Optics MCQs with Answers – Part 2 (NEET Physics)

Explanation: The incident ray initially travels through the surrounding liquid, which has the greater refractive index. The slab has a smaller refractive index than the liquid. Therefore, light travels faster inside the slab than in the surrounding medium. At the first interface, the ray passes from the optically denser liquid into the optically rarer slab material. For an obliquely incident transmitted ray, Snell's law requires the angle of refraction inside the slab to be greater than the incidence angle. The ray therefore bends away from the normal at entry. The two slab faces are plane and parallel, so their normals are parallel. When the ray reaches the second face, it passes from the optically rarer slab back into the optically denser liquid. It therefore bends towards the normal at exit. Because the external medium is the same on both sides, the second refraction restores the original angle with the normal. The emergent ray is consequently parallel to the incident ray. However, the ray does not generally return to the straight-line continuation of its original path. Inside the slab, its greater angle with the normal makes it travel farther along the interface direction than it would have travelled without refraction. The emergent ray is therefore laterally displaced towards the side corresponding to this larger internal transverse displacement. This is opposite to the displacement produced when a ray enters a slab of greater refractive index than its surroundings. A plane-parallel slab does not universally displace a ray in only one transverse direction; the displacement direction depends on whether its refractive index is greater or smaller than that of the external medium. The emergent ray remains parallel because the initial and final media are identical, not because the refractive-index contrast is zero. Thus, the correct result is a parallel emergent ray displaced away from the normal relative to the unrefracted continuation.

Explanation: \(\textbf{Initial arrangement:}\) The object is initially at a depth \[ d_1=36\,\text{cm} \] The liquid refractive index is \[ n=1.5 \] The apparent depth for near-normal observation from air is \[ d'_1=\frac{d_1}{n} \] Substituting, \[ d'_1=\frac{36}{1.5} \] \[ d'_1=24\,\text{cm} \] \(\textbf{Choose a fixed coordinate system:}\) Let the original liquid surface be at vertical coordinate \(y=0\). Take upward as positive. The actual initial object position is \[ y_{o1}=-36\,\text{cm} \] Its initial apparent position is \[ y'_{o1}=-24\,\text{cm} \] \(\textbf{Move the actual object:}\) The object moves upward through \[ \Delta y_o=+9\,\text{cm} \] Therefore, \[ y_{o2}=-36+9 \] \[ y_{o2}=-27\,\text{cm} \] \(\textbf{Raise the liquid surface:}\) The surface rises by \[ \Delta y_s=+6\,\text{cm} \] Its new position is \[ y_{s2}=+6\,\text{cm} \] \(\textbf{Determine the new actual depth:}\) The object's depth below the new surface is \[ d_2=y_{s2}-y_{o2} \] Substituting, \[ d_2=6-(-27) \] \[ d_2=33\,\text{cm} \] \(\textbf{Calculate the new apparent depth:}\) \[ d'_2=\frac{d_2}{n} \] \[ d'_2=\frac{33}{1.5} \] \[ d'_2=22\,\text{cm} \] \(\textbf{Convert to the fixed coordinate system:}\) The new surface is at \(y=+6\,\text{cm}\). The apparent image lies \(22\,\text{cm}\) below this surface. Therefore, \[ y'_{o2}=6-22 \] \[ y'_{o2}=-16\,\text{cm} \] \(\textbf{Find the apparent-image displacement:}\) \[ \Delta y'=y'_{o2}-y'_{o1} \] \[ \Delta y'=-16-(-24) \] \[ \Delta y'=+8\,\text{cm} \] \(\textbf{Separate the two physical contributions:}\) For a fixed liquid surface, upward movement of the actual object through \(9\,\text{cm}\) would move its apparent image upward by \[ \Delta y'_{\text{object}}=\frac9{1.5} \] \[ \Delta y'_{\text{object}}=6\,\text{cm} \] For a stationary object, raising the liquid surface through \(6\,\text{cm}\) would move its apparent image upward by \[ \Delta y'_{\text{surface}}=6\left(1-\frac1{1.5}\right) \] \[ \Delta y'_{\text{surface}}=2\,\text{cm} \] The two contributions add. \[ \Delta y'=6+2 \] \[ \Delta y'=8\,\text{cm} \] \(\textbf{Physical interpretation:}\) The actual object moves upward by \(9\,\text{cm}\), but its apparent image moves upward by only \(8\,\text{cm}\). The difference arises because the apparent position depends on both the object coordinate and the liquid-surface coordinate. The actual depth below the surface decreases from \(36\,\text{cm}\) to \(33\,\text{cm}\), despite the liquid level rising. The apparent depth correspondingly decreases from \(24\,\text{cm}\) to \(22\,\text{cm}\). The image displacement relative to the vessel must account for the movement of the reference surface as well as the change in apparent depth. \(\textbf{Final answer:}\) \[ \boxed{\Delta y'=8\,\text{cm}\text{ upward}} \]

Explanation: Statement I is correct because the lateral displacement through a plane-parallel slab is directly proportional to its perpendicular thickness when the incidence and refraction angles remain fixed. The relevant relation is \[ d=\frac{t\sin(i-r)}{\cos r} \] Here, \(t\) is the slab thickness, \(i\) is the incidence angle and \(r\) is the refraction angle. For the same material and unchanged incidence angle, Snell's law gives the same refraction angle. Therefore, doubling \(t\) doubles \(d\). Statement II is incorrect because the angle of emergence is determined by Snell's law and the refractive indices of the media, not by the slab thickness. For a plane-parallel slab surrounded by air on both sides, the emergence angle equals the incidence angle. Doubling the thickness changes the lateral displacement but does not double the emergence angle. Statement III is correct because a normally incident ray travels along the common normal to the slab faces. Snell's law gives zero refraction angle at each surface. Consequently, the ray experiences no lateral displacement. Nevertheless, the propagation speed inside the glass is lower than in air because the glass has refractive index greater than unity. Thus, zero bending does not imply unchanged speed. Statement IV is correct because the two slab faces are parallel and the ray returns to the same external medium. The angular change at the second surface reverses the angular change at the first surface. The emergent ray is therefore parallel to the incident ray. It may still be laterally displaced at oblique incidence. The distinction between angular deviation and lateral displacement is essential. A slab can have zero net angular deviation without producing zero lateral displacement. Therefore, Statements I, III and IV are correct.

