Spherical Mirror Questions in English

(B) The magnification $m$ is given by the ratio of the height of the image $(h_i)$ to the height of the object $(h_o)$.
Here,$h_i = 2 \ m$ and $h_o = 6 \ m$.
Magnification $m = \frac{h_i}{h_o} = \frac{2}{6} = \frac{1}{3}$.
Since the image is erect and diminished $(|m| < 1)$,the mirror must be a convex mirror. $A$ plane mirror always produces an image of the same size as the object,and a concave mirror produces a diminished erect image only when the object is placed between the pole and the focus,but it is not the standard case for general observation.

(C) convex mirror always forms a virtual,erect,and diminished image for all positions of the object placed in front of it. $A$ plane mirror always forms an image of the same size as the object. $A$ concave mirror can form a virtual image,but it is always magnified. Therefore,the correct option is $C$.

(C) Given: Object height $O = 5 \ cm$,Object distance $u = -100 \ cm$,Radius of curvature $R = -20 \ cm$.
Focal length $f = R/2 = -10 \ cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} + \frac{1}{-100} = \frac{1}{-10} \Rightarrow \frac{1}{v} = -\frac{1}{10} + \frac{1}{100} = \frac{-10+1}{100} = -\frac{9}{100}$.
So,$v = -\frac{100}{9} \ cm$.
Magnification $m = \frac{I}{O} = -\frac{v}{u}$.
$I = O \times (-\frac{v}{u}) = 5 \times (-\frac{-100/9}{-100}) = 5 \times (-\frac{1}{9}) = -\frac{5}{9} \approx -0.55 \ cm$.
The size of the image is $0.55 \ cm$.

(A) For a concave mirror,the focal length $f = -50 \, cm$.
Since the image is inverted and magnified,it must be a real image.
For a real image,the magnification $m = -2$.
Using the magnification formula $m = \frac{f}{f - u}$:
$-2 = \frac{-50}{-50 - u}$
$-2(-50 - u) = -50$
$100 + 2u = -50$
$2u = -150$
$u = -75 \, cm$.
Thus,the object should be placed at a distance of $75 \, cm$ in front of the mirror.

(B) Given: Object size $O = 7.5 \ cm$,Radius of curvature $R = 25 \ cm$,Object distance $u = -40 \ cm$.
For a convex mirror,the focal length $f = R/2 = 25/2 = 12.5 \ cm$.
Using the magnification formula for a mirror: $m = \frac{I}{O} = -\frac{v}{u}$.
Also,from the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we have $v = \frac{uf}{f - u}$.
Substituting $v$ into the magnification formula: $\frac{I}{O} = -\left(\frac{uf}{f - u}\right) \cdot \frac{1}{u} = \frac{f}{f - u}$.
Substituting the values: $\frac{I}{7.5} = \frac{12.5}{12.5 - (-40)} = \frac{12.5}{52.5}$.
$I = 7.5 \times \frac{12.5}{52.5} = 7.5 \times \frac{1}{4.2} \approx 1.78 \ cm$.

(C) convex mirror is a spherical mirror that curves outward.
Because of its outward curvature,it diverges the incident rays of light.
This divergence allows the mirror to capture light from a much wider area compared to plane or concave mirrors of the same size.
Consequently,the field of view is maximum for a convex mirror.

(B) For a concave mirror,the magnification $m$ is given by $m = \frac{I}{O} = \frac{f}{f - u}$.
Here,$u$ is the object distance from the pole. The distance of the object from the principal focus is $x$.
Since the object is placed in front of the mirror,the distance from the pole is $u = -(f + x)$.
Substituting this into the magnification formula:
$m = \frac{f}{f - (-(f + x))} = \frac{f}{f + f + x} = \frac{f}{2f + x}$.
However,using Newton's form of the mirror equation,$x_1 x_2 = f^2$,where $x_1$ and $x_2$ are distances of object and image from the focus respectively.
Here $x_1 = x$. So,$x_2 = \frac{f^2}{x}$.
The magnification $m = \frac{f}{x_1} = \frac{f}{x}$ (magnitude).
Thus,the ratio of the size of the image to the size of the object is $\frac{f}{x}$.

