Optics MCQs with Answers – Part 1 (NEET Physics)

Explanation: The laws of reflection specify both the angular relationship and the geometrical arrangement of the rays. The angle of incidence is measured between the incident ray and the normal at the point of incidence. Similarly, the angle of reflection is measured between the reflected ray and the same normal. The first law gives \(i=r\), where \(i\) and \(r\) are the angles of incidence and reflection. The second law states that the incident ray, reflected ray and normal at the point of incidence lie in the same plane. The normal is perpendicular to the reflecting surface, so an angle measured from the surface is complementary to the corresponding angle measured from the normal. Although the incident and reflected rays make equal angles with the mirror surface as well, those surface-based angles are not the conventional definitions of \(i\) and \(r\). Both reflection laws must hold simultaneously, irrespective of the angle at which the light ray strikes the mirror.

Explanation: \(\textbf{Given:}\) Angle between the incident ray and mirror surface \(=25^\circ\). \(\textbf{Angle of incidence:}\) The normal is perpendicular to the mirror surface. Therefore, the angle of incidence is \[ i=90^\circ-25^\circ \] \[ i=65^\circ \] \(\textbf{Law of reflection:}\) The angle of reflection equals the angle of incidence. \[ r=i=65^\circ \] \(\textbf{Angle between the rays:}\) The incident and reflected ray segments extend away from the point of incidence on opposite sides of the normal. Hence, the angle between these two segments is \[ \theta=i+r \] \[ \theta=65^\circ+65^\circ \] \[ \theta=130^\circ \] \(\textbf{Physical check:}\) Both ray segments make \(25^\circ\) with the mirror surface, so the angle between them is also \(180^\circ-2(25^\circ)=130^\circ\). The smaller angle between two intersecting ray segments must be at most \(180^\circ\), which is satisfied here. \(\textbf{Final answer:}\) \[ \boxed{\theta=130^\circ} \]

Explanation: \(\textbf{Given:}\) Mirror rotation \(=12^\circ\), while the incident ray remains fixed. \(\textbf{Change in the normal:}\) The normal is rigidly attached to the mirror surface. When the mirror rotates through an angle \(\theta\), its normal also rotates through \(\theta\). The direction of the incident ray is unchanged, but its angle relative to the normal changes. \(\textbf{Reflection geometry:}\) At every mirror orientation, the incident and reflected ray segments make equal angles with the normal. If the normal rotates through \(\theta\), the reflected direction changes through twice that angle. \[ \Delta\phi=2\theta \] \(\textbf{Substitution:}\) \[ \Delta\phi=2(12^\circ) \] \[ \Delta\phi=24^\circ \] \(\textbf{Physical check:}\) The reflected-ray rotation is measured relative to a fixed external direction, not relative to the rotating mirror. The factor of two arises because the reflected direction depends on the normal's orientation on both sides of the reflection geometry. This result applies while the incident ray remains fixed and the reflected path is geometrically possible. \(\textbf{Final answer:}\) \[ \boxed{\Delta\phi=24^\circ} \]

Explanation: \(\textbf{Given ray directions:}\) Take anticlockwise angular displacement from the positive horizontal direction as positive. The direction of the incident ray, measured along its propagation, is \[ \alpha=-20^\circ \] The direction of the reflected ray is \[ \beta=+70^\circ \] \(\textbf{Reflection geometry:}\) Reflection from a plane mirror reverses the component of the propagation direction perpendicular to the mirror while preserving the parallel component. Consequently, the incident and reflected propagation directions are symmetric about the mirror surface. If the mirror surface makes an angle \(\theta\) with the positive horizontal, then \[ \beta-\theta=\theta-\alpha \] Rearranging, \[ 2\theta=\alpha+\beta \] \(\textbf{Substitution:}\) \[ 2\theta=-20^\circ+70^\circ \] \[ 2\theta=50^\circ \] \[ \theta=25^\circ \] \(\textbf{Consistency check:}\) The incident propagation direction makes \(45^\circ\) with the mirror surface. The reflected propagation direction also makes \(45^\circ\) with that surface. Thus, the corresponding angles with the normal are equal. Both rays lie on the physically appropriate side of the reflecting surface. \(\textbf{Final answer:}\) \[ \boxed{\theta=25^\circ} \]

