Mirror Formula and Magnification Questions in English

(A) The magnification $(m)$ produced by a spherical mirror is given by the formula:
$m = -\frac{v}{u}$
Since the image is real and magnified three times,the magnification $m = -3$ (as real images are inverted).
Given,object distance $u = -10 \, cm$ (according to sign convention).
Substituting the values in the formula:
$-3 = -\frac{v}{-10}$
$-3 = \frac{v}{10}$
$v = -3 \times 10 = -30 \, cm$.
The negative sign indicates that the image is formed in front of the mirror at a distance of $30 \, cm$.

(A) Focal length of convex mirror,$f = +15\, cm$.
Object distance,$u = -10\, cm$.
According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values,$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - (\frac{1}{-10}) = \frac{1}{15} + \frac{1}{10}$.
$\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$.
Therefore,$v = +6\, cm$.
The positive value of $v$ indicates that the image is formed $6\, cm$ behind the mirror.
Magnification,$m = -\frac{v}{u} = -(\frac{6}{-10}) = +0.6$.
The positive value of magnification indicates that the image formed is virtual and erect.

(B) The magnification $(m)$ produced by a mirror is defined as the ratio of the height of the image $(H_i)$ to the height of the object $(H_o)$:
$m = \frac{H_i}{H_o}$
Given that the magnification produced by a plane mirror is $+1$:
$1$. The magnitude $1$ indicates that the height of the image is equal to the height of the object $(H_i = H_o)$,meaning the image is of the same size as the object.
$2$. The positive sign $(+)$ indicates that the image formed is virtual and erect.

(N/A) Object distance,$u = -20\, cm$.
Object height,$h = 5\, cm$.
Radius of curvature,$R = 30\, cm$.
Focal length,$f = R/2 = 15\, cm$.
According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - (\frac{1}{-20}) = \frac{1}{15} + \frac{1}{20} = \frac{4+3}{60} = \frac{7}{60}$.
$v = \frac{60}{7} \approx 8.57\, cm$.
The positive value of $v$ indicates that the image is formed behind the mirror.
Magnification,$m = -\frac{v}{u} = -\frac{8.57}{-20} = 0.428$.
The positive value of magnification indicates that the image is virtual and erect.
Height of the image,$h' = m \times h = 0.428 \times 5 = 2.14\, cm$.
Thus,the image is virtual,erect,and $2.14\, cm$ in size,formed $8.57\, cm$ behind the mirror.

(A) Object distance,$u = -27\, cm$.
Object height,$h = 7\, cm$.
Focal length,$f = -18\, cm$.
According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27} = \frac{-3 + 2}{54} = \frac{-1}{54}$.
Thus,$v = -54\, cm$.
The screen should be placed at a distance of $54\, cm$ in front of the mirror.
Magnification,$m = -\frac{v}{u} = -\frac{-54}{-27} = -2$.
The negative sign of magnification indicates that the image is real and inverted.
Magnification,$m = \frac{h'}{h} \implies h' = m \times h = -2 \times 7 = -14\, cm$.
The size of the image is $14\, cm$.

(B) Given: Height of object $(h_o)$ = $10\, mm = 1\, cm$. Height of image $(h_i)$ = $-5\, mm = -0.5\, cm$ (since the image is real and inverted,it is formed in front of the mirror). Image distance $(v)$ = $-30\, cm$.
Magnification $(m)$ = $h_i / h_o = -v / u$.
Substituting the values: $-0.5 / 1 = -(-30) / u$.
$-0.5 = 30 / u$.
$u = 30 / -0.5 = -60\, cm$.
Using the mirror formula: $1/f = 1/v + 1/u$.
$1/f = 1/(-30) + 1/(-60)$.
$1/f = (-2 - 1) / 60 = -3 / 60$.
$1/f = -1 / 20$.
Therefore,$f = -20\, cm$.

(A) rear-view mirror in vehicles is a convex mirror.
Convex mirrors always form virtual,erect,and diminished images of objects placed in front of them.
Since the image is always diminished (smaller than the object),the height of the image $(h')$ is always less than the height of the object $(h)$.
Magnification $(m)$ is defined as the ratio of the height of the image to the height of the object,i.e.,$m = h'/h$.
Because $h' < h$,the magnification $m$ is always less than $1$.

