence that the transmitted wavefront cannot be treated as an unchanged expanding sphere centred on S. \(\textbf{Final answer:}\) \[ \boxed{t_{\mathrm{near}}=0.45\,\text{ns},\quad t_{\mathrm{far}}=0.75\,\text{ns},\quad s_{\mathrm{outside}}=3.75\,\text{cm}} \]

Explanation: \(\textbf{Determine the two propagation speeds:}\) The red refractive index is \[ n_R=1.5 \] Therefore, \[ v_R=\frac c{n_R} \] \[ v_R=\frac{3.0\times10^8}{1.5} \] \[ v_R=2.0\times10^8\,\text{m s}^{-1} \] The violet refractive index is \[ n_V=1.6 \] Hence, \[ v_V=\frac{3.0\times10^8}{1.6} \] \[ v_V=1.875\times10^8\,\text{m s}^{-1} \] \(\textbf{Calculate reflected-wavelet expansion:}\) The common time interval is \[ \Delta t=2.0\,\text{fs} \] \[ \Delta t=2.0\times10^{-15}\,\text{s} \] Each secondary wavelet expands with the light speed appropriate to its colour in the medium. For red, \[ r_R=v_R\Delta t \] \[ r_R=(2.0\times10^8)(2.0\times10^{-15}) \] \[ r_R=4.0\times10^{-7}\,\text{m} \] \[ r_R=400\,\text{nm} \] For violet, \[ r_V=v_V\Delta t \] \[ r_V=(1.875\times10^8)(2.0\times10^{-15}) \] \[ r_V=3.75\times10^{-7}\,\text{m} \] \[ r_V=375\,\text{nm} \] \(\textbf{Determine the wavelengths in the medium:}\) For light of vacuum wavelength \(\lambda_0\), \[ \lambda=\frac{\lambda_0}{n} \] For red, \[ \lambda_R=\frac{600}{1.5} \] \[ \lambda_R=400\,\text{nm} \] For violet, \[ \lambda_V=\frac{400}{1.6} \] \[ \lambda_V=250\,\text{nm} \] \(\textbf{Compare each wavelet radius with its wavelength:}\) For red, \[ \frac{r_R}{\lambda_R}=\frac{400}{400} \] \[ \frac{r_R}{\lambda_R}=1 \] For violet, \[ \frac{r_V}{\lambda_V}=\frac{375}{250} \] \[ \frac{r_V}{\lambda_V}=\frac32 \] Thus, the same observation time corresponds to one red period but one and a half violet periods. \(\textbf{Apply the reflection law:}\) The plane mirror is stationary. The incident angles are equal for both colours. The law of reflection gives \[ i_R=r_R^{(\mathrm{angle})} \] and \[ i_V=r_V^{(\mathrm{angle})} \] Therefore, the two reflected-ray angles are equal. The different wave speeds do not alter the angular equality of reflection. \(\textbf{Physical interpretation:}\) In a dispersive medium, red and violet waves can have different speeds and wavelengths. Their Huygens secondary wavelets consequently expand through different radii during the same elapsed time. However, each reflected wave remains in its original medium and retains its frequency and wavelength. The geometry of reflection still imposes equality of incidence and reflection angles for each colour. It is therefore possible for the two wavelet radii to differ while the reflected ray directions obey the same angular law. The wavelet radius after a chosen time interval must not automatically be identified with one wavelength for every frequency. \(\textbf{Final answer:}\) \[ \boxed{r_R=400\,\text{nm},\quad r_V=375\,\text{nm};\quad\lambda_R=400\,\text{nm},\quad\lambda_V=250\,\text{nm}} \]

Explanation: \(\textbf{Identify the incident convergence point:}\) The incoming wave travels to the right. Its rays would meet at \[ x_F=40\,\text{cm} \] if the mirror were absent. Thus, the wavefront is converging towards F. \(\textbf{Interpret F relative to the first mirror:}\) The initial mirror coordinate is \[ x_{m1}=25\,\text{cm} \] The would-be incident focus lies beyond the mirror. Its distance behind the mirror is \[ 40-25=15\,\text{cm} \] The rays incident on the mirror are already converging towards this point. Therefore, F acts as a virtual object for the plane mirror. \(\textbf{Locate the reflected focus:}\) A plane mirror reflects the converging bundle so that its reflected rays meet at the mirror-symmetric position of F. The image coordinate is \[ x'_F=2x_m-x_F \] For the initial mirror position, \[ x'_{F1}=2(25)-40 \] \[ x'_{F1}=10\,\text{cm} \] The reflected rays propagate to the left and converge towards this real point. The reflected wavefront is therefore converging rather than diverging. \(\textbf{Translate the mirror:}\) The new mirror coordinate is \[ x_{m2}=30\,\text{cm} \] The incoming focus remains at \[ x_F=40\,\text{cm} \] The virtual-object distance behind the translated mirror is now \[ 40-30=10\,\text{cm} \] \(\textbf{Locate the new reflected focus:}\) \[ x'_{F2}=2x_{m2}-x_F \] \[ x'_{F2}=2(30)-40 \] \[ x'_{F2}=20\,\text{cm} \] The reflected rays again converge to a real point on the incident side of the mirror. \(\textbf{Calculate focus displacement:}\) \[ \Delta x'_F=x'_{F2}-x'_{F1} \] \[ \Delta x'_F=20-10 \] \[ \Delta x'_F=10\,\text{cm} \] \(\textbf{Interpret the wavefront shapes:}\) Before reaching either mirror, the incident wavefront is converging. At reflection, the propagation direction reverses. The reflected rays remain directed towards their real meeting point on the left. Consequently, the reflected wavefront stays converging, with its centre of curvature at the corresponding reflected focus. \(\textbf{Physical interpretation:}\) A plane mirror does not necessarily reflect every incident wave into a diverging wave. Its reflected-wavefront geometry depends on whether the incoming wave is converging, diverging or plane. Here the incoming rays are already converging towards a point behind the mirror. The mirror reverses their propagation and redirects their convergence towards the mirror-symmetric real point on the incident side. Translating the mirror shifts that reflected focus by twice the mirror displacement because the incoming convergence point remains fixed. The wavefront description is consistent with the geometrical treatment of a plane mirror receiving a virtual object. \(\textbf{Final answer:}\) \[ \boxed{x'_{F1}=10\,\text{cm},\quad x'_{F2}=20\,\text{cm},\quad\Delta x'_F=10\,\text{cm}} \]

