Optics MCQs with Answers – Part 4 (NEET Physics)

Explanation: \(\textbf{Use the exact minimum-deviation relation:}\) For a prism in air, \[ n=\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin(A/2)} \] Both prisms have \[ \delta_m=30^\circ \] However, their prism angles differ. Therefore, equal minimum deviations do not necessarily imply equal refractive indices. \(\textbf{Determine the index of prism P:}\) For P, \[ A_P=60^\circ \] Substituting, \[ n_P=\frac{\sin45^\circ}{\sin30^\circ} \] \[ n_P=\frac{1/\sqrt2}{1/2} \] \[ n_P=\sqrt2 \] \(\textbf{Determine the index of prism Q:}\) For Q, \[ A_Q=30^\circ \] Therefore, \[ n_Q=\frac{\sin30^\circ}{\sin15^\circ} \] \[ n_Q=\frac{1/2}{(\sqrt6-\sqrt2)/4} \] \[ n_Q=\frac2{\sqrt6-\sqrt2} \] Rationalising, \[ n_Q=\frac{\sqrt6+\sqrt2}{2} \] \(\textbf{Compare the material indices:}\) \[ n_P\approx1.414 \] \[ n_Q\approx1.932 \] Thus, \[ n_Q\gt n_P \] Prism Q has the greater absolute refractive index, despite having the smaller refracting angle. \(\textbf{Immerse both prisms in the specified liquid:}\) The surrounding-liquid index is \[ n_l=n_P=\sqrt2 \] For prism P, the relative refractive index becomes \[ \mu_P=\frac{n_P}{n_l} \] \[ \mu_P=1 \] There is no refractive-index contrast at either face. Consequently, P produces no angular deviation under ideal geometrical optics. \(\textbf{Check prism Q in the same liquid:}\) Its relative index is \[ \mu_Q=\frac{n_Q}{n_l} \] \[ \mu_Q=\frac{\sqrt6+\sqrt2}{2\sqrt2} \] \[ \mu_Q=\frac{\sqrt3+1}{2} \] Since \[ \mu_Q\gt 1 \] Q remains optically denser than the liquid. It can still deviate transmitted rays towards its base. \(\textbf{Physical interpretation:}\) Minimum deviation depends jointly on prism angle and relative refractive index. A prism with a smaller refracting angle can produce the same minimum deviation as another prism by having a greater refractive index. Immersion distinguishes the materials because only the prism whose index matches the liquid becomes optically ineffective. Equal measured minimum deviations in air therefore do not establish equivalence under a changed surrounding medium. The inference requires both the exact prism formula and the refractive-index matching principle. \(\textbf{Final answer:}\) The index of Q is \(\frac{\sqrt6+\sqrt2}{2}\), greater than \(n_P=\sqrt2\); P becomes index-matched, while Q remains deviating towards its base.

Explanation: \(\textbf{Infer the common glass index:}\) For the original prism, \[ A_1=60^\circ \] \[ \delta_{m1}=30^\circ \] The exact minimum-deviation relation is \[ n=\frac{\sin[(A+\delta_m)/2]}{\sin(A/2)} \] Therefore, \[ n=\frac{\sin45^\circ}{\sin30^\circ} \] \[ n=\sqrt2 \] Both prisms have this refractive index because they are made from the same glass and tested at the same wavelength. \(\textbf{Identify the first minimum-deviation incidence:}\) At minimum deviation, \[ i_m=e_m \] The incidence angle is \[ i_{m1}=\frac{A_1+\delta_{m1}}2 \] \[ i_{m1}=\frac{60^\circ+30^\circ}{2} \] \[ i_{m1}=45^\circ \] \(\textbf{Apply symmetry to the second prism:}\) The second prism has \[ A_2=30^\circ \] At minimum deviation, \[ r_1=r_2=\frac{A_2}{2} \] \[ r_1=r_2=15^\circ \] \(\textbf{Find its required external incidence:}\) Snell's law at the first air-glass face gives \[ \sin i_{m2}=n\sin15^\circ \] Substituting, \[ \sin i_{m2}=\sqrt2\sin15^\circ \] Therefore, \[ i_{m2}=\sin^{-1}(\sqrt2\sin15^\circ) \] Since \[ \sqrt2\sin15^\circ\lt \sin45^\circ \] we obtain \[ i_{m2}\lt 45^\circ \] \(\textbf{Determine the second minimum deviation:}\) Using \[ \delta_{m2}=2i_{m2}-A_2 \] gives \[ \delta_{m2}=2\sin^{-1}(\sqrt2\sin15^\circ)-30^\circ \] Numerically, \[ \sin15^\circ\approx0.2588 \] \[ i_{m2}\approx21.47^\circ \] Therefore, \[ \delta_{m2}\approx12.94^\circ \] \(\textbf{Physical interpretation:}\) Both prisms possess the same material refractive index. Nevertheless, changing the prism angle changes the internal refraction angles of the symmetric minimum-deviation path. The external incidence angles must consequently change according to Snell's law. The minimum deviation is not proportional to prism angle according to an exact linear law for arbitrary prism angles. Here the smaller prism angle gives a smaller minimum-deviation incidence and a smaller minimum deviation. The calculation must use the actual symmetric geometry of each prism rather than transferring the first prism's minimum-deviation angle unchanged. \(\textbf{Final answer:}\) \[ \boxed{\delta_{m2}=2\sin^{-1}(\sqrt2\sin15^\circ)-30^\circ\approx12.94^\circ,\quad i_{m2}\approx21.47^\circ} \]

