Spherical Mirror Questions in English

(A) For a concave mirror,the radius of curvature $R$ is equal to $2f$. The object distance $u = -R = -2f$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we substitute $u = -2f$:
$\frac{1}{v} + \frac{1}{-2f} = \frac{1}{-f}$
$\frac{1}{v} = \frac{1}{2f} - \frac{1}{f} = -\frac{1}{2f}$
Thus,$v = -2f = -R$.
The image is formed at the same position as the object (the centre of curvature).
Since the image is formed in front of the mirror,it is real and inverted,and its size is equal to the size of the object.

(D) For any spherical mirror,the line joining the object point and the image point always passes through the centre of curvature $(C)$ of the mirror.
This is because any light ray passing through the centre of curvature strikes the mirror surface normally (at an angle of incidence $i = 0^{\circ}$),and according to the laws of reflection,it retraces its path back.
Since the normal to the mirror at any point on a spherical surface always passes through the centre of curvature,the line connecting the object and its image is always perpendicular to the mirror surface at the point of incidence,regardless of whether the mirror is plane,concave,or convex.

(A) For the first case,magnification $m_1 = -3$ (real image). So,$v_1 = 3u_1$. Let $u_1 = u$,then $v_1 = 3u$.
Using mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{3u} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{4}{3u} = \frac{1}{f} \Rightarrow f = \frac{3u}{4}$.
For the second case,magnification $m_2 = -2$. The object is shifted by $6 \ cm$,so $u_2 = u + 6$. Then $v_2 = 2(u + 6)$.
Using mirror formula $\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f} \Rightarrow \frac{1}{2(u+6)} + \frac{1}{u+6} = \frac{1}{f} \Rightarrow \frac{3}{2(u+6)} = \frac{1}{f}$.
Equating the two expressions for $f$: $\frac{3u}{4} = \frac{2(u+6)}{3} \Rightarrow 9u = 8u + 48 \Rightarrow u = 48 \ cm$.
Then $f = \frac{3(48)}{4} = 36 \ cm$.
Initial screen position $v_1 = 3(48) = 144 \ cm$. Final screen position $v_2 = 2(48+6) = 108 \ cm$.
Shift of screen = $v_1 - v_2 = 144 - 108 = 36 \ cm$.

(D) For a convex mirror,the focal length $f = +\frac{a}{2}$.
The object is placed at a distance $b$ from the focus. Since the object is real,it is placed in front of the mirror. The distance of the object from the pole is $u = -(b + \frac{a}{2})$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} + \frac{1}{-(b + a/2)} = \frac{1}{a/2}$
$\frac{1}{v} = \frac{2}{a} + \frac{1}{b + a/2} = \frac{2}{a} + \frac{2}{2b + a}$
$\frac{1}{v} = \frac{2(2b + a) + 2a}{a(2b + a)} = \frac{4b + 4a}{a(2b + a)}$
$v = \frac{a(2b + a)}{4(b + a)}$
The distance of the image from the focus is $|v - f|$:
$|v - f| = |\frac{2ab + a^2}{4(b + a)} - \frac{a}{2}| = |\frac{2ab + a^2 - 2a(b + a)}{4(b + a)}|$
$= |\frac{2ab + a^2 - 2ab - 2a^2}{4(b + a)}| = |\frac{-a^2}{4(b + a)}| = \frac{a^2}{4(b + a)}$
Since this result is not among the options,the correct answer is $D$.

(B) When a concave mirror forms an image of a distant object like the Sun,every part of the mirror contributes to the formation of the complete image.
If the lower half of the mirror is covered,the light rays from the Sun that would have reflected off that portion are blocked.
However,the remaining upper half of the mirror still receives light from all parts of the Sun and focuses it to form the complete image.
Since the total amount of light reaching the screen is reduced,the intensity (brightness) of the image decreases,but the image remains complete.

(A) For a concave mirror,the mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Differentiating with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$.
Thus,the velocity of the image $v_i = \frac{dv}{dt} = -(\frac{v}{u})^2 \frac{du}{dt}$.
Here,the object is between the Pole $(P)$ and Focus $(F)$,so $u < f$. The image formed is virtual,erect,and magnified,located behind the mirror ($v$ is positive).
As the object moves away from the mirror,$u$ increases (becomes less negative),and the image also moves away from the mirror ($v$ increases).
The magnification $m = \frac{v}{|u|} = \frac{f}{f-u}$. As $u$ increases towards $f$,the magnitude of $m$ increases rapidly.
Since $v_i = -m^2 v_o$,and $m$ increases as the object moves away from the mirror,the magnitude of the image velocity $|v_i| = m^2 |v_o|$ increases.
Therefore,the image moves away from the mirror with an increasing magnitude of velocity.