Explanation: The Assertion is true because the intersection between an advancing plane wavefront and a fixed plane surface is a geometrical feature. Its motion along the surface need not equal the actual propagation speed of the light. The light speed in the medium is \[ v=\frac c{1.5} \] \[ v=2.0\times10^8\,\text{m s}^{-1} \] For a plane wavefront incident at angle \(i\) to the interface normal, the intersection pattern advances along the interface with speed \[ u_\parallel=\frac v{\sin i} \] The specified incidence condition is \[ \sin i=\frac13 \] Therefore, \[ u_\parallel=\frac{2.0\times10^8}{1/3} \] \[ u_\parallel=6.0\times10^8\,\text{m s}^{-1} \] This is twice the vacuum speed of light. However, the wave disturbance itself continues propagating through the medium along its ray normals at \[ v=2.0\times10^8\,\text{m s}^{-1} \] The intersection point is not a physical object moving along the interface. It also does not represent a signal or packet of energy independently travelling along the surface at the calculated speed. Different points of the interface are reached by different portions of the same advancing plane wavefront. Their successive geometrical intersection positions create the apparently rapid surface motion. The Reason correctly gives both the geometrical relation and the distinction between pattern motion and physical propagation. It therefore explains why the calculated intersection speed may exceed \(c\) without implying superluminal transport of information or energy. Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Convert wavefront inclinations into ray angles:}\) In a homogeneous isotropic medium, the direction of propagation is normal to the wavefront. The normal to the plane interface is also perpendicular to the interface. Therefore, the acute angle between a wavefront and the interface equals the corresponding ray angle with the interface normal. For the incident wave, \[ i=37^\circ \] For the transmitted wave, \[ r=53^\circ \] \(\textbf{Infer the unknown refractive index:}\) Huygens' construction gives Snell's law: \[ n_P\sin i=n_Q\sin r \] Rearranging, \[ n_Q=n_P\frac{\sin i}{\sin r} \] Substituting, \[ n_Q=1.6\frac{3/5}{4/5} \] \[ n_Q=1.6\times\frac34 \] \[ n_Q=1.2 \] \(\textbf{Find the wavelength in P:}\) The vacuum wavelength is \[ \lambda_0=600\,\text{nm} \] The frequency remains unchanged upon transmission at the stationary boundary. Therefore, \[ \lambda_P=\frac{\lambda_0}{n_P} \] \[ \lambda_P=\frac{600}{1.6} \] \[ \lambda_P=375\,\text{nm} \] \(\textbf{Find the wavelength in Q:}\) \[ \lambda_Q=\frac{\lambda_0}{n_Q} \] \[ \lambda_Q=\frac{600}{1.2} \] \[ \lambda_Q=500\,\text{nm} \] \(\textbf{Check the bending direction:}\) The transmitted ray angle is larger: \[ 53^\circ\gt 37^\circ \] Thus, the ray bends away from the normal. This agrees with \[ n_Q\lt n_P \] because the wave propagates faster in Q. \(\textbf{Check the wavelength and angle ratios:}\) \[ \frac{\lambda_Q}{\lambda_P}=\frac{500}{375} \] \[ \frac{\lambda_Q}{\lambda_P}=\frac43 \] From Huygens' refraction relation, \[ \frac{\sin r}{\sin i}=\frac{4/5}{3/5} \] \[ \frac{\sin r}{\sin i}=\frac43 \] The two ratios agree. \(\textbf{Physical interpretation:}\) The question supplies wavefront orientations rather than ray angles, so the wavefront-normal relationship must be applied before using refraction laws. The transmitted wavefront becomes more steeply inclined to the interface because the ray normal bends away from the interface normal. The transmitted wavelength also increases because the wave speed increases while frequency remains fixed. The independent geometrical and wavelength predictions are consistent with the same refractive-index ratio. \(\textbf{Final answer:}\) \[ \boxed{n_Q=1.2,\quad\lambda_P=375\,\text{nm},\quad\lambda_Q=500\,\text{nm},\quad r=53^\circ} \]