Explanation: The Assertion is true because the original path satisfies the symmetry condition for minimum deviation. The prism has fixed refracting angle \(60^\circ\), and the wavelength and surrounding medium remain unchanged. At minimum deviation, incidence and emergence are equal. Their common value follows from \[ \delta_m=i_m+e_m-A \] and \[ i_m=e_m \] Therefore, \[ i_m=\frac{A+\delta_m}{2} \] Using the measured deviation, \[ i_m=\frac{60^\circ+40^\circ}{2} \] \[ i_m=50^\circ \] The two internal refraction angles are equal to \(30^\circ\). This establishes the original ray as the symmetric minimum-deviation ray for the specified prism. When the incidence angle is decreased slightly, the first internal refraction angle also decreases. The sum of the two internal angles must remain equal to the prism angle. Consequently, the second internal incidence angle increases. The resulting path is no longer symmetric. For a small change that still permits ordinary transmission, its deviation is greater than the minimum value. The Reason is therefore true and correctly explains the Assertion. The conclusion is not merely that incidence and emergence cease to be equal. It is that their equality identifies a minimum of the deviation-versus-incidence relation for this fixed prism and wavelength. A sufficiently small change to either side of that minimum increases deviation. The restriction to ordinary transmission is important because total internal reflection would remove the usual emergent-ray deviation from consideration. Both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Interpret the measured graph slopes:}\) For a thin prism in air, \[ \delta\approx(n_g-1)A \] Therefore, the slope of a deviation-versus-prism-angle graph is \[ S=n_g-1 \] \(\textbf{Recover the two glass refractive indices:}\) The red slope is \[ S_R=0.50 \] Hence, \[ n_R=1+0.50 \] \[ n_R=1.50 \] For violet, \[ S_V=0.65 \] Therefore, \[ n_V=1+0.65 \] \[ n_V=1.65 \] \(\textbf{Determine the relative indices in the liquid:}\) The surrounding liquid index is \[ n_l=1.55 \] For red light, \[ \mu_R=\frac{1.50}{1.55} \] \[ \mu_R=\frac{30}{31} \] For violet, \[ \mu_V=\frac{1.65}{1.55} \] \[ \mu_V=\frac{33}{31} \] \(\textbf{Calculate the red deviation:}\) The new prism angle is \[ A=0.031\,\text{rad} \] Using the thin-prism relation, \[ \delta_R=(\mu_R-1)A \] \[ \delta_R=\left(\frac{30}{31}-1\right)(0.031) \] \[ \delta_R=-0.001\,\text{rad} \] Red light deviates towards the prism apex. \(\textbf{Calculate the violet deviation:}\) \[ \delta_V=\left(\frac{33}{31}-1\right)(0.031) \] \[ \delta_V=+0.002\,\text{rad} \] Violet light deviates towards the prism base. \(\textbf{Convert angular deviations into screen positions:}\) The screen distance is \[ L=2.0\,\text{m} \] For small angles, \[ y\approx L\delta \] Therefore, \[ y_R=2(-0.001) \] \[ y_R=-0.002\,\text{m}=-2\,\text{mm} \] Similarly, \[ y_V=2(+0.002) \] \[ y_V=+0.004\,\text{m}=+4\,\text{mm} \] \(\textbf{Determine the spot separation:}\) \[ s=|y_V-y_R| \] \[ s=|4-(-2)| \] \[ s=6\,\text{mm} \] \(\textbf{Physical interpretation:}\) The graph slopes determine the absolute red and violet indices of the prism glass. After immersion, the liquid index lies between those values. Red light consequently sees a relative refractive index below unity, while violet sees one above unity. Their deviations are therefore towards opposite sides of the undeviated direction. The angular and screen separations must include both signed deviations. It would be incorrect to subtract their positive magnitudes as though both rays bent towards the same side. \(\textbf{Final answer:}\) \[ \boxed{y_R=-2\,\text{mm},\quad y_V=+4\,\text{mm},\quad s=6\,\text{mm}} \]

Explanation: \(\textbf{Internal ray geometry at normal entrance:}\) The entrance incidence angle is \[ i=0^\circ \] Therefore, the internal refraction angle at the first face is \[ r_1=0^\circ \] The prism relation gives \[ r_1+r_2=A \] With \[ A=30^\circ \] we obtain \[ r_2=30^\circ \] This internal second-face incidence remains unchanged between the two arrangements. \(\textbf{Original emergence into air:}\) The glass index is \[ n_g=\sqrt2 \] The air index is unity. At the second face, Snell's law gives \[ n_g\sin r_2=\sin e_1 \] Substituting, \[ \sin e_1=\sqrt2\sin30^\circ \] \[ \sin e_1=\frac1{\sqrt2} \] Therefore, \[ e_1=45^\circ \] \(\textbf{Determine the original deviation:}\) The prism deviation relation is \[ \delta=i+e-A \] Therefore, \[ \delta_1=0^\circ+45^\circ-30^\circ \] \[ \delta_1=+15^\circ \] The positive sign corresponds to deviation towards the base. \(\textbf{Immerse the prism in the higher-index liquid:}\) The liquid refractive index is \[ n_l=2 \] The ray still enters normally, so the internal second-face incidence remains \(30^\circ\). Snell's law at the glass-liquid exit gives \[ n_g\sin30^\circ=n_l\sin e_2 \] Substituting, \[ \sqrt2\left(\frac12\right)=2\sin e_2 \] \[ \sin e_2=\frac1{2\sqrt2} \] Therefore, \[ e_2=\sin^{-1}\!\left(\frac1{2\sqrt2}\right) \] \(\textbf{Determine the immersed-prism deviation:}\) \[ \delta_2=i+e_2-A \] \[ \delta_2=\sin^{-1}\!\left(\frac1{2\sqrt2}\right)-30^\circ \] Since \[ \sin^{-1}\!\left(\frac1{2\sqrt2}\right)\lt 30^\circ \] the signed deviation is negative. Its magnitude is \[ |\delta_2|=30^\circ-\sin^{-1}\!\left(\frac1{2\sqrt2}\right) \] \[ |\delta_2|\approx9.30^\circ \] \(\textbf{Physical interpretation:}\) In air, the glass is optically denser than its surroundings, so the emerging ray bends away from the exit normal and the prism deviation is towards its base. In the liquid, the glass is optically rarer than its surroundings. The emerging ray bends towards the exit normal. Its emergence angle becomes smaller than the prism angle, reversing the signed deviation. The direction reversal follows from exact Snell-law geometry rather than from a thin-prism approximation. The physical prism shape and internal incidence angle remain unchanged. \(\textbf{Final answer:}\) \[ \boxed{\delta_{\mathrm{air}}=+15^\circ,\quad\delta_{\mathrm{liquid}}=\sin^{-1}\!\left(\frac1{2\sqrt2}\right)-30^\circ\approx-9.30^\circ} \]

