Spherical Mirror Questions in English

(A) For a concave mirror,a real image is formed when the object is placed beyond the focus $F$.
When the object is placed at the center of curvature $C$ (i.e.,at a distance of $2f$ from the pole),the image is also formed at the center of curvature $C$.
In this specific case,the object and the image coincide at the same point.
Therefore,the distance between the real object and the real image is $0$.

(A) Given: Focal length $f = -10 \ cm$,object distance $u = -25 \ cm$,side of square $s = 3 \ cm$.
Area of the object $A_0 = s^2 = 3^2 = 9 \ cm^2$.
The magnification $m$ is given by $m = \frac{f}{f - u} = \frac{-10}{-10 - (-25)} = \frac{-10}{15} = -\frac{2}{3}$.
The area of the image $A_i$ is given by $A_i = m^2 \times A_0$.
$A_i = \left(-\frac{2}{3}\right)^2 \times 9 = \frac{4}{9} \times 9 = 4 \ cm^2$.

(B) For a convex mirror,the focal length $f = +30 \ cm$.
The magnification $m$ is given as $m = +1/4$ (since the image formed by a convex mirror is always virtual and erect).
The formula for magnification is $m = \frac{f}{f - u}$.
Substituting the values: $\frac{1}{4} = \frac{30}{30 - u}$.
Cross-multiplying: $30 - u = 120$.
Solving for $u$: $u = 30 - 120 = -90 \ cm$.
The distance of the object from the mirror is the magnitude of $u$,which is $90 \ cm$.

(B) Given: Focal length $f = -10 \, cm$,Object distance $u = -25 \, cm$,Area of object $A_{obj} = (3.0 \, cm)^2 = 9 \, cm^2$.
Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we find the magnification $m = \frac{f}{f - u}$.
$m = \frac{-10}{-10 - (-25)} = \frac{-10}{15} = -\frac{2}{3}$.
The area of the image is given by $A_{image} = m^2 \times A_{obj}$.
$A_{image} = \left( -\frac{2}{3} \right)^2 \times 9 = \frac{4}{9} \times 9 = 4 \, cm^2$.

(C) Given: Focal length of the concave mirror $f = -12\; cm$ (sign convention).
Object height $h = 4\; cm$.
Image height $h' = -1\; cm$ (since the image is real,it is inverted).
Magnification $m = \frac{h'}{h} = \frac{-1}{4} = -0.25$.
We know that magnification $m = \frac{-v}{u}$,so $\frac{-v}{u} = -0.25$,which gives $v = 0.25u$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-12} = \frac{1}{0.25u} + \frac{1}{u}$.
$\frac{1}{-12} = \frac{4}{u} + \frac{1}{u} = \frac{5}{u}$.
$u = -12 \times 5 = -60\; cm$.
Thus,the object should be placed at a distance of $60\; cm$ in front of the mirror.

(A) For a concave mirror,the focal length $f = -50 \ cm$.
Since the image is real and inverted,the magnification $m = -2$.
The formula for magnification is $m = \frac{f}{f - u}$.
Substituting the values: $-2 = \frac{-50}{-50 - u}$.
$-2(-50 - u) = -50$.
$100 + 2u = -50$.
$2u = -150$.
$u = -75 \ cm$.
The object should be placed at a distance of $75 \ cm$ from the mirror.

(A) Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
For a concave mirror,$f$ is negative,so $\frac{1}{-f} = \frac{1}{v} + \frac{1}{-4f}$.
Rearranging gives: $\frac{1}{v} = \frac{1}{4f} - \frac{1}{f} = \frac{1-4}{4f} = -\frac{3}{4f}$.
Thus,$v = -\frac{4f}{3}$.
The magnification $m$ is given by $m = \frac{I}{O} = -\frac{v}{u}$.
Substituting the values: $\frac{I}{6} = -\frac{(-4f/3)}{-4f}$.
$\frac{I}{6} = -\frac{1}{3}$.
$I = -\frac{6}{3} = -2 \ cm$.
The magnitude of the length of the image is $2 \ cm$ (the negative sign indicates an inverted image).