Explanation: \(\textbf{Identify the common construction interval:}\) Let \(\Delta t\) be the time between the wavefront reaching A and its later point B reaching C. The reflected secondary wavelet propagates in medium P. Its radius is \[ r_P=v_P\Delta t=12\,\text{cm} \] The transmitted secondary wavelet propagates in Q. Its radius is \[ r_Q=v_Q\Delta t=9\,\text{cm} \] The two wavelets expand for the same duration. \(\textbf{Recover the incident angle from wavefront geometry:}\) In the standard Huygens construction, triangle ABC is right-angled at B because the incident ray BC is perpendicular to the incident wavefront AB. Thus, \[ BC=AC\sin i \] The distance travelled by B in medium P equals the reflected wavelet's radius over the same interval: \[ BC=v_P\Delta t=12\,\text{cm} \] Therefore, \[ \sin i=\frac{BC}{AC} \] \[ \sin i=\frac{12}{20} \] \[ \sin i=\frac35 \] \(\textbf{Determine the reflected direction:}\) The reflected wavefront is the tangent from C to the reflected secondary wavelet centred at A. The corresponding right-triangle construction gives equality of incidence and reflection angles. Hence, \[ r_{\mathrm{reflection}}=i \] and \[ \sin r_{\mathrm{reflection}}=\frac35 \] \(\textbf{Determine the transmitted direction:}\) The transmitted wavefront is the tangent drawn from C to the secondary wavelet in Q. Its angle of refraction satisfies \[ \sin r=\frac{r_Q}{AC} \] Substituting, \[ \sin r=\frac9{20} \] \(\textbf{Compare the two propagation speeds:}\) \[ \frac{v_Q}{v_P}=\frac{r_Q}{r_P} \] \[ \frac{v_Q}{v_P}=\frac9{12} \] \[ \frac{v_Q}{v_P}=\frac34 \] The transmitted medium supports a smaller wave speed. Accordingly, \[ \sin r\lt \sin i \] and the transmitted ray bends towards the interface normal. \(\textbf{Physical interpretation:}\) The same initial disturbance can produce both reflected and transmitted wavefronts at a boundary. The reflected secondary wavelets propagate in the original medium, while the transmitted wavelets propagate in the second medium. Their unequal radii after the same time reveal the difference in wave speeds. The common boundary-contact geometry determines the incidence angle, and the two tangent constructions determine the two outgoing directions. The reflected and transmitted wavefronts therefore obey different angular laws while sharing one incident wavefront. \(\textbf{Final answer:}\) \[ \boxed{\sin i=\sin r_{\mathrm{reflection}}=\frac35,\quad\sin r_{\mathrm{transmission}}=\frac9{20}} \]

Explanation: \(\textbf{Determine the speed in each layer:}\) The propagation speed in a medium is \[ v=\frac cn \] For P, \[ v_P=\frac{3.0\times10^8}{1.5} \] \[ v_P=2.0\times10^8\,\text{m s}^{-1} \] For Q, \[ v_Q=\frac{3.0\times10^8}{2.0} \] \[ v_Q=1.5\times10^8\,\text{m s}^{-1} \] For R, \[ v_R=\frac{3.0\times10^8}{1.25} \] \[ v_R=2.4\times10^8\,\text{m s}^{-1} \] \(\textbf{Find the time spent crossing P:}\) The thickness of P is \[ d_P=0.20\,\text{m} \] Therefore, \[ t_P=\frac{d_P}{v_P} \] \[ t_P=\frac{0.20}{2.0\times10^8} \] \[ t_P=1.0\,\text{ns} \] \(\textbf{Track the wavefront at }3.0\,\text{ns}\textbf{:}\) The wavefront entered Q at \(t=1.0\,\text{ns}\). Thus, by \(t=3.0\,\text{ns}\), it has travelled within Q for \[ \Delta t_Q=3.0-1.0 \] \[ \Delta t_Q=2.0\,\text{ns} \] The distance travelled in Q is \[ s_Q=v_Q\Delta t_Q \] \[ s_Q=(1.5\times10^8)(2.0\times10^{-9}) \] \[ s_Q=0.30\,\text{m} \] \[ s_Q=30\,\text{cm} \] The wavefront is therefore exactly at the Q-R boundary, not \(15\,\text{cm}\) inside Q. \(\textbf{Determine the remaining travel time:}\) Only layer R remains. Its thickness is \[ d_R=0.10\,\text{m} \] Therefore, \[ t_R=\frac{d_R}{v_R} \] \[ t_R=\frac{0.10}{2.4\times10^8} \] \[ t_R=\frac5{12}\,\text{ns} \] \(\textbf{Interpret the result:}\) At \(t=3.0\,\text{ns}\), the wavefront has reached the Q-R boundary and needs only another \(\frac5{12}\,\text{ns}\) to emerge. This matches the Q-R boundary position and remaining travel time given in option D. \(\textbf{Physical interpretation:}\) Normal transmission preserves the plane orientation of a wavefront but changes its speed and wavelength in each layer. The time spent in each layer depends on the layer thickness divided by its local wave speed. A wavefront's location cannot be inferred by multiplying the total elapsed time by one average speed chosen without justification. The entry time into each successive layer must be tracked. The calculation independently confirms that the wavefront is at the Q-R boundary at the specified observation instant. \(\textbf{Final answer:}\) \[ \boxed{\text{At Q-R boundary at }t=3.0\,\text{ns};\quad t_{\mathrm{remaining}}=\frac5{12}\,\text{ns}} \]

