Ray Optics and Optical Instruments MCQs with Answers – Part 3 (Class 12 Physics)

Explanation: In the usual sign convention, the direction of incident light is taken as positive. If light travels from left to right, distances measured to the right of the pole are positive. The centre of curvature of the described convex refracting surface lies on the right side of the pole. Therefore the radius of curvature \(R\) is positive. The sign comes from the position of the centre of curvature, not from the word convex alone.

Explanation: The pole of the spherical surface is the origin for measuring distances. With incident light travelling from left to right, the right side is positive and the left side is negative. If the centre of curvature lies to the left of the pole, \(R\) is negative. The sign does not depend directly on whether the second medium is glass or water. In formula work, a wrong sign of \(R\) can change both the image position and image nature.

Explanation: The object is on the side from which light is incident. If light travels from left to right, the positive direction is to the right. A real object placed on the left of the refracting surface lies opposite to this positive direction. Hence its object distance \(u\) is negative. This is the same general sign idea used in lens and mirror calculations, though the formula for a refracting surface is different.

Explanation: A mirror reflects light back into the same medium, so its formula involves only \(u\), \(v\), and \(f\). At a spherical refracting surface, light passes from one medium to another. The refractive indices \(n_1\) and \(n_2\) must therefore appear in the relation. The standard formula is \(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\). The radius of curvature \(R\) is still essential because the bending depends on the curvature of the surface.

Explanation: The standard formula for refraction at a spherical surface is derived using the paraxial approximation. Paraxial rays remain close to the principal axis and make small angles with it. This allows the small-angle geometrical relations to be used. Rays far from the axis may produce aberrations and do not follow the simple ideal image-forming relation exactly. The approximation is a controlled simplification, not a statement that non-paraxial rays cannot refract.

Explanation: \( \textbf{Given:} \) Incident medium is air, so \(n_1=1.0\). Refracting medium is glass, so \(n_2=1.5\). The object is in front of the surface, so \(u=-45\,cm\). The centre of curvature is on the refracted side, so \(R=+30\,cm\). \( \textbf{Formula:} \) \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] \( \textbf{Substitution:} \) \[ \frac{1.5}{v}-\frac{1.0}{-45}=\frac{1.5-1.0}{30} \] \[ \frac{1.5}{v}+\frac{1}{45}=\frac{0.5}{30} \] \[ \frac{1.5}{v}+\frac{1}{45}=\frac{1}{60} \] \[ \frac{1.5}{v}=\frac{1}{60}-\frac{1}{45} \] \[ \frac{1.5}{v}=\frac{3-4}{180}=-\frac{1}{180} \] \[ v=1.5\times(-180)=-270\,cm \] The negative sign shows that the image is on the incident side of the refracting surface. \( \textbf{Final answer:} \) \(-270\,cm\).

Explanation: \( \textbf{Given data:} \) The incident medium is air, so \(n_1=1.0\). The refracting medium is glass, so \(n_2=1.5\). The object is in front of the surface, so \(u=-30\,cm\). The convex surface has its centre to the right, so \(R=+20\,cm\). Use the spherical refracting surface formula: \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Substitute the values: \[ \frac{1.5}{v}-\frac{1.0}{-30}=\frac{1.5-1.0}{20} \] \[ \frac{1.5}{v}+\frac{1}{30}=\frac{0.5}{20} \] \[ \frac{1.5}{v}+\frac{1}{30}=\frac{1}{40} \] \[ \frac{1.5}{v}=\frac{1}{40}-\frac{1}{30}=-\frac{1}{120} \] \[ v=1.5\times(-120)=-180\,cm \] \( \textbf{Final answer:} \) \(-180\,cm\), so the image lies on the incident side and is virtual.

Explanation: \( \textbf{Given data:} \) \(n_1=1.0\), \(n_2=1.5\), \(u=-90\,cm\), and \(R=+30\,cm\). Use \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Substitute the signed values: \[ \frac{1.5}{v}-\frac{1.0}{-90}=\frac{1.5-1.0}{30} \] \[ \frac{1.5}{v}+\frac{1}{90}=\frac{0.5}{30} \] \[ \frac{1.5}{v}+\frac{1}{90}=\frac{1}{60} \] \[ \frac{1.5}{v}=\frac{1}{60}-\frac{1}{90} \] \[ \frac{1.5}{v}=\frac{3-2}{180}=\frac{1}{180} \] \[ v=1.5\times180=270\,cm \] \( \textbf{Final answer:} \) \(+270\,cm\), placing the real image in the second medium.

Explanation: Statement I is correct because \(n_1\) is the refractive index of the incident medium. Statement II is correct because \(n_2\) is the refractive index of the medium into which the ray enters after refraction. Statement III is also correct because \(R\) is an algebraic distance from the pole to the centre of curvature. Statement IV is not correct because using \(cm\) or \(m\) does not remove sign convention. In this formula, the side of the object, image, and centre of curvature must be represented by signs.

Explanation: In the Cartesian convention, the direction of incident light is taken as positive. If light travels from left to right, a centre of curvature on the right gives \(R\gt 0\), so row P is correct. A centre of curvature on the left gives \(R\lt 0\), so row Q is correct. A plane surface can be treated as a spherical surface with infinite radius of curvature, so row R is acceptable. Row S is wrong because a point on the right side of the pole is in the positive direction. The sign of \(R\) is fixed by the position of the centre of curvature, not by the medium alone.