(B) Given: Radius of curvature $R = 22 \; cm$.
For a convex mirror,the focal length $f$ is positive.
$f = \frac{R}{2} = \frac{22}{2} = 11 \; cm$.
The object distance $u$ is always taken as negative in the sign convention: $u = -14 \; cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{11} = \frac{1}{v} + \frac{1}{-14}$.
Rearranging for $v$: $\frac{1}{v} = \frac{1}{11} + \frac{1}{14}$.
$\frac{1}{v} = \frac{14 + 11}{154} = \frac{25}{154}$.
$v = \frac{154}{25} = 6.16 \; cm \approx 6.2 \; cm$.
Since $v$ is positive,the image is formed at $6.2 \; cm$ on the back side of the mirror.

(D) Given: Object distance $u = -30 \, cm$ (by sign convention).
Focal length of convex mirror $f = +30 \, cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{30} = \frac{1}{v} + \frac{1}{-30}$.
Rearranging the terms: $\frac{1}{v} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$.
Therefore,$v = +15 \, cm$.
The positive sign indicates that the image is formed $15 \, cm$ behind the mirror.

(D) The sun is at an effectively infinite distance from the concave mirror,so the object distance $u = \infty$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting $u = \infty$,we get $\frac{1}{v} + 0 = \frac{1}{f}$,which implies $v = f$.
Thus,the image of the sun is formed at the focus $F$ of the concave mirror.
Let $D_i$ be the diameter of the image formed at the focal plane.
Since the angle subtended by the sun at the pole $P$ is $\theta$,the image formed at the focus also subtends the same angle $\theta$ at the pole.
Using the definition of angle in radians,$\theta = \frac{\text{arc length}}{\text{radius}} = \frac{D_i}{f}$.
Therefore,the diameter of the image is $D_i = f \theta$.

(C) For a concave mirror,the focal length $f = -10 \, cm$. The magnification $m$ can be $2$ (virtual image) or $-2$ (real image).
Using the magnification formula $m = \frac{f}{f - u}$:
Case $1$: For $m = 2$ (virtual image),
$2 = \frac{-10}{-10 - u} \implies -20 - 2u = -10 \implies 2u = -10 \implies u = -5 \, cm$.
Case $2$: For $m = -2$ (real image),
$-2 = \frac{-10}{-10 - u} \implies 20 + 2u = -10 \implies 2u = -30 \implies u = -15 \, cm$.
Thus,the object can be placed at $5 \, cm$ or $15 \, cm$ from the mirror.

(D) From the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Differentiating both sides with respect to $u$,we get $-\frac{1}{v^2} dv - \frac{1}{u^2} du = 0$.
This implies $dv = -\frac{v^2}{u^2} du$. The magnitude of the image length $|dv|$ is given by $|dv| = \left( \frac{v}{u} \right)^2 |du|$.
Since the object is small,$|du| = b$,so the image length is $L = \left( \frac{v}{u} \right)^2 b$.
From the mirror formula,$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{fu}$,which gives $\frac{v}{u} = \frac{f}{u - f}$.
Substituting this into the expression for $L$,we get $L = b \left( \frac{f}{u - f} \right)^2$.

(A) Given: Focal length $f = -20 \,cm$ (for a concave mirror),Object distance $u = -40 \,cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-40}$.
Rearranging: $\frac{1}{v} = \frac{1}{40} - \frac{1}{20} = \frac{1 - 2}{40} = -\frac{1}{40}$.
Thus,$v = -40 \,cm$.
The magnification $m = -\frac{v}{u} = -\frac{-40}{-40} = -1$.
Since $v$ is negative,the image is real. Since $m = -1$,the image is inverted and of the same size as the object.