Explanation: \(\textbf{Compare the two wavelet expansion speeds:}\) For a secondary wavelet expanding in a homogeneous medium, \[ r_w=v\Delta t \] The observation intervals are identical. Therefore, \[ \frac{v_Q}{v_P}=\frac{r_Q}{r_P} \] Using the measured radii, \[ \frac{v_Q}{v_P}=\frac{18}{12} \] \[ \frac{v_Q}{v_P}=\frac32 \] \(\textbf{Use the common frequency:}\) The waves have the same frequency. Because \[ v=f\lambda \] the wavelength ratio equals the wave-speed ratio. Thus, \[ \frac{\lambda_Q}{\lambda_P}=\frac32 \] The wavelength in P is \[ \lambda_P=480\,\text{nm} \] Therefore, \[ \lambda_Q=\frac32(480) \] \[ \lambda_Q=720\,\text{nm} \] \(\textbf{Apply reflection at stationary mirrors:}\) A reflected wave remains in its incident medium. Its frequency does not change when it reflects from a stationary mirror. Its speed and wavelength consequently remain unchanged. Therefore, after reflection, \[ \lambda'_P=480\,\text{nm} \] \[ \lambda'_Q=720\,\text{nm} \] \(\textbf{Determine the effect of mirror rotation:}\) For a fixed incident-ray direction, rotating a plane mirror through a small angle \(\alpha\) changes the reflected-ray direction through \[ \Delta\theta=2\alpha \] This geometrical result follows from equality of incidence and reflection angles about the rotated normal. It does not contain the wave speed or wavelength. Therefore, both reflected rays undergo the same angular change for equal mirror rotations. \(\textbf{Compare wavefront orientation and spacing:}\) The reflected plane wavefront is perpendicular to its reflected ray. The mirror rotation therefore changes its orientation. However, the perpendicular spacing between successive reflected wavefronts remains the wavelength characteristic of each medium. \(\textbf{Physical interpretation:}\) The larger secondary-wavelet radius in Q indicates a larger propagation speed. At the same frequency, that produces a longer wavelength. Reflection changes the direction of wavefront propagation but not the medium's wave speed. The angular response to mirror rotation is determined by reflection geometry and is identical for the two media. Thus, the wavelet-size difference affects the wavelengths but not the factor of two relating mirror rotation to reflected-ray rotation. \(\textbf{Final answer:}\) \[ \boxed{\lambda_Q=720\,\text{nm},\quad\Delta\theta_P=\Delta\theta_Q=2\alpha} \]

Explanation: \(\textbf{Relate the reflected and transmitted directions:}\) The reflected ray makes angle \(i\) with the interface normal. The transmitted ray makes angle \(r\) with the normal on the other side. The angle between their propagation directions is \[ 180^\circ-i-r \] The two rays are stipulated to be perpendicular. Therefore, \[ 180^\circ-i-r=90^\circ \] \[ i+r=90^\circ \] Thus, \[ r=90^\circ-i \] \(\textbf{Combine the condition with Snell's law:}\) At the stationary interface, \[ n_P\sin i=n_Q\sin r \] Substituting \[ r=90^\circ-i \] gives \[ n_P\sin i=n_Q\cos i \] Therefore, \[ \tan i=\frac{n_Q}{n_P} \] \[ \tan i=\frac1\mu \] \(\textbf{Interpret the angular condition:}\) Because \[ \mu\gt 1 \] the result requires \[ \tan i\lt 1 \] Hence, \[ i\lt 45^\circ \] and \[ r\gt 45^\circ \] This agrees with a ray travelling from the optically denser medium into the optically rarer one and bending away from the normal. \(\textbf{Check reversibility:}\) In the reversed experiment, the original transmitted ray becomes the new incident ray in Q. Its incidence angle is \(r\). Applying Snell's law in reverse, \[ n_Q\sin r=n_P\sin i \] The new transmitted ray consequently follows the original incident direction backward. The original ordinary transmitted path guarantees that this exact reversed path is subcritical. \(\textbf{Physical interpretation:}\) The perpendicularity of reflected and transmitted rays adds a geometrical constraint to the normal reflection and refraction laws. Combining that constraint with Snell's law determines the incidence angle from the refractive-index ratio. Optical reversibility then predicts the complete reversed transmitted path. The incident angle is smaller than \(45^\circ\) when light travels from higher to lower refractive index under this perpendicular-ray condition. \(\textbf{Final answer:}\) \[ \boxed{\tan i=\frac1\mu,\quad r=90^\circ-i} \]