Explanation: \(\textbf{Use the relative-index form of prism deviation:}\) For a thin prism immersed in a medium, \[ \delta\approx\left(\frac{n_g}{n_l}-1\right)A \] The sign depends on the refractive-index contrast at the particular wavelength. A positive sign corresponds to deviation towards the base. A negative sign corresponds to deviation towards the apex. \(\textbf{Red light in liquid P:}\) The glass red index is \[ n_{gR}=1.50 \] The liquid P red index is \[ n_{PR}=1.40 \] Therefore, \[ \mu_{PR}=\frac{1.50}{1.40} \] \[ \mu_{PR}=\frac{15}{14}\gt 1 \] The red deviation is positive. \[ \delta_{PR}=\left(\frac{15}{14}-1\right)(0.050) \] \[ \delta_{PR}=+\frac{0.050}{14} \] Thus, red light deviates towards the base. \(\textbf{Violet light in liquid P:}\) For violet light, \[ n_{gV}=1.60 \] \[ n_{PV}=1.60 \] The two refractive indices match. \[ \mu_{PV}=1 \] Therefore, \[ \delta_{PV}=0 \] Violet light experiences no prism angular deviation in the ideal model. \(\textbf{Red light in liquid Q:}\) The Q liquid red index is \[ n_{QR}=1.55 \] Therefore, \[ \mu_{QR}=\frac{1.50}{1.55} \] \[ \mu_{QR}=\frac{30}{31}\lt 1 \] Hence, \[ \delta_{QR}=\left(\frac{30}{31}-1\right)(0.050)\lt 0 \] Red light deviates towards the apex. \(\textbf{Violet light in liquid Q:}\) The Q liquid violet index is \[ n_{QV}=1.65 \] Therefore, \[ \mu_{QV}=\frac{1.60}{1.65} \] \[ \mu_{QV}=\frac{32}{33}\lt 1 \] Thus, \[ \delta_{QV}=\left(\frac{32}{33}-1\right)(0.050)\lt 0 \] Violet light also deviates towards the apex. \(\textbf{Physical interpretation:}\) A surrounding liquid may have its own wavelength-dependent refractive index. The prism's behaviour must therefore be determined from the relative index at each colour, rather than from the glass indices alone. Liquid P matches the glass for violet but not for red, making violet undeviated while red bends towards the base. Liquid Q has a greater index than the glass at both wavelengths. Both colours then bend towards the apex. The wavelength ordering of absolute refractive indices is not by itself sufficient to determine the direction of a colour's prism deviation. The appropriate comparison is always between the prism material and the surrounding medium at that same wavelength. \(\textbf{Final answer:}\) In liquid P, red deviates towards the base and violet is undeviated; in liquid Q, both colours deviate towards the apex.

Explanation: \(\textbf{Determine the prism's minimum deviation:}\) The refracting angle is \[ A=60^\circ \] The refractive index is \[ n=\sqrt2 \] At minimum deviation, the internal ray angles are \[ r_1=r_2=\frac A2 \] \[ r_1=r_2=30^\circ \] \(\textbf{Find the external incidence angle:}\) Snell's law gives \[ \sin i_m=n\sin30^\circ \] Substituting, \[ \sin i_m=\frac1{\sqrt2} \] Therefore, \[ i_m=45^\circ \] \(\textbf{Calculate minimum deviation:}\) At minimum deviation, \[ i_m=e_m \] Therefore, \[ \delta_m=2i_m-A \] \[ \delta_m=2(45^\circ)-60^\circ \] \[ \delta_m=30^\circ \] \(\textbf{Interpret the two prism orientations:}\) In experiment P, the prism deviates the ray towards the upper base. Its final direction makes \[ +30^\circ \] with the original reference direction. In experiment Q, reversing the base orientation reverses the direction of deviation. The final direction makes \[ -30^\circ \] with the reference direction. \(\textbf{Convert direction into screen position:}\) The screen is perpendicular to the original direction, not to either deviated ray. Therefore, the transverse coordinate is obtained from the tangent of the deviation angle. If the screen distance is \(L\), \[ y=L\tan\delta \] The screen distance is \[ L=2.0\,\text{m} \] For P, \[ y_P=2\tan30^\circ \] \[ y_P=\frac2{\sqrt3}\,\text{m} \] For Q, \[ y_Q=-2\tan30^\circ \] \[ y_Q=-\frac2{\sqrt3}\,\text{m} \] \(\textbf{Calculate the physical separation:}\) \[ s=|y_P-y_Q| \] Substituting, \[ s=\left|\frac2{\sqrt3}-\left(-\frac2{\sqrt3}\right)\right| \] \[ s=\frac4{\sqrt3}\,\text{m} \] \(\textbf{Physical interpretation:}\) Reversing a prism's base direction reverses the transverse direction of its deviation for an otherwise equivalent ray path. The minimum deviation magnitude remains unchanged because the prism's angle, material and wavelength remain unchanged. The two emergent rays lie on opposite sides of the original direction. Since the deviation angle is \(30^\circ\), it is not appropriate to replace \(\tan30^\circ\) by the numerical angle in radians without assessing the approximation. The exact straight-line screen geometry gives the separation through the tangent relation. \(\textbf{Final answer:}\) \[ \boxed{s=\frac4{\sqrt3}\,\text{m}} \]

