Spherical Mirror Questions in English

(N/A) Height of the needle,$h_{1} = 4.5 \; cm$.
Object distance,$u = -12 \; cm$.
Focal length of the convex mirror,$f = 15 \; cm$.
Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - (\frac{1}{-12}) = \frac{1}{15} + \frac{1}{12} = \frac{4+5}{60} = \frac{9}{60}$.
Thus,$v = \frac{60}{9} \approx 6.7 \; cm$.
The image is formed $6.7 \; cm$ behind the mirror.
Magnification $m = -\frac{v}{u} = -\frac{6.7}{-12} \approx 0.56$.
The height of the image $h_{2} = m \times h_{1} = 0.56 \times 4.5 \approx 2.5 \; cm$.
The image is virtual,erect,and diminished.
As the needle is moved farther from the mirror,the object distance $u$ increases,causing the image distance $v$ to increase towards the focus $f$,and the size of the image decreases.

(N/A) For a concave mirror,the focal length $(f)$ is negative,$f < 0$. When the object is placed on the left side of the mirror,the object distance $(u)$ is negative,$u < 0$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$. Since the object lies between $f$ and $2f$,we have $2f < u < f$ (considering magnitudes). This leads to $\frac{1}{2f} > \frac{1}{u} > \frac{1}{f}$,which implies $\frac{1}{v} < 0$. Thus,$v$ is negative,and the image is real and formed beyond $2f$.
$(b)$ For a convex mirror,$f > 0$ and $u < 0$. From $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$,since $f > 0$ and $u < 0$,$\frac{1}{v}$ is always positive. Thus,$v > 0$,meaning the image is always virtual and formed behind the mirror,regardless of $u$.
$(c)$ For a convex mirror,$f > 0$ and $u < 0$. From $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$,we see $\frac{1}{v} > \frac{1}{f}$,which implies $v < f$. Thus,the image is always located between the pole and the focus. Since magnification $m = -\frac{v}{u}$ and $|v| < |u|$,the image is diminished.
$(d)$ For a concave mirror,$f < 0$. When $0 < |u| < |f|$,then $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ results in $\frac{1}{v} > 0$,so $v > 0$. The image is virtual. Since $|v| > |u|$,the magnification $m = |\frac{v}{u}| > 1$,so the image is enlarged.

(N/A) $(1)$ Radius of Curvature $(R)$: The radius of the spherical shell of which the mirror is a part is called the radius of curvature.
$(2)$ Centre of Curvature $(C)$: The centre of the spherical shell of which the mirror is a part is called the centre of curvature.
$(3)$ Pole $(P)$: The geometric centre of the reflecting surface of a spherical mirror is called its pole.
$(4)$ Principal Axis: The straight line passing through the pole $(P)$ and the centre of curvature $(C)$ is called the principal axis.
$(5)$ Aperture: The diameter of the reflecting surface of a spherical mirror is called its aperture.
$(6)$ Principal Focus $(F)$: The point on the principal axis where rays parallel to the principal axis converge (concave mirror) or appear to diverge (convex mirror) after reflection is called the principal focus.
$(7)$ Focal Plane: $A$ plane passing through the principal focus and perpendicular to the principal axis is called the focal plane.
$(8)$ Focal Length $(f)$: The distance between the pole $(P)$ and the principal focus $(F)$ is called the focal length.
$(9)$ Paraxial Rays: Rays of light that are close to the principal axis and make small angles with it are called paraxial rays.

(N/A) Let $C$ be the centre of curvature and $P$ be the pole of the spherical mirror.
Consider a light ray parallel to the principal axis striking the mirror at point $M$. The line $CM$ is normal to the mirror at $M$.
Let $\theta$ be the angle of incidence. According to the law of reflection,the angle of reflection is also $\theta$.
From the geometry,$\angle MCP = \theta$ (alternate interior angles) and $\angle MFP = 2\theta$ (exterior angle of $\triangle MCF$ is equal to the sum of opposite interior angles).
Let $MD$ be the perpendicular from $M$ to the principal axis. For paraxial rays,$D$ is very close to $P$,so $CD \approx CP = R$ and $FD \approx FP = f$.
In $\triangle MDC$,$\tan \theta = \frac{MD}{CD} \approx \frac{MD}{R}$. For small $\theta$,$\tan \theta \approx \theta$,so $\theta \approx \frac{MD}{R}$.
In $\triangle MDF$,$\tan 2\theta = \frac{MD}{FD} \approx \frac{MD}{f}$. For small $\theta$,$\tan 2\theta \approx 2\theta$,so $2\theta \approx \frac{MD}{f}$.
Substituting $\theta$ from the first equation into the second: $2(\frac{MD}{R}) = \frac{MD}{f}$.
Therefore,$R = 2f$ or $f = \frac{R}{2}$.