Explanation: \(\textbf{Recover the incident direction from the wavefront graph:}\) The incident equal-phase line has slope \[ m_i=\frac43 \] A vector along that wavefront is \[ \mathbf{t}_i=(3,4) \] The ray direction must be perpendicular to the wavefront. A perpendicular vector is \[ \mathbf{k}_i=(4,-3) \] because \[ (3,4)\cdot(4,-3)=12-12 \] \[ (3,4)\cdot(4,-3)=0 \] This normal points downwards and to the right, as required. \(\textbf{Determine the incidence angle:}\) The boundary is horizontal, so its normal is vertical. The incident propagation vector has horizontal and vertical components proportional to 4 and 3. Therefore, \[ \sin i=\frac4{\sqrt{4^2+3^2}} \] \[ \sin i=\frac45 \] \(\textbf{Apply Snell's law at the boundary:}\) The refractive indices are \[ n_P=1.5,\qquad n_Q=2.0 \] Using \[ n_P\sin i=n_Q\sin r \] gives \[ \sin r=\frac{1.5}{2.0}\times\frac45 \] \[ \sin r=\frac35 \] For the acute transmitted angle, \[ \cos r=\frac45 \] \(\textbf{Construct the transmitted ray vector:}\) The transmitted ray continues downwards and to the right. Its horizontal component is proportional to \(\sin r\), and its downward vertical component to \(\cos r\). Therefore, \[ \mathbf{k}_r\parallel(3,-4) \] \(\textbf{Recover the transmitted wavefront slope:}\) A direction perpendicular to \((3,-4)\) is \[ \mathbf{t}_r=(4,3) \] Thus, the transmitted equal-phase line has slope \[ m_r=\frac34 \] \(\textbf{Check the physical bending:}\) The ray changes from direction \((4,-3)\) to \((3,-4)\). Its horizontal component becomes smaller relative to its vertical component. Thus, the ray bends towards the boundary normal, consistent with entering the optically denser medium Q. \(\textbf{Physical interpretation:}\) A graph of a wavefront must not be read as though it were a graph of a ray trajectory. The incident ray is first obtained by constructing a normal to the given equal-phase line. Snell's law then changes the ray orientation according to the refractive-index ratio. Only after finding the transmitted direction can the transmitted wavefront be constructed as its perpendicular. The transmitted wavefront is less steeply inclined to the horizontal boundary because the transmitted ray bends towards the vertical normal. This connects graphical wavefront interpretation with the geometrical law of refraction. \(\textbf{Final answer:}\) \[ \boxed{m_r=\frac34,\quad\mathbf{k}_i\parallel(4,-3),\quad\mathbf{k}_r\parallel(3,-4)} \]

Explanation: \(\textbf{Identify the symmetry of the original wavefronts:}\) An ideal coherent line source has translational symmetry along its length. In the stipulated homogeneous isotropic medium, its outgoing wavefronts are cylindrical surfaces centred on the source line. Their ray normals lie in planes perpendicular to the line source. \(\textbf{Reflect the cylindrical wavefronts:}\) The reflecting plane is \[ x=0 \] For a plane mirror, reflected rays have backward extensions that originate from the mirror image of the source. An extended straight line parallel to the mirror is imaged as a corresponding straight virtual line behind it. The reflected wavefronts therefore remain cylindrical rather than becoming spherical or plane. \(\textbf{Locate the initial virtual line:}\) The original source line lies at \[ x_{S1}=-15\,\text{cm} \] Its plane-mirror image lies at \[ x_{S'_1}=+15\,\text{cm} \] The detector coordinate is \[ x_D=-5\,\text{cm} \] \(\textbf{Determine the original radius of curvature:}\) At the detector, the reflected cylindrical wavefront is centred on the virtual line at \(x=+15\,\text{cm}\). Its transverse radius is \[ R_1=|x_D-x_{S'_1}| \] \[ R_1=|-5-15| \] \[ R_1=20\,\text{cm} \] \(\textbf{Translate the actual line source:}\) The source moves \(5\,\text{cm}\) towards the mirror. Its new position is \[ x_{S2}=-15+5 \] \[ x_{S2}=-10\,\text{cm} \] \(\textbf{Locate the new virtual line:}\) The plane-mirror image is at \[ x_{S'_2}=+10\,\text{cm} \] The reflected cylindrical wavefronts now have their virtual axis along that image line. \(\textbf{Determine the new detector radius:}\) \[ R_2=|x_D-x_{S'_2}| \] \[ R_2=|-5-10| \] \[ R_2=15\,\text{cm} \] \(\textbf{Compare the radii:}\) \[ R_2-R_1=15-20 \] \[ R_2-R_1=-5\,\text{cm} \] The reflected wavefront curvature increases in magnitude because the detector is closer to the virtual line. \(\textbf{Physical interpretation:}\) The wavefront shape reflects the dimensional symmetry of its source. A plane mirror changes the apparent location of that source without changing a line source into a point source. Consequently, cylindrical incident wavefronts remain cylindrical after ideal reflection. Their virtual axes shift when the actual source moves. At a fixed detector, the radius of curvature follows from the distance to the corresponding apparent source line. The problem combines plane-mirror imaging with cylindrical wavefront geometry and distinguishes wavefront curvature from the shape of the reflecting surface. \(\textbf{Final answer:}\) \[ \boxed{x_{S'_1}=15\,\text{cm},\quad x_{S'_2}=10\,\text{cm},\quad R_1=20\,\text{cm},\quad R_2=15\,\text{cm}} \]