(C) For a convex mirror,the magnification $m$ is given by $m = \frac{f}{f - u}$,where $f$ is the focal length and $u$ is the object distance (which is negative).
Given that the image size is $(1/n)$ times the object size,the magnification $m = +\frac{1}{n}$ (since the image is virtual and erect).
Substituting the values: $\frac{1}{n} = \frac{f}{f - u}$.
Rearranging the equation: $f - u = nf$.
Solving for $u$: $u = f - nf = -(n - 1)f$.
The distance of the object from the mirror is the magnitude of $u$,which is $|u| = (n - 1)f$.

(B) For a concave mirror,the focal length $f = -30\; cm$.
Since the image is real,the magnification $m = -3$.
The formula for magnification is $m = \frac{f}{f - u}$.
Substituting the values: $-3 = \frac{-30}{-30 - u}$.
$-3(-30 - u) = -30$.
$90 + 3u = -30$.
$3u = -120$.
$u = -40\; cm$.
The object should be placed at a distance of $40\; cm$ in front of the mirror.

(D) For a concave mirror,the focal length $f = -30 \, cm$.
Given that the image is erect (virtual),the magnification $m$ must be positive.
Since the image is three times the size of the object,$m = +3$.
The formula for magnification in terms of focal length and object distance $u$ is $m = \frac{f}{f - u}$.
Substituting the given values:
$3 = \frac{-30}{-30 - u}$
$3(-30 - u) = -30$
$-90 - 3u = -30$
$-3u = 60$
$u = -20 \, cm$.
The negative sign indicates that the object is placed $20 \, cm$ in front of the mirror.

(C) For an erect image,the magnification $m = +3$.
Since $m = -v/u$,we have $3 = -v/u$,which implies $v = -3u$.
The distance between the object and the image is given as $|v - u| = 80 \ cm$.
Substituting $v = -3u$,we get $|-3u - u| = 80$,so $|-4u| = 80$,which gives $u = -20 \ cm$.
Then,$v = -3(-20) = 60 \ cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{60} + \frac{1}{-20} = \frac{1 - 3}{60} = \frac{-2}{60} = -\frac{1}{30}$.
Therefore,$f = -30 \ cm$.

(B) The mirror formula is given by: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Rearranging this equation in the form of a linear equation $y = mx + c$,we get: $\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$.
Here,$y = \frac{1}{v}$,$x = \frac{1}{u}$,slope $m = -1$,and intercept $c = \frac{1}{f}$.
Since the intercept $c = \frac{1}{f}$ is non-zero for a spherical mirror,the graph of $\frac{1}{v}$ versus $\frac{1}{u}$ is a straight line that does not pass through the origin.

(C) The mirror formula is given by: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
Rearranging this equation to express $1/v$ in terms of $1/u$,we get:
$\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$
This equation is of the form $y = mx + c$,where:
$y = \frac{1}{v}$
$x = \frac{1}{u}$
$m = -1$ (slope is negative)
$c = \frac{1}{f}$ (positive y-intercept)
This represents a straight line with a negative slope and a positive y-intercept. Comparing this with the given options,graph $C$ represents a straight line with a negative slope and a positive intercept on the $1/v$ axis.

(C) When an object is placed between the pole $(P)$ and the focus $(F)$ of a concave mirror,the light rays diverge after reflection.
By extending these rays backward,a virtual image is formed behind the mirror.
Properties of this image are:
$1$. It is virtual (not real).
$2$. It is magnified (larger than the object).
$3$. It is erect (upright).
Therefore,statements $II$ and $III$ are correct,while statement $I$ is incorrect.

(B) For a convex mirror,the focal length $f$ is positive $(f = +10 \ cm)$.
The mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
As the object distance $u$ varies from $-\infty$ to $0$,the image distance $v$ varies from $f$ to $0$.
Specifically,when the object is at infinity $(u = -\infty)$,the image is formed at the focus $(v = f = 10 \ cm)$.
When the object is at the pole $(u = 0)$,the image is formed at the pole $(v = 0)$.
Therefore,the image always forms between the pole and the focus of the convex mirror.
The maximum distance of the image from the mirror is equal to the focal length,which is $10 \ cm$.