(A) convex mirror always forms a virtual,erect,and diminished image for any real object placed in front of it.
Since the reflected rays appear to diverge from a point behind the mirror,the image is virtual.

(B) According to Newton's formula for mirrors,when distances are measured from the focus,the relationship is given by $x_1 x_2 = f^2$,where $x_1$ is the distance of the object from the focus and $x_2$ is the distance of the image from the focus.
Alternatively,using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
Let the object distance be $u = -(f + x_1)$ and the image distance be $v = -(f + x_2)$.
Substituting these into the mirror formula:
$\frac{1}{-(f + x_2)} + \frac{1}{-(f + x_1)} = -\frac{1}{f}$
$\frac{1}{f + x_2} + \frac{1}{f + x_1} = \frac{1}{f}$
$\frac{(f + x_1) + (f + x_2)}{(f + x_1)(f + x_2)} = \frac{1}{f}$
$f(2f + x_1 + x_2) = (f + x_1)(f + x_2)$
$2f^2 + f x_1 + f x_2 = f^2 + f x_1 + f x_2 + x_1 x_2$
$f^2 = x_1 x_2$
$f = \sqrt{x_1 x_2}$

(D) convex mirror always forms a virtual,erect,and diminished image for any real object placed in front of it. The image is always formed between the pole $(P)$ and the focus $(F)$ behind the mirror. Therefore,the statement that the image is real is incorrect.

(B) When a point source of light is placed at the principal focus $(F)$ of a concave mirror,the light rays originating from the source strike the mirror and are reflected parallel to the principal axis.
This is because,according to the laws of reflection,any ray passing through the focus of a concave mirror becomes parallel to the principal axis after reflection.
Therefore,a concave mirror can produce a parallel beam of light from a point source.

(B) For a convex mirror,the focal length $f$ is positive,so $f = +30 \ cm$.
The magnification $m$ for a virtual and erect image formed by a convex mirror is positive. Given that the image size is a quarter of the object size,$m = +\frac{1}{4}$.
The formula for magnification in terms of focal length and object distance $u$ is $m = \frac{f}{f - u}$.
Substituting the values: $\frac{1}{4} = \frac{30}{30 - u}$.
Cross-multiplying gives: $30 - u = 120$.
Solving for $u$: $u = 30 - 120 = -90 \ cm$.
The negative sign indicates that the object is placed in front of the mirror at a distance of $90 \ cm$.

(B) The magnification $m$ is given by the ratio of the height of the image to the height of the object,which is $m = \frac{h_i}{h_o} = \frac{1}{5}$.
Since the image is erect,the magnification is positive $(m = +0.2)$.
$A$ plane mirror always produces an image of the same size $(m = 1)$,so it is not a plane mirror.
$A$ concave mirror can produce an erect image only when the object is placed between the pole and the focus,but in that case,the image is always magnified $(m > 1)$.
$A$ convex mirror always forms a virtual,erect,and diminished image for all positions of the object.
Therefore,the mirror must be a convex mirror.

(C) plane mirror always forms a virtual and erect image regardless of the distance.
$A$ convex mirror also always forms a virtual and erect image for all positions of the object.
$A$ concave mirror forms a virtual and erect image when the object is placed between the pole $(P)$ and the focus $(F)$. As the object moves beyond the focus $(F)$,the image becomes real and inverted.
Since the image changes from virtual to inverted as the mirror is moved away,the mirror must be a concave mirror.

(B) The linear magnification $m$ for a spherical mirror is given by $m = -\frac{v}{u}$.
From the mirror formula,we have $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Multiplying both sides by $u$,we get $\frac{u}{f} = \frac{u}{v} + 1$.
Rearranging the terms,we get $\frac{u}{v} = \frac{u}{f} - 1 = \frac{u - f}{f}$.
Taking the reciprocal,we get $\frac{v}{u} = \frac{f}{u - f}$.
Since $m = -\frac{v}{u}$,we have $m = -\frac{f}{u - f} = \frac{f}{f - u}$.
Therefore,the correct relation is $m = \frac{f}{f - u}$.