Explanation: \(\textbf{Calculate the two propagation speeds:}\) Inside the spherical region, \[ n_1=1.5 \] Therefore, \[ v_1=\frac{3.0\times10^8}{1.5} \] \[ v_1=2.0\times10^8\,\text{m s}^{-1} \] This is equivalent to \[ v_1=20\,\text{cm ns}^{-1} \] Outside the region, \[ n_2=1.2 \] Thus, \[ v_2=\frac{3.0\times10^8}{1.2} \] \[ v_2=2.5\times10^8\,\text{m s}^{-1} \] \[ v_2=25\,\text{cm ns}^{-1} \] \(\textbf{Determine the boundary-crossing time:}\) The spherical boundary is \[ R=30\,\text{cm} \] from the central source. Each crest requires \[ t_b=\frac R{v_1} \] \[ t_b=\frac{30}{20} \] \[ t_b=1.5\,\text{ns} \] after its own emission to reach the boundary. Therefore, the first crest crosses at \[ t_{b1}=1.5\,\text{ns} \] The second crest crosses at \[ t_{b2}=1.0+1.5 \] \[ t_{b2}=2.5\,\text{ns} \] \(\textbf{Locate the first crest at }2.0\,\text{ns}\textbf{:}\) It has spent \[ 2.0-1.5=0.5\,\text{ns} \] outside the spherical region. Its radius is \[ r_{1,2}=30+25(0.5) \] \[ r_{1,2}=42.5\,\text{cm} \] \(\textbf{Locate the second crest at }2.0\,\text{ns}\textbf{:}\) It was emitted at \(t=1.0\,\text{ns}\). Its age is \(1.0\,\text{ns}\), and it is still inside the central medium. Therefore, \[ r_{2,2}=20(1.0) \] \[ r_{2,2}=20\,\text{cm} \] \(\textbf{Calculate the earlier separation:}\) \[ \Delta r_{2.0}=42.5-20 \] \[ \Delta r_{2.0}=22.5\,\text{cm} \] \(\textbf{Locate both crests at }3.0\,\text{ns}\textbf{:}\) The first crest has been outside for \[ 3.0-1.5=1.5\,\text{ns} \] Its radius is \[ r_{1,3}=30+25(1.5) \] \[ r_{1,3}=67.5\,\text{cm} \] The second crest has been outside for \[ 3.0-2.5=0.5\,\text{ns} \] Its radius is \[ r_{2,3}=30+25(0.5) \] \[ r_{2,3}=42.5\,\text{cm} \] \(\textbf{Calculate the later separation:}\) \[ \Delta r_{3.0}=67.5-42.5 \] \[ \Delta r_{3.0}=25\,\text{cm} \] \(\textbf{Physical interpretation:}\) A central source produces concentric spherical wavefronts that cross the concentric spherical interface normally. The two crests retain their original emission-time separation. However, their instantaneous radial spacing depends on the speeds of the regions they currently occupy. When one crest has entered the faster outer medium but the next crest remains inside, the spacing takes an intermediate value. Once both crests propagate outside, their separation equals the outer-medium speed multiplied by their emission-time difference. Thus, an unchanged temporal period can correspond to different spatial wavefront spacings in different media. \(\textbf{Final answer:}\) \[ \boxed{\Delta r(2.0\,\text{ns})=22.5\,\text{cm},\quad\Delta r(3.0\,\text{ns})=25\,\text{cm}} \]

Explanation: \(\textbf{Identify the incident and reflected ray directions:}\) Take the positive horizontal direction as the reference. The incident-ray direction is \[ \theta_i=0^\circ \] The reflected-ray direction is \[ \theta_r=120^\circ \] \(\textbf{Recover the reflecting-surface orientation:}\) Reflection from a straight mirror line of orientation \(\alpha\) gives \[ \theta_r=2\alpha-\theta_i \] This relation expresses the symmetry of the incident and reflected directions about the reflecting surface. Therefore, \[ 2\alpha=\theta_r+\theta_i \] Substituting, \[ 2\alpha=120^\circ+0^\circ \] \[ \alpha=60^\circ \] The mirror surface makes \(60^\circ\) with the positive horizontal direction. \(\textbf{Determine the incidence angle:}\) The mirror normal is perpendicular to its surface. One normal orientation is \[ \theta_n=60^\circ+90^\circ \] \[ \theta_n=150^\circ \] Because a normal defines a line rather than a unique arrow, the corresponding acute angle between the incident ray and the normal line is \[ i=30^\circ \] The reflected ray also makes \(30^\circ\) with that normal line. \(\textbf{Determine the incident wavefront:}\) The incident ray travels horizontally. The incident plane wavefront is therefore vertical because its normal is horizontal. Its straight-line orientation is \[ 90^\circ \] to the positive horizontal. \(\textbf{Determine the reflected wavefront:}\) The reflected ray direction is \[ 120^\circ \] The reflected wavefront must be perpendicular to it. A wavefront-line orientation is therefore \[ 120^\circ-90^\circ=30^\circ \] Thus, the reflected wavefront makes an acute angle of \(30^\circ\) with the horizontal. \(\textbf{Check the law of reflection:}\) The incidence angle is \(30^\circ\). The reflected-ray angle relative to the normal is also \(30^\circ\). The inferred mirror orientation is therefore consistent with specular reflection. \(\textbf{Physical interpretation:}\) A reflected ray direction can be used to reconstruct an unknown mirror orientation when the incident direction is known. The reflecting surface is the angular bisector of the incident and reflected direction lines in the appropriate geometrical sense. Once the mirror orientation is fixed, its normal determines the incidence angle. The reflected wavefront is then constructed perpendicular to the reflected ray, not parallel to the mirror. The result links inverse ray geometry with Huygens' wavefront-normal description. \(\textbf{Final answer:}\) \[ \boxed{\alpha_{\mathrm{mirror}}=60^\circ,\quad i=30^\circ,\quad\alpha_{\mathrm{wavefront}}=30^\circ} \]