Explanation: Statement I is correct because a plane mirror reverses the position coordinate perpendicular to its surface but preserves coordinates parallel to the surface. Hence, upward movement of the object produces an equal upward movement of its image. Statement II is correct because the virtual image lies as far behind the mirror as the object lies in front of it. If the object-mirror distance is \(d\), the object-image separation is \(2d\). A decrease of \(3\,\text{cm}\) in \(d\) therefore decreases the separation by \(6\,\text{cm}\). Statement III is incorrect because the virtual-image position is determined by the object and mirror geometry, not by the position of the observer. Moving sideways changes the observer's line of sight and may change which rays enter the eye. It does not shift the geometrical image while the object and mirror remain fixed. Statement IV is correct because plane-mirror magnification is \(+1\). The positive sign represents an erect image, and its magnitude represents equal object and image sizes. Lateral inversion should not be confused with vertical inversion. A vertically displaced object does not produce image motion in the opposite vertical direction. The correct combination is therefore Statements I, II and IV.

Explanation: \(\textbf{Initial configuration:}\) The mirror is at \(x_m=0\), and the object is at \(x_o=-30\,\text{cm}\). A plane-mirror image is the reflection of the object position across the mirror plane. In the chosen coordinate system, \[ x_i-x_m=x_m-x_o \] Therefore, \[ x_i=2x_m-x_o \] \(\textbf{Initial image position:}\) \[ x_{i1}=2(0)-(-30) \] \[ x_{i1}=+30\,\text{cm} \] \(\textbf{Changed configuration:}\) The mirror moves \(6\,\text{cm}\) along positive \(x\). \[ x_{m2}=+6\,\text{cm} \] The object moves \(4\,\text{cm}\) along positive \(x\). \[ x_{o2}=-30+4=-26\,\text{cm} \] \(\textbf{Final image position:}\) \[ x_{i2}=2(6)-(-26) \] \[ x_{i2}=+38\,\text{cm} \] \(\textbf{Image displacement:}\) \[ \Delta x_i=x_{i2}-x_{i1} \] \[ \Delta x_i=38-30=+8\,\text{cm} \] \(\textbf{Physical check:}\) For a stationary object, the mirror's \(6\,\text{cm}\) movement would shift its image by \(12\,\text{cm}\). The object's additional \(4\,\text{cm}\) movement in the positive direction contributes a \(4\,\text{cm}\) image shift in the negative direction. The resulting image displacement is therefore \(12-4=8\,\text{cm}\) along positive \(x\). \(\textbf{Final answer:}\) \[ \boxed{\Delta x_i=+8\,\text{cm}} \]

Explanation: \(\textbf{Initial reflection:}\) The incident ray is vertically downward, so it initially strikes the horizontal mirror normally. The reflected ray initially travels vertically upward and meets the screen directly above the point of incidence. \(\textbf{Effect of mirror rotation:}\) For a fixed incident ray, a mirror rotation through \(\theta\) rotates the reflected ray through \(2\theta\). \[ \phi=2\theta \] \[ \phi=2(15^\circ)=30^\circ \] Because the mirror rotates anticlockwise, the reflected ray rotates anticlockwise from its initially upward direction. It therefore travels \(30^\circ\) to the left of the vertical. \(\textbf{Screen geometry:}\) The screen is horizontal, and its vertical separation from the point of incidence is \[ h=\sqrt{3}\,\text{m} \] Let \(s\) be the horizontal displacement of the spot from its initial position. The reflected ray, vertical separation and horizontal displacement form a right-angled triangle. \[ \tan30^\circ=\frac{s}{h} \] \(\textbf{Substitution:}\) \[ s=h\tan30^\circ \] \[ s=\sqrt{3}\times\frac{1}{\sqrt{3}} \] \[ s=1\,\text{m} \] \(\textbf{Direction check:}\) The reflected ray tilts leftward from the vertical, so the spot shifts to the left. Using \(15^\circ\) instead of \(30^\circ\) would ignore the doubling of reflected-ray rotation. \(\textbf{Final answer:}\) \[ \boxed{s=1\,\text{m}\text{ to the left}} \]