(B) Given: Object distance $u = -30 \,cm$. Magnification $m = +1/5$ (since the image is erect).
For a spherical mirror,magnification $m = -v/u$.
Substituting the values: $1/5 = -v / (-30)$.
$1/5 = v / 30 \implies v = 6 \,cm$.
Since the image distance $v$ is positive,the image is formed behind the mirror.
Using the mirror formula: $1/f = 1/v + 1/u$.
$1/f = 1/6 + 1/(-30) = 5/30 - 1/30 = 4/30 = 2/15$.
$f = 15/2 = +7.5 \,cm$.
Since the focal length $f$ is positive,the mirror is a convex mirror.

(B) Let $f$ be the focal length of the concave mirror.
According to Newton's formula for mirrors,the distance of the object from the focus is $x_1$ and the distance of the image from the focus is $x_2$.
The relationship between these distances and the focal length $f$ is given by $f^2 = x_1 x_2$.
Therefore,the focal length is $f = \sqrt{x_1 x_2}$.

(D) Let the rod be $AB$ with length $L = f/3$. The end $B$ is at distance $u_B = 5f/3$ from the pole,and end $A$ is at $u_A = 2f$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with sign convention (all distances negative):
For end $B$ $(u_B = -5f/3)$: $\frac{1}{v_B} + \frac{1}{-5f/3} = \frac{1}{-f} \Rightarrow \frac{1}{v_B} = -\frac{1}{f} + \frac{3}{5f} = -\frac{2}{5f} \Rightarrow v_B = -2.5f$.
For end $A$ $(u_A = -2f)$: $\frac{1}{v_A} + \frac{1}{-2f} = \frac{1}{-f} \Rightarrow \frac{1}{v_A} = -\frac{1}{f} + \frac{1}{2f} = -\frac{1}{2f} \Rightarrow v_A = -2f$.
The length of the image is $|v_B - v_A| = |-2.5f - (-2f)| = 0.5f = f/2$.
The longitudinal magnification $m = \frac{\text{length of image}}{\text{length of object}} = \frac{f/2}{f/3} = 1.5$.
Since the image is inverted,the magnification is $m = -3/2$.

(A) The sun is at a very large distance from the mirror,so the rays coming from the sun are parallel to the principal axis.
These rays will focus at the focal plane of the mirror.
The image of the sun is formed at the focal plane.
From the geometry of the setup,the angle subtended by the sun at the pole is $\theta$.
Since the image is formed at the focal plane at a distance $f$ from the pole,we can use the relation: $\text{Angle} = \frac{\text{Arc}}{\text{Radius}}$.
Here,the arc corresponds to the diameter of the image $(d)$ and the radius corresponds to the focal length $(f)$.
Therefore,$\theta = \frac{d}{f}$.
Thus,the diameter of the image is $d = f \theta$.

(D) The focal length of a spherical mirror is determined solely by its radius of curvature $(f = R/2)$.
Unlike lenses,where the refractive index depends on the wavelength of light (causing chromatic aberration),the reflection of light from a mirror surface is independent of the wavelength or color of the incident light.
Therefore,the focal length remains the same for all colors of light.

(D) For this situation,the incident rays are converging towards a point $O$ behind the mirror,which acts as a virtual object.
Here,the object distance $u = +10 \, cm$ (since it is behind the mirror) and the focal length of the convex mirror $f = +20 \, cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
Substituting the values: $\frac{1}{v} + \frac{1}{10} = \frac{1}{20}$
$\frac{1}{v} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = -\frac{1}{20}$
Therefore,$v = -20 \, cm$.
The negative sign indicates that the image is formed in front of the mirror at a distance of $20 \, cm$ from the pole.

(B) According to the rules of reflection for a concave mirror,a light ray passing through the principal focus $(F)$ of the mirror will become parallel to the principal axis after reflection.
In the given diagram,the incident ray $PQ$ passes through the focus $(F)$.
Therefore,after reflection from the concave mirror,the ray must travel parallel to the principal axis.
Among the given options,ray $2$ is the only one that is parallel to the principal axis.
Thus,ray $2$ correctly represents the reflected ray.