Explanation: \( \textbf{Starting formula:} \) \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Rearrange it: \[ \frac{n_2}{v}=\frac{n_1}{u}+\frac{n_2-n_1}{R} \] Let \[ Y=\frac{n_2}{v} \] and \[ X=\frac{n_1}{u} \] Then the equation becomes \[ Y=X+\frac{n_2-n_1}{R} \] This has the straight-line form \(Y=X+\text{constant}\). \( \textbf{Final answer:} \) slope \(+1\), with vertical intercept \(\frac{n_2-n_1}{R}\).

Explanation: The formula for a spherical refracting surface is \(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\). For a plane surface, the radius of curvature is infinite. Therefore the term \(\frac{n_2-n_1}{R}\) becomes zero. The formula then reduces to \(\frac{n_2}{v}-\frac{n_1}{u}=0\). Rearranging gives \(\frac{n_2}{v}=\frac{n_1}{u}\). A plane surface has no curvature term, but the refractive indices still affect the apparent position.

Explanation: \( \textbf{Required condition:} \) Image at infinity means \(v=\infty\), so \(\frac{n_2}{v}=0\). Use the spherical surface formula: \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Substitute \(\frac{n_2}{v}=0\): \[ -\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Insert \(n_1=1.0\), \(n_2=1.5\), and \(R=+30\,cm\): \[ -\frac{1}{u}=\frac{0.5}{30} \] \[ -\frac{1}{u}=\frac{1}{60} \] \[ u=-60\,cm \] \( \textbf{Final answer:} \) \(-60\,cm\), meaning the real object is \(60\,cm\) in front of the surface.

Explanation: \( \textbf{Known values:} \) \(n_1=1.0\), \(n_2=1.5\), \(u=-90\,cm\), and \(v=+270\,cm\). For refraction at a spherical surface, lateral magnification is \[ m=\frac{n_1v}{n_2u} \] Substitute the values: \[ m=\frac{1.0\times270}{1.5\times(-90)} \] \[ m=\frac{270}{-135} \] \[ m=-2 \] The magnitude \(2\) means the image is twice the object size. The negative sign shows that the image is inverted relative to the object. \( \textbf{Final answer:} \) \(-2\).

Explanation: \( \textbf{Given data:} \) Object height is \(h_o=+3\,cm\). \(n_1=1.0\), \(n_2=1.5\), \(u=-90\,cm\), and \(v=+270\,cm\). First calculate the magnification: \[ m=\frac{n_1v}{n_2u} \] \[ m=\frac{1.0\times270}{1.5\times(-90)} \] \[ m=-2 \] Now use \[ m=\frac{h_i}{h_o} \] Therefore, \[ h_i=mh_o \] \[ h_i=(-2)(3\,cm) \] \[ h_i=-6\,cm \] \( \textbf{Final answer:} \) \(-6\,cm\), where the negative sign shows an inverted image.

Explanation: Mirror magnification and refracting-surface magnification are not the same relation. A mirror reflects light back into the same medium, so its formula can be written as \(m=-\frac{v}{u}\). At a spherical refracting surface, light crosses from a medium of refractive index \(n_1\) into a medium of refractive index \(n_2\). The refractive indices therefore appear in the magnification relation. The correct relation is \(m=\frac{n_1v}{n_2u}\). Omitting \(n_1\) and \(n_2\) treats refraction as if it were reflection in one medium.

Explanation: The assertion is true because \(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\) is not symmetric under a careless interchange of \(n_1\) and \(n_2\). The incident medium and refracted medium play different roles in the formula. The reason is also true because \(n_1\) belongs to the side from which light comes, while \(n_2\) belongs to the side into which light enters. Reversing the ray path can be handled, but the object side, image side, and signs must then be reassigned consistently. The formula is reliable only when the medium labels and sign convention describe the actual ray direction.

Explanation: \( \textbf{Given data:} \) Incident medium is glass, so \(n_1=1.5\). Refracted medium is air, so \(n_2=1.0\). The real object is on the incident side, so \(u=-60\,cm\). The centre of curvature lies on the incident side, so \(R=-30\,cm\). Use \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Substitute: \[ \frac{1.0}{v}-\frac{1.5}{-60}=\frac{1.0-1.5}{-30} \] \[ \frac{1}{v}+\frac{1.5}{60}=\frac{-0.5}{-30} \] \[ \frac{1}{v}+\frac{1}{40}=\frac{1}{60} \] \[ \frac{1}{v}=\frac{1}{60}-\frac{1}{40} \] \[ \frac{1}{v}=-\frac{1}{120} \] \[ v=-120\,cm \] \( \textbf{Final answer:} \) \(-120\,cm\), so the image lies on the incident side.

Explanation: Row P is acceptable because a positive image distance places the image to the right of the pole when light travels from left to right. Row Q is acceptable because a negative image distance places the image on the left side. Row R is acceptable because a negative magnification represents an inverted image. Row S is not acceptable because \(|m|\lt1\) means the image is diminished, not enlarged. The sign of \(v\) locates the image, while the sign and magnitude of \(m\) describe orientation and relative size.

Explanation: Refraction occurs because light changes speed when it crosses into a medium of different refractive index. If \(n_1=n_2\), there is no optical contrast at the boundary. In the formula \(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\), the right-hand side becomes zero. This indicates that the curved boundary does not act as a refracting surface in the usual sense. The geometry may still exist physically, but optically the boundary is ineffective when the refractive indices are equal.