(N/A) We can take any two rays emanating from a point on an object,trace their paths,find their point of intersection,and thus,obtain the image of the point due to reflection at a spherical mirror.
In practice,however,it is convenient to choose any two of the following rays:
$(i)$ The ray from the point which is parallel to the principal axis. The reflected ray goes through the focus of the mirror.
$(ii)$ The ray passing through the centre of curvature of a concave mirror or appearing to pass through it for a convex mirror. The reflected ray simply retraces the path.
$(iii)$ The ray directed towards the pole of the mirror. The reflected ray follows the laws of reflection,making equal angles with the principal axis.
$(iv)$ The ray passing through the focus of a concave mirror or directed towards the focus of a convex mirror. The reflected ray becomes parallel to the principal axis.

(N/A) The figure shows the ray diagram considering three rays originating from point $A$ of an object $AB$.
$1$. $A$ ray parallel to the principal axis,after reflection,passes through the principal focus $F$.
$2$. $A$ ray passing through the center of curvature $C$ is reflected back along the same path.
$3$. $A$ ray incident at the pole $P$ is reflected at an equal angle with the principal axis.
These three rays intersect at point $A^{\prime}$,which is the image of point $A$. By dropping a perpendicular from $A^{\prime}$ to the principal axis,we get the image $A^{\prime}B^{\prime}$ of the object $AB$. Since rays actually intersect at $A^{\prime}$,the image is real and inverted.

(N/A) Consider an object $AB$ placed perpendicular to the principal axis beyond the center of curvature $C$ of a concave mirror.
$A$ ray $AM$ from point $A$ incident on the mirror at $M$ reflects through the principal focus $F$.
$A$ ray $AP$ from point $A$ incident on the pole $P$ reflects back following the law of reflection,such that $\angle APB = \angle A'PB'$.
These reflected rays intersect at $A'$,forming a real image $A'B'$.
Let $FP = f$ (focal length),$CP = R$ (radius of curvature),$BP = u$ (object distance),and $B'P = v$ (image distance).
For paraxial rays,$MP$ can be considered a straight line perpendicular to the principal axis.
The right-angled triangles $\triangle A'B'F$ and $\triangle MPF$ are similar.
Therefore,$\frac{B'A'}{PM} = \frac{B'F}{FP}$. Since $PM = AB$,we have $\frac{B'A'}{AB} = \frac{B'F}{FP} = \frac{B'P - FP}{FP} = \frac{v - f}{f} \quad \dots (1)$
Similarly,$\triangle A'B'P$ and $\triangle ABP$ are similar.
Therefore,$\frac{B'A'}{AB} = \frac{B'P}{BP} = \frac{v}{u} \quad \dots (2)$
Equating $(1)$ and $(2)$,we get $\frac{v - f}{f} = \frac{v}{u}$.
Dividing by $v$,we get $\frac{1}{f} - \frac{1}{v} = \frac{1}{u}$,which simplifies to the mirror equation: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.

(N/A) $(1)$ Object position: $(u = \infty)$
Image position: At principal focus $F$
Image type: Real and inverted
Image size: Highly diminished (point-sized)
Magnification: $m \approx 0$
$(2)$ Object position: Beyond $C$ $(\infty > u > R)$
Image position: Between $F$ and $C$
Image type: Real and inverted
Image size: Diminished
Magnification: $-1 < m < 0$
$(3)$ Object position: At $C$ $(u = 2f = R)$
Image position: At $C$
Image type: Real and inverted
Image size: Same as the object
Magnification: $m = -1$

(N/A) Linear magnification $(m)$ is defined as the ratio of the height of the image $(h^{\prime})$ to the height of the object $(h)$.
$\therefore m = \frac{h^{\prime}}{h} \quad \dots (1)$
Consider an object $AB$ of height $h$ placed on the principal axis in front of a concave mirror as shown in the figure.
Two rays $AQ$ and $AP$ intersect at $A^{\prime}$ after reflection from the mirror,where $A^{\prime}$ is the image of $A$. Rays from object $AB$ form an image $A^{\prime}B^{\prime}$ after reflection. Let $A^{\prime}B^{\prime} = h^{\prime}$.
From the figure,triangles $\triangle A^{\prime}B^{\prime}P$ and $\triangle ABP$ are similar.
Therefore,$\frac{B^{\prime}A^{\prime}}{BA} = \frac{B^{\prime}P}{BP}$.
According to the sign convention,$B^{\prime}A^{\prime} = -h^{\prime}$ (downward),$BA = h$ (upward),$B^{\prime}P = -v$,and $BP = -u$.
Substituting these values:
$\frac{-h^{\prime}}{h} = \frac{-v}{-u}$
$\therefore \frac{h^{\prime}}{h} = -\frac{v}{u}$
$\therefore m = -\frac{v}{u} \quad (\text{From equation } 1)$.
For a real image,the image formed is inverted,hence its magnification is negative. For a virtual image,the image formed is erect,hence its magnification is positive.

(N/A) The centre of curvature of a convex mirror is the centre of the sphere of which the mirror is a part. For a convex mirror,the centre of curvature lies behind the mirror surface.

(N/A) The focal plane of a concave mirror is defined as a plane passing through the focus $(F)$ of the mirror and oriented perpendicular to the principal axis.
Any parallel beam of light rays that is incident on the mirror at an angle to the principal axis will converge at a point on this focal plane after reflection.