Explanation: \(\textbf{First slab — oblique incidence:}\) The first slab has thickness \[ t_1=8\,\text{cm} \] The incidence and refraction angles are \[ i_1=45^\circ,\qquad r_1=30^\circ \] The lateral displacement is \[ d_1=\frac{t_1\sin(i_1-r_1)}{\cos r_1} \] Substituting, \[ d_1=\frac{8\sin(45^\circ-30^\circ)}{\cos30^\circ} \] \[ d_1=\frac{8\sin15^\circ}{\cos30^\circ} \] Since \(\sin15^\circ\) is positive, \[ d_1\gt 0 \] Thus, the first slab produces a non-zero lateral displacement. \(\textbf{Second slab — normal incidence:}\) The second slab is thicker. \[ t_2=12\,\text{cm} \] However, the incident ray is normal to the surface. \[ i_2=0^\circ \] Snell's law gives \[ n_{\text{air}}\sin i_2=n_{\text{glass}}\sin r_2 \] Therefore, \[ r_2=0^\circ \] \(\textbf{Lateral displacement in the second slab:}\) Using the slab relation, \[ d_2=\frac{t_2\sin(i_2-r_2)}{\cos r_2} \] Substituting, \[ d_2=\frac{12\sin0^\circ}{\cos0^\circ} \] \[ d_2=0 \] \(\textbf{Interpret the comparison:}\) The first slab produces a non-zero displacement because the ray is oblique and changes direction inside the glass. The second slab produces zero lateral displacement despite its greater thickness. At normal incidence, the ray travels along the normal through both surfaces. There is no transverse component of refraction-induced displacement. The thickness of a slab affects lateral displacement only when the incidence and refraction angles permit a non-zero shift. Increasing thickness cannot create lateral displacement for a ray that remains normally incident. Thus, the incident-ray geometry must be examined before using thickness proportionality. \(\textbf{Final answer:}\) The first slab produces a non-zero lateral displacement, whereas the second produces zero displacement.

Explanation: \(\textbf{Original slab:}\) The original thickness is \[ t_1=20\,\text{cm} \] Its refractive index is \[ n_1=1.5 \] The mark lies on the lower surface. For near-normal observation from air, the apparent depth is \[ d_1=\frac{t_1}{n_1} \] Substituting, \[ d_1=\frac{20}{1.5} \] \[ d_1=\frac{40}{3}\,\text{cm} \] \(\textbf{Replacement slab condition:}\) The replacement slab has refractive index \[ n_2=1.25=\frac54 \] Its unknown thickness is \(t_2\). The apparent depth must remain equal to the original apparent depth. Therefore, \[ \frac{t_2}{n_2}=\frac{t_1}{n_1} \] Rearranging, \[ t_2=t_1\frac{n_2}{n_1} \] Substituting, \[ t_2=20\frac{1.25}{1.5} \] \[ t_2=20\left(\frac56\right) \] \[ t_2=\frac{50}{3}\,\text{cm} \] \[ t_2\approx16.67\,\text{cm} \] \(\textbf{Compare light speeds:}\) The speed of light in a transparent medium is \[ v=\frac{c}{n} \] Therefore, \[ \frac{v_2}{v_1}=\frac{c/n_2}{c/n_1} \] \[ \frac{v_2}{v_1}=\frac{n_1}{n_2} \] Substituting, \[ \frac{v_2}{v_1}=\frac{1.5}{1.25} \] \[ \frac{v_2}{v_1}=\frac65 \] \(\textbf{Physical check:}\) The replacement material has a lower refractive index. Light therefore travels faster in the replacement slab. Its speed is \(1.2\) times the speed in the original slab. To preserve the same apparent depth, the replacement slab must also be physically thinner. The calculated thickness of approximately \(16.67\,\text{cm}\) is smaller than the original \(20\,\text{cm}\). Both changes are consistent with the relation \(d'=t/n\). The equality of apparent depths does not require the physical thicknesses or light speeds to be equal. \(\textbf{Final answer:}\) \[ \boxed{t_2=\frac{50}{3}\,\text{cm}\approx16.67\,\text{cm},\quad\frac{v_2}{v_1}=\frac65} \]

Explanation: A plane-parallel slab has two reflecting-free refracting interfaces whose normals are parallel. The incident ray enters from air and refracts into the slab according to Snell's law. For a fixed angle of incidence, fixed refractive index and unchanged surrounding medium, the internal refraction angle remains fixed. The slab is translated parallel to its faces without rotation. Such a translation does not change the orientation of either refracting surface. It also does not change the perpendicular distance between the two faces. Therefore, the incidence angle and refraction angle remain the same. The ray still undergoes equal and opposite angular changes at the entry and exit surfaces. Since the external medium is air on both sides, the emergent ray remains parallel to the original incident ray. Its lateral displacement is \[ d=\frac{t\sin(i-r)}{\cos r} \] This relation depends on the slab thickness and the two ray angles. All three quantities remain unchanged during the translation. Consequently, the lateral displacement remains unchanged. The exact points at which the fixed incident line intersects the physical material may change relative to the slab's edges, but the infinite-plane optical geometry is unchanged. The assumption that the slab is sufficiently large ensures that edge effects do not prevent transmission. A translation parallel to the faces does not create an additional refracting angle or alter the normal separation of the faces. Thus, neither the final propagation direction nor the lateral displacement changes. This illustrates that optical ray geometry depends on the positions and orientations of the actual interfaces, not on the arbitrary location of the finite slab's material boundaries parallel to those interfaces.