(B) For street lighting, the goal is to spread the light over a wide area. A $Convex$ mirror is a diverging mirror, which spreads the light rays falling on it. Therefore, it is used as a reflector in street lamps to ensure the light covers a larger region on the road.

(A) Let the distance of the flower from the mirror be $u$. Since the image is formed on a wall,the image is real,so the magnification $m = -16$.
By definition,magnification $m = -\frac{v}{u}$. Thus,$-\frac{v}{u} = -16$,which gives $v = 16u$.
The distance between the object (flower) and the image (wall) is given as $120 \ cm$. Since the image is real and formed by a concave mirror,the image is formed on the same side as the object. The distance between them is $|v - u| = 120$.
Substituting $v = 16u$,we get $|16u - u| = 120$,which simplifies to $15u = 120$.
Therefore,$u = \frac{120}{15} = 8 \ cm$.

(A) concave mirror forms a virtual,erect,and magnified (larger than the object) image when the object is placed between the pole $(P)$ and the principal focus $(F)$ of the mirror.
In contrast,a convex mirror and a concave lens always form virtual images that are smaller than the object.
$A$ plane mirror always forms a virtual image that is equal in size to the object.
Therefore,the correct option is $A$.

(A) Given: Focal length of the concave mirror $f = -20\;cm$. Object distance $u = -40\;cm$.
Since the object is placed at a distance of $40\;cm$ from the mirror,and the radius of curvature $R = 2f = 2 \times 20 = 40\;cm$,the object is placed at the centre of curvature $(C)$.
According to the rules of image formation by a concave mirror,when an object is placed at the centre of curvature,the image is formed at the same position.
The image formed is real,inverted,and of the same size as the object.

(B) For a concave mirror,the radius of curvature $R = -36 \ cm$ (by sign convention).
Therefore,the focal length $f = R/2 = -18 \ cm$.
Since the image is virtual and magnified three times,the magnification $m = +3$.
The formula for magnification in terms of focal length and object distance $u$ is $m = f / (f - u)$.
Substituting the values: $3 = -18 / (-18 - u)$.
Multiplying both sides: $3(-18 - u) = -18$.
$-54 - 3u = -18$.
$-3u = 54 - 18 = 36$.
$u = -12 \ cm$.
The distance of the object from the mirror is $12 \ cm$.

(D) Given: Radius of curvature $R = -40 \ cm$ (for concave mirror),so focal length $f = R/2 = -20 \ cm$. Magnification $m = \pm 2$.
Case $1$: For a real image,$m = -2$.
Using the formula $m = f / (f - u)$:
$-2 = -20 / (-20 - u)$
$-2(-20 - u) = -20$
$40 + 2u = -20$
$2u = -60 \Rightarrow u = -30 \ cm$.
Case $2$: For a virtual image,$m = +2$.
Using the formula $m = f / (f - u)$:
$2 = -20 / (-20 - u)$
$2(-20 - u) = -20$
$-40 - 2u = -20$
$-2u = 20 \Rightarrow u = -10 \ cm$.
Since $30 \ cm$ is the only option provided that matches the real image case,the correct answer is $30 \ cm$.

(D) convex mirror always forms a virtual,erect,and diminished (smaller) image for any real object placed in front of it.
Therefore,the statement that a real,inverted,same-sized image can be formed using a convex mirror is incorrect.
Thus,the correct option is $D$.

(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,$v = +20 \ cm$.
The positive sign of $v$ indicates that 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 upright. Since $|m| > 1$,the image is enlarged.
Therefore,the image is enlarged,upright,and virtual.

(C) For a convex mirror,the focal length is taken as positive,so $f_{mirror} = +f$.
Since the object is placed in front of the mirror,the object distance is $u = -f$.
Using the mirror formula: $\frac{1}{f_{mirror}} = \frac{1}{v} + \frac{1}{u}$.
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,the image distance is $v = f/2$.