(N/A) For a spherical mirror of small aperture,the relationship between the radius of curvature $(R)$ and the focal length $(f)$ is given by the formula:
$R = 2f$
Alternatively,it can be expressed as:
$f = \frac{R}{2}$
This indicates that the focal length is half of the radius of curvature.

(N/A) The mirror equation relates the object distance $(u)$,the image distance $(v)$,and the focal length $(f)$ of a spherical mirror. It is given by the formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
Where:
$f$ is the focal length of the mirror.
$v$ is the distance of the image from the pole of the mirror.
$u$ is the distance of the object from the pole of the mirror.

(B) When an object is placed between the focus $(F)$ and the pole $(P)$ of a concave mirror,the light rays diverge after reflection.
$1$. $A$ ray parallel to the principal axis passes through the focus $(F)$ after reflection.
$2$. $A$ ray passing through the center of curvature $(C)$ reflects back along the same path.
$3$. By extending these reflected rays backward,they appear to meet behind the mirror.
$4$. The image formed is virtual,erect,and magnified,located behind the mirror.

(N/A) The mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Differentiating both sides with respect to $u$,we get $-\frac{1}{v^2} \frac{dv}{du} - \frac{1}{u^2} = 0$.
Therefore,$\frac{dv}{du} = -\frac{v^2}{u^2}$.
Since $v = \frac{fu}{u-f}$,we have $\frac{dv}{du} = -\left(\frac{fu}{u-f}\right)^2 \cdot \frac{1}{u^2} = -\left(\frac{f}{u-f}\right)^2$.
The length of the image $L'$ is given by $|dv| = |\frac{dv}{du}| \cdot L$.
Substituting the value of $\frac{dv}{du}$,we get $L' = \left(\frac{f}{u-f}\right)^2 L$.

(A) From the figure,the radius of curvature $R = 16 \ cm$. Since the inner surface is reflecting,it is a concave mirror.
Focal length $f = \frac{R}{2} = \frac{-16}{2} = -8 \ cm$.
The object is placed at $u = -10 \ cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} + \frac{1}{-10} = \frac{1}{-8}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = \frac{-1}{40}$
$v = -40 \ cm$.
Magnification $m = \frac{-v}{u} = \frac{-(-40)}{-10} = -4$.
Since $m$ is negative,the image is real and inverted. Since $|m| = 4 > 1$,the image is magnified.

(B) Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
For a concave mirror,the focal length $f$ is taken as negative,so $f_{mirror} = -f$.
The object distance $u$ is also negative,so $u = -1.5 f = -\frac{3 f}{2}$.
Substituting these values into the mirror formula:
$\frac{1}{v} + \frac{1}{-1.5 f} = \frac{1}{-f}$
$\frac{1}{v} - \frac{2}{3 f} = -\frac{1}{f}$
$\frac{1}{v} = -\frac{1}{f} + \frac{2}{3 f}$
$\frac{1}{v} = \frac{-3 + 2}{3 f}$
$\frac{1}{v} = -\frac{1}{3 f}$
Therefore,$v = -3 f$. The image is formed at a distance of $3 f$ in front of the mirror.

(A) For a spherical mirror,the relationship between focal length $f$ and radius of curvature $R$ is given by $f = \frac{R}{2}$.
For a convex mirror,the focus $F$ and the center of curvature $C$ are located behind the mirror.
According to the sign convention,distances measured in the direction of incident light (behind the mirror) are taken as positive.
Therefore,for a convex mirror,both $f$ and $R$ are positive.
Thus,the relationship is $f = +\frac{R}{2}$.

(D) $1$. The magnification $m$ is given by the ratio of image height to object height: $m = \frac{h_i}{h_o} = \frac{25\, cm}{100\, cm} = +0.25$.
$2$. Since the magnification is positive $(m > 0)$,the image is erect (same orientation as the object) and virtual.
$3$. For a spherical mirror,a virtual and erect image is formed by both concave and convex mirrors. However,a concave mirror produces a magnified virtual image $(|m| > 1)$,whereas a convex mirror produces a diminished virtual image $(|m| < 1)$.
$4$. Since the image height $(25\, cm)$ is less than the object height $(100\, cm)$,the magnification is less than $1$. Therefore,the mirror must be a convex mirror.
$5$. $A$ virtual image formed by a mirror is always located on the opposite side of the mirror relative to the object.

(A) For a spherical mirror,Newton's formula relates the distances of the object and image from the focus $F$. Let $x_{1}$ and $x_{2}$ be the distances of the object and image from the focus,respectively. The formula is $x_{1} x_{2} = f^{2}$.
In this problem,the distances are measured from the centre of curvature $C$. The distance of the focus $F$ from $C$ is the focal length $f$.
Thus,the distance of the object from the focus is $x_{1} = d_{1} + f$,and the distance of the image from the focus is $x_{2} = f - d_{2}$.
Substituting these into Newton's formula: $(f + d_{1})(f - d_{2}) = f^{2}$.
Expanding the equation: $f^{2} - f d_{2} + f d_{1} - d_{1} d_{2} = f^{2}$.
Simplifying: $f(d_{1} - d_{2}) = d_{1} d_{2}$.
Therefore,the focal length is $f = \frac{d_{1} d_{2}}{d_{1} - d_{2}}$.
The radius of curvature $R$ is given by $R = 2f$.
Hence,$R = \frac{2 d_{1} d_{2}}{d_{1} - d_{2}}$.