(B) Given: Focal length $f = -20 \, cm$ (for a concave mirror),Object distance $u = -10 \, cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-10}$.
$\frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{2-1}{20} = \frac{1}{20}$.
So,$v = +20 \, cm$.
Since $v$ is positive,the image is formed behind the mirror,which means it is virtual.
The magnification $m = -\frac{v}{u} = -\frac{20}{-10} = +2$.
Since $m$ is positive,the image is erect.
Since $|m| > 1$,the image is magnified.
Therefore,the image is magnified,erect,and virtual.

(C) For a concave mirror, the magnification $m$ is given by $m = -v/u$. Given that the image is real and $n$ times the size of the object, the magnification $m = -n$.
Thus, $-n = -v/u$, which implies $v = nu$.
Using the mirror formula: $1/f = 1/v + 1/u$.
Substituting $v = nu$: $1/f = 1/(nu) + 1/u$.
$1/f = (1 + n) / (nu)$.
Rearranging for $u$: $u = f(n + 1) / n = [(n + 1) / n] f$.
Therefore, the distance of the object from the mirror is $[(n + 1) / n] f$.

(B) Given: Focal length $f = -15 \ cm$ (for a concave mirror).
Magnification $m = +2$ (since the image is virtual and erect,magnification is positive).
The formula for magnification in terms of focal length and object distance $u$ is $m = \frac{f}{f - u}$.
Substituting the values: $2 = \frac{-15}{-15 - u}$.
$2(-15 - u) = -15$.
$-30 - 2u = -15$.
$-2u = -15 + 30$.
$-2u = 15$.
$u = -7.5 \ cm$.
The negative sign indicates that the object is placed in front of the mirror at a distance of $7.5 \ cm$.

(C) For a concave mirror,the radius of curvature is $R = 2f$.
Given that the nearest end of the object is at a distance $u > R = 2f$.
Since the entire object is placed beyond the center of curvature $(C)$,the image of every point of the object will be formed between the focus $(F)$ and the center of curvature $(C)$.
For an object placed beyond $C$,the image formed by a concave mirror is always real,inverted,and diminished (smaller than the object).
Therefore,the correct option is $C$.

(C) For an erect image,the magnification $m = +3$.
Since $m = -v/u$,we have $v = -3u$.
Let the object distance be $u = -x$ (where $x > 0$). Then $v = 3x$.
The distance between the object and the image is given as $|v - u| = 80 \, cm$.
Substituting the values: $|3x - (-x)| = 80 \implies 4x = 80 \implies x = 20 \, cm$.
Thus,$u = -20 \, cm$ and $v = 60 \, cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{60} + \frac{1}{-20} = \frac{1 - 3}{60} = \frac{-2}{60} = \frac{-1}{30}$.
Therefore,$f = -30 \, cm$.

(C) Given: Height of object $O = 5 \, cm$,Radius of curvature $R = 20 \, cm$,Object distance $u = -1 \, m = -100 \, cm$.
For a concave mirror,focal length $f = -R/2 = -10 \, cm$.
The magnification formula is given by $m = I/O = f/(f - u)$.
Substituting the values: $I/5 = -10 / (-10 - (-100))$.
$I/5 = -10 / (-10 + 100) = -10 / 90 = -1/9$.
$I = -5/9 \, cm \approx -0.55 \, cm$.
The magnitude of the height of the image is $0.55 \, cm$.

(C) Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
For a convex mirror,the focal length $f$ is positive $(+f)$ and the object distance $u$ is negative $(-f)$.
Substituting these values into the formula: $\frac{1}{+f} = \frac{1}{v} + \frac{1}{-f}$.
Rearranging the terms: $\frac{1}{v} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$.
Therefore,$v = \frac{f}{2}$.
The image is formed at a distance of $f/2$ behind the mirror.

(A) Given: Focal length $f = -10 \, cm$,Object distance $u = -25 \, cm$,Side of square object $s_o = 3 \, cm$.
Area of the object $A_o = (s_o)^2 = (3)^2 = 9 \, cm^2$.
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 relationship between the area of the image $A_i$ and the area of the object $A_o$ is $A_i = m^2 \times A_o$.
$A_i = (-\frac{2}{3})^2 \times 9 = \frac{4}{9} \times 9 = 4 \, cm^2$.