(A) For a concave mirror, the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
For a concave mirror, $f < 0$. Let $f = -|f|$.
If the object is virtual, $u > 0$. Let $u = |u|$.
Then $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{|f|} - \frac{1}{|u|} = -(\frac{1}{|f|} + \frac{1}{|u|})$.
Since $v$ is negative, the image formed is always real for a virtual object.
Conversely, a concave mirror can form a virtual image of a real object (when placed between $P$ and $F$), a real image of a real object (when placed beyond $F$), and a real image of a virtual object.
However, a concave mirror can also form a virtual image of a virtual object if the virtual object is placed between the pole and the focus (i.e., $0 < u < |f|$).
Actually, the statement "$A$ concave mirror cannot form a virtual image of a virtual object" is incorrect because it can form one if the virtual object is placed between the pole and the focus.
Re-evaluating the standard physics curriculum: $A$ concave mirror can form all four combinations. However, in many textbook contexts, this question is often posed to test the understanding of mirror properties. Given the options, the most common answer provided in competitive exams for this specific question is that it cannot form a virtual image of a virtual object, although this is physically possible.

(A) The object $AB$ is placed such that point $A$ is at the centre of curvature $(C = 2f)$ of the concave mirror.
Since $A$ is at the centre of curvature, its image $A'$ will also be formed at the centre of curvature $A' = A$.
Point $B$ is placed at a distance greater than $2f$ from the mirror. For an object placed beyond $C$, the image is formed between $C$ and $f$, is real, inverted, and diminished.
Since $B$ is at an angle of $45^{\circ}$ with the principal axis, the image $B'$ will also be formed below the principal axis.
As the object $AB$ is tilted away from the mirror, the image $B'$ will be formed such that it is inverted and points towards the principal axis.
Comparing this with the given options, option $A$ correctly represents the inverted image $A'B'$ where $A'$ coincides with $A$ and $B'$ is formed between $C$ and $f$ pointing downwards.

(D) In a reflecting telescope,a spherical mirror suffers from spherical aberration,which causes parallel rays incident on the mirror to focus at different points rather than a single point,resulting in a blurred image.
By using a parabolic mirror,all parallel rays incident on the mirror are reflected to a single point (the focus) regardless of their distance from the principal axis.
Therefore,the primary purpose of using a parabolic mirror is to eliminate spherical aberration and produce a sharp,clear image.

(B) Let the incident vector be $\vec{v}_i = 3\hat i + 4\hat j + 12\hat k$. The normal vector is $\vec{n} = 3\hat i + 4\hat j$.
First,find the unit normal vector $\hat{n} = \frac{3\hat i + 4\hat j}{\sqrt{3^2 + 4^2}} = \frac{3\hat i + 4\hat j}{5}$.
The incident ray can be decomposed into components parallel and perpendicular to the normal.
The component of $\vec{v}_i$ parallel to $\hat{n}$ is $\vec{v}_{i\parallel} = (\vec{v}_i \cdot \hat{n})\hat{n} = \left( \frac{3(3) + 4(4) + 12(0)}{5} \right) \hat{n} = \frac{25}{5} \hat{n} = 5 \hat{n} = 3\hat i + 4\hat j$.
The component of $\vec{v}_i$ perpendicular to $\hat{n}$ is $\vec{v}_{i\perp} = \vec{v}_i - \vec{v}_{i\parallel} = (3\hat i + 4\hat j + 12\hat k) - (3\hat i + 4\hat j) = 12\hat k$.
Upon reflection,the parallel component reverses direction,while the perpendicular component remains unchanged.
Thus,the reflected vector $\vec{v}_r = -\vec{v}_{i\parallel} + \vec{v}_{i\perp} = -(3\hat i + 4\hat j) + 12\hat k = -3\hat i - 4\hat j + 12\hat k$.
The unit vector along the reflected ray is $\hat{v}_r = \frac{-3\hat i - 4\hat j + 12\hat k}{|\vec{v}_r|} = \frac{-3\hat i - 4\hat j + 12\hat k}{\sqrt{(-3)^2 + (-4)^2 + 12^2}} = \frac{-3\hat i - 4\hat j + 12\hat k}{\sqrt{9 + 16 + 144}} = \frac{1}{13}(-3\hat i - 4\hat j + 12\hat k)$.