Explanation: \(\textbf{Incident ray and surface normal:}\) The forward normal from air into the transparent medium has orientation \[ \theta_n=20^\circ \] The direction of incident propagation is \[ \theta_i=80^\circ \] Therefore, the angle of incidence, measured from the normal, is \[ i=\theta_i-\theta_n \] \[ i=80^\circ-20^\circ \] \[ i=60^\circ \] \(\textbf{Refraction into the second medium:}\) The refractive indices are \[ n_1=1,\qquad n_2=\sqrt3 \] Snell's law gives \[ n_1\sin i=n_2\sin r \] Substituting, \[ 1\times\sin60^\circ=\sqrt3\sin r \] \[ \frac{\sqrt3}{2}=\sqrt3\sin r \] \[ \sin r=\frac12 \] \[ r=30^\circ \] \(\textbf{Determine the absolute refracted direction:}\) The refracted ray bends towards the normal because the second medium has the greater refractive index. It remains on the same side of the forward normal as the incident propagation direction. Therefore, \[ \theta_t=\theta_n+r \] \[ \theta_t=20^\circ+30^\circ \] \[ \theta_t=50^\circ \] \(\textbf{Determine the reflected direction:}\) Reflection reverses the component of propagation perpendicular to the interface but preserves the component parallel to it. The reflecting surface is perpendicular to the specified normal. Therefore, its orientation is \[ \theta_s=20^\circ+90^\circ \] \[ \theta_s=110^\circ \] For a reflected propagation direction \(\theta_r\), \[ \theta_r=2\theta_s-\theta_i \] Substituting, \[ \theta_r=2(110^\circ)-80^\circ \] \[ \theta_r=140^\circ \] \(\textbf{Physical consistency:}\) The refracted ray makes \(30^\circ\) with the forward normal, while the incident ray makes \(60^\circ\). Thus, the transmitted ray bends towards the normal. The reflected ray travels back into the incident medium, which is required for reflection. Its direction of \(140^\circ\) makes an angle of \(60^\circ\) with the backward normal at \(200^\circ\). This equals the angle of incidence and satisfies the reflection law. The two outgoing rays are therefore distinguished by both their propagation regions and their angles. \(\textbf{Final answer:}\) \[ \boxed{\theta_t=50^\circ,\quad\theta_r=140^\circ} \]

Explanation: \(\textbf{Identify the colour-dependent refractive indices:}\) For red light, \[ n_R=\frac43 \] For violet light, \[ n_V=\frac32 \] Both rays have the same incidence angle. \[ i=30^\circ \] \(\textbf{Apply Snell's law to red light:}\) Taking the refractive index of air as unity, \[ \sin i=n_R\sin r_R \] Therefore, \[ \sin r_R=\frac{\sin30^\circ}{\frac43} \] \[ \sin r_R=\frac12\times\frac34 \] \[ \sin r_R=\frac38 \] \(\textbf{Apply Snell's law to violet light:}\) Similarly, \[ \sin r_V=\frac{\sin30^\circ}{\frac32} \] \[ \sin r_V=\frac12\times\frac23 \] \[ \sin r_V=\frac13 \] \(\textbf{Compare the ray directions:}\) Since \[ \frac13\lt \frac38 \] the violet refraction angle is smaller than the red refraction angle. Violet light therefore bends more towards the normal. \(\textbf{Determine the red wavelength:}\) The incident red wavelength is \[ \lambda_{R,\text{air}}=720\,\text{nm} \] Frequency remains unchanged at the stationary boundary. Thus, \[ \lambda_{R,\text{medium}}=\frac{\lambda_{R,\text{air}}}{n_R} \] \[ \lambda_{R,\text{medium}}=\frac{720}{\frac43} \] \[ \lambda_{R,\text{medium}}=720\times\frac34 \] \[ \lambda_{R,\text{medium}}=540\,\text{nm} \] \(\textbf{Determine the violet wavelength:}\) \[ \lambda_{V,\text{medium}}=\frac{\lambda_{V,\text{air}}}{n_V} \] \[ \lambda_{V,\text{medium}}=\frac{480}{\frac32} \] \[ \lambda_{V,\text{medium}}=480\times\frac23 \] \[ \lambda_{V,\text{medium}}=320\,\text{nm} \] \(\textbf{Physical interpretation:}\) The refractive index is larger for violet light in the specified material. Consequently, violet light has a lower speed and undergoes greater angular bending towards the normal. Red and violet light do not have identical frequencies because they originate with different wavelengths in air. However, the frequency of each individual colour remains unchanged during its own refraction. The transmitted wavelengths decrease according to the respective colour-dependent refractive indices. These conditions produce different refraction angles even though both rays have the same incident direction. \(\textbf{Final conclusion:}\) Violet light bends closer to the normal; the wavelengths inside the material are \(540\,\text{nm}\) for red and \(320\,\text{nm}\) for violet.