(D) Given: Initial distance of object $u_0 = -100\,cm$. Radius of curvature $R = -200\,cm$. Focal length $f = R/2 = -100\,cm$.
Speed of object $v_{obj} = 2\,cm/s$ towards the mirror.
After time $t = 10\,s$,the distance moved by the object is $d = v_{obj} \times t = 2\,cm/s \times 10\,s = 20\,cm$.
The new position of the object from the mirror is $u = u_0 + d = -100\,cm + 20\,cm = -80\,cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} + \frac{1}{-80} = \frac{1}{-100}$.
$\frac{1}{v} = \frac{1}{80} - \frac{1}{100} = \frac{5 - 4}{400} = \frac{1}{400}$.
Therefore,$v = 400\,cm$.

(A) For a large aperture mirror,rays parallel to the axis do not converge at a single point (focus $F$) due to spherical aberration. Instead,they form a circular patch at the focal plane.
Let $N$ be a point on the periphery of the mirror at a distance $r$ from the axis. Let the ray reflected from $N$ intersect the axis at $Q$. In $\triangle NQC$,where $C$ is the center of curvature,the angle of incidence equals the angle of reflection,both being $\theta$. Thus,$\angle N C P = \theta$ and $\angle N Q C = \theta$.
In $\triangle NQC$,by the law of sines or geometry,$QC = \frac{R}{2 \cos \theta}$.
The distance of $Q$ from the pole $P$ is $PQ = R - QC = R - \frac{R}{2 \cos \theta}$.
The distance of $Q$ from the focus $F$ is $QF = PF - PQ = \frac{R}{2} - (R - \frac{R}{2 \cos \theta}) = \frac{R}{2 \cos \theta} - \frac{R}{2} = \frac{R}{2} (\sec \theta - 1)$.
Let $d$ be the radius of the disc at the focus. From similar triangles formed by the reflected ray,the radius $d$ is given by $d = QF \tan(2\theta)$.
For small $\theta$,$\sin \theta \approx \tan \theta \approx \frac{r}{R}$. Also,$\sec \theta = (1 - \sin^2 \theta)^{-1/2} \approx 1 + \frac{\theta^2}{2} = 1 + \frac{r^2}{2R^2}$.
Substituting these,$QF \approx \frac{R}{2} (1 + \frac{r^2}{2R^2} - 1) = \frac{r^2}{4R}$.
Since $\tan(2\theta) \approx 2\theta \approx 2(\frac{r}{R})$,we have $d = QF \cdot 2\theta = (\frac{r^2}{4R}) \cdot (\frac{2r}{R}) = \frac{r^3}{2R^2}$.
The area of the disc is $A = \pi d^2 = \pi (\frac{r^3}{2R^2})^2 = \frac{\pi r^6}{4R^4}$.

(B) For a convex mirror,the image formed is always virtual,erect,and diminished.
Consider the line $PQ$. Point $Q$ is farther from the mirror than point $P$. For a convex mirror,the magnification $m = -v/u$ is positive and less than $1$. Since $Q$ is farther from the mirror $(u_Q > u_P)$,the image of $Q$ $(Q')$ will be closer to the mirror than the image of $P$ $(P')$,and the distance between $P'$ and $Q'$ will be smaller than the distance between $P$ and $Q$.
Furthermore,because the mirror is convex,the rays from points farther from the pole appear to originate from points closer to the focus. Thus,the image $P'Q'$ will appear tilted such that the end closer to the mirror $(P')$ is further from the principal axis than the end farther from the mirror $(Q')$.
Applying this logic to both lines $PQ$ and $RS$,we find that the correct schematic representation is given by option $B$.

(B) rear-view mirror is a convex mirror.
Given: Radius of curvature $R = 1.50 \, m$,Object distance $u = -10.0 \, m$.
The focal length $f = \frac{R}{2} = \frac{1.5}{2} = 0.75 \, m$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{0.75} - \frac{1}{-10} = \frac{4}{3} + \frac{1}{10} = \frac{40 + 3}{30} = \frac{43}{30}$.
Thus,$v = \frac{30}{43} \, m$.
The magnification $m$ is given by $m = -\frac{v}{u}$.
$m = -\frac{(30/43)}{-10} = \frac{30}{430} = \frac{3}{43} \approx 0.0697$.
Rounding to two decimal places,the magnification factor is approximately $0.07$.

(A) For commercial solar energy applications,parabolic mirrors are preferred.
These mirrors are designed to focus parallel rays of sunlight coming from infinity onto a single focal point.
This concentration of solar radiation generates high temperatures,which are necessary for efficient energy conversion processes.