Explanation: \(\textbf{First reflection:}\) Reflection from a plane mirror reverses the component of ray propagation perpendicular to the reflecting surface. The component parallel to the surface remains unchanged. For the vertical mirror \(M_1\), the horizontal component of propagation reverses direction. Thus, the ray changes from leftward-downward propagation to rightward-downward propagation. \(\textbf{Second reflection:}\) The ray next strikes the horizontal mirror \(M_2\). Reflection from this mirror reverses the vertical component but preserves the horizontal component. The final ray therefore travels rightward and upward. \(\textbf{Combined effect:}\) Take the initial propagation components as \((-a,-b)\), where \(a\) and \(b\) are positive. After reflection from \(M_1\), the components become \[ (-a,-b)\longrightarrow(a,-b) \] After reflection from \(M_2\), \[ (a,-b)\longrightarrow(a,b) \] The final propagation vector is the negative of the initial propagation vector. The two propagation directions are therefore antiparallel. \(\textbf{Physical interpretation:}\) The ray undergoes two reflections, reversing two mutually perpendicular directional components. Although the final direction is opposite to the initial direction, the outgoing ray need not retrace the same spatial path. This distinction separates directional reversal from exact path reversal. \(\textbf{Final answer:}\) \[ \boxed{\phi=180^\circ} \]

Explanation: \(\textbf{Mirror geometry:}\) The pole is at the origin, and the centre of curvature is \(36\,\text{cm}\) to its left. In the Cartesian sign convention, distances measured opposite to the incident-light direction are negative. \[ R=-36\,\text{cm} \] The mirror is concave towards the approaching light because its centre of curvature lies on the incident side. \(\textbf{Focal-length relation:}\) For a spherical mirror under the paraxial approximation, \[ f=\frac{R}{2} \] Substituting the signed radius, \[ f=\frac{-36}{2} \] \[ f=-18\,\text{cm} \] \(\textbf{Focus position:}\) The principal focus lies at a distance equal to the focal length from the pole. \[ x_F=-18\,\text{cm} \] \(\textbf{Image-forming interpretation:}\) Paraxial rays incident parallel to the principal axis actually converge at this focus after reflection. Their physical intersection makes the focus real. The focus lies between the pole and centre of curvature, as required for a concave spherical mirror. The negative focal length specifies position along the chosen axis; it does not imply an imaginary or unphysical distance. \(\textbf{Final answer:}\) \[ \boxed{x_F=-18\,\text{cm}} \] The focus is real and lies in front of the mirror.

Explanation: A convex spherical mirror is a diverging mirror for light arriving from a real object in front of it. Its reflected rays diverge, and their backward extensions meet behind the mirror. Consequently, the image remains virtual and erect for every finite real-object position. When the object is extremely distant, the incident rays reaching the mirror are nearly parallel. Their reflected rays appear to originate from the principal focus \(F\). As the object approaches the mirror, its image shifts from the focal region towards the pole \(P\). Using Cartesian signs, a convex mirror has \(f\gt 0\), a real object has \(u\lt 0\), and its virtual image has \(v\gt 0\). The mirror equation gives \(1/v=1/f+1/|u|\), showing that \(v\) decreases as the object approaches. The positive magnification is \(m=-v/u=v/|u|\). Equivalently, \(m=f/(|u|+f)\), which remains smaller than unity for every non-zero real-object distance. As \(|u|\) decreases, this magnification increases, so the image becomes larger than it was previously but never larger than the object. A real image cannot suddenly appear merely because the real object approaches a convex mirror. The correct behaviour is therefore motion from \(F\) towards \(P\), with an increasingly large yet diminished virtual image.