(D) For a concave mirror,a real image is formed when the object is placed at a distance greater than the focal length $f$ from the mirror.
When the object is placed at the centre of curvature $(u = 2f)$,the image is also formed at the centre of curvature $(v = 2f)$.
In this specific case,the object and the image coincide at the same point.
Therefore,the distance between the object and its real image is $Zero$.

(C) Given: Object distance $u = -20 \, cm$,Focal length $f = +10 \, cm$ (for a convex mirror).
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 for $v$: $\frac{1}{v} = \frac{1}{10} + \frac{1}{20} = \frac{2+1}{20} = \frac{3}{20}$.
Thus,$v = +\frac{20}{3} \, cm$.
Since $v$ is positive,the image is virtual and formed behind the mirror at a distance of $\frac{20}{3} \, cm$.

(D) The relationship between the focal length $(f)$ and the radius of curvature $(R)$ of a spherical mirror is given by the formula: $f = \frac{R}{2}$.
Given,the focal length of the convex mirror is $f = 20\, cm$.
Substituting the value into the formula: $20 = \frac{R}{2}$.
Therefore,the radius of curvature is $R = 20 \times 2 = 40\, cm$.
Thus,the correct option is $D$.

(B) For a concave mirror,the focal length $f = -15\, cm$.
Since the image is virtual,the magnification $m = +2$.
Using the magnification formula $m = -\frac{v}{u}$,we get $2 = -\frac{v}{u}$,which implies $v = -2u$.
Applying the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{-15} = \frac{1}{-2u} + \frac{1}{u}$
$\frac{1}{-15} = \frac{-1 + 2}{2u}$
$\frac{1}{-15} = \frac{1}{2u}$
$2u = -15$
$u = -7.5\, cm$.
Thus,the object is placed at a distance of $7.5\, cm$ from the mirror.

(D) Given: Object height $O = +2.5 \, cm$,object distance $u = -10 \, cm$,and radius of curvature $R = -30 \, cm$ (for a concave mirror).
Focal length $f = R/2 = -30/2 = -15 \, cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-15} = \frac{1}{v} + \frac{1}{-10}$.
$\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$.
So,$v = +30 \, cm$.
Magnification $m = \frac{I}{O} = -\frac{v}{u}$.
$\frac{I}{2.5} = -\frac{30}{-10} = 3$.
$I = 3 \times 2.5 = 7.5 \, cm$.

(D) concave mirror has a reflecting surface that bulges inwards,which causes incident light rays from real objects to converge and meet at a point after reflection.
This convergence of reflected rays results in the formation of a real image,provided the real object is not placed between the focus $(F)$ and the pole $(P)$ of the mirror.

(A) For a concave mirror,the focal length $f$ is taken as negative,so $f_{mirror} = -f$. The object distance $u$ is also negative,so $u = -4f$. The height of the object $h_o = 6 \ cm$.
Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} + \frac{1}{-4f} = \frac{1}{-f}$.
$\frac{1}{v} = \frac{1}{4f} - \frac{1}{f} = \frac{1 - 4}{4f} = -\frac{3}{4f}$.
Thus,$v = -\frac{4f}{3}$.
The magnification $m$ is given by $m = -\frac{v}{u} = \frac{h_i}{h_o}$.
$m = -\frac{(-4f/3)}{(-4f)} = -\frac{1}{3}$.
Therefore,$h_i = m \times h_o = -\frac{1}{3} \times 6 \ cm = -2 \ cm$.
The length (magnitude) of the image is $2 \ cm$.

(D) The convergence (or power) of a mirror is defined by its focal length,which is given by $f = R/2$. This focal length depends only on the radius of curvature $R$ of the mirror surface. Unlike lenses,the reflection of light by a mirror does not depend on the refractive index of the surrounding medium. Therefore,dipping a concave mirror in any medium like water or oil does not change its focal length or its convergence. Thus,the correct answer is $D$.