(C) For a concave mirror,the focal length $f = -0.15 \,m = -15 \,cm$.
Since the image is virtual and magnified,the magnification $m = +2$.
The formula for magnification is $m = \frac{f}{f-u}$.
Substituting the values: $2 = \frac{-15}{-15 - u}$.
$2(-15 - u) = -15$.
$-30 - 2u = -15$.
$-2u = 15$.
$u = -7.5 \,cm$.
The negative sign indicates that the object is placed in front of the mirror at a distance of $7.5 \,cm$.

(A) The radius of curvature of the cylindrical mirror is $R = 10 \,cm$.
The focal length $f$ of the mirror is given by $f = \frac{R}{2} = \frac{10 \,cm}{2} = 5 \,cm$.
The field of view $\theta$ (in radians) for an observer at a large distance is determined by the arc length $s$ and the focal length $f$ as $\theta = \frac{s}{f}$.
Given the arc length $s = 10 \,cm$ and $f = 5 \,cm$,we have:
$\theta = \frac{10 \,cm}{5 \,cm} = 2 \,rad$.
Thus,the correct option is $A$.

(A) The mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ is derived using the paraxial approximation,which assumes that the rays are close to the principal axis and the mirror aperture is small compared to the radius of curvature. Thus,the Assertion $(A)$ is true.
The laws of reflection (angle of incidence equals angle of reflection) are fundamental laws of optics and are valid for any reflecting surface,whether plane or spherical,regardless of size. Therefore,the Reason $(R)$ is false.

(B) Given: Focal length $f = -20\,cm$. The midpoint of the rod is at $40\,cm$ from the pole. Since the rod length is $10\,cm$,the ends $A$ and $B$ are at distances $u_A = -(40 + 5) = -45\,cm$ and $u_B = -(40 - 5) = -35\,cm$ from the pole.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we have $v = \frac{uf}{u-f}$.
For end $A$: $v_A = \frac{(-45)(-20)}{-45 - (-20)} = \frac{900}{-25} = -36\,cm$.
For end $B$: $v_B = \frac{(-35)(-20)}{-35 - (-20)} = \frac{700}{-15} = -\frac{140}{3}\,cm$.
The length of the image is $|v_A - v_B| = |-36 - (-140/3)| = |-108/3 + 140/3| = |32/3|\,cm$.
Comparing this with $\frac{x}{3}\,cm$,we get $x = 32$.

(C) Given: Radius of curvature $R = -40\,cm$ (for concave mirror). Focal length $f = R/2 = -20\,cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
For object $A$ at $u_1 = -15\,cm$:
$\frac{1}{v_1} + \frac{1}{-15} = \frac{1}{-20} \implies \frac{1}{v_1} = -\frac{1}{20} + \frac{1}{15} = \frac{-3 + 4}{60} = \frac{1}{60}$.
So,$v_1 = 60\,cm$ (image is virtual and behind the mirror).
For object $B$ at $u_2 = -25\,cm$:
$\frac{1}{v_2} + \frac{1}{-25} = \frac{1}{-20} \implies \frac{1}{v_2} = -\frac{1}{20} + \frac{1}{25} = \frac{-5 + 4}{100} = -\frac{1}{100}$.
So,$v_2 = -100\,cm$ (image is real and in front of the mirror).
The distance between the images is $d = |v_1 - v_2| = |60 - (-100)| = 160\,cm$.

(A) Given: Radius of curvature $R = 30 \,cm$.
Focal length $f = R / 2 = +15 \,cm$ (for a convex mirror).
Magnification $m = +1/2$ (since a convex mirror always forms a virtual, erect image for a real object).
Using the magnification formula for mirrors: $m = f / (f - u)$.
Substituting the values: $1/2 = 15 / (15 - u)$.
$15 - u = 30$.
$-u = 30 - 15$.
$-u = 15$.
$u = -15 \,cm$.
Thus, the object distance is $-15 \,cm$.

(C) For a virtual image, magnification $m = +2$.
Using the magnification formula for a spherical mirror, $m = -v/u$.
Since the image is virtual, it is formed behind the mirror, so $v$ is positive. Let the object distance be $u$ (where $u$ is negative by sign convention).
The distance between the object and the image is $|v - u| = 15 \,cm$.
Since the object is in front of the mirror $(u < 0)$ and the virtual image is behind the mirror $(v > 0)$, the distance is $v - u = 15$.
Substituting $v = -mu = -2u$:
$-2u - u = 15
-3u = 15
u = -5 \,cm$.
Then, $v = -2(-5) = 10 \,cm$.
Using the mirror formula, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{10} + \frac{1}{-5} = \frac{1 - 2}{10} = -\frac{1}{10}$.
Therefore, $f = -10 \,cm$.