(D) Given: Focal length $f = -20\, cm$ (for a concave mirror),Magnification $m = -1/3$ (for a real and inverted image).
Using the magnification formula $m = -v/u$,we get $-1/3 = -v/u$,which implies $v = u/3$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,substitute $v = u/3$:
$\frac{1}{u/3} + \frac{1}{u} = \frac{1}{-20}$
$\frac{3}{u} + \frac{1}{u} = -\frac{1}{20}$
$\frac{4}{u} = -\frac{1}{20}$
$u = -80\, cm$.
The object is placed at a distance of $80\, cm$ from the mirror.
The image is real and inverted,and the mirror is concave.

(B) Magnification $(m)$ is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
$m = h'/h$.
If the magnification is positive $(m > 0)$,it indicates that the image is formed above the principal axis.
This implies that the image is virtual and erect.
Therefore,the correct option is $B$.

(B) In the context of spherical mirrors, magnification $(m)$ is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
If the magnification is negative $(m < 0)$, it indicates that the height of the image is negative relative to the height of the object.
Since the object is typically placed upright above the principal axis (positive height), a negative image height implies that the image is formed below the principal axis.
Therefore, a negative magnification signifies that the image is $real$ and $inverted$.

(N/A) The magnification $m$ of a spherical mirror is defined as the ratio of the height of the image to the height of the object,which is also given by $m = -v/u$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,we can derive the relationship between magnification,focal length $(f)$,and object distance $(u)$ or image distance $(v)$.
By substituting $v = \frac{fu}{u-f}$ into $m = -v/u$,we get $m = \frac{f}{f-u}$.
Similarly,by substituting $u = \frac{fv}{v-f}$ into $m = -v/u$,we get $m = \frac{f-v}{f}$.
Thus,the relationship is $m = \frac{f-v}{f} = \frac{f}{f-u}$.

(B) The magnification $m$ is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$,i.e.,$m = h'/h$.
Given $m = +1$,this implies $h'/h = +1$,which means $h' = h$.
Since the magnification is positive $(+)$,the image is virtual and erect.
Since the magnitude of magnification is $1$,the size of the image is exactly equal to the size of the object.
Therefore,the image is virtual,erect,and of the same size as the object.

(N/A) Lateral magnification is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
Mathematically,it is expressed as $m = \frac{h'}{h}$.
It indicates how much larger or smaller the image is compared to the object.

(B) The relationship between object distance $(u)$,image distance $(v)$,and focal length $(f)$ for a spherical mirror is known as the mirror formula.
This formula is given by: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Here,$(u)$ is the distance of the object from the pole,$(v)$ is the distance of the image from the pole,and $(f)$ is the focal length of the mirror.

(B) The magnification $(m)$ for any spherical lens is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
Mathematically,this is expressed as $m = h'/h = v/u$,where $v$ is the image distance and $u$ is the object distance.
This formula is universal and applies to both convex and concave lenses.
While the sign conventions for $v$ and $u$ differ depending on the type of lens and the position of the object,the fundamental equation for magnification remains the same.

(B) Given: Magnification $(m)$ = $2$,Size of the object $(O)$ = $1\, m$.
The formula for magnification is $m = \frac{I}{O}$,where $I$ is the size of the image.
Rearranging the formula to find the image size: $I = m \times O$.
Substituting the values: $I = 2 \times 1\, m = 2\, m$.
Therefore,the size of the image is $2\, m$.

(N/A) Linear magnification $(m)$ produced by a spherical mirror is defined as the ratio of the height of the image $(h^{\prime})$ to the height of the object $(h)$.
Mathematically,it is expressed as:
$m = \frac{\text{height of the image}}{\text{height of the object}} = \frac{h^{\prime}}{h}$
Additionally,it is related to the object distance $(u)$ and image distance $(v)$ as:
$m = \frac{h^{\prime}}{h} = -\frac{v}{u}$
Here,$u$ is the distance of the object from the mirror,and $v$ is the distance of the image from the mirror.