Explanation: The Assertion is true within the stated thin-prism approximation. For a prism in a surrounding medium of refractive index \(n_m\), the small-angle deviation is \[ \delta=\left(\frac{n_g}{n_m}-1\right)A \] Because the bases of P and Q are opposed, their signed deviations subtract. Therefore, \[ \delta_{\mathrm{net}}=\left(\frac{n_P}{n_m}-1\right)A_P-\left(\frac{n_Q}{n_m}-1\right)A_Q \] Rearranging, \[ \delta_{\mathrm{net}}=\frac{n_PA_P-n_QA_Q}{n_m}-(A_P-A_Q) \] Cancellation in air requires \[ n_PA_P-n_QA_Q=A_P-A_Q \] Cancellation in the liquid requires \[ n_PA_P-n_QA_Q=n_l(A_P-A_Q) \] Both equations must hold for the unchanged prism pair. Subtracting them gives \[ (n_l-1)(A_P-A_Q)=0 \] The liquid index differs from unity. Therefore, \[ A_P=A_Q \] Substituting this result into either cancellation condition gives \[ n_PA_P=n_QA_Q \] Both prism angles are stipulated to be non-zero and are now equal. Consequently, \[ n_P=n_Q \] The Reason is true because the two surrounding-media conditions impose independent restrictions that must be satisfied simultaneously. The first is equality of material-index–angle products, and the second is equality of angles. Together they establish the equality of the material indices. This result is more restrictive than cancellation in one surrounding medium, where different prism angles and different material indices may still compensate. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Recover the original internal refraction angle:}\) At minimum deviation, the incident and emergent angles are equal. The two internal refraction angles are also equal: \[ r_1=r_2=\frac A2 \] The glass refractive index is \[ n=1.5 \] The minimum-deviation incidence angle is \[ i_m=55^\circ \] Snell's law gives \[ \sin55^\circ=1.5\sin r_1 \] Therefore, \[ \sin r_1=\frac{\sin55^\circ}{1.5} \] \(\textbf{Compare with the critical angle:}\) For glass into air, \[ \sin i_c=\frac1{1.5} \] Since \[ \sin55^\circ\lt 1 \] it follows that \[ \sin r_1\lt \sin i_c \] Therefore, \[ r_1\lt i_c \] \(\textbf{Recover the original prism angle:}\) At minimum deviation, \[ A=2r_1 \] Therefore, \[ A\lt 2i_c \] This confirms that the original symmetric transmitted path is physically possible. \(\textbf{Double the prism angle:}\) The reshaped prism has \[ A'=2A \] Therefore, \[ A'=4r_1 \] To decide whether any ordinarily transmitted path is possible, compare this value with \(2i_c\). Using the given numerical data, \[ r_1=\sin^{-1}\left(\frac{\sin55^\circ}{1.5}\right) \] \[ r_1\approx33.09^\circ \] Hence, \[ A\approx66.18^\circ \] and \[ A'\approx132.36^\circ \] \(\textbf{Determine the maximum permissible prism angle:}\) The glass-air critical angle is \[ i_c=\sin^{-1}\left(\frac23\right) \] \[ i_c\approx41.81^\circ \] For ordinary entry from air and ordinary emergence into air, the internal angles must satisfy \[ r_1\leq i_c \] \[ r_2\leq i_c \] Since \[ r_1+r_2=A' \] a transmitted path requires \[ A'\leq2i_c \] But \[ 2i_c\approx83.62^\circ \] whereas \[ A'\approx132.36^\circ \] Therefore, ordinary transmission through both faces is impossible in the reshaped prism. \(\textbf{Physical interpretation:}\) The original minimum-deviation ray establishes the refracting angle of the first prism. Doubling that angle makes the required sum of the two internal refraction angles too large. No ordinary entrance ray from air can produce an internal angle greater than the glass-air critical angle. The second face would consequently be supercritical for any ray entering through the first face. The outcome follows from combining an experimental minimum-deviation measurement, Snell's law and the limiting transmission geometry. \(\textbf{Final answer:}\) No ordinary two-face transmitted path exists for the reshaped prism.