(B) The phenomenon described is known as parallax. When the eye is moved to the left,if the image appears to move to the right relative to the object,it indicates that the image is located behind the object pin (further from the mirror).
For a concave mirror,an inverted image is formed when the object is placed beyond the focal point $f$.
If the object is placed between $f$ and $2f$,the image is formed beyond $2f$.
Since the image is formed at a distance greater than the object distance,the image will appear to shift in the opposite direction to the eye's movement relative to the object.
Therefore,the object must be placed in the region $f < x < 2f$.

(C) $Statement-1$ is True. The mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ is derived using the paraxial approximation, which assumes that the rays are close to the principal axis and the aperture of the mirror is small compared to the radius of curvature $(R)$.
$Statement-2$ is False. The laws of reflection (angle of incidence equals angle of reflection) are fundamental and hold true for any reflecting surface, whether plane or spherical, regardless of size. The deviation in large spherical mirrors is due to spherical aberration, not a violation of the laws of reflection.

(C) For a concave mirror,the focal length $f = -24 \ cm$. The mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,which gives $v = \frac{uf}{u-f}$.
We check the recorded values against the theoretical values calculated using the mirror formula:
$1$. For $u = -42 \ cm$: $v = \frac{(-42)(-24)}{-42+24} = \frac{1008}{-18} = -56 \ cm$. (Matches)
$2$. For $u = -48 \ cm$: $v = \frac{(-48)(-24)}{-48+24} = \frac{1152}{-24} = -48 \ cm$. (Matches)
$3$. For $u = -60 \ cm$: $v = \frac{(-60)(-24)}{-60+24} = \frac{1440}{-36} = -40 \ cm$. (Matches)
$4$. For $u = -66 \ cm$: $v = \frac{(-66)(-24)}{-66+24} = \frac{1584}{-42} \approx -37.71 \ cm$. The recorded value is $33 \ cm$. The difference $|37.71 - 33| = 4.71 \ cm$,which is much greater than the error limit of $0.2 \ cm$.
$5$. For $u = -78 \ cm$: $v = \frac{(-78)(-24)}{-78+24} = \frac{1872}{-54} \approx -34.66 \ cm$. The recorded value is $39 \ cm$. The difference $|34.66 - 39| = 4.34 \ cm$,which is much greater than the error limit of $0.2 \ cm$.
Thus,the data sets $(66, 33)$ and $(78, 39)$ are incorrectly recorded.

(A) Given: Focal length $f = -10 \text{ cm}$ (for a concave mirror),object distance $u = -30 \text{ cm}$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{-10} - \frac{1}{-30} = \frac{-3+1}{30} = -\frac{2}{30} = -\frac{1}{15}$.
Thus,$v = -15 \text{ cm}$.
The velocity of the image $v_I$ with respect to the mirror $v_m$ is given by $v_{I/m} = -\left(\frac{v}{u}\right)^2 v_{o/m}$.
Here,$v_{o/m} = v_o - v_m$ and $v_{I/m} = v_I - v_m$.
Given that the image is at rest with respect to the laboratory frame,$v_I = 0$.
So,$0 - v_m = -\left(\frac{-15}{-30}\right)^2 (v_o - v_m)$.
$-v_m = -\left(\frac{1}{2}\right)^2 (v_o - v_m) = -\frac{1}{4} (v_o - v_m)$.
$v_m = \frac{1}{4} v_o - \frac{1}{4} v_m$.
$\frac{5}{4} v_m = \frac{1}{4} v_o$.
$v_m = \frac{v_o}{5} = \frac{15}{5} = 3 \text{ cm s}^{-1}$.
Therefore,the magnitude of $V_m$ is $3 \text{ cm s}^{-1}$.

(B) Given: $R = 2 \ m$,so focal length $f = R/2 = 1 \ m$. The speed of the object $v_0 = 90 \ km/h = 90 \times (5/18) = 25 \ m/s$. Since the object is approaching,$u = -24 \ m$ and $du/dt = v_0 = 25 \ m/s$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we differentiate with respect to time: $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$,which gives $v_I = \frac{dv}{dt} = -(\frac{v}{u})^2 v_0 = -m^2 v_0$.
For $u = -24 \ m$ and $f = 1 \ m$,$v = \frac{uf}{u-f} = \frac{(-24)(1)}{-24-1} = \frac{24}{25} \ m$.
Magnification $m = -v/u = -(\frac{24/25}{-24}) = \frac{1}{25}$.
Velocity of image $v_I = -m^2 v_0 = -(\frac{1}{25})^2 (25) = -\frac{1}{25} \ m/s$.
Differentiating the velocity equation $\frac{dv}{dt} = -(\frac{v}{u})^2 \frac{du}{dt}$ again with respect to time:
$a_I = \frac{d^2v}{dt^2} = -[2(\frac{v}{u})(\frac{u \frac{dv}{dt} - v \frac{du}{dt}}{u^2})] v_0 - (\frac{v}{u})^2 a_0$. Since $a_0 = 0$,$a_I = -2(\frac{v}{u})(\frac{u v_I - v v_0}{u^2}) v_0$.
Substituting values: $a_I = -2(\frac{24/25}{-24})(\frac{(-24)(-1/25) - (24/25)(25)}{(-24)^2}) (25) = -2(-\frac{1}{25})(\frac{24/25 - 24}{576}) (25) = 2(\frac{1}{25})(\frac{24 - 600}{25 \times 576}) (25) = 2(\frac{-576}{25 \times 576}) = -\frac{2}{25} \ m/s^2$.
The magnitude $a = |-\frac{2}{25}| = 0.08 \ m/s^2$. Thus,$100a = 100 \times 0.08 = 8$.