Explanation: At the centre of curvature \(C\), a real object placed in front of a concave mirror forms its image at \(C\). The reflected principal rays actually converge, so the initial image is real. The image is inverted, and its size equals the object size because the magnitude of magnification is unity. As the object moves from \(C\) towards the principal focus \(F\), the real image moves away from the mirror and becomes progressively enlarged. At the focal position, the reflected paraxial rays are parallel, so the image is formed at infinity rather than at a finite location. When the object crosses \(F\) towards the pole, the reflected rays become divergent. Their backward extensions meet behind the mirror, producing a virtual image. For an object between \(F\) and \(P\), this virtual image is erect and magnified. The specified final object distance is only half the focal-length magnitude, so the object definitely lies between \(F\) and \(P\). In fact, at this distance, the magnitude of mirror magnification is \(2\), confirming that the final image is enlarged. The transition from real to virtual imaging is controlled by the object's position relative to the focus, not merely by whether the mirror is concave. Therefore, the initial image is real, inverted and equal-sized, whereas the final image is virtual, erect and enlarged.

Explanation: \(\textbf{Given:}\) The concave mirror has a negative focal length under the Cartesian sign convention. \[ f=-12\,\text{cm},\qquad h_o=3\,\text{cm} \] Both object positions are in front of the mirror. \[ u_1=-36\,\text{cm},\qquad u_2=-18\,\text{cm} \] \(\textbf{Initial image position:}\) Apply the mirror formula. \[ \frac{1}{f}=\frac{1}{v_1}+\frac{1}{u_1} \] Rearranging, \[ \frac{1}{v_1}=\frac{1}{f}-\frac{1}{u_1} \] Substituting, \[ \frac{1}{v_1}=-\frac{1}{12}+\frac{1}{36} \] \[ \frac{1}{v_1}=-\frac{2}{36}=-\frac{1}{18} \] \[ v_1=-18\,\text{cm} \] \(\textbf{Initial magnification:}\) For a spherical mirror, \[ m_1=-\frac{v_1}{u_1} \] \[ m_1=-\frac{-18}{-36}=-\frac12 \] Hence, the initial image height is \[ h_{i1}=m_1h_o=-\frac12(3)=-1.5\,\text{cm} \] \(\textbf{Final image position:}\) For the changed object position, \[ \frac{1}{v_2}=\frac{1}{f}-\frac{1}{u_2} \] \[ \frac{1}{v_2}=-\frac{1}{12}+\frac{1}{18} \] \[ \frac{1}{v_2}=-\frac{1}{36} \] \[ v_2=-36\,\text{cm} \] \(\textbf{Final magnification:}\) \[ m_2=-\frac{v_2}{u_2} \] \[ m_2=-\frac{-36}{-18}=-2 \] Thus, \[ h_{i2}=m_2h_o=-2(3)=-6\,\text{cm} \] \(\textbf{Comparison:}\) The question asks for the ratio of image-height magnitudes, so the negative orientation signs are not included in the ratio. \[ \frac{|h_{i2}|}{|h_{i1}|}=\frac{6}{1.5}=4 \] \(\textbf{Physical check:}\) Both object positions lie outside the focal point, so both images must be real and inverted. The second object is closer to the focus, producing a larger real image, consistent with the calculation. \(\textbf{Final answer:}\) \[ \boxed{|h_{i2}|:|h_{i1}|=4:1} \]

Explanation: \(\textbf{Given image geometry:}\) The object is real and lies in front of the mirror. Therefore, \[ u=-30\,\text{cm} \] The image is located behind the mirror. Therefore, \[ v=+15\,\text{cm} \] \(\textbf{Determine the focal length:}\) The spherical-mirror equation is \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \] Substituting the signed distances, \[ \frac{1}{f}=\frac{1}{15}-\frac{1}{30} \] \[ \frac{1}{f}=\frac{2-1}{30} \] \[ \frac{1}{f}=\frac{1}{30} \] \[ f=+30\,\text{cm} \] \(\textbf{Identify the mirror:}\) A positive focal length indicates a convex spherical mirror when light is incident from the front. Its focus lies behind the mirror. \(\textbf{Determine the magnification:}\) \[ m=-\frac{v}{u} \] \[ m=-\frac{+15}{-30} \] \[ m=+\frac12 \] The positive sign indicates an erect image. The magnitude of magnification is smaller than unity, so the image is diminished. \(\textbf{Physical check:}\) A convex mirror forms virtual, erect and diminished images of real objects. Its virtual image lies between the pole and focus; here \(0\lt v\lt f\), since \(15\lt 30\). Thus, the calculated image position and magnification are both consistent with a convex mirror. \(\textbf{Final answer:}\) The mirror is convex, and the image is virtual, erect and diminished.