(N/A) The mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
For a plane mirror,the focal length $f$ is considered to be infinity $(\infty)$.
Substituting $f = \infty$ into the mirror formula,we get:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{\infty}$
Since $\frac{1}{\infty} = 0$,the equation becomes:
$\frac{1}{v} + \frac{1}{u} = 0$
$\frac{1}{v} = -\frac{1}{u}$
$v = -u$
This result shows that for a plane mirror,the image distance $v$ is equal in magnitude to the object distance $u$ but opposite in sign,which is consistent with the properties of plane mirrors (virtual image formed at the same distance behind the mirror).

(A) Given: Object height $h_{1} = 4.5 \, cm$,Object distance $u = -12 \, cm$,Focal length $f = +15 \, cm$ (for a convex mirror).
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{15} = \frac{1}{v} + \frac{1}{-12}$.
Rearranging: $\frac{1}{v} = \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60} = \frac{3}{20}$.
Thus,$v = \frac{20}{3} \approx +6.67 \, cm$.
The image is formed $6.67 \, cm$ behind the mirror.
Magnification $m = -\frac{v}{u} = -\frac{6.67}{-12} \approx +0.56$.
As the needle is moved farther from the mirror,the object distance $u$ increases. Consequently,the image moves closer to the focus $F$ and its size decreases.

(A) Since the image is formed on a screen,it must be a real image. Only a concave mirror can form a real image.
$(b)$ Given: Object distance $u = -15\, cm$,Image distance $v = -60\, cm$ (as it is in front of the mirror).
Linear magnification $m = \frac{v}{u} = \frac{-60}{-15} = -4$.
$(c)$ The object is at $15\, cm$ from the pole and the image is at $60\, cm$ from the pole on the same side.
Distance between object and image $= |v - u| = |-60 - (-15)| = |-60 + 15| = |-45| = 45\, cm$.
$(d)$ The ray diagram shows the object placed between the focus $F$ and the pole $P$ is incorrect based on the values; however,for a real image,the object must be placed between $C$ and $F$. The provided diagram shows the object between $C$ and $F$ forming a real,inverted,and magnified image beyond $C$.

(A) The mirror formula expressing the relationship between object distance $(u)$,image distance $(v)$,and focal length $(f)$ is given by: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Given that the image is real and magnified by $2$ times,the magnification $(m)$ is $-2$ (since real images are inverted).
The object distance $(u)$ is $-10 \ cm$ (using sign convention).
Using the magnification formula: $m = -\frac{v}{u}$.
Substituting the values: $-2 = -\frac{v}{-10}$.
Solving for $v$: $-2 = \frac{v}{10}$,which gives $v = -20 \ cm$.
Thus,the image is formed at a distance of $20 \ cm$ in front of the mirror.

(A) $(i)$ If the image is real and inverted:
$m = -v/u = -3$,which implies $v = 3u$. Given $f = -30 \, cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{3u} + \frac{1}{u} = \frac{1}{-30} \Rightarrow \frac{1+3}{3u} = \frac{1}{-30} \Rightarrow \frac{4}{3u} = \frac{1}{-30} \Rightarrow u = -40 \, cm$.
Thus,the object is placed at $40 \, cm$ in front of the mirror.
$(ii)$ If the image is virtual and erect:
$m = -v/u = +3$,which implies $v = -3u$. Given $f = -30 \, cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{-3u} + \frac{1}{u} = \frac{1}{-30} \Rightarrow \frac{-1+3}{3u} = \frac{1}{-30} \Rightarrow \frac{2}{3u} = \frac{1}{-30} \Rightarrow u = -20 \, cm$.
Thus,the object is placed at $20 \, cm$ in front of the mirror.

(A) Given: Object height $O = 4.5\, cm$,Object distance $u = -12\, cm$,Focal length $f = +15\, cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} + \frac{1}{-12} = \frac{1}{15}$.
$\frac{1}{v} = \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60}$.
$v = \frac{60}{9} \approx 6.67\, cm \approx 6.7\, cm$.
The positive sign indicates that the image is formed $6.7\, cm$ behind the mirror.
Magnification $m = -\frac{v}{u} = -\frac{6.67}{-12} \approx 0.56$.
Since $m$ is positive and less than $1$,the image is virtual,erect,and diminished.