(A) The focal length of a spherical mirror is given by $f = R/2$, where $R$ is the radius of curvature.
This formula depends only on the geometry (radius of curvature) of the mirror and is independent of the refractive index of the surrounding medium.
Therefore, when the concave mirror is immersed in water, its focal length remains unchanged.
Thus, the new focal length is $10 \, cm$.

(C) Given: Object height $O = +3\, cm$. Image height $I = -9\, cm$ (since the image is formed on the wall,it is real and inverted).
Distance of the object from the mirror $u = -(x - 3)\, m = -100(x - 3)\, cm$.
Distance of the image from the mirror $v = -x\, m = -100x\, cm$.
Using the magnification formula $m = \frac{I}{O} = -\frac{v}{u}$:
$\frac{-9}{3} = -\frac{-100x}{-100(x - 3)}$
$-3 = -\frac{x}{x - 3}$
$3(x - 3) = x$
$3x - 9 = x$
$2x = 9$
$x = 4.5\, m = 450\, cm$.

(B) The object is placed such that its farthest end is at the center of curvature $C$ (distance $2f$ from the mirror). Since the image of the farthest end is formed at the same position,the farthest end is at $u = -2f$,and its image is at $v = -2f$.
The nearest end of the object is at a distance $u' = -(2f - f/3) = -5f/3$ from the mirror.
Using the mirror formula $\frac{1}{f} = \frac{1}{v'} + \frac{1}{u'}$:
$\frac{1}{-f} = \frac{1}{v'} + \frac{1}{-5f/3}$
$\frac{1}{v'} = \frac{3}{5f} - \frac{1}{f} = \frac{3-5}{5f} = \frac{-2}{5f}$
$v' = -2.5f = -5f/2$.
The length of the image is the distance between the image of the farthest end (at $2f$) and the image of the nearest end (at $2.5f$):
Length $= |v'| - |v| = 2.5f - 2f = 0.5f = f/2$.

(D) Given,focal length of the concave mirror $f = -10\, cm$.
For end $A$ of the rod,the object distance is $u_A = -20\, cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v_A} + \frac{1}{-20} = \frac{1}{-10}$
$\frac{1}{v_A} = -\frac{1}{10} + \frac{1}{20} = -\frac{1}{20}$
$v_A = -20\, cm$.
For end $B$ of the rod,the object distance is $u_B = -(20 + 10) = -30\, cm$.
Using the mirror formula:
$\frac{1}{v_B} + \frac{1}{-30} = \frac{1}{-10}$
$\frac{1}{v_B} = -\frac{1}{10} + \frac{1}{30} = -\frac{2}{30} = -\frac{1}{15}$
$v_B = -15\, cm$.
The length of the image is the distance between the image positions of the two ends:
$L = |v_A - v_B| = |-20 - (-15)| = |-5| = 5\, cm$.

(D) The magnification for a spherical mirror is given by $m = -v/u$.
$1$. For $m = -2$: Since $m$ is negative,the image is real. $A$ real image is always inverted and formed by a concave mirror. Thus,$(1-b, c)$.
$2$. For $m = -1/2$: Since $m$ is negative,the image is real. $A$ real image is formed by a concave mirror. Thus,$(2-b, c)$.
$3$. For $m = +2$: Since $m$ is positive,the image is virtual. $A$ virtual image with magnification greater than $1$ is formed by a concave mirror. Thus,$(3-b, d)$.
$4$. For $m = +1/2$: Since $m$ is positive,the image is virtual. $A$ virtual image with magnification less than $1$ is formed by a convex mirror. Thus,$(4-a, d)$.
Therefore,the correct matching is $(1-b, c), (2-b, c), (3-b, d), (4-a, d)$.