Explanation: \(\textbf{Determine the unmodified lens image plane:}\) The focal length is \[ f=+20\,\text{cm} \] The real object is at \[ u=-60\,\text{cm} \] Using the thin-lens equation, \[ \frac1v=\frac1f+\frac1u \] \[ \frac1v=\frac1{20}-\frac1{60} \] \[ \frac1v=\frac1{30} \] Therefore, \[ v=+30\,\text{cm} \] The lens would form its real image \(30\,\text{cm}\) to its right without the prism. \(\textbf{Use angular dispersion as the source of transverse separation:}\) The prism introduces a difference between the red and violet ray directions: \[ \Delta\delta=0.006\,\text{rad} \] For a prism placed within a converging paraxial bundle, this angular difference produces transverse separation after the rays travel from the prism to the image plane. For a small angle, \[ s\approx L\Delta\delta \] where \(L\) is the axial distance from the effective prism plane to the image plane. \(\textbf{Arrangement P — prism immediately before the lens:}\) The prism is effectively in the lens plane. Therefore, \[ L_P=v \] \[ L_P=30\,\text{cm} \] The colour-image separation is \[ s_P=L_P\Delta\delta \] \[ s_P=30(0.006) \] \[ s_P=0.18\,\text{cm} \] \[ s_P=1.8\,\text{mm} \] \(\textbf{Arrangement Q — prism after the lens:}\) The prism is moved to \[ x_Q=+10\,\text{cm} \] relative to the lens. The real image plane remains at \[ x_i=+30\,\text{cm} \] The remaining propagation distance is \[ L_Q=x_i-x_Q \] \[ L_Q=30-10 \] \[ L_Q=20\,\text{cm} \] \(\textbf{Calculate the new colour-image separation:}\) \[ s_Q=L_Q\Delta\delta \] \[ s_Q=20(0.006) \] \[ s_Q=0.12\,\text{cm} \] \[ s_Q=1.2\,\text{mm} \] \(\textbf{Compare the two placements:}\) \[ \frac{s_Q}{s_P}=\frac{1.2}{1.8} \] \[ \frac{s_Q}{s_P}=\frac23 \] \(\textbf{Physical interpretation:}\) The prism's colour-dependent angular deviation is unchanged between the arrangements. However, the distance over which that angular difference develops into a transverse separation changes. The prism placed nearer the real image plane gives the rays less distance to separate. Thus, equal angular dispersion does not imply equal separation of colour images. The final observed effect depends on the prism's axial location relative to the imaging lens and its conjugate image plane. \(\textbf{Final answer:}\) \[ \boxed{s_P=1.8\,\text{mm},\quad s_Q=1.2\,\text{mm}} \]

Explanation: Modification P reverses the orientation of the prism base relative to the reference incident direction. A prism with the same angle and refractive index retains the same magnitude of minimum deviation when its orientation is reversed. However, the transverse direction of deviation changes. Therefore, P matches (3). Modification Q makes the refractive index of the glass equal to that of the surrounding liquid. Snell's law then gives no refractive bending at either face. The angular deviation becomes zero in the ideal geometrical-optics treatment. Therefore, Q matches (1). Modification R reduces the material refractive index while keeping it greater than the surrounding air index. The minimum-deviation relation is \[ n=\frac{\sin[(A+\delta_m)/2]}{\sin(A/2)} \] For the same prism angle, reducing \(n\) reduces the permitted minimum-deviation angle. The prism continues to deviate light towards its base, but by a smaller angle. Therefore, R matches (4). Modification S uses a shorter wavelength for which the refractive index is greater. The stronger refraction at the prism faces produces a larger minimum deviation at that wavelength. Therefore, S matches (2). The four modifications affect different physical features of prism behaviour. Reversing the base changes the sign of deviation without changing its magnitude. Index matching eliminates refraction. Reducing refractive-index contrast weakens deviation. Increasing the material refractive index at a new wavelength strengthens minimum deviation. The correct matching is P–(3), Q–(1), R–(4), S–(2).

Explanation: For a fixed prism and wavelength, Snell's law at the first face relates the external incidence angle \(i\) to the first internal refraction angle \(r_1\). As \(i\) increases within the ordinary-transmission range, \(r_1\) also increases. The internal prism-angle relation is \[ r_1+r_2=A \] The prism angle \(A\) remains constant. Therefore, increasing \(r_1\) necessarily decreases the internal incidence angle \(r_2\) at the second face. Snell's law at that face relates \(r_2\) to the emergence angle \(e\). As \(r_2\) decreases, the emergence angle also decreases within the permitted transmitted-ray range. Thus, the emergence-angle graph decreases continuously as the entrance incidence increases. The total prism deviation is \[ \delta=i+e-A \] Two competing changes now occur: \(i\) increases while \(e\) decreases. The total deviation therefore need not vary monotonically. Initially, as the entrance incidence approaches its minimum-deviation value from below, the decrease in emergence angle dominates and the deviation falls. At the symmetric minimum-deviation path, \[ i=e \] and \[ r_1=r_2=\frac A2 \] The deviation reaches its minimum. Beyond this condition, the increase in entrance incidence dominates and the deviation rises again. Consequently, the two graphs have different qualitative shapes. The emergence angle decreases throughout the transmitted range, whereas the deviation has a minimum. Optical reversibility explains why two distinct entrance angles on opposite sides of that minimum can produce the same deviation, even though their emergence angles differ. The correct interpretation is a monotonically decreasing emergence angle and a deviation that first decreases and then increases.

Explanation: \(\textbf{Determine the internal ray geometry:}\) The entrance incidence is normal. Therefore, \[ i=0^\circ \] The first internal refraction angle is \[ r_1=0^\circ \] For a prism, \[ r_1+r_2=A \] The prism angle is \[ A=45^\circ \] Thus, \[ r_2=45^\circ \] \(\textbf{Check the initial glass-air condition:}\) The glass index is \[ n_g=\sqrt2 \] The air index is unity. The critical-angle relation gives \[ \sin i_c=\frac1{\sqrt2} \] \[ i_c=45^\circ \] The second-face incidence is therefore exactly critical. \(\textbf{Determine original deviation:}\) At critical incidence, the limiting refracted direction lies along the exit face. Therefore, \[ e_1=90^\circ \] The deviation formula gives \[ \delta_1=i+e_1-A \] \[ \delta_1=0^\circ+90^\circ-45^\circ \] \[ \delta_1=45^\circ \] \(\textbf{Immerse the prism:}\) The surrounding-liquid index becomes \[ n_l=\sqrt{\frac32} \] The ray still enters normally. Hence, its internal second-face incidence remains \(45^\circ\). \(\textbf{Calculate the immersed exit angle:}\) Snell's law at the glass-liquid boundary gives \[ n_g\sin45^\circ=n_l\sin e_2 \] Substituting, \[ \sqrt2\left(\frac1{\sqrt2}\right)=\sqrt{\frac32}\sin e_2 \] \[ 1=\sqrt{\frac32}\sin e_2 \] Therefore, \[ \sin e_2=\sqrt{\frac23} \] \[ e_2=\sin^{-1}\!\sqrt{\frac23} \] This is smaller than \(90^\circ\), confirming ordinary refraction after immersion. \(\textbf{Determine the new deviation:}\) \[ \delta_2=i+e_2-A \] \[ \delta_2=\sin^{-1}\!\sqrt{\frac23}-45^\circ \] Numerically, \[ e_2\approx54.74^\circ \] Therefore, \[ \delta_2\approx9.74^\circ \] The deviation remains towards the prism base. \(\textbf{Physical interpretation:}\) The initial ray is critically incident at the second face rather than supercritical. Replacing air with a higher-index liquid raises the critical angle and makes the same internal ray subcritical. The ray can then emerge at an angle smaller than \(90^\circ\). The total prism deviation decreases considerably even though its refracting angle and internal ray path remain unchanged. The surrounding refractive index changes the outgoing direction without changing the first-face normal-entry condition. \(\textbf{Final answer:}\) \[ \boxed{\delta_{\mathrm{air}}=45^\circ,\quad\delta_{\mathrm{liquid}}=\sin^{-1}\!\sqrt{\frac23}-45^\circ\approx9.74^\circ} \]