Explanation: \(\textbf{Angular measurements at the interface:}\) The incidence and refraction angles are \[ i=30^\circ,\qquad r=45^\circ \] The absolute refractive index of Q is \[ n_Q=\sqrt2 \] The refractive index of P is unknown. \(\textbf{Apply Snell's law:}\) \[ n_P\sin i=n_Q\sin r \] Rearranging, \[ n_P=n_Q\frac{\sin r}{\sin i} \] Substituting, \[ n_P=\sqrt2\frac{\sin45^\circ}{\sin30^\circ} \] \[ n_P=\sqrt2\frac{\frac1{\sqrt2}}{\frac12} \] \[ n_P=2 \] \(\textbf{Interpret the original refraction:}\) The refractive index decreases from P to Q. Therefore, the ray bends away from the normal. This agrees with the increase in angle from \(30^\circ\) to \(45^\circ\). \(\textbf{Apparent-distance observation:}\) A small object is placed in medium P at real depth \[ d=30\,\text{cm} \] The observer is in medium Q. For near-normal viewing across a plane interface, \[ d'=d\frac{n_{\text{observer}}}{n_{\text{object}}} \] Here, \[ n_{\text{observer}}=n_Q=\sqrt2 \] and \[ n_{\text{object}}=n_P=2 \] \(\textbf{Substitution:}\) \[ d'=30\frac{\sqrt2}{2} \] \[ d'=15\sqrt2\,\text{cm} \] \(\textbf{Check the apparent-image direction:}\) The observer's medium has a smaller refractive index than the object medium. Therefore, the object appears closer to the interface than its actual position. The calculated apparent distance is approximately \(21.2\,\text{cm}\), smaller than \(30\,\text{cm}\). This agrees with the expected apparent raising of an object viewed from an optically rarer medium. \(\textbf{Important distinction:}\) The angles \(30^\circ\) and \(45^\circ\) are used to determine the refractive-index contrast. The apparent-depth relation is then applied only to a separate near-normal observation. It would be incorrect to insert the original oblique angles directly into the near-normal apparent-depth formula. The two parts of the problem are linked through the material refractive indices, not through identical viewing-ray geometry. \(\textbf{Final answer:}\) \[ \boxed{n_P=2,\quad d'=15\sqrt2\,\text{cm}} \]

Explanation: \(\textbf{Define the unknown layer thicknesses:}\) Let the thickness of the upper liquid be \(t_1\). Let the thickness of the lower liquid be \(t_2\). The actual total depth is \[ t_1+t_2=60\,\text{cm} \] The two refractive indices are \[ n_1=\frac43,\qquad n_2=\frac32 \] \(\textbf{Near-normal apparent-depth relation:}\) For an object viewed from air through plane-parallel liquid layers, \[ d'=\frac{t_1}{n_1}+\frac{t_2}{n_2} \] The measured apparent depth is \[ d'=42\,\text{cm} \] Therefore, \[ \frac{t_1}{\frac43}+\frac{t_2}{\frac32}=42 \] \[ \frac34t_1+\frac23t_2=42 \] \(\textbf{Use the known total depth:}\) From \[ t_1+t_2=60 \] we obtain \[ t_2=60-t_1 \] Substitute into the apparent-depth equation. \[ \frac34t_1+\frac23(60-t_1)=42 \] \[ \frac34t_1+40-\frac23t_1=42 \] \[ \left(\frac34-\frac23\right)t_1=2 \] \[ \frac1{12}t_1=2 \] \[ t_1=24\,\text{cm} \] The lower-layer thickness is \[ t_2=60-24 \] \[ t_2=36\,\text{cm} \] \(\textbf{Verify the original observation:}\) \[ d'=\frac{24}{\frac43}+\frac{36}{\frac32} \] \[ d'=18+24 \] \[ d'=42\,\text{cm} \] The measured apparent depth is reproduced. \(\textbf{Effect of interchanging the layer order:}\) The hypothetical rearrangement preserves the physical thickness belonging to each liquid. The upper and lower positions are interchanged, but the optical contribution of each material remains determined by its own thickness and refractive index. For near-normal viewing from air, the new apparent depth is therefore \[ d'_{\text{new}}=\frac{36}{\frac32}+\frac{24}{\frac43} \] \[ d'_{\text{new}}=24+18 \] \[ d'_{\text{new}}=42\,\text{cm} \] \(\textbf{Physical interpretation:}\) The apparent-depth contributions add because the liquid interfaces are plane and parallel and the observation is paraxial. Changing the order of these layers does not change the sum of their optical thickness contributions. The internal ray directions may be considered successively using Snell's law, but the final near-normal apparent depth depends on the thickness-index pairs. The real object remains at the same physical depth of \(60\,\text{cm}\). The apparent upward shift consequently remains \(60-42=18\,\text{cm}\). This invariance should not be extended to arbitrary non-parallel interfaces or strongly oblique viewing. \(\textbf{Final answer:}\) \[ \boxed{t_1=24\,\text{cm},\quad d'_{\text{new}}=42\,\text{cm}} \]