Explanation: \(\textbf{Identify the double-pass geometry:}\) The beams are normally incident on the mirrors. In arrangement Q, the light crosses the slab on the way to the mirror and again after reflection. Therefore, the total distance travelled within the slab is \[ 2t \] where \[ t=300\,\text{nm} \] \(\textbf{Compare corresponding travel paths:}\) In arrangement P, the corresponding total distance is travelled through air. In Q, the same geometrical segments are occupied by refractive-index \(n=1.5\) material. For light travelling through a thickness \(t\), replacing air by glass adds optical path \[ (n-1)t \] The two passages therefore add \[ \Delta=2(n-1)t \] \(\textbf{Calculate the additional optical path:}\) Substituting, \[ \Delta=2(1.5-1)(300\,\text{nm}) \] \[ \Delta=2(0.5)(300\,\text{nm}) \] \[ \Delta=300\,\text{nm} \] \(\textbf{Relate optical path to phase:}\) The vacuum wavelength is \[ \lambda_0=600\,\text{nm} \] The additional phase lag has magnitude \[ |\Delta\phi|=\frac{2\pi}{\lambda_0}\Delta \] Substituting, \[ |\Delta\phi|=\frac{2\pi}{600}(300) \] \[ |\Delta\phi|=\pi \] \(\textbf{Distinguish the common mirror-reflection effect:}\) Both arrangements contain one reflection from an equivalent stationary mirror. Any phase change common to those mirror reflections cancels when comparing the two returned waves. The calculated additional phase difference is caused by the slab's unequal optical path, not by counting one mirror reflection as a second independent difference. \(\textbf{Check direction and frequency:}\) The beam is normally incident throughout. The slab faces are parallel, and the mirror is perpendicular to the beam. Thus, the reflected beam returns along the original line in both arrangements. The stationary optical components do not change the monochromatic frequency. The spatial phase of the returned wave can nevertheless differ. \(\textbf{Physical interpretation:}\) A plane-parallel slab may change phase without changing the final direction of a normally incident beam. Placing it immediately before a mirror causes the light to traverse it twice. Consequently, its optical-path contribution is doubled compared with a single transmission. A path difference of half a vacuum wavelength produces a relative phase difference of \(\pi\). The geometry illustrates why propagation phase and ray deviation are distinct characteristics of a light wave. \(\textbf{Final answer:}\) \[ \boxed{\Delta=300\,\text{nm},\quad|\Delta\phi|=\pi} \]

Explanation: Statement I is correct because the two portions of an obliquely incident plane wavefront do not reach the interface simultaneously. The later point travels through medium P for a definite time before reaching the boundary. That interval must be calculated using the speed of the incident wave in P. Statement II is correct because the part of the disturbance that first reaches the interface gives rise to a transmitted secondary wavelet propagating in Q. The transmitted wavelet expands with the speed appropriate to Q, not the speed in P. Its radius after the common interval is therefore \(v_Q\Delta t\). Statement III is correct because the forward envelope of the transmitted secondary wavelets represents the advancing transmitted wavefront. For the standard two-dimensional construction, that envelope is represented by the appropriate tangent drawn from the later boundary-contact point to the transmitted wavelet. The normal to that transmitted wavefront gives the direction of the transmitted ray. The wavefront orientation then yields Snell's law through the ratio of the wave speeds. Statement IV is incorrect because the physically transmitted wavefront is selected by forward propagation into Q. An arbitrary tangent selected to make the propagation normal point backwards into P would not describe the ordinary transmitted wave crossing into Q. The elementary geometric construction requires the propagation direction to be physically identified; drawing every formal tangent does not make every envelope a transmitted wavefront. The distinction is especially important when the refractive-index change alters the wavefront inclination. The construction therefore requires both correct propagation distances and correct selection of the outgoing envelope. Statements I, II and III are correct.

Explanation: \(\textbf{Identify the centre of reflected wavefront curvature:}\) The point source emits diverging spherical wavefronts. After reflection from a plane mirror, the rays appear to diverge from the virtual image S' of the source. The reflected spherical wavefronts therefore have their centre of curvature at S'. \(\textbf{Express the virtual-image coordinate:}\) The source is at \[ x_S=-10\,\text{cm} \] For a mirror at coordinate \(x_m\), \[ x_{S'}=2x_m-x_S \] Therefore, \[ x_{S'}=2x_m+10 \] \(\textbf{Relate the measured radius to image position:}\) The detector is at \[ x_D=-20\,\text{cm} \] The mirror and its virtual image lie to the detector's right. Thus, the radius of curvature of the reflected wavefront at the detector is \[ R=x_{S'}-x_D \] Substituting the source-image relation, \[ R=(2x_m+10)-(-20) \] \[ R=2x_m+30 \] \(\textbf{Recover the first mirror position:}\) In experiment P, \[ R_1=50\,\text{cm} \] Therefore, \[ 50=2x_{m1}+30 \] \[ 2x_{m1}=20 \] \[ x_{m1}=10\,\text{cm} \] \(\textbf{Recover the second mirror position:}\) The new radius is \[ R_2=70\,\text{cm} \] Therefore, \[ 70=2x_{m2}+30 \] \[ 2x_{m2}=40 \] \[ x_{m2}=20\,\text{cm} \] \(\textbf{Calculate the mirror displacement:}\) \[ \Delta x_m=x_{m2}-x_{m1} \] \[ \Delta x_m=20-10 \] \[ \Delta x_m=+10\,\text{cm} \] The positive sign denotes movement to the right. \(\textbf{Check the virtual centres:}\) Initially, \[ x_{S'_1}=2(10)+10 \] \[ x_{S'_1}=30\,\text{cm} \] Finally, \[ x_{S'_2}=2(20)+10 \] \[ x_{S'_2}=50\,\text{cm} \] The detector distances to these centres are \(50\,\text{cm}\) and \(70\,\text{cm}\), respectively. \(\textbf{Physical interpretation:}\) A reflected spherical wavefront is curved about the apparent source from which its reflected rays seem to diverge. For a plane mirror, that apparent source is the virtual mirror image of the real point source. Moving the mirror shifts its virtual image through twice the mirror displacement when the real source stays fixed. The change in reflected-wavefront curvature measured at a stationary detector can therefore be used to infer the mirror's movement. This reverses the usual problem of calculating a virtual image from a known mirror position. \(\textbf{Final answer:}\) \[ \boxed{x_{m1}=10\,\text{cm},\quad x_{m2}=20\,\text{cm},\quad\Delta x_m=+10\,\text{cm}} \]