Explanation: \(\textbf{Viewing conditions:}\) The focal length of the magnifying lens is \[ f=+10\,\text{cm} \] The observer's least distance of distinct vision is \[ D=25\,\text{cm} \] The eye is sufficiently close to the lens for the standard simple-microscope angular magnification relations to apply. \(\textbf{Experiment P — relaxed-eye observation:}\) For the final image at infinity, the object must lie at the first principal focus. Therefore, \[ u_P=-f \] \[ u_P=-10\,\text{cm} \] The angular magnifying power is \[ M_P=\frac Df \] Substituting, \[ M_P=\frac{25}{10} \] \[ M_P=2.5 \] \(\textbf{Experiment Q — final image at the near point:}\) The desired virtual image distance is \[ v_Q=-25\,\text{cm} \] The thin-lens equation is \[ \frac1v-\frac1u=\frac1f \] Rearranging, \[ \frac1u=\frac1v-\frac1f \] Substituting, \[ \frac1{u_Q}=-\frac1{25}-\frac1{10} \] \[ \frac1{u_Q}=-\frac{2+5}{50} \] \[ u_Q=-\frac{50}{7}\,\text{cm} \] The object is now approximately \(7.14\,\text{cm}\) from the lens, inside its focal length. \(\textbf{Near-point angular magnifying power:}\) For the final image at the near point, \[ M_Q=1+\frac Df \] \[ M_Q=1+\frac{25}{10} \] \[ M_Q=3.5 \] \(\textbf{Check the change in object position:}\) The object moves from \(10\,\text{cm}\) to approximately \(7.14\,\text{cm}\) from the lens. Thus, it moves closer to the lens. Both arrangements produce virtual, erect images, but the final image is located differently. \(\textbf{Physical interpretation:}\) The simple microscope increases the angular size of a nearby object rather than merely supplying a large linear image. For relaxed-eye viewing, the emerging rays are parallel and the final image is at infinity. For maximum conventional near-point magnification, the object is placed slightly nearer the lens so that the final virtual image appears at the near point. The latter condition gives the larger angular magnifying power but requires accommodation by the observer. The object distance must be determined consistently with the intended final image position. \(\textbf{Final answer:}\) \[ \boxed{|u_P|=10\,\text{cm},\ M_P=2.5;\quad |u_Q|=\frac{50}{7}\,\text{cm},\ M_Q=3.5} \]

Explanation: \(\textbf{Initial microscope and near-point image:}\) The initial focal length is \[ f_1=+5\,\text{cm} \] The least distance of distinct vision is \[ D=25\,\text{cm} \] The desired virtual-image position is \[ v_1=-25\,\text{cm} \] \(\textbf{Determine the initial object distance:}\) Using the thin-lens equation, \[ \frac1{v_1}-\frac1{u_1}=\frac1{f_1} \] Rearranging, \[ \frac1{u_1}=\frac1{v_1}-\frac1{f_1} \] Substituting, \[ \frac1{u_1}=-\frac1{25}-\frac15 \] \[ \frac1{u_1}=-\frac6{25} \] Therefore, \[ u_1=-\frac{25}{6}\,\text{cm} \] \(\textbf{Initial angular magnifying power:}\) For near-point viewing, \[ M_1=1+\frac D{f_1} \] \[ M_1=1+\frac{25}{5} \] \[ M_1=6 \] \(\textbf{Replacement lens and relaxed-eye viewing:}\) The replacement focal length is \[ f_2=+10\,\text{cm} \] For a final image at infinity, the object must be at its first focus. Therefore, \[ u_2=-10\,\text{cm} \] \(\textbf{New angular magnifying power:}\) For relaxed-eye viewing, \[ M_2=\frac D{f_2} \] \[ M_2=\frac{25}{10} \] \[ M_2=2.5 \] \(\textbf{Find the object displacement:}\) The object's initial distance magnitude is \(\frac{25}{6}\,\text{cm}\), and its final distance is \(10\,\text{cm}\). Therefore, the required movement away from the lens is \[ \Delta s=10-\frac{25}{6} \] \[ \Delta s=\frac{60-25}{6} \] \[ \Delta s=\frac{35}{6}\,\text{cm} \] \(\textbf{Physical interpretation:}\) Two different changes occur simultaneously. The replacement lens has a longer focal length, reducing its angular magnifying ability. The final viewing condition also changes from near-point accommodation to relaxed-eye observation. Both changes contribute to the reduction in magnifying power. The new object is placed at the replacement lens's first focus rather than at the previous near-point object position. It would therefore be incorrect to compare only the focal lengths without recalculating the object positions and viewing conditions. \(\textbf{Final answer:}\) \[ \boxed{M_1=6,\quad M_2=2.5,\quad\Delta s=\frac{35}{6}\,\text{cm}\text{ away}} \]