(N/A) Given: Radius of curvature $R = +3\, m$,Focal length $f = R/2 = +1.5\, m$.
Object distance $u = -5\, m$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{1.5} = \frac{1}{v} + \frac{1}{-5}$.
$\frac{1}{v} = \frac{1}{1.5} + \frac{1}{5} = \frac{5 + 1.5}{7.5} = \frac{6.5}{7.5}$.
$v = \frac{7.5}{6.5} \approx +1.15\, m$.
The image is formed at $1.15\, m$ behind the mirror.
Magnification $m = -\frac{v}{u} = -\frac{1.15}{-5} = +0.23$.
Since $v$ is positive,the image is virtual and erect. Since $m < 1$,the image is diminished (smaller in size) by a factor of $0.23$.

(A) Given:
Object size,$h = +4.0 \, cm$
Object distance,$u = -25.0 \, cm$
Focal length,$f = -15.0 \, cm$
Using the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-25} = \frac{-5 + 3}{75} = \frac{-2}{75}$
$v = -37.5 \, cm$
The negative sign indicates that the image is formed in front of the mirror,so a screen must be placed at $37.5 \, cm$ from the mirror on the same side as the object. The image is real.
Magnification,$m = \frac{h'}{h} = -\frac{v}{u}$
$h' = -\frac{v \cdot h}{u} = -\frac{(-37.5) \cdot (4.0)}{-25.0} = -6.0 \, cm$
The image size is $6.0 \, cm$. The negative sign indicates that the image is inverted.

(A) Given: Object height $h = 3 \, cm$,Object distance $u = -9 \, cm$,Focal length $f = -18 \, cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-18} = \frac{1}{v} + \frac{1}{-9}$.
$\frac{1}{v} = \frac{1}{9} - \frac{1}{18} = \frac{2-1}{18} = \frac{1}{18}$.
Therefore,$v = +18 \, cm$. Since $v$ is positive,the image is formed $18 \, cm$ behind the mirror,which means it is virtual and erect.
Now,magnification $m = \frac{h'}{h} = -\frac{v}{u}$.
$h' = -\frac{v \times h}{u} = -\frac{18 \times 3}{-9} = 6 \, cm$.
Thus,the image is virtual,erect,$18 \, cm$ behind the mirror,and $6 \, cm$ in size.

(A) Given: Object distance $u = -12 \ cm$. Magnification $m = -4$ (since the image is real and inverted).
Using the magnification formula for a spherical mirror: $m = -\frac{v}{u}$.
Substituting the values: $-4 = -\frac{v}{-12}$.
Solving for $v$: $-4 = \frac{v}{12} \implies v = -48 \ cm$.
The negative sign indicates that the image is formed in front of the mirror,which is consistent with a real image. Thus,the distance of the image from the mirror is $48 \ cm$.

(A) Given: Object height $O = 1 \, cm$,Object distance $u = -15 \, cm$,Focal length $f = -10 \, cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = \frac{-3 + 2}{30} = \frac{-1}{30}$.
Thus,$v = -30 \, cm$. The negative sign indicates the image is formed in front of the mirror.
Magnification $m = \frac{I}{O} = -\frac{v}{u}$.
$I = -\frac{v}{u} \times O = -\frac{-30}{-15} \times 1 = -2 \, cm$.
The image is $30 \, cm$ in front of the mirror,$2 \, cm$ tall,and is real and inverted.

(B) For a convex mirror,the magnification $m$ is given by $m = -v/u$. Given that the image size is one-third of the object size,$m = 1/3$.
Since $m = -v/u = 1/3$,we have $u = -3v$.
The mirror formula is $1/f = 1/v + 1/u$.
Given $f = +12 \, cm$ (for a convex mirror),we substitute the values: $1/12 = 1/v + 1/(-3v)$.
$1/12 = (3 - 1) / 3v = 2 / 3v$.
$3v = 24$,which gives $v = +8 \, cm$.
$A$ positive value for $v$ indicates that the image is formed $8 \, cm$ behind the mirror.

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

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

(B) The mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,where $f$ is the focal length,$v$ is the image distance,and $u$ is the object distance.
It relates the focal length of a spherical mirror to the object distance and the image distance.
The relationship between the height of the image $(h')$ and the height of the object $(h)$ is defined by the magnification formula,which is $m = \frac{h'}{h} = -\frac{v}{u}$.
Therefore,the statement is false.