(B) Using the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
For a concave mirror,$f = -15 \, cm$.
Initially,the object distance $u_1 = -40 \, cm$.
$\frac{1}{-15} = \frac{1}{v_1} + \frac{1}{-40} \Rightarrow \frac{1}{v_1} = \frac{1}{40} - \frac{1}{15} = \frac{3 - 8}{120} = \frac{-5}{120} = -\frac{1}{24}$.
So,$v_1 = -24 \, cm$ (image is $24 \, cm$ in front of the mirror).
When the object is displaced by $20 \, cm$ towards the mirror,the new object distance $u_2 = -(40 - 20) = -20 \, cm$.
$\frac{1}{-15} = \frac{1}{v_2} + \frac{1}{-20} \Rightarrow \frac{1}{v_2} = \frac{1}{20} - \frac{1}{15} = \frac{3 - 4}{60} = -\frac{1}{60}$.
So,$v_2 = -60 \, cm$ (image is $60 \, cm$ in front of the mirror).
The displacement of the image is $|v_2 - v_1| = |-60 - (-24)| = |-36| = 36 \, cm$.
Since the magnitude of the image distance increased,the image shifts $36 \, cm$ away from the mirror.

(C) The incident light rays are converging towards a point $12 \, cm$ behind the convex mirror. This point acts as a virtual object for the mirror.
Since the image is formed at the same position as the virtual object,the light rays must be striking the mirror normally (along the radius of curvature).
This implies that the virtual object is located at the center of curvature $(C)$ of the convex mirror.
Therefore,the radius of curvature $R = 12 \, cm$.
The focal length $f$ of the mirror is given by $f = \frac{R}{2}$.
Substituting the value of $R$,we get $f = \frac{12 \, cm}{2} = 6 \, cm$.

(B) Given: Focal length $f = -10 \, cm$,length of the rod $l = 5 \, cm$.
One end of the rod is at $2f = 20 \, cm$ from the mirror. Let this end be $A$. The other end is at $u = 20 - 5 = 15 \, cm$ from the mirror.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
For end $A$ at $u_A = 20 \, cm$,the image $v_A = 20 \, cm$ (at the center of curvature).
For the other end at $u_B = 15 \, cm$:
$\frac{1}{-10} = \frac{1}{v_B} + \frac{1}{-15}$
$\frac{1}{v_B} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} = -\frac{1}{30}$
So,$v_B = -30 \, cm$.
The length of the image is $|v_A - v_B| = |20 - 30| = 10 \, cm$.
Magnification $m = \frac{\text{Length of image}}{\text{Length of object}} = \frac{10 \, cm}{5 \, cm} = 2$.

(B) For a concave mirror,when an object is placed at the focus $(u = -f)$,the image is formed at infinity.
However,for a convex mirror,if an object is placed at a distance equal to the focal length $(u = -f)$,the image is formed behind the mirror at a distance of $f/2$ from the pole.
Since the question specifies a 'spherical mirror' without distinguishing between concave and convex,the image formation depends on the type of mirror.
Therefore,the image may be at infinity (if the mirror is concave) or at another position (if the mirror is convex).
Thus,the correct option is that the image 'may be at infinity'.

(B) For a ray incident parallel to the principal axis to undergo multiple reflections,it must strike the mirror at a point such that after the first reflection,it is directed towards another point on the mirror.
Consider a ray incident at the edge of the mirror. After reflection,the angle of incidence equals the angle of reflection. The normal at any point on the spherical mirror passes through the centre of curvature $C$.
For the ray to hit the other end of the mirror after the first reflection,the reflected ray must make an angle of $90^{\circ}$ with the incident ray (which is parallel to the principal axis).
From the geometry of the reflection,the angle between the normal and the reflected ray is $\theta$. The angle between the incident ray and the normal is also $\theta$.
Thus,the angle between the incident ray and the reflected ray is $2\theta$. For the reflected ray to be perpendicular to the incident ray,we must have $2\theta = 90^{\circ}$.
Therefore,$\theta = 45^{\circ}$.