(D) For a concave mirror,the magnification $m$ is given by $m = \frac{f}{f-u}$.
For the first position,$u_1 = -25 \ cm$,so $m_1 = \frac{f}{f - (-25)} = \frac{f}{f+25}$.
For the second position,the object is moved $15 \ cm$ farther,so $u_2 = -(25 + 15) = -40 \ cm$. Thus,$m_2 = \frac{f}{f - (-40)} = \frac{f}{f+40}$.
Given $\frac{m_1}{m_2} = 4$,we substitute the expressions: $\frac{f/(f+25)}{f/(f+40)} = 4$.
This simplifies to $\frac{f+40}{f+25} = 4$.
$f + 40 = 4(f + 25) \implies f + 40 = 4f + 100$.
$3f = -60 \implies f = -20 \ cm$.
The focal length of a concave mirror is $20 \ cm$ (magnitude).

(C) For a concave mirror,magnification $m = \frac{f}{f-u}$. Given $m = 1.8$ and $u = -10 \ cm$ (since the object is in front of the mirror).
$1.8 = \frac{f}{f - (-10)} = \frac{f}{f + 10}$.
$1.8(f + 10) = f \Rightarrow 1.8f + 18 = f \Rightarrow 0.8f = -18 \Rightarrow f = -22.5 \ cm$.
Since $f$ is negative,the mirror is concave with focal length $22.5 \ cm$ and radius of curvature $R = 2f = 45 \ cm$.
In the second case,the object distance $u = -50 \ cm$.
Since $|u| > |R|$ (i.e.,$50 \ cm > 45 \ cm$),the object is placed beyond the center of curvature $C$.
For a concave mirror,when the object is placed beyond $C$,the image formed is real,inverted,and diminished.

(A) The lateral magnification $m$ is given by $m = \frac{h_i}{h_o} = -2$.
Since $m$ is negative,the image is inverted with respect to the object.
For a virtual object (where the object distance $u > 0$),if the magnification is negative,the image formed is real (where the image distance $v < 0$).
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,for a virtual object and a real image,the mirror must be a convex mirror.
Specifically,for a convex mirror,the focal length $f$ is positive.
Thus,the correct configuration is a convex mirror forming a real image.

(A) For a concave mirror,the mirror formula is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Given $u = -10 \ cm$ and $v = -20 \ cm$,we find the focal length $f$:
$\frac{1}{f} = \frac{1}{-20} + \frac{1}{-10} = \frac{-1-2}{20} = -\frac{3}{20} \implies f = -\frac{20}{3} \ cm$.
Differentiating the mirror formula with respect to $v$ and $u$:
$-\frac{1}{v^2} dv - \frac{1}{u^2} du = 0 \implies \frac{dv}{du} = -\frac{v^2}{u^2}$.
Here,$du = -0.1 \ cm$ (as the object moves towards the mirror).
Substituting the values:
$dv = -\left(\frac{-20}{-10}\right)^2 \times (-0.1) = -(2)^2 \times (-0.1) = -4 \times (-0.1) = 0.4 \ cm$.
The positive sign indicates that the image shifts away from the mirror.

(C) For a concave mirror, the magnification $m$ is given by $m = -\frac{v}{u}$.
Since the image is real and $n$ times the size of the object, the magnification is $m = -n$.
Using the mirror formula $m = \frac{f}{f-u}$, we substitute the values:
$-n = \frac{f}{f-u}$
$-n(f-u) = f$
$-nf + nu = f$
$nu = f + nf$
$nu = f(n+1)$
$u = \frac{(n+1)f}{n}$
Thus, the magnitude of the object distance is $\left( \frac{n+1}{n} \right)f$.

(D) For a convex mirror,the magnification $m$ is given by $m = \frac{f}{f-u}$. Since for a real object,$u$ is negative (let $u = -|u|$),the magnification becomes $m = \frac{f}{f+|u|}$. Because $f$ and $|u|$ are positive,the denominator $(f+|u|)$ is always greater than the numerator $f$. Therefore,the magnification $m$ is always less than $1$ $(m < 1)$. This implies that a convex mirror always forms a diminished image of a real object. Thus,it can never form an image larger than the object.

(B) For a mirror,the nature of the image formed depends on the nature of the object and the mirror type.
$1$. If the object is real $(R.O.)$,the image is real $(R.I.)$ if it is inverted,and virtual $(V.I.)$ if it is erect.
$2$. If the object is virtual $(V.O.)$,the image is real $(R.I.)$ if it is erect,and virtual $(V.I.)$ if it is inverted.
Specifically:
- $R.O. \to R.I.$ (Inverted)
- $V.O. \to V.I.$ (Inverted)
- $R.O. \to V.I.$ (Erect)
- $V.O. \to R.I.$ (Erect)
Thus,the image is erect if the object is real and the image is virtual (case $c$),or if the object is virtual and the image is real (case $d$).
Therefore,the correct option is $(b)$.