Explanation: The Assertion is true because a plane wave has parallel ray normals throughout a homogeneous isotropic medium. When such a wave encounters a stationary plane boundary, all its incident rays have the same angle relative to the common interface normal. Snell's law therefore gives the same transmitted angle for every ray in the homogeneous second medium. The transmitted rays remain parallel, so their common equal-phase surfaces remain plane. For a spherical wave emitted by a nearby point source, the incident rays diverge in different directions. They consequently encounter the plane boundary at different points and generally at different incidence angles. Their transmitted angles vary according to their individual applications of Snell's law. The portions of the disturbance also reach the interface at different times. Huygens' construction must account for these arrival times and the secondary-wavelet speed in the transmitted medium. The resulting forward envelope is not generally an exact sphere centred on a single geometrical point. A paraxial approximation may allow the transmitted rays to be treated as though they originate from an apparent point source, but that approximation is not an exact statement about every ray. The Reason correctly identifies the difference between uniform incidence of plane wavefronts and varying incidence of spherical wavefronts. It also explains the resulting distinction in transmitted wavefront geometry. A plane interface does not by itself guarantee a plane transmitted wavefront regardless of the incident wavefront shape. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Determine the critical-angle condition:}\) The refractive index of the sphere is \[ n_1=1.5 \] The surrounding air index is \[ n_2=1 \] For rays travelling from the solid into air, \[ \sin i_c=\frac{n_2}{n_1} \] \[ \sin i_c=\frac1{1.5} \] \[ \sin i_c=\frac23 \] Total internal reflection requires \[ \sin i\gt \frac23 \] \(\textbf{Relate boundary incidence to source displacement:}\) Let the source displacement from the sphere centre be \[ OS=d \] Consider a ray emitted from S and meeting the sphere at a point A. The boundary normal at A is along OA. Let the shortest perpendicular distance from O to the incident-ray line SA be \(b\). The geometrical relation between this distance and the incidence angle at the spherical surface is \[ b=R\sin i \] where \[ R=10\,\text{cm} \] \(\textbf{Determine the maximum possible incidence:}\) All rays originate from the fixed point S. The perpendicular distance from O to any line through S cannot exceed OS. Thus, \[ b_{\max}=d \] Therefore, \[ R\sin i_{\max}=d \] \[ \sin i_{\max}=\frac dR \] \(\textbf{First source position:}\) Initially, \[ d_1=5\,\text{cm} \] Therefore, \[ \sin i_{\max,1}=\frac5{10} \] \[ \sin i_{\max,1}=\frac12 \] Compare with the critical condition: \[ \frac12\lt \frac23 \] Even the largest incidence angle is below the critical angle. Thus, every ray can undergo ordinary refraction through the boundary, and total internal reflection is impossible. \(\textbf{Second source position:}\) The source moves to \[ d_2=8\,\text{cm} \] Therefore, \[ \sin i_{\max,2}=\frac8{10} \] \[ \sin i_{\max,2}=\frac45 \] Now, \[ \frac45\gt \frac23 \] Some rays can reach the spherical boundary above the critical angle. These rays undergo total internal reflection. Other rays with smaller incidence angles can still transmit ordinarily. \(\textbf{Determine the displacement threshold:}\) The first possibility of critical incidence occurs when \[ \sin i_{\max}=\sin i_c \] Therefore, \[ \frac{d_c}{R}=\frac23 \] \[ d_c=\frac{20}{3}\,\text{cm} \] The first source position is below this threshold, while the second is above it. \(\textbf{Physical interpretation:}\) A point source at the centre of a sphere sends rays normally towards the entire spherical boundary. Displacing the source changes the relationship between its radial emission directions and the local boundary normals. The incidence angles are then not all zero. Their maximum value grows as the source moves farther from the centre. If the displacement becomes sufficiently large, some rays reach the surface above the critical angle even though others remain subcritical. This connects spherical wavefront geometry, local ray normals and total internal reflection in a genuinely changed-condition problem. \(\textbf{Final answer:}\) \[ \boxed{\sin i_{\max,1}=\frac12\lt \frac23,\quad\sin i_{\max,2}=\frac45\gt \frac23,\quad d_c=\frac{20}{3}\,\text{cm}} \]