(D) The focal length of a spherical mirror is given by the formula $f = \frac{R}{2}$,where $R$ is the radius of curvature of the mirror.
This formula depends only on the geometry of the mirror surface.
Unlike lenses,the focal length of a mirror does not depend on the refractive index of the surrounding medium.
Therefore,when the mirror is dipped in a liquid of refractive index $\mu$,its focal length remains unchanged.
Thus,the focal length in the liquid will be $f$.

(D) Given magnification $m = -3$. We know that $m = -v/u$,so $-v/u = -3$,which implies $v = 3u$.
Since the image is real and formed by a concave mirror,both $u$ and $v$ are negative. The distance between the object and the image is $|v - u| = 20\ cm$.
Since the image is formed at a greater distance than the object for $m = -3$,we have $|v| > |u|$. Thus,$v - u = -20\ cm$ (using sign convention where $v$ and $u$ are negative).
Substituting $v = 3u$,we get $3u - u = -20$,so $2u = -20$,which gives $u = -10\ cm$.
Then $v = 3(-10) = -30\ cm$.
Using the mirror formula $1/f = 1/v + 1/u = 1/(-30) + 1/(-10) = (-1 - 3)/30 = -4/30 = -2/15$.
So,$f = -7.5\ cm$.
The radius of curvature $R = 2|f| = 2 \times 7.5 = 15\ cm$.

(A) Given,radius of curvature $R = 12 \ cm$. The focal length $f = R / 2 = 6 \ cm$.
For a concave mirror,$f = -6 \ cm$ and object distance $u = -18 \ cm$. Using the magnification formula $m = f / (f - u)$:
$m_{concave} = -6 / (-6 - (-18)) = -6 / 12 = -1 / 2$. The size of the image is $|m_{concave}| = 1 / 2$.
For a convex mirror,$f = +6 \ cm$ and object distance $u = -18 \ cm$. Using the magnification formula $m = f / (f - u)$:
$m_{convex} = 6 / (6 - (-18)) = 6 / 24 = 1 / 4$. The size of the image is $|m_{convex}| = 1 / 4$.
The ratio of the size of the image formed by the convex mirror to that formed by the concave mirror is:
Ratio $= |m_{convex}| / |m_{concave}| = (1 / 4) / (1 / 2) = 1 / 2$.

(A) The magnification formula for a spherical mirror is $m = \frac{f}{f-u}$.
Given $m = \frac{1}{2}$ and $u = -40\ cm$,we have $\frac{1}{2} = \frac{f}{f - (-40)}$.
$f + 40 = 2f$,which gives $f = 40\ cm$.
Now,to obtain a magnification of $m = \frac{1}{3}$,we use the same formula: $\frac{1}{3} = \frac{40}{40 - u'}$.
$40 - u' = 120$,so $u' = -80\ cm$.
The object was initially at $40\ cm$ and is now at $80\ cm$ from the mirror.
Therefore,the object must be moved $80 - 40 = 40\ cm$ away from the mirror.

(A) Given magnification $M = -\frac{1}{3}$.
Since $M = -\frac{v}{u}$,we have $-\frac{v}{u} = -\frac{1}{3}$,which implies $v = \frac{u}{3}$.
The distance between the object and the image is given by $|u - v| = 30 \ cm$. Since the image is real and inverted (magnification is negative),both object and image are on the same side of the mirror. Thus,$|u| - |v| = 30 \ cm$. Let $u = -u_0$ and $v = -v_0$,then $u_0 - v_0 = 30$.
Substituting $v_0 = \frac{u_0}{3}$,we get $u_0 - \frac{u_0}{3} = 30 \Rightarrow \frac{2u_0}{3} = 30 \Rightarrow u_0 = 45 \ cm$.
So,$u = -45 \ cm$ and $v = -15 \ cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{-15} + \frac{1}{-45} = \frac{-3 - 1}{45} = -\frac{4}{45}$.
Therefore,$f = -\frac{45}{4} \ cm$.
The magnitude of the focal length is $\frac{45}{4} \ cm$. Comparing this with $\frac{x}{4} \ cm$,we get $x = 45$.