(A) The mirror formula is given by the expression $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,where $f$ is the focal length,$v$ is the image distance,and $u$ is the object distance.
This formula is a universal relationship derived from the laws of reflection and geometry.
It applies to both spherical mirrors,i.e.,concave mirrors and convex mirrors,provided that the sign convention is applied correctly during calculation.

(A) The magnification $(m)$ of a spherical mirror is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
It is also related to the object distance $(u)$ and image distance $(v)$ by the formula: $m = \frac{h'}{h} = -\frac{v}{u}$.
Therefore,the magnification equation establishes a direct relationship between the image distance and the object distance.

(A) The magnification $(m)$ produced by a spherical mirror is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
The formula is given by $m = \frac{h'}{h} = -\frac{v}{u}$,where $v$ is the image distance and $u$ is the object distance.
This formula is derived from the geometry of reflection and is universally applicable to all spherical mirrors,whether they are concave or convex,provided that the sign convention is followed correctly.

(B) The magnification $(m)$ of a mirror is defined as the ratio of the height of the image $(h_i)$ to the height of the object $(h_o)$,expressed as $m = h_i / h_o$.
In a plane mirror,the image formed is always virtual,erect,and of the same size as the object.
Since the height of the image is equal to the height of the object $(h_i = h_o)$,the magnification is $m = h_i / h_o = 1$.
Additionally,the image is erect,so the magnification is positive,resulting in $m = +1$.

(B) For a concave mirror,magnification $m = -v/u$. Given $m = -2$.
Substituting this,$-2 = -v/u$,which implies $v = 2u$.
According to the mirror formula,$1/f = 1/v + 1/u$.
Substituting $v = 2u$,we get $1/f = 1/(2u) + 1/u = 3/(2u)$.
This gives $u = 1.5f$.
Since the focal length $f$ is between the pole and the center of curvature $(2f)$,an object placed at $1.5f$ lies between the principal focus $(F)$ and the center of curvature $(C)$.
Thus,the correct position is between the principal focus and the center of curvature.

(A) The magnification $m$ of a spherical mirror is given by $m = -v/u = h'/h$.
$1.$ Magnification $m = -1$: The negative sign indicates a real and inverted image. The magnitude $1$ indicates the image size is equal to the object size. Thus, $(1-b)$.
$2.$ Magnification $m = +5$: The positive sign indicates a virtual and erect image. The magnitude $5$ $( > 1)$ indicates an enlarged image. Thus, $(2-d)$.
$3.$ Magnification $m = -5$: The negative sign indicates a real and inverted image. The magnitude $5$ $( > 1)$ indicates an enlarged image. Thus, $(3-a)$.
$4.$ Magnification $m = -0.5$: The negative sign indicates a real and inverted image. The magnitude $0.5$ $( < 1)$ indicates a diminished image. Thus, $(4-c)$.
Therefore, the correct matching is $(1-b), (2-d), (3-a), (4-c)$.

(B) For a concave mirror,a real image is always inverted.
According to the sign convention,the height of an inverted image is taken as negative $(h_i < 0)$ and the height of the object is taken as positive $(h_o > 0)$.
Magnification $(m)$ is defined as the ratio of the height of the image to the height of the object: $m = h_i / h_o$.
Since $h_i$ is negative and $h_o$ is positive,the magnification $m$ must be negative $(m < 0)$.

(B) According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Given that the object is at infinity,$u = \infty$.
Substituting this into the formula: $\frac{1}{v} + \frac{1}{\infty} = \frac{1}{f}$.
Since $\frac{1}{\infty} = 0$,we get $\frac{1}{v} + 0 = \frac{1}{f}$.
Therefore,$v = f$.
When an object is placed at infinity in front of a concave mirror,the rays of light are parallel to the principal axis,and the image is formed at the principal focus $(F)$.

(B) The magnification $(m)$ of a mirror is given by the ratio of the height of the image $(h')$ to the height of the object $(h)$,which is also equal to $(-\frac{v}{u})$.
Given $m = -\frac{1}{3}$.
$1$. The negative sign $(-)$ indicates that the image is real and inverted.
$2$. The magnitude of magnification is $\frac{1}{3}$,which is less than $1$,indicating that the image is diminished.
$3$. $A$ plane mirror always produces a virtual,erect image with a magnification of $+1$.
$4$. $A$ convex mirror always produces a virtual,erect,and diminished image with a magnification between $0$ and $+1$.
$5$. $A$ concave mirror can produce real,inverted,and diminished images when the object is placed beyond the center of curvature $(C)$.
Therefore,the mirror must be a concave mirror.