Explanation: \(\textbf{Actual object geometry:}\) The two luminous points are at equal depth below the liquid surface. Their horizontal separation is \[ s=3\,\text{cm} \] Their real depth is \[ d=24\,\text{cm} \] The liquid refractive index is \[ n=1.5 \] \(\textbf{Determine the apparent depth:}\) For near-normal viewing from air, \[ d'=\frac{d}{n} \] Substituting, \[ d'=\frac{24}{1.5} \] \[ d'=16\,\text{cm} \] Both points appear at this same depth. \(\textbf{Horizontal apparent separation:}\) A plane refracting interface changes the apparent axial depth under paraxial observation. For the stated small transverse separation, the horizontal separation of the apparent points remains \[ s'=3\,\text{cm} \] \(\textbf{Initial eye position:}\) The observer is initially \(18\,\text{cm}\) above the surface. The perpendicular distance from the eye to the plane containing the apparent points is \[ L_1=18+d' \] \[ L_1=18+16 \] \[ L_1=34\,\text{cm} \] \(\textbf{Initial angular separation:}\) In the paraxial approximation, \[ \alpha_1\approx\frac{s'}{L_1} \] \[ \alpha_1\approx\frac3{34}\,\text{rad} \] \(\textbf{Changed eye position:}\) The observer lowers the eye to \(8\,\text{cm}\) above the surface. The new distance to the apparent-point plane is \[ L_2=8+d' \] \[ L_2=8+16 \] \[ L_2=24\,\text{cm} \] \(\textbf{New angular separation:}\) \[ \alpha_2\approx\frac{s'}{L_2} \] \[ \alpha_2\approx\frac3{24}\,\text{rad} \] \(\textbf{Calculate the increase factor:}\) \[ \frac{\alpha_2}{\alpha_1}=\frac{\frac3{24}}{\frac3{34}} \] \[ \frac{\alpha_2}{\alpha_1}=\frac{34}{24} \] \[ \frac{\alpha_2}{\alpha_1}=\frac{17}{12} \] \(\textbf{Physical interpretation:}\) The luminous points and liquid surface remain fixed. Consequently, their apparent depths do not change during the observer's movement. However, the eye approaches the plane containing their apparent positions. The unchanged apparent transverse separation therefore subtends a greater angle. The increase is governed by the ratio of the initial and final eye-to-apparent-image distances. Using the actual depth of \(24\,\text{cm}\) instead of the apparent depth of \(16\,\text{cm}\) would give an incorrect ratio. This problem combines the axial effect of plane-surface refraction with the angular geometry of observation. \(\textbf{Final answer:}\) \[ \boxed{\frac{\alpha_2}{\alpha_1}=\frac{17}{12}} \]

Explanation: Both colours are incident at the same angle \(i\). The refractive indices of the slab are different for the two colours. The given ordering is \(n_V\gt n_R\), so violet light travels more slowly inside the slab than red light. Snell's law at the first air-slab interface gives \[ \sin r=\frac{\sin i}{n} \] For the same positive incidence angle, increasing the refractive index decreases the refraction angle. Therefore, violet light makes the smaller angle with the normal inside the slab. Both rays enter a material having refractive index greater than that of air, so both bend towards the normal. For a plane-parallel slab, the lateral displacement is \[ d=\frac{t\sin(i-r)}{\cos r} \] Here, \(t\) and \(i\) are fixed for both colours. Using the trigonometric identity for the difference of angles, \[ d=t(\sin i-\cos i\tan r) \] For a fixed oblique incidence angle, a smaller positive refraction angle gives a smaller value of \(\tan r\). Consequently, the lateral displacement is greater for the colour having the smaller refraction angle. Thus, violet light undergoes a greater lateral displacement than red light. At the second face, each colour passes from the slab back into air. Since the slab faces are parallel, the exit refraction restores the original external propagation direction for each colour. Both emergent rays are therefore parallel to the common incident direction. However, their different lateral displacements mean that they need not emerge along the same straight line. The difference in lateral shift is caused by the wavelength dependence of refractive index, not by a permanent angular deviation through the parallel-sided slab. Therefore, violet light has the smaller internal angle and greater lateral displacement.

Explanation: \(\textbf{Interpret the measured slope:}\) The lateral-displacement relation for a plane-parallel slab is \[ d=\frac{t\sin(i-r)}{\cos r} \] The graph plots \(d\) vertically against \(t\) horizontally. Therefore, its slope is \[ S=\frac{d}{t}=\frac{\sin(i-r)}{\cos r} \] The measured slope is \[ S=\frac1{\sqrt3} \] The fixed incidence angle is \[ i=60^\circ \] \(\textbf{Rewrite the slope relation:}\) Use \[ \sin(i-r)=\sin i\cos r-\cos i\sin r \] Dividing by \(\cos r\), \[ S=\sin i-\cos i\tan r \] Substitute the known quantities. \[ \frac1{\sqrt3}=\sin60^\circ-\cos60^\circ\tan r \] \[ \frac1{\sqrt3}=\frac{\sqrt3}{2}-\frac12\tan r \] Multiply by two. \[ \frac2{\sqrt3}=\sqrt3-\tan r \] Therefore, \[ \tan r=\sqrt3-\frac2{\sqrt3} \] \[ \tan r=\frac1{\sqrt3} \] Hence, \[ r=30^\circ \] \(\textbf{Determine the refractive index:}\) Snell's law gives \[ n_{\text{air}}\sin i=n\sin r \] With \(n_{\text{air}}=1\), \[ n=\frac{\sin60^\circ}{\sin30^\circ} \] \[ n=\frac{\frac{\sqrt3}{2}}{\frac12} \] \[ n=\sqrt3 \] \(\textbf{Determine the light speed:}\) Using the definition of absolute refractive index, \[ v=\frac{c}{n} \] Substituting, \[ v=\frac{3\times10^8}{\sqrt3} \] Rationalising, \[ v=\sqrt3\times10^8\,\text{m s}^{-1} \] \(\textbf{Physical consistency:}\) The inferred refractive index is greater than unity. The ray therefore bends towards the normal when entering the slab from air. The calculated internal angle of \(30^\circ\) is smaller than the external incidence angle of \(60^\circ\). For those angles, the slab formula predicts \[ \frac{d}{t}=\frac{\sin30^\circ}{\cos30^\circ} \] \[ \frac{d}{t}=\frac1{\sqrt3} \] This reproduces the measured graph slope. The graph thus allows the refractive index to be inferred indirectly from the displacement rather than from a direct measurement of the internal angle. \(\textbf{Final answer:}\) \[ \boxed{n=\sqrt3,\quad v=\sqrt3\times10^8\,\text{m s}^{-1}} \]