Explanation: \(\textbf{Objective imaging stage:}\) The objective focal length is \[ f_o=+2\,\text{cm} \] The real object is placed at \[ u_o=-2.5\,\text{cm} \] The objective forms the intermediate image. Using the thin-lens equation, \[ \frac1{v_o}-\frac1{u_o}=\frac1{f_o} \] Rearranging, \[ \frac1{v_o}=\frac1{f_o}+\frac1{u_o} \] Substituting, \[ \frac1{v_o}=\frac12-\frac1{2.5} \] \[ \frac1{v_o}=\frac12-\frac25 \] \[ \frac1{v_o}=\frac1{10} \] Therefore, \[ v_o=+10\,\text{cm} \] \(\textbf{Objective transverse magnification:}\) \[ m_o=\frac{v_o}{u_o} \] \[ m_o=\frac{10}{-2.5} \] \[ m_o=-4 \] The intermediate image is real, inverted and enlarged four times. \(\textbf{Position the eyepiece for relaxed-eye viewing:}\) The eyepiece focal length is \[ f_e=+5\,\text{cm} \] For a final virtual image at infinity, the intermediate image must lie at the eyepiece's first focal plane. Therefore, the intermediate image must be \(5\,\text{cm}\) to the left of the eyepiece. \(\textbf{Determine the lens separation:}\) The intermediate image lies \(10\,\text{cm}\) to the right of the objective. The eyepiece must lie a further \(5\,\text{cm}\) beyond it. Thus, \[ L=v_o+f_e \] \[ L=10+5 \] \[ L=15\,\text{cm} \] \(\textbf{Eyepiece angular magnification:}\) The least distance of distinct vision is \[ D=25\,\text{cm} \] For final imaging at infinity, \[ M_e=\frac D{f_e} \] \[ M_e=\frac{25}{5} \] \[ M_e=5 \] \(\textbf{Combine the two imaging stages:}\) The magnitude of total angular magnifying power is \[ |M|=|m_o|M_e \] Substituting, \[ |M|=4(5) \] \[ |M|=20 \] The final image remains inverted relative to the original object because the objective introduces the inversion. \(\textbf{Physical interpretation:}\) The microscope's objective produces a real enlarged intermediate image. The eyepiece acts as a simple magnifier for that intermediate image. Normal adjustment requires the intermediate image to coincide with the eyepiece's first focal plane. Thus, the lens separation is obtained from the actual objective-image distance, not merely by adding the focal lengths of objective and eyepiece. The angular magnifying power combines the objective's transverse magnification with the eyepiece's angular magnification. These two roles are physically distinct and must be evaluated in sequence. \(\textbf{Final answer:}\) \[ \boxed{L=15\,\text{cm},\quad |M|=20} \]

Explanation: \(\textbf{Find the objective's intermediate image:}\) The objective parameters are \[ f_o=+2\,\text{cm} \] \[ u_o=-2.5\,\text{cm} \] Using the lens formula, \[ \frac1{v_o}=\frac1{f_o}+\frac1{u_o} \] \[ \frac1{v_o}=\frac12-\frac1{2.5} \] \[ \frac1{v_o}=\frac1{10} \] \[ v_o=+10\,\text{cm} \] The objective magnification is \[ m_o=\frac{v_o}{u_o} \] \[ m_o=-4 \] The intermediate image remains fixed because the object and objective do not move. \(\textbf{Initial eyepiece position:}\) The eyepiece focal length is \[ f_e=+5\,\text{cm} \] For the final image at infinity, the intermediate image lies at the eyepiece's first focus. Therefore, \[ |u_{e1}|=5\,\text{cm} \] The initial lens separation is \[ L_1=v_o+|u_{e1}| \] \[ L_1=10+5 \] \[ L_1=15\,\text{cm} \] \(\textbf{Initial angular magnifying power:}\) For relaxed-eye viewing, \[ |M_1|=|m_o|\frac D{f_e} \] \[ |M_1|=4\left(\frac{25}{5}\right) \] \[ |M_1|=20 \] \(\textbf{Eyepiece condition for near-point viewing:}\) The final virtual image must be at \[ v_{e2}=-25\,\text{cm} \] Using the eyepiece lens formula, \[ \frac1{v_{e2}}-\frac1{u_{e2}}=\frac1{f_e} \] Rearranging, \[ \frac1{u_{e2}}=\frac1{v_{e2}}-\frac1{f_e} \] \[ \frac1{u_{e2}}=-\frac1{25}-\frac15 \] \[ \frac1{u_{e2}}=-\frac6{25} \] Therefore, \[ u_{e2}=-\frac{25}{6}\,\text{cm} \] \(\textbf{New lens separation:}\) \[ L_2=v_o+|u_{e2}| \] \[ L_2=10+\frac{25}{6} \] \[ L_2=\frac{85}{6}\,\text{cm} \] \(\textbf{Required eyepiece movement:}\) \[ \Delta L=L_2-L_1 \] \[ \Delta L=\frac{85}{6}-15 \] \[ \Delta L=-\frac56\,\text{cm} \] The negative change means the eyepiece moves towards the objective. \(\textbf{New angular magnifying power:}\) For final near-point viewing, \[ |M_2|=|m_o|\left(1+\frac D{f_e}\right) \] \[ |M_2|=4\left(1+\frac{25}{5}\right) \] \[ |M_2|=24 \] \(\textbf{Physical interpretation:}\) The objective image and its linear magnification remain unchanged. Only the eyepiece viewing condition is altered. To move its final image from infinity to the near point, the intermediate image must lie slightly inside the eyepiece focal length. The eyepiece is therefore brought closer to the objective. This adjustment increases the angular magnifying power but requires the observer to accommodate for a near-point image. The tube-length change and magnification change follow from the same altered eyepiece conjugate condition. \(\textbf{Final answer:}\) \[ \boxed{\Delta L=-\frac56\,\text{cm},\quad |M_1|=20,\quad |M_2|=24} \]