(B) For a convex mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Given the reflecting surface is towards the negative $x-$ axis,the focal length is positive,so $f = +f_0$ (where $f_0 > 0$).
For real objects,the object distance $u$ is negative,so let $u = -x$ where $x > 0$.
The mirror formula becomes $\frac{1}{v} + \frac{1}{-x} = \frac{1}{f_0} \Rightarrow \frac{1}{v} = \frac{1}{f_0} + \frac{1}{x} = \frac{x + f_0}{x f_0}$.
Thus,$v = \frac{x f_0}{x + f_0}$.
As $x \to 0$,$v \to 0$.
As $x \to \infty$,$v \to f_0$.
Since $v$ is always positive and less than $f_0$,the graph represents a curve starting from the origin $(0,0)$ and asymptotically approaching $v = f_0$ as $u$ becomes more negative. This corresponds to the curve shown in option $B$.

(C) For a concave mirror, the magnification $m$ can be either $+2$ (virtual image) or $-2$ (real image).
Given focal length $f = -20\,cm$.
The magnification formula is $m = \frac{f}{f - u}$.
Case $1$: For a virtual image, $m = +2$.
$+2 = \frac{-20}{-20 - u_1} \implies -40 - 2u_1 = -20 \implies 2u_1 = -20 \implies u_1 = -10\,cm$.
Case $2$: For a real image, $m = -2$.
$-2 = \frac{-20}{-20 - u_2} \implies 40 + 2u_2 = -20 \implies 2u_2 = -60 \implies u_2 = -30\,cm$.
Thus, the object can be placed at $10\,cm$ or $30\,cm$ in front of the mirror.
Therefore, the correct option is $(C)$.

(C) Given: Focal length $f = -20 \, cm$,object distance $u = -30 \, cm$,and object length $L_o = 2 \, mm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-20} - \frac{1}{-30} = \frac{-3 + 2}{60} = -\frac{1}{60}$.
So,$v = -60 \, cm$.
The transverse magnification is $m = -\frac{v}{u} = -\frac{-60}{-30} = -2$.
The longitudinal magnification $m_L$ for a small object is given by $m_L = -m^2$.
$m_L = -(-2)^2 = -4$.
The length of the image $L_I = |m_L| \times L_o = |-4| \times 2 \, mm = 8 \, mm$.

(B) For a convex mirror,the focal length $f = +10\,cm$. The object $AB$ has length $5\,cm$. Point $B$ is at $u_B = -20\,cm$. Since $AB = 5\,cm$,point $A$ is at $u_A = -20 - 5 = -25\,cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $v = \frac{uf}{u-f}$.
For point $A$ $(u_A = -25\,cm)$:
$v_A = \frac{(-25) \times 10}{-25 - 10} = \frac{-250}{-35} = \frac{50}{7}\,cm$.
For point $B$ $(u_B = -20\,cm)$:
$v_B = \frac{(-20) \times 10}{-20 - 10} = \frac{-200}{-30} = \frac{20}{3}\,cm$.
The size of the image is the difference between the image positions: $|v_A - v_B| = |\frac{50}{7} - \frac{20}{3}| = |\frac{150 - 140}{21}| = \frac{10}{21}\,cm$.

(A) For a convex mirror,the magnification $m$ is given by $m = \frac{f}{f-u}$.
Given that the image size is $\frac{1}{n}$ times the object size,the magnification $m = \frac{1}{n}$.
Substituting this into the formula: $\frac{1}{n} = \frac{f}{f-u}$.
Cross-multiplying gives: $f - u = nf$.
Rearranging for $u$: $-u = nf - f = (n-1)f$.
Therefore,the distance of the object from the mirror is $|u| = (n-1)f$.

(A) Given: Object side $s = 3\, cm$,Object area $A_0 = s^2 = 9\, cm^2$. Object distance $u = -25\, cm$,Focal length $f = -10\, cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{-10} - \frac{1}{-25} = \frac{-5 + 2}{50} = \frac{-3}{50}$.
So,$v = -\frac{50}{3}\, cm$.
Linear magnification $m = -\frac{v}{u} = -\left(\frac{-50/3}{-25}\right) = -\frac{2}{3}$.
Areal magnification $m_A = m^2 = \left(-\frac{2}{3}\right)^2 = \frac{4}{9}$.
Image area $A_I = m_A \times A_0 = \frac{4}{9} \times 9 = 4\, cm^2$.

(C) concave mirror is used as a shaving mirror to obtain a magnified,virtual image.
Given:
Object distance,$u = -10\,cm$ (placed in front of the mirror).
The image is virtual and formed at the closest comfortable distance of distinct vision,so $v = -25\,cm$.
Using the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
Substituting the values:
$\frac{1}{f} = \frac{1}{-25} + \frac{1}{-10} = \frac{-2 - 5}{50} = \frac{-7}{50}$
$f = -\frac{50}{7}\,cm$
The radius of curvature $R$ is given by $R = 2f$:
$R = 2 \times (-\frac{50}{7}) = -\frac{100}{7} \approx -14.28\,cm$.
The magnitude of the radius of curvature is $14.28\,cm$.