Explanation: \(\textbf{Actual rod geometry:}\) The endpoints have different depths below the liquid surface. \[ d_A=12\,\text{cm} \] \[ d_B=27\,\text{cm} \] Their horizontal separation is \[ x=12\,\text{cm} \] The actual vertical separation is \[ \Delta d=d_B-d_A \] \[ \Delta d=27-12 \] \[ \Delta d=15\,\text{cm} \] \(\textbf{Apparent depth of endpoint A:}\) For near-normal viewing from air, \[ d'=\frac{d}{n} \] The liquid refractive index is \[ n=1.5 \] Therefore, \[ d'_A=\frac{12}{1.5} \] \[ d'_A=8\,\text{cm} \] \(\textbf{Apparent depth of endpoint B:}\) \[ d'_B=\frac{27}{1.5} \] \[ d'_B=18\,\text{cm} \] \(\textbf{Apparent vertical separation:}\) \[ \Delta d'=d'_B-d'_A \] \[ \Delta d'=18-8 \] \[ \Delta d'=10\,\text{cm} \] \(\textbf{Horizontal separation:}\) For the stated paraxial observation through a plane interface, the apparent transverse coordinates are unchanged. Thus, \[ x'=12\,\text{cm} \] \(\textbf{Apparent inclination of the rod:}\) Let the apparent image make an angle \(\theta'\) with the horizontal. Its tangent is the apparent vertical separation divided by the horizontal separation. \[ \tan\theta'=\frac{\Delta d'}{x'} \] Substituting, \[ \tan\theta'=\frac{10}{12} \] \[ \tan\theta'=\frac56 \] \(\textbf{Apparent length of the rod:}\) The apparent endpoint separation follows from the Pythagorean relation. \[ L'=\sqrt{(x')^2+(\Delta d')^2} \] \[ L'=\sqrt{12^2+10^2} \] \[ L'=\sqrt{144+100} \] \[ L'=\sqrt{244} \] \[ L'=2\sqrt{61}\,\text{cm} \] \(\textbf{Physical interpretation:}\) A plane liquid-air interface makes submerged points appear closer to the surface when viewed from air. Both endpoints are shifted upward, but the deeper endpoint has the greater upward shift. Consequently, the apparent vertical separation of the endpoints is reduced. The horizontal separation is unchanged in the paraxial treatment. The rod therefore appears less steeply inclined to the horizontal and shorter than its actual length. The actual inclination would satisfy \(\tan\theta=15/12=5/4\), which exceeds the apparent value \(5/6\). Thus, the image is not obtained by translating the entire rod upward through one common distance. Different object depths produce different apparent shifts. \(\textbf{Final answer:}\) \[ \boxed{\tan\theta'=\frac56,\quad L'=2\sqrt{61}\,\text{cm}} \]

Explanation: Statement I is correct because the plate and surrounding liquid have equal refractive indices. Snell's law at either interface is \[ n\sin i=n\sin r \] Therefore, for the transmitted ray, \[ r=i \] The propagation direction does not change at either boundary. Statement II is incorrect because the absence of refractive-index contrast means that the plate does not bend the ray. A ray travelling through the matched plate follows the same straight-line path it would follow through the surrounding liquid. Therefore, the plate produces zero lateral displacement relative to that uninterrupted path, irrespective of its thickness. Increasing the thickness of an optically matched plate does not create a lateral displacement. Statement III is correct because the ray remains in media having the same refractive index. The phase speed is therefore unchanged. The frequency also remains unchanged across the stationary interfaces. Using \(v=f\lambda\), the wavelength must remain unchanged as well. Statement IV is correct because replacing a region of liquid with an optically matched plate does not change the ray's geometrical propagation or phase speed under the stated conditions. For the specified monochromatic light, the plate has no refractive-index contrast relative to the liquid. Its plane boundaries therefore do not alter the transmitted ray path. The conclusion does not require the plate material and the liquid to have identical mass densities or chemical compositions. The relevant property is their equality of refractive indices at the wavelength being considered. The distinction between matching refractive indices and merely having parallel faces is important. An ordinary unmatched parallel slab can produce lateral displacement even when its emergent ray is parallel to the incident ray. An index-matched plate produces neither angular deviation nor lateral displacement. Thus, Statements I, III and IV are correct.