Explanation: \(\textbf{Interpret the graph:}\) The spherical-mirror formula is \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \] Rearranging, \[ \frac{1}{v}=-\frac{1}{u}+\frac{1}{f} \] Comparing this relation with \(y=mx+c\), where \(y=1/v\) and \(x=1/u\), gives \[ m=-1,\qquad c=\frac{1}{f} \] The given vertical intercept is \[ \frac{1}{f}=-\frac{1}{30}\,\text{cm}^{-1} \] Therefore, \[ f=-30\,\text{cm} \] \(\textbf{Identify the mirror:}\) The negative focal length identifies a concave mirror facing the incident light. \(\textbf{Calculate the new image distance:}\) The real object is \(45\,\text{cm}\) in front of the mirror. \[ u=-45\,\text{cm} \] Using the mirror equation, \[ \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \] \[ \frac{1}{v}=-\frac{1}{30}+\frac{1}{45} \] \[ \frac{1}{v}=\frac{-3+2}{90} \] \[ \frac{1}{v}=-\frac{1}{90} \] \[ v=-90\,\text{cm} \] \(\textbf{Image nature:}\) The negative image distance places the image in front of the mirror. The reflected rays actually converge there, so the image is real. \(\textbf{Check orientation and size:}\) \[ m=-\frac{v}{u} \] \[ m=-\frac{-90}{-45}=-2 \] The negative magnification indicates an inverted image. Its magnitude is \(2\), indicating enlargement. \(\textbf{Consistency check:}\) The object is between \(F\) and \(C\), since its distance is \(45\,\text{cm}\), while the focal-length magnitude is \(30\,\text{cm}\). A concave mirror must produce a real, inverted and enlarged image for an object in this region. This agrees with the graph-derived calculation. \(\textbf{Final answer:}\) The image is real, inverted and magnified.

Explanation: \(\textbf{Original arrangement:}\) The object and its real image are both in front of the concave mirror. Their signed distances are \[ u_1=-30\,\text{cm},\qquad v_1=-60\,\text{cm} \] \(\textbf{Determine the focal length:}\) The mirror formula gives \[ \frac{1}{f}=\frac{1}{u_1}+\frac{1}{v_1} \] Substituting, \[ \frac{1}{f}=-\frac{1}{30}-\frac{1}{60} \] \[ \frac{1}{f}=-\frac{2+1}{60} \] \[ \frac{1}{f}=-\frac{1}{20} \] \[ f=-20\,\text{cm} \] \(\textbf{New object position:}\) The object is now placed at the previous real-image position. \[ u_2=-60\,\text{cm} \] \(\textbf{New image position:}\) The focal length remains unchanged. \[ \frac{1}{v_2}=\frac{1}{f}-\frac{1}{u_2} \] \[ \frac{1}{v_2}=-\frac{1}{20}+\frac{1}{60} \] \[ \frac{1}{v_2}=-\frac{3-1}{60} \] \[ \frac{1}{v_2}=-\frac{1}{30} \] \[ v_2=-30\,\text{cm} \] The new image forms \(30\,\text{cm}\) in front of the mirror. \(\textbf{Image magnification:}\) \[ m_2=-\frac{v_2}{u_2} \] \[ m_2=-\frac{-30}{-60} \] \[ m_2=-\frac12 \] The negative sign indicates inversion, and the magnitude indicates an image half the object's size. \(\textbf{Physical interpretation:}\) This result illustrates the reversibility of light paths in mirror imaging. When an object and its real-image positions are interchanged, the image forms at the original object position. The object was initially between \(F\) and \(C\), producing an enlarged real image beyond \(C\). After the positions are exchanged, the object lies beyond \(C\), producing a diminished real image between \(F\) and \(C\). The calculated image distance and magnification agree with this reciprocal geometry. \(\textbf{Final answer:}\) The new image forms \(30\,\text{cm}\) in front of the mirror and is real, inverted and half-sized.