(B) First,we find the image formed by the convex lens using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Given $u = -30\, cm$ and $f = +20\, cm$,we have $\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$.
Thus,$v = +60\, cm$ to the right of the lens.
Since the image position does not change when the mirror is removed,the light rays must be striking the mirror normally,meaning the image formed by the lens is at the center of curvature $(C)$ of the concave mirror.
The distance between the lens and the mirror is $80\, cm$. The image is at $60\, cm$ from the lens,so the distance of the image from the mirror is $80 - 60 = 20\, cm$.
Therefore,the radius of curvature $R = 20\, cm$,which implies the focal length $f_m = \frac{R}{2} = 10\, cm$.
$A$ concave mirror produces a virtual image only when the object is placed between the pole $(P)$ and the focus $(F)$.
Thus,the maximum distance of the object from the mirror for which a virtual image is formed is equal to the focal length,which is $10\, cm$.

(D) For a concave mirror,the focal length $f = -0.4\,m = -40\,cm$.
For an upright (virtual) image,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: $5 = \frac{-40}{-40 - u}$.
Cross-multiplying: $5(-40 - u) = -40$.
$-200 - 5u = -40$.
$-5u = 160$.
$u = -32\,cm$.
Converting back to meters: $u = -0.32\,m$.
The distance at which you hold the mirror is the magnitude of $u$,which is $0.32\,m$.

(B) Given: Object height $h_o = 1 \, cm$,Image height $h_i = 3 \, cm$,Object distance $u = -4 \, cm$.
Since the image is upright,the magnification $m = \frac{h_i}{h_o} = \frac{+3}{+1} = +3$.
Using the magnification formula for a spherical mirror: $m = \frac{f}{f - u}$.
Substituting the values: $3 = \frac{f}{f - (-4)} = \frac{f}{f + 4}$.
$3(f + 4) = f \Rightarrow 3f + 12 = f \Rightarrow 2f = -12 \Rightarrow f = -6 \, cm$.
$A$ negative focal length indicates a concave mirror.
The radius of curvature $R = 2|f| = 2 \times 6 = 12 \, cm$.
Thus,a concave mirror of radius of curvature $12 \, cm$ is required.

(B) Given: Focal length $f = -40 \, cm$ (concave mirror),object distance $u = -60 \, cm$,and object height $h_o = 6 \, cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\frac{1}{-40} = \frac{1}{v} + \frac{1}{-60}$
$\frac{1}{v} = \frac{1}{60} - \frac{1}{40} = \frac{2 - 3}{120} = -\frac{1}{120}$
So,$v = -120 \, cm$.
The magnification $m$ is given by $m = -\frac{v}{u} = -\frac{-120}{-60} = -2$.
The height of the image $h_i = m \times h_o = -2 \times 6 \, cm = -12 \, cm$.
Since the image is formed at $v = -120 \, cm$ and its height is $-12 \, cm$,the coordinates of the image are $(-120 \, cm, -12 \, cm)$.

(A) For a convex mirror,the radius of curvature $R = 10\,cm$. The focal length $f = R/2 = 5\,cm$.
For a real object placed at a distance $u$ (where $u < 0$),the image distance $v$ is given by the mirror formula: $1/v + 1/u = 1/f$.
Since $f = 5\,cm$,we have $1/v = 1/5 - 1/u$.
As the object distance $u$ varies from $-\infty$ to $0$,the image distance $v$ varies from $0$ to $f$.
Specifically,as $u \to -\infty$,$v \to 0$,and as $u \to 0$,$v \to 0$. The maximum value of $v$ occurs as the object approaches the pole,where $v$ approaches $f = 5\,cm$.
Therefore,the image is always formed between the pole $P$ and the focus $F$,and its distance from the mirror cannot exceed the focal length $5\,cm$.

(A) For a convex mirror,the focal length is $f = +f$ and the object distance is $u = -f/3$.
Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} - \frac{1}{-f/3} = \frac{1}{f} + \frac{3}{f} = \frac{4}{f}$.
Thus,$v = f/4$.
The transverse magnification $m_T$ is given by $m_T = -v/u = -(f/4) / (-f/3) = 3/4$.
The longitudinal magnification $m_L$ for a small object is given by $m_L = -m_T^2$ (or $|m_L| = m_T^2$).
The ratio of transverse magnification to longitudinal magnification is $\frac{m_T}{|m_L|} = \frac{m_T}{m_T^2} = \frac{1}{m_T}$.
Substituting the value of $m_T$,we get $\frac{1}{3/4} = 4/3$.