(C) Given magnification $m = \frac{1}{4}$. Since the image is smaller than the object and formed by a mirror,we assume a concave mirror forming a real image.
For a real image,$m = -\frac{v}{u} = -\frac{1}{4}$,so $u = 4v$.
The distance between the object and the image is $|u - v| = 40 \ cm$.
Since $u$ and $v$ are on the same side for a real image,$u - v = 40 \ cm$.
Substituting $u = 4v$,we get $4v - v = 40 \implies 3v = 40 \implies v = \frac{40}{3} \ cm$.
Then $u = 4 \times \frac{40}{3} = \frac{160}{3} \ cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ with sign convention ($u = -\frac{160}{3} \ cm$,$v = -\frac{40}{3} \ cm$):
$\frac{1}{f} = -\frac{3}{40} - \frac{3}{160} = \frac{-12 - 3}{160} = -\frac{15}{160} = -\frac{3}{32}$.
$f = -\frac{32}{3} \approx -10.7 \ cm$.
The magnitude of the focal length is $10.7 \ cm$.

(B) In the parallax method,if the eye is shifted to the left and the image appears to shift to the right relative to the object pin,it implies that the image is further away from the eye than the object pin.
Let the position of the object pin be $O$ and the position of the image be $I$. If the image $I$ is further from the eye than the object $O$,then the image $I$ is behind the object $O$ (further from the pole $\text{P}$).
For a concave mirror,an inverted image is formed behind the object when the object is placed between the focus $f$ and the center of curvature $2f$. In this region,the image is formed beyond $2f$.
Therefore,the condition for the object distance $x$ is $f < x < 2f$.

(D) Given: Object height $h_o = 2 \ mm = 2 \times 10^{-3} \ m$. Radius of curvature $R = +40 \ cm$ (for convex mirror). Focal length $f = R/2 = +20 \ cm$. Object distance $u = -20 \ cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{20} = \frac{1}{20}$.
$\frac{1}{v} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \Rightarrow v = +10 \ cm$.
Magnification $m = -\frac{v}{u} = \frac{h_i}{h_o}$.
$m = -\frac{10}{-20} = +0.5$.
$h_i = m \times h_o = 0.5 \times 2 \ mm = 1 \ mm$.

(A) For a concave mirror,the focal length $f = -20 \ cm$.
Since the image is real and magnified two times,the magnification $m = -2$.
The formula for magnification in terms of focal length and object distance is $m = \frac{f}{f - u}$.
Substituting the values: $-2 = \frac{-20}{-20 - u}$.
$-2(-20 - u) = -20$.
$40 + 2u = -20$.
$2u = -60$.
$u = -30 \ cm$.
Thus,the object must be placed at a distance of $30 \ cm$ in front of the mirror.

(A) For a concave mirror,the focal length is taken as $-f$.
The distance of the object from the principal focus is $x$. Since the object is placed in front of the mirror,the object distance $u = -(f + x)$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f_{mirror}}$
$\frac{1}{v} + \frac{1}{-(f+x)} = \frac{1}{-f}$
$\frac{1}{v} = \frac{1}{f+x} - \frac{1}{f} = \frac{f - (f+x)}{f(f+x)} = \frac{-x}{f(f+x)}$
$v = -\frac{f(f+x)}{x}$
Magnification $m = -\frac{v}{u} = -\left( \frac{-\frac{f(f+x)}{x}}{-(f+x)} \right) = -\frac{f}{x}$
The magnitude of magnification is $|m| = \frac{f}{x}$.

(B) For a concave mirror,parallel rays incident at an angle $\theta$ with the principal axis focus on the focal plane at a distance $y$ from the axis.
Using the small angle approximation,the distance $y$ from the axis is given by $y = f \tan \theta$.
Given $f = 20 \ cm$ and $\theta = 5^{\circ}$.
Converting the angle to radians: $\theta = 5 \times \frac{\pi}{180} \approx 0.0873 \ rad$.
Since $\tan \theta \approx \theta$ for small angles,$y = 20 \times (5 \times \frac{\pi}{180})$.
$y = 20 \times 0.08726 = 1.745 \ cm$.
Rounding to the nearest value,$y \approx 1.75 \ cm$.

(D) For a concave mirror,the focal length $f = -60 \ cm$.
Since the image is real,the magnification $m = -5$.
The formula for magnification in terms of focal length and object distance $u$ is $m = \frac{f}{f - u}$.
Substituting the given values: $-5 = \frac{-60}{-60 - u}$.
Multiplying both sides by $(-60 - u)$,we get: $-5(-60 - u) = -60$.
$300 + 5u = -60$.
$5u = -60 - 300$.
$5u = -360$.
$u = -72 \ cm$.
The distance between the mirror and the object is the magnitude of $u$,which is $72 \ cm$.

(B) Given: Focal length $f = -10 \ cm$,Object distance $u = -25 \ cm$,Area of object $A_{\text{obj}} = (3 \ cm)^2 = 9 \ cm^2$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,the magnification $m$ is given by $m = \frac{f}{f-u}$.
Substituting the values: $m = \frac{-10}{-10 - (-25)} = \frac{-10}{15} = -\frac{2}{3}$.
The area of the image $A_{\text{image}}$ is related to the area of the object by $A_{\text{image}} = m^2 \times A_{\text{obj}}$.
$A_{\text{image}} = \left(-\frac{2}{3}\right)^2 \times 9 = \frac{4}{9} \times 9 = 4 \ cm^2$.