(D) In the context of spherical mirrors,the magnification $m$ is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$,i.e.,$m = h'/h$.
For real images,the image is formed below the principal axis,making the height of the image $(h')$ negative,while the object height $(h)$ is taken as positive.
Therefore,a negative magnification $(m < 0)$ indicates that the image is real and inverted.
Since all real images formed by spherical mirrors are inverted,the correct nature of the image is inverted.

(A) convex mirror always forms a virtual,erect,and diminished image for all positions of the object placed in front of it.
For an erect image,the height of the image $(h')$ and the height of the object $(h)$ are both taken as positive.
Magnification $(m)$ is defined as the ratio of the height of the image to the height of the object: $m = h'/h$.
Since both $h'$ and $h$ are positive for an erect image,the magnification $(m)$ is always positive.
Additionally,for a convex mirror,the image is always diminished,meaning the size of the image is smaller than the size of the object $(h' < h)$,so the magnification is always less than $1$ $(m < 1)$.

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

(A) Magnification $(m)$ produced by a spherical mirror or lens is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
Mathematically,it is expressed as: $m = \frac{h'}{h}$.
Therefore,magnification is the ratio of the height of the image to the height of the object.

(D) Given: Object height $(h_o)$ = $4 \, cm$,Object distance $(u)$ = $-12 \, cm$ (by sign convention).
Since the image is formed behind the mirror,it is a virtual image. Thus,Image distance $(v)$ = $+24 \, cm$.
The magnification $(m)$ of a mirror is given by the formula: $m = \frac{h_i}{h_o} = -\frac{v}{u}$.
Substituting the values: $\frac{h_i}{4} = -\frac{24}{-12}$.
$\frac{h_i}{4} = 2$.
$h_i = 2 \times 4 = 8 \, cm$.
Therefore,the height of the image is $8 \, cm$.

(B) Given: Focal length $(f)$ = $-12 \ cm$ (for a concave mirror),Object distance $(u)$ = $-18 \ cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-12} = \frac{1}{v} + \frac{1}{-18}$.
Rearranging: $\frac{1}{v} = \frac{1}{18} - \frac{1}{12}$.
Calculating the common denominator $(36)$: $\frac{1}{v} = \frac{2 - 3}{36} = -\frac{1}{36}$.
Therefore,$v = -36 \ cm$.
The center of curvature $(C)$ is at $2f = 2 \times 12 = 24 \ cm$.
Since the image distance $(36 \ cm)$ is greater than the center of curvature $(24 \ cm)$,the image is formed beyond the center of curvature.

(D) Given: Focal length $f = -10 \text{ cm}$ (for a concave mirror),Object distance $u = -20 \text{ cm}$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-10} = \frac{1}{v} + \frac{1}{-20}$.
Rearranging: $\frac{1}{v} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = -\frac{1}{20}$.
Therefore,$v = -20 \text{ cm}$.
Since the object is placed at $2f$ $(2 \times 10 \text{ cm} = 20 \text{ cm})$,the image is also formed at $2f$,which is the center of curvature.

(C) Given: Focal length $(f)$ = $-20 \text{ cm}$ (for a concave mirror),Object distance $(u)$ = $-10 \text{ cm}$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-10}$.
Rearranging: $\frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{2-1}{20} = \frac{1}{20}$.
So,$v = +20 \text{ cm}$.
Since the image distance $(v)$ is positive,the image is formed behind the mirror,which means it is virtual.
Magnification $(m)$ = $-\frac{v}{u} = -(\frac{20}{-10}) = +2$.
Since $m$ is positive and greater than $1$,the image is virtual and magnified.

(D) The magnification $m$ is given by the ratio of the height of the image $(h')$ to the height of the object $(h)$,i.e.,$m = h'/h$.
Since the value of $m$ is $+0.5$,the positive sign indicates that the image is virtual and erect.
The magnitude $0.5$ (which is less than $1$) indicates that the image is diminished.
$A$ convex mirror always forms a virtual,erect,and diminished image for all positions of the object.
Therefore,a convex mirror can produce a magnification of $+0.5$.