(D) Using the mirror formula:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
Given:
Object distance $u = -30 \, cm$ (by sign convention)
Focal length $f = +30 \, cm$ (for a convex mirror)
Substituting the values:
$\frac{1}{v} + \frac{1}{-30} = \frac{1}{30}$
$\frac{1}{v} = \frac{1}{30} + \frac{1}{30}$
$\frac{1}{v} = \frac{2}{30}$
$\frac{1}{v} = \frac{1}{15}$
$v = +15 \, cm$
Since the value of $v$ is positive,the image is formed at $15 \, cm$ behind the mirror.

(D) convex mirror is a diverging mirror. For any real object placed in front of a convex mirror,the light rays diverge after reflection. The reflected rays appear to originate from a point behind the mirror. Consequently,the image formed by a convex mirror is always erect,diminished,and virtual,regardless of the object distance $u$ (where $u > 0$).

(A) The lateral magnification $m$ is defined as the ratio of the height of the image $h_i$ to the height of the object $h_o$,i.e.,$m = \frac{h_i}{h_o}$.
Given the lateral magnification $m = 0.4$.
For a two-dimensional object placed perpendicular to the principal axis,the area magnification $m_A$ is the square of the lateral magnification.
$m_A = \frac{A_i}{A_o} = m^2$.
Given the area of the object $A_o = 100\,cm^2$.
Substituting the values: $A_i = A_o \times m^2 = 100 \times (0.4)^2$.
$A_i = 100 \times 0.16 = 16\,cm^2$.
Therefore,the area of the image is $16\,cm^2$.

(C) The height of the object is $h_o = 3 \, cm$ and the height of the real image formed on the wall is $h_i = -9 \, cm$ (since it is inverted).
Magnification $m = \frac{h_i}{h_o} = \frac{-9}{3} = -3$.
We know that $m = -\frac{v}{u}$,so $-3 = -\frac{v}{u} \implies v = 3u$.
From the geometry of the setup,the distance of the image from the mirror is $v = x$,and the distance of the object from the mirror is $u = x - 300$.
Substituting these into the magnification equation: $x = 3(x - 300)$.
$x = 3x - 900$.
$2x = 900$.
$x = 450 \, cm$.

(A) Given,object distance $u = -20\,cm$.
Since the image is real,the magnification $m$ must be negative. Given $|m| = 3$,so $m = -3$.
Using the magnification formula $m = -v/u$,we have $-3 = -v / (-20)$,which gives $v = -60\,cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,we substitute the values:
$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = \frac{-4}{60} = \frac{-1}{15}$.
Therefore,$f = -15\,cm$.
The magnitude of the focal length is $15\,cm$.

(D) The incident ray $XY$ is directed towards the pole $P$ of the concave mirror,but it strikes the mirror at point $Y$ between the focus $F$ and the pole $P$.
According to the laws of reflection,the angle of incidence equals the angle of reflection with respect to the normal at the point of incidence.
For a concave mirror,if an object is placed between the focus $F$ and the pole $P$,the image formed is virtual,erect,and magnified.
This implies that the reflected rays must appear to diverge from a point behind the mirror.
Looking at the geometry of the rays,ray $4$ is the only ray that is diverging away from the principal axis in a manner consistent with the reflection of a ray originating from an object placed between $F$ and $P$.
Therefore,ray $4$ is the correct reflected ray.

(B) Since the image is formed on a screen, it must be a real image.
For a real image, the magnification $m$ is negative.
Given that the image is $2$ times magnified, the magnitude of magnification is $|m| = 2$.
Therefore, $m = -2$.
A convex mirror always forms a virtual, diminished image, so it cannot form an image on a screen.
A concave mirror can form a real, magnified image when the object is placed between $F$ and $2F$.
Thus, the mirror is concave and $m = -2$.

(B) Given: Magnification $m = -3$ (since the image is real,it must be inverted).
Object distance $u = -20 \ cm$.
Using the magnification formula: $m = -v/u$.
$-3 = -v / (-20) \implies v = -60 \ cm$.
Using the mirror formula: $1/v + 1/u = 1/f$.
$1/(-60) + 1/(-20) = 1/f$.
$(-1 - 3) / 60 = 1/f$.
$-4 / 60 = 1/f$.
$f = -15 \ cm$.

(C) For an erect image,the magnification $m = +3$. Since the image is erect,the mirror must be a concave mirror forming a virtual image.
Given $m = -v/u = 3$,so $v = -3u$.
Let the object distance be $u = -x$ (where $x > 0$). Then the image distance $v = 3x$.
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 $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{60} + \frac{1}{-20} = \frac{1}{f}$
$\frac{1 - 3}{60} = \frac{1}{f} \implies \frac{-2}{60} = \frac{1}{f}$
$f = -30 \, cm$.
The magnitude of the focal length is $30 \, cm$.