Explanation: \(\textbf{Given:}\) The mirror is concave, so its focal length is negative. \[ f=-15\,\text{cm} \] The image is erect and three times the object height. Therefore, the signed magnification is \[ m=+3 \] \(\textbf{Use the magnification condition:}\) For a spherical mirror, \[ m=-\frac{v}{u} \] Substituting the known magnification, \[ 3=-\frac{v}{u} \] Rearranging, \[ v=-3u \] \(\textbf{Combine with the mirror formula:}\) The mirror equation is \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \] Using \(v=-3u\), \[ \frac{1}{f}=-\frac{1}{3u}+\frac{1}{u} \] Taking the common denominator, \[ \frac{1}{f}=\frac{-1+3}{3u} \] \[ \frac{1}{f}=\frac{2}{3u} \] Therefore, \[ 3u=2f \] \[ u=\frac{2f}{3} \] \(\textbf{Calculate the object distance:}\) \[ u=\frac{2(-15)}{3} \] \[ u=-10\,\text{cm} \] \(\textbf{Calculate the image distance:}\) Using the magnification relation, \[ v=-3u \] \[ v=-3(-10) \] \[ v=+30\,\text{cm} \] \(\textbf{Physical consistency:}\) The object lies \(10\,\text{cm}\) in front of the mirror, which is within the focal-length magnitude of \(15\,\text{cm}\). A concave mirror forms a virtual, erect and magnified image when a real object is placed between its pole and principal focus. The positive image distance places the image behind the mirror, as required. A direct check gives \[ m=-\frac{+30}{-10}=+3 \] The calculated magnification agrees exactly with the stated image-height ratio and orientation. \(\textbf{Final answer:}\) \[ \boxed{(u,v)=(-10\,\text{cm},+30\,\text{cm})} \]

Explanation: \(\textbf{Given dimensions:}\) Student's height: \[ H=180\,\text{cm} \] Eye height above the floor: \[ h_e=160\,\text{cm} \] The student and the eyes are assumed to lie in the same vertical plane perpendicular to the mirror. \(\textbf{Reflection geometry:}\) A point of the body is visible when light from that point reflects from the mirror into the student's eye. For a vertical plane mirror, the required reflection point lies midway in height between the body point and the eye. This follows from constructing the virtual image of the body point behind the mirror and drawing a straight line from that image to the eye. \(\textbf{Lower edge of the mirror:}\) To see the feet, the reflected ray must connect a point at floor level to the eye. The minimum height of the required reflecting point is \[ h_{\text{lower}}=\frac{0+h_e}{2} \] Substituting, \[ h_{\text{lower}}=\frac{160}{2} \] \[ h_{\text{lower}}=80\,\text{cm} \] \(\textbf{Upper edge of the mirror:}\) To see the top of the head, the reflecting point must lie midway between the eye and the head's uppermost point. \[ h_{\text{upper}}=\frac{H+h_e}{2} \] Substituting, \[ h_{\text{upper}}=\frac{180+160}{2} \] \[ h_{\text{upper}}=170\,\text{cm} \] \(\textbf{Minimum mirror length:}\) \[ L=h_{\text{upper}}-h_{\text{lower}} \] \[ L=170-80 \] \[ L=90\,\text{cm} \] \(\textbf{Physical check:}\) The minimum mirror length is half the student's total height. The eye position determines where the mirror must be placed vertically, but it does not change the minimum required length. Moving nearer to or farther from the mirror does not reduce this minimum length for the stated geometry. A mirror extending from \(80\,\text{cm}\) to \(170\,\text{cm}\) provides the required reflection points for both extreme body positions. \(\textbf{Final answer:}\) \[ \boxed{L=90\,\text{cm},\quad h_{\text{lower}}=80\,\text{cm},\quad h_{\text{upper}}=170\,\text{cm}} \]