Explanation: \(\textbf{Normal-incidence condition:}\) The beam is normally incident on each plate. Its direction remains unchanged when it enters or emerges from a plane-parallel plate. Therefore, the actual geometrical path length through each plate equals its perpendicular thickness. \(\textbf{Additional travel-time relation:}\) For a plate of thickness \(t\) and refractive index \(n\), the light speed is \[ v=\frac{c}{n} \] The time taken to cross the plate is \[ T_{\text{plate}}=\frac{t}{v} \] Substituting \(v=c/n\), \[ T_{\text{plate}}=\frac{nt}{c} \] The time required to cross the same thickness of air is \[ T_{\text{air}}=\frac{t}{c} \] Therefore, the additional delay is \[ \Delta T=T_{\text{plate}}-T_{\text{air}} \] \[ \Delta T=\frac{(n-1)t}{c} \] \(\textbf{Delay produced by plate P:}\) For plate P, \[ t_P=6\,\text{cm}=0.06\,\text{m} \] \[ n_P=1.5 \] Substituting, \[ \Delta T_P=\frac{(1.5-1)(0.06)}{3\times10^8} \] \[ \Delta T_P=\frac{0.5(0.06)}{3\times10^8} \] \[ \Delta T_P=\frac{0.03}{3\times10^8} \] \[ \Delta T_P=1.0\times10^{-10}\,\text{s} \] \(\textbf{Delay produced by plate Q:}\) For plate Q, \[ t_Q=9\,\text{cm}=0.09\,\text{m} \] \[ n_Q=\frac43 \] Therefore, \[ \Delta T_Q=\frac{\left(\frac43-1\right)(0.09)}{3\times10^8} \] \[ \Delta T_Q=\frac{\frac13(0.09)}{3\times10^8} \] \[ \Delta T_Q=\frac{0.03}{3\times10^8} \] \[ \Delta T_Q=1.0\times10^{-10}\,\text{s} \] \(\textbf{Compare the two plates:}\) The plates have different refractive indices and different thicknesses. However, their products \((n-1)t\) are equal. Therefore, both produce the same additional travel time relative to the corresponding air path. \(\textbf{Physical interpretation:}\) Light travels more slowly in plate P because its refractive index is greater. Nevertheless, plate P is physically thinner than plate Q. The greater refractive-index contrast of P is exactly compensated by the smaller thickness. Normal incidence ensures that changes in oblique geometrical path length do not enter this comparison. The result concerns additional travel time, not equality of the total times taken through the two plates. The plates replace different lengths of air, so their total plate traversal times are not equal. The additional delays are equal because the excess optical path lengths are equal. \(\textbf{Final answer:}\) \[ \boxed{\Delta T_P=\Delta T_Q=1.0\times10^{-10}\,\text{s}} \]

Explanation: \(\textbf{Identify the media and curvature sign:}\) The ray travels from air into glass. Therefore, \[ n_1=1,\qquad n_2=1.5 \] The pole is at the origin, and the centre of curvature lies to its right. With the positive direction chosen along incident propagation, \[ R=+20\,\text{cm} \] The real object lies to the left of the surface. Hence, \[ u=-30\,\text{cm} \] \(\textbf{Apply the spherical-refraction equation:}\) For a single spherical refracting surface, \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Substituting, \[ \frac{1.5}{v}-\frac1{-30}=\frac{1.5-1}{20} \] \[ \frac{1.5}{v}+\frac1{30}=\frac{0.5}{20} \] \[ \frac{1.5}{v}=\frac1{40}-\frac1{30} \] \[ \frac{1.5}{v}=\frac{3-4}{120} \] \[ \frac{1.5}{v}=-\frac1{120} \] Therefore, \[ v=-180\,\text{cm} \] \(\textbf{Interpret the negative image distance:}\) The negative sign places the image to the left of the surface, in the incident medium. The transmitted rays are divergent, and their backward extensions meet at this image position. The image is therefore virtual. \(\textbf{Determine transverse magnification:}\) For refraction at a spherical surface, \[ m=\frac{n_1v}{n_2u} \] Substituting, \[ m=\frac{1(-180)}{1.5(-30)} \] \[ m=\frac{-180}{-45} \] \[ m=+4 \] \(\textbf{Physical check:}\) The surface bends paraxial rays towards the normal because light enters a higher-index medium. However, the object lies inside the first focal distance of the refracting surface, so the transmitted rays remain divergent. A converging refracting surface need not form a real image of every real object. The positive magnification indicates that the virtual image is erect. Its magnitude of \(4\) shows that the image is enlarged. The radius sign follows from the position of the centre of curvature, not merely from describing the surface as convex. \(\textbf{Final answer:}\) \[ \boxed{v=-180\,\text{cm},\quad m=+4} \]

Explanation: \(\textbf{Common object and medium data:}\) The incident medium is air. \[ n_1=1 \] The transmitted medium is glass. \[ n_2=1.5 \] The real object is on the incident side. \[ u=-90\,\text{cm} \] \(\textbf{Assign curvature signs:}\) The positive direction is from left to right. For interface P, the centre of curvature lies to the right. \[ R_P=+30\,\text{cm} \] For interface Q, the centre lies to the left. \[ R_Q=-30\,\text{cm} \] \(\textbf{Image formed by interface P:}\) The spherical-refraction formula is \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Substitute the data for P. \[ \frac{1.5}{v_P}-\frac1{-90}=\frac{0.5}{30} \] \[ \frac{1.5}{v_P}+\frac1{90}=\frac1{60} \] \[ \frac{1.5}{v_P}=\frac1{60}-\frac1{90} \] \[ \frac{1.5}{v_P}=\frac1{180} \] Hence, \[ v_P=+270\,\text{cm} \] The positive value means that transmitted rays actually converge in the glass. The image is real. \(\textbf{Image formed by interface Q:}\) For Q, the curvature radius is negative. \[ \frac{1.5}{v_Q}-\frac1{-90}=\frac{0.5}{-30} \] \[ \frac{1.5}{v_Q}+\frac1{90}=-\frac1{60} \] \[ \frac{1.5}{v_Q}=-\frac1{60}-\frac1{90} \] \[ \frac{1.5}{v_Q}=-\frac{3+2}{180} \] \[ \frac{1.5}{v_Q}=-\frac1{36} \] Therefore, \[ v_Q=-54\,\text{cm} \] The negative image distance indicates a virtual image to the left of the interface. \(\textbf{Compare image formation:}\) The two surfaces have equal curvature-radius magnitudes and identical refractive-index contrasts. However, their curvature signs are opposite. For P, the positive surface power bends the transmitted rays towards convergence. For Q, the negative surface power increases their divergence. The change in curvature orientation therefore changes the nature of image formation. \(\textbf{Final answer:}\) Interface P forms a real image \(270\,\text{cm}\) to the right, while Q forms a virtual image \(54\,\text{cm}\) to the left.