(A) For a concave mirror,the magnification $m = -v/u$. Given $|m| = 2$,so $v/u = 2$ or $v/u = -2$.
Case $1$: $v = 2u$. Using the mirror formula $1/v + 1/u = 1/f$ with $f = -20\,cm$:
$1/(2u) + 1/u = -1/20 \implies 3/(2u) = -1/20 \implies u_1 = -30\,cm$.
Case $2$: $v = -2u$. Using the mirror formula:
$1/(-2u) + 1/u = -1/20 \implies 1/(2u) = -1/20 \implies u_2 = -10\,cm$.
The distance between the two positions is $|u_1 - u_2| = |-30 - (-10)| = |-20| = 20\,cm$.

(C) Given: Radius of curvature $R = 35.0 \, cm$. The focal length $f = R / 2 = 35.0 / 2 = 17.5 \, cm$. Since it is a concave mirror,$f = -17.5 \, cm$.
For an upright (virtual) image,the magnification $m = +2.5$.
The magnification formula for a mirror is $m = -v / u$,so $v = -m \cdot u = -2.5 \cdot u$.
Using the mirror formula: $1/f = 1/v + 1/u$.
Substituting the values: $1 / (-17.5) = 1 / (-2.5 \cdot u) + 1 / u$.
$1 / (-17.5) = (-1 + 2.5) / (2.5 \cdot u) = 1.5 / (2.5 \cdot u)$.
$2.5 \cdot u = -17.5 \cdot 1.5$.
$u = -(17.5 \cdot 1.5) / 2.5 = -7 \cdot 1.5 = -10.5 \, cm$.
The distance of the mirror from the face is $|u| = 10.5 \, cm$.

(A) In the first case,the image is virtual and magnified,so magnification $m = +2$. Let the object distance be $u_1 = -x$.
Using the magnification formula $m = -v/u$,we get $2 = -v_1 / (-x)$,which implies $v_1 = 2x$.
Using the mirror formula $1/v + 1/u = 1/f$:
$1/(2x) - 1/x = -1/f$
$-1/(2x) = -1/f$,so $x = f/2$.
In the second case,the image is real and magnified,so magnification $m = -2$. Let the object distance be $u_2 = -y$.
Using $m = -v/u$,we get $-2 = -v_2 / (-y)$,which implies $v_2 = -2y$.
Using the mirror formula $1/v + 1/u = 1/f$:
$1/(-2y) - 1/y = -1/f$
$-3/(2y) = -1/f$,so $y = 3f/2$.
The distance the object must be moved is $\Delta d = y - x = 3f/2 - f/2 = f$.

(C) 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 mirror aperture is small compared to the radius of curvature. Thus,the Assertion is correct.
The laws of reflection (angle of incidence equals angle of reflection) are universal and hold true for any reflecting surface,whether plane or spherical,regardless of size. Therefore,the Reason is incorrect.

(C) For a convex mirror,the focal length $f$ is taken as positive $(f = +26 \ cm)$. The object distance $u$ is always taken as negative $(u = -26 \ cm)$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} + \frac{1}{-26} = \frac{1}{26}$.
$\frac{1}{v} = \frac{1}{26} + \frac{1}{26} = \frac{2}{26} = \frac{1}{13}$.
Thus,$v = +13 \ cm$.
The image is formed at $13 \ cm$ behind the mirror,not at infinity. Therefore,the Assertion is correct.
The Reason states that the equation gives $v = \infty$,which is mathematically incorrect based on the sign convention. Thus,the Reason is incorrect.

(C) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Using the sign convention,$u = -x$,where $x$ is the distance of the object from the mirror $(x > 0)$.
Thus,$-\frac{1}{v} - \frac{1}{x} = -\frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{x} = \frac{x-f}{fx} \implies v = \frac{fx}{x-f}$.
The linear magnification $m$ is given by $m = -\frac{v}{u} = -\frac{fx/(x-f)}{-x} = \frac{f}{x-f}$.
The magnitude of linear magnification is $|m| = \left| \frac{f}{x-f} \right|$.
As the object moves away from the focal point $(x = f)$ towards infinity $(x \to \infty)$:
$1$. At $x = f$,$|m| \to \infty$.
$2$. At $x = 2f$,$|m| = |f / (2f - f)| = 1$.
$3$. As $x \to \infty$,$|m| \to 0$.
Therefore,the graph must show $|m| \to \infty$ at $x = f$,$|m| = 1$ at $x = 2f$,and $|m| \to 0$ as $x \to \infty$. This corresponds to the graph shown in option $C$.