(A) Let $C$ be the center of curvature and $F$ be the focus of the concave mirror. Let the incident ray strike the mirror at point $A$. The height of the incident ray from the principal axis is $AD = f$,where $f$ is the focal length.
In $\Delta ADC$,where $D$ is the projection of $A$ on the principal axis,$\sin i = \frac{AD}{AC} = \frac{f}{R} = \frac{f}{2f} = \frac{1}{2}$.
Thus,the angle of incidence $i = 30^{\circ}$.
Since the reflected ray $AB$ makes an angle $2i$ with the principal axis,in $\Delta ADB$,$\sin 2i = \frac{AD}{AB} = \frac{f}{AB}$.
Therefore,$AB = \frac{f}{\sin 60^{\circ}} = \frac{f}{\sqrt{3}/2} = \frac{2f}{\sqrt{3}}$.
In $\Delta ABC$,since $CA$ is the normal,$\angle CAB = i = 30^{\circ}$. Also,$\angle ACB = i = 30^{\circ}$ (alternate interior angles). Thus,$\Delta ABC$ is isosceles with $CB = AB = \frac{2f}{\sqrt{3}}$.
The distance of the focus from the center of curvature is $FC = R - f = 2f - f = f$.
The required ratio is $\frac{CB}{FC} = \frac{2f/\sqrt{3}}{f} = \frac{2}{\sqrt{3}}$.

(C) For a concave mirror,the mirror formula is given by $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.
Using sign convention,$u = -x$ and $v = -v_{i}$ (where $v_{i}$ is the image distance,and we are interested in $|v_{i}|$).
Substituting these,we get $\frac{1}{f} = -\frac{1}{x} - \frac{1}{v_{i}}$,which simplifies to $\frac{1}{v_{i}} = -(\frac{1}{f} + \frac{1}{x}) = -(\frac{x+f}{xf})$.
Thus,$|v_{i}| = \frac{xf}{f-x}$.
As $x \to 0$,$|v_{i}| \to 0$.
As $x \to f$,$|v_{i}| \to \infty$.
For $x > f$,the image becomes real and inverted. The mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$. With $u = -x$,we have $\frac{1}{v} = \frac{1}{f} - \frac{1}{x} = \frac{x-f}{xf}$,so $v = \frac{xf}{x-f}$.
Thus,$|v| = \frac{xf}{x-f}$.
As $x \to f^{+}$,$|v| \to \infty$.
As $x \to \infty$,$|v| \to f$.
Comparing these behaviors,the graph in option $C$ correctly shows the asymptotic behavior at $x = f$ and the limits as $x \to 0$ and $x \to \infty$.

(A) $1$. The object $O$ is above the principal axis $AB$,and the image $I$ is below the principal axis $AB$. This indicates that the image is inverted.
$2$. The distance of the object from the principal axis is $d_1$,and the distance of the image from the principal axis is $d_2$. Since $d_2 > d_1$,the image is magnified.
$3$. $A$ convex mirror always forms a virtual,erect,and diminished image. Therefore,the mirror cannot be convex.
$4$. $A$ concave mirror can form a real,inverted,and magnified image when the object is placed between the focus $F$ and the center of curvature $C$.
$5$. For a real object to form a real image,the mirror must be placed on the side of the object and image. Since the light rays must travel from the object to the mirror and reflect back to form the image at $I$,the mirror must be placed to the right of both $O$ and $I$ (or specifically,to the right of $I$ to intercept the rays). Looking at the geometry,the mirror must be placed to the right of $I$ to reflect the rays from $O$ to $I$.

(D) For a real object,a convex mirror always forms a virtual,erect,and diminished image.
$1$. The magnification $m$ for a convex mirror is always $0 < m < 1$,so statement $A$ is correct.
$2$. $A$ plane mirror forms a virtual,erect,and same-sized image,so statement $B$ is correct.
$3$. $A$ concave mirror forms a virtual,erect,and magnified image when the object is placed between the pole and the focus,so statement $C$ is correct.
$4$. $A$ convex mirror can never form a real image for a real object. Therefore,statement $D$ is incorrect.