Explanation: \(\textbf{Angular changes:}\) Take anticlockwise rotations as positive and clockwise rotations as negative. The mirror rotates through \[ \Delta\theta=+7^\circ \] The incident propagation direction rotates through \[ \Delta\alpha=-4^\circ \] \(\textbf{Reflection-direction relation:}\) Let \(\theta\) be the orientation of the reflecting surface and \(\alpha\) the direction of incident propagation, measured from the same fixed reference axis. If \(\beta\) is the reflected propagation direction, the reflection law gives \[ \beta=2\theta-\alpha \] This relation expresses the symmetry of the incident and reflected propagation directions about the reflecting surface. It also allows the mirror and incident-ray directions to change independently. \(\textbf{Effect of simultaneous rotations:}\) The change in reflected direction is \[ \Delta\beta=2\Delta\theta-\Delta\alpha \] Substituting the signed rotations, \[ \Delta\beta=2(+7^\circ)-(-4^\circ) \] \[ \Delta\beta=14^\circ+4^\circ \] \[ \Delta\beta=+18^\circ \] \(\textbf{Interpretation:}\) The positive sign represents an anticlockwise rotation. The mirror's rotation alone would produce a \(14^\circ\) anticlockwise change in the reflected ray. The clockwise rotation of the incident propagation direction contributes another \(4^\circ\) anticlockwise change in the reflected direction. These effects add because they act through different terms in the reflection relation. \(\textbf{Physical check:}\) If the incident ray were fixed, the reflected-ray rotation would be twice the mirror rotation. If the mirror were fixed, rotating the incident propagation direction clockwise would rotate the reflected propagation direction anticlockwise through the same magnitude. The combined result is therefore consistent with both limiting cases. \(\textbf{Final answer:}\) \[ \boxed{\Delta\beta=18^\circ\text{ anticlockwise}} \]

Explanation: Statement I is correct because a concave spherical mirror converges paraxial incident rays that are parallel to its principal axis. After reflection, these rays pass through the real principal focus situated in front of the mirror. Statement II is correct because the radius drawn from the centre of curvature to the point of incidence is normal to the spherical reflecting surface. A ray directed towards the centre of curvature therefore strikes the mirror normally. Its angle of incidence is zero, so it returns along its original path. Statement III is incorrect because the principal focus of a convex mirror is virtual and lies behind the reflecting surface. A ray directed towards this virtual focus is reflected parallel to the principal axis. It does not physically pass through the virtual focus after reflection. Statement IV is correct because a convex mirror causes a paraxial beam parallel to its principal axis to diverge after reflection. The backward extensions of the reflected rays meet at the virtual principal focus behind the mirror. These rules follow from the reflection law and spherical-mirror geometry. A real focus is a location where reflected rays actually converge, whereas a virtual focus is the apparent point from which divergent reflected rays originate. The principal-ray rules are applied using the paraxial approximation, under which a definite principal focus is obtained. A frequent error is to treat the virtual focus of a convex mirror as though reflected rays physically pass through it. Another error is to assume that the principal focus and centre of curvature play identical roles in ray construction. The centre-of-curvature ray retraces its path because of normal incidence, while a focal ray is associated with reflection parallel to the principal axis. Therefore, Statements I, II and IV are correct.

Explanation: A concave spherical mirror possesses a real principal focus in front of its reflecting surface. When incident paraxial rays travel parallel to the principal axis, reflection makes them converge at this focus. The reversibility of light states that a light ray can traverse the same geometrical path in the reverse direction when the optical arrangement is unchanged. Consequently, when a point source is placed at the principal focus, its rays incident on the concave mirror are reflected parallel to the principal axis. The first reflected beam is therefore a parallel beam rather than a convergent or divergent one. In the second situation, the directions of these reflected rays are reversed. The reversed rays are now parallel incident rays approaching the same concave mirror. According to the principal-ray rule, these rays converge at the principal focus after reflection. That focus coincides with the original point-source position. The rays do not converge at the centre of curvature because the centre of curvature is associated with normal-incidence rays that retrace their paths. The pole is also not the focal point of a spherical mirror with finite non-zero focal length. The result depends on retaining the same mirror geometry and working within the paraxial approximation. Thus, the first process produces a parallel beam, and the reversed process reconstructs convergence at the original source position.