(N/A) You may think that the image will now show only half of the object,but taking the laws of reflection to be true for all points of the remaining part of the mirror,the image will be that of the whole object.
However,as the area of the reflecting surface has been reduced,the intensity of the image will be low (in this case,half).

(N/A) The ray diagram for the formation of the image of the phone is shown in the figure.
Since the mobile phone is placed along the principal axis,different parts of the phone are at different distances from the pole of the mirror.
The magnification $m$ for a spherical mirror is given by $m = -v/u$.
As the object distance $u$ varies for different parts of the phone,the image distance $v$ and consequently the magnification $m$ also vary for different parts.
This leads to non-uniform magnification,causing the image to be distorted.
Yes,the distortion of the image depends on the location of the phone with respect to the mirror. If the phone is moved closer to or further from the mirror,the range of $u$ changes,which changes the range of $v$ and the degree of distortion.

(N/A) The focal length $f = -15 / 2 \; cm = -7.5 \; cm$.
$(i)$ The object distance $u = -10 \; cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{7.5} = \frac{7.5 - 10}{75} = \frac{-2.5}{75} = -\frac{1}{30}$
So,$v = -30 \; cm$.
The image is $30 \; cm$ from the mirror on the same side as the object. Magnification $m = -\frac{v}{u} = -\frac{(-30)}{(-10)} = -3$.
The image is magnified,real,and inverted.
$(ii)$ The object distance $u = -5 \; cm$.
$\frac{1}{v} + \frac{1}{-5} = \frac{1}{-7.5}$
$\frac{1}{v} = \frac{1}{5} - \frac{1}{7.5} = \frac{7.5 - 5}{37.5} = \frac{2.5}{37.5} = \frac{1}{15}$
So,$v = 15 \; cm$.
The image is formed at $15 \; cm$ behind the mirror. Magnification $m = -\frac{v}{u} = -\frac{15}{(-5)} = 3$.
The image is magnified,virtual,and erect.

(N/A) The mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,which implies $v = \frac{fu}{u-f}$.
For a convex mirror,$R = 2 \; m$,so the focal length $f = \frac{R}{2} = 1 \; m$.
Since the jogger is moving at a constant speed of $5 \; m s^{-1}$,we calculate the image position $v_1$ at distance $u$ and $v_2$ at distance $u' = u + 5$ (since the jogger moves $5 \; m$ towards the mirror in $1 \; s$). The speed of the image is $|v_1 - v_2| / 1 \; s$.
$(a)$ For $u = -39 \; m$,$v_1 = \frac{1(-39)}{-39-1} = \frac{39}{40} \; m$. For $u' = -34 \; m$,$v_2 = \frac{1(-34)}{-34-1} = \frac{34}{35} \; m$. Speed $= |\frac{39}{40} - \frac{34}{35}| = |\frac{1365 - 1360}{1400}| = \frac{5}{1400} = \frac{1}{280} \; m s^{-1}$.
$(b)$ For $u = -29 \; m$,$v_1 = \frac{29}{30} \; m$. For $u' = -24 \; m$,$v_2 = \frac{24}{25} \; m$. Speed $= |\frac{29}{30} - \frac{24}{25}| = |\frac{145 - 144}{150}| = \frac{1}{150} \; m s^{-1}$.
$(c)$ For $u = -19 \; m$,$v_1 = \frac{19}{20} \; m$. For $u' = -14 \; m$,$v_2 = \frac{14}{15} \; m$. Speed $= |\frac{19}{20} - \frac{14}{15}| = |\frac{57 - 56}{60}| = \frac{1}{60} \; m s^{-1}$.
$(d)$ For $u = -9 \; m$,$v_1 = \frac{9}{10} \; m$. For $u' = -4 \; m$,$v_2 = \frac{4}{5} \; m$. Speed $= |\frac{9}{10} - \frac{4}{5}| = |\frac{9 - 8}{10}| = \frac{1}{10} \; m s^{-1}$.

(D) Given: Size of the candle,$h = 2.5\; cm$. Object distance,$u = -27\; cm$. Radius of curvature,$R = -36\; cm$.
Focal length,$f = R/2 = -18\; cm$.
Using 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 $54\; cm$ in front of the mirror.
Magnification $m = \frac{h'}{h} = -\frac{v}{u}$.
$h' = -\frac{v}{u} \times h = -\left(\frac{-54}{-27}\right) \times 2.5 = -5\; cm$.
The image is real,inverted,and $5\; cm$ in size.
If the candle is moved closer to the mirror (i.e.,$u$ decreases),the image distance $v$ increases,so the screen must be moved away from the mirror.