Explanation: \(\textbf{Image of parallel rays at a spherical interface:}\) For paraxial refraction at a spherical surface, \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] An incident parallel beam corresponds to an object at infinity. Therefore, \[ \frac{n_1}{u}\rightarrow0 \] The image distance then satisfies \[ \frac{n_2}{v}=\frac{n_2-n_1}{R} \] Hence, \[ v=\frac{n_2R}{n_2-n_1} \] \(\textbf{Experiment P — air to glass:}\) The incident direction is from left to right. The centre of curvature lies to the right of the pole. Therefore, \[ R_P=+30\,\text{cm} \] The medium indices are \[ n_{1P}=1,\qquad n_{2P}=1.5 \] Substituting, \[ v_P=\frac{1.5(30)}{1.5-1} \] \[ v_P=\frac{45}{0.5} \] \[ v_P=+90\,\text{cm} \] The positive image distance indicates convergence in the glass to the right of the interface. \(\textbf{Experiment Q — glass to air:}\) The incident direction is now from right to left. The positive Cartesian direction is redefined along this reversed propagation direction. The physical centre of curvature remains to the right of the pole. It is now opposite to the positive direction. Therefore, \[ R_Q=-30\,\text{cm} \] The incident and transmitted media are also interchanged. \[ n_{1Q}=1.5,\qquad n_{2Q}=1 \] \(\textbf{Calculate the second image distance:}\) \[ v_Q=\frac{n_{2Q}R_Q}{n_{2Q}-n_{1Q}} \] \[ v_Q=\frac{1(-30)}{1-1.5} \] \[ v_Q=\frac{-30}{-0.5} \] \[ v_Q=+60\,\text{cm} \] The positive value indicates a real image in air, to the left of the interface. \(\textbf{Physical interpretation:}\) Reversing the direction of illumination changes both the medium assignments and the signed curvature radius. It does not change the physical shape of the spherical surface. The two parallel beams converge at different distances because the refractive-index coefficient of the transmitted medium differs between the experiments. The resulting image distances are not required to be equal. Both images are real because the corresponding signed image distances are positive in their separately defined incident-direction coordinate systems. The distance ratio is \[ v_P:v_Q=90:60 \] \[ v_P:v_Q=3:2 \] \(\textbf{Final answer:}\) \[ \boxed{v_P=+90\,\text{cm},\quad v_Q=+60\,\text{cm}} \]

Explanation: \(\textbf{Initial interface parameters:}\) The incident light travels from glass into air. \[ n_1=1.5,\qquad n_2=1 \] The object is real and lies to the left of the interface. \[ u=-30\,\text{cm} \] The centre of curvature is initially to the right. \[ R=+30\,\text{cm} \] \(\textbf{Initial spherical-surface image:}\) Use \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Substituting, \[ \frac1v-\frac{1.5}{-30}=\frac{1-1.5}{30} \] \[ \frac1v+\frac1{20}=-\frac1{60} \] \[ \frac1v=-\frac1{60}-\frac1{20} \] \[ \frac1v=-\frac{1+3}{60} \] \[ \frac1v=-\frac1{15} \] Therefore, \[ v=-15\,\text{cm} \] The image is virtual and lies to the left of the pole. \(\textbf{Change the surface curvature:}\) The magnitude of the positive radius of curvature is gradually increased. The refractive power of the surface is proportional to \[ \frac{n_2-n_1}{R} \] As \(R\) becomes extremely large, this quantity approaches zero. The surface approaches a plane interface. \(\textbf{Plane-interface limiting equation:}\) In the limit, \[ R\rightarrow\infty \] and \[ \frac{n_2-n_1}{R}\rightarrow0 \] Therefore, \[ \frac{n_2}{v}-\frac{n_1}{u}=0 \] Rearranging, \[ v=\frac{n_2}{n_1}u \] Substituting, \[ v=\frac1{1.5}(-30) \] \[ v=-20\,\text{cm} \] \(\textbf{Compare the positions:}\) The initial virtual image is \(15\,\text{cm}\) to the left of the pole. In the plane-surface limit, the virtual image is \(20\,\text{cm}\) to the left. Therefore, the apparent image moves farther to the left as the curvature decreases. Its distance from the surface increases by \(5\,\text{cm}\). \(\textbf{Physical consistency:}\) The limiting position agrees with the ordinary apparent-depth relation for a real object in glass viewed from air. A plane interface makes the object appear at a depth reduced by the factor \(n_2/n_1\). The resulting apparent distance is \(20\,\text{cm}\), smaller than the actual distance of \(30\,\text{cm}\). The finite positive curvature produces an additional refractive effect that shifts the image closer to the pole. Flattening the surface removes that curvature-dependent contribution. The radius-of-curvature limit therefore connects spherical-surface refraction continuously with plane-surface refraction. \(\textbf{Final answer:}\) \[ \boxed{v_{\text{initial}}=-15\,\text{cm},\quad v_{\text{limit}}=-20\,\text{cm}} \]