Explanation: Statement I is correct because an object placed at the first focus of a convex lens produces parallel emerging rays. In the compound microscope, the objective's intermediate image acts as the object for the eyepiece. When that intermediate image is at the eyepiece focal plane, the final image is at infinity. This is the relaxed-eye or normal-adjustment condition. Statement II is correct because a convex eyepiece forms a virtual image when its object lies inside its focal length. For a fixed intermediate image, moving the eyepiece slightly towards it reduces the separation between them. The intermediate image consequently moves inside the eyepiece's first focal distance. An appropriate adjustment then places the final virtual image at the observer's near point. Statement III is incorrect because an object lying beyond twice the focal length of an ordinary convex lens produces a real image on its transmitted side. Such a condition does not give the virtual final image required for the normal compound-microscope viewing arrangement. The eyepiece cannot be treated as a magnifying glass irrespective of its object distance. Its image nature must be determined from the lens formula or corresponding ray geometry. Statement IV is correct because the objective initially forms a real inverted intermediate image. In its usual microscope operating condition, the eyepiece produces a virtual erect image relative to that intermediate image. Therefore, the final image remains inverted relative to the original specimen. The eyepiece increases the apparent angular size; it does not ordinarily restore the specimen's original orientation. The intermediate image position determines both the required eyepiece adjustment and the type of final image. The correct statements are I, II and IV.

Explanation: \(\textbf{Locate the required intermediate image:}\) The eyepiece focal length is \[ f_e=+5\,\text{cm} \] For relaxed-eye viewing, the objective's intermediate image must lie at the first focal plane of the eyepiece. The separation between objective and eyepiece is \[ L=20\,\text{cm} \] Therefore, the intermediate image must lie \[ v_o=L-f_e \] \[ v_o=20-5 \] \[ v_o=+15\,\text{cm} \] to the right of the objective. \(\textbf{Find the specimen distance:}\) The objective focal length is \[ f_o=+1.5\,\text{cm} \] The objective lens equation gives \[ \frac1{v_o}-\frac1{u_o}=\frac1{f_o} \] Rearranging, \[ \frac1{u_o}=\frac1{v_o}-\frac1{f_o} \] Substituting, \[ \frac1{u_o}=\frac1{15}-\frac1{1.5} \] \[ \frac1{u_o}=\frac1{15}-\frac23 \] \[ \frac1{u_o}=\frac{1-10}{15} \] \[ \frac1{u_o}=-\frac35 \] Therefore, \[ u_o=-\frac53\,\text{cm} \] The specimen lies slightly beyond the objective's first focal point. \(\textbf{Calculate objective magnification:}\) \[ m_o=\frac{v_o}{u_o} \] \[ m_o=\frac{15}{-5/3} \] \[ m_o=-9 \] The intermediate image is real, inverted and enlarged nine times. \(\textbf{Calculate eyepiece angular magnification:}\) The observer's near-point distance is \[ D=25\,\text{cm} \] For relaxed-eye viewing, \[ M_e=\frac D{f_e} \] \[ M_e=\frac{25}{5} \] \[ M_e=5 \] \(\textbf{Combine the magnifications:}\) The total angular magnifying-power magnitude is \[ |M|=|m_o|M_e \] \[ |M|=9(5) \] \[ |M|=45 \] \(\textbf{Physical interpretation:}\) The fixed objective-eyepiece separation and the requirement of a final image at infinity determine where the intermediate image must form. The objective lens equation then determines the specimen position rather than allowing it to be selected independently. The objective provides substantial transverse enlargement, while the eyepiece provides angular magnification of that enlarged intermediate image. The final magnifying power therefore depends on both optical stages. It would be incorrect to use only the eyepiece focal length or the approximate microscope tube length without first locating the actual intermediate image. \(\textbf{Final answer:}\) \[ \boxed{|u_o|=\frac53\,\text{cm},\quad |M|=45} \]

Explanation: The objective and eyepiece play different roles in a compound microscope. The objective forms a real intermediate image with transverse enlargement. The eyepiece then allows the observer to view that intermediate image under a suitable angular-magnification condition. Modification P changes the focal length of the objective. For an appropriately chosen real-object distance, a shorter objective focal length can yield greater linear enlargement of the intermediate image. The actual conjugate distances must remain consistent with the objective lens formula. Therefore, P matches (4). Modification Q decreases the eyepiece focal length while the microscope is readjusted so that the intermediate image remains at the eyepiece's first focus. For relaxed-eye viewing, its angular magnification is \[ M_e=\frac D{f_e} \] Reducing \(f_e\) increases this factor. Therefore, Q matches (2). Modification R increases the aperture of the objective. A wider aperture can collect more light from the specimen. If the focal length and imaging conjugates remain unchanged, the geometrical magnification is not automatically increased merely because the aperture becomes larger. Therefore, R matches (1). Modification S moves the intermediate image beyond the eyepiece focal distance. An ordinary converging eyepiece with a real object beyond its first focus produces a real image rather than the virtual final image desired in normal microscope viewing. It no longer operates in the intended simple-magnifier mode. Therefore, S matches (3). The distinctions are important because focal length, aperture and image position control different optical properties. The objective largely determines transverse enlargement and light collection. The eyepiece determines the angular viewing factor and final image condition. The correct matching is P–(4), Q–(2), R–(1), S–(3).