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

(A) For a concave mirror,the focal length $f = -6 \, cm$. The magnification $m = \pm 3$.
Using the formula $m = \frac{f}{f - u}$,we have $\pm 3 = \frac{-6}{-6 - u}$.
Case $1$: For a real image,$m = -3$.
$-3 = \frac{-6}{-6 - u} \implies -3(-6 - u) = -6 \implies 18 + 3u = -6 \implies 3u = -24 \implies u = -8 \, cm$.
Case $2$: For a virtual image,$m = +3$.
$3 = \frac{-6}{-6 - u} \implies 3(-6 - u) = -6 \implies -18 - 3u = -6 \implies -3u = 12 \implies u = -4 \, cm$.
Thus,the possible distances of the object from the mirror are $4 \, cm$ or $8 \, cm$.

(A) The focal length of a spherical mirror is given by $f = R/2$, where $R$ is the radius of curvature of the mirror.
This formula depends only on the geometry of the mirror surface.
Unlike a lens, the reflection of light by a mirror does not depend on the refractive index of the surrounding medium.
Therefore, when the mirror is immersed in water, its focal length remains unchanged.
Thus, the focal length in water is $f$.

(B) The sun is at a very large distance,so its image is formed at the focus of the concave mirror. The distance of the image from the pole of the mirror is equal to the focal length,$f = 100 \ cm$.
Let $x$ be the diameter of the image of the sun.
The angle subtended by the sun at the mirror is $\theta = 30' = (30/60)^\circ = 0.5^\circ$.
Converting the angle into radians:
$\theta = 0.5 \times (\pi / 180) \ rad = \pi / 360 \ rad$.
Using the relation $\text{Angle} = \text{Arc} / \text{Radius}$,where the arc is the diameter $x$ and the radius is the focal length $f$:
$\theta = x / f$
$x = f \times \theta = 100 \times (\pi / 360) \ cm$.
$x = 100 \times 3.14159 / 360 \approx 0.872 \ cm$.
Thus,the diameter of the image is approximately $0.87 \ cm$.

(D) From the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ .....$(i)$
Differentiating equation $(i)$ with respect to $u$,we get:
$0 = - \frac{1}{v^2} \frac{dv}{du} - \frac{1}{u^2}$
$\frac{dv}{du} = - \left( \frac{v}{u} \right)^2$
Since the object length $l = du$ is small,the image length $dv$ is given by $dv = - \left( \frac{v}{u} \right)^2 du$ .....$(ii)$
From the mirror formula,we can express the magnification $m = \frac{v}{u}$ as:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{fu}$
$\frac{v}{u} = \frac{f}{u - f}$ .....$(iii)$
Substituting equation $(iii)$ into equation $(ii)$,we get:
$dv = - \left( \frac{f}{u - f} \right)^2 l$
The magnitude of the size of the image is $l' = |dv| = l \left( \frac{f}{u - f} \right)^2$.

(B) Let the rod be placed between $u = 2f - f/3 = 5f/3$ and $u = 2f$ from the concave mirror. The end of the rod at $u = 2f$ forms an image at $v = 2f$.
For the other end of the rod at $u = 5f/3$,we use the mirror formula $\frac{1}{f_{mirror}} = \frac{1}{v} + \frac{1}{u}$.
Using sign convention,$f_{mirror} = -f$ and $u = -5f/3$.
$\frac{1}{-f} = \frac{1}{v} + \frac{1}{-5f/3} \Rightarrow \frac{1}{v} = \frac{3}{5f} - \frac{1}{f} = \frac{3-5}{5f} = -\frac{2}{5f}$.
Thus,$v = -2.5f$.
The length of the image is the distance between the two image positions: $|2.5f - 2f| = 0.5f = f/2$.

(B) $1$. Without the mirror,the illuminance $I_1$ on the screen due to the point source at a distance $r$ is given by $I_1 = \frac{L}{r^2}$,where $L$ is the luminous intensity of the source.
$2$. With the mirror,the source is placed at the center of curvature $(u = r)$. For a concave mirror,an object placed at the center of curvature forms a real image at the same position $(v = r)$.
$3$. The light reaching the screen now consists of two parts: the direct light from the source and the light reflected from the mirror,which appears to come from the image formed at the same location as the source.
$4$. The illuminance due to the reflected light is also $I_{reflected} = \frac{L}{r^2}$.
$5$. The total illuminance on the screen with the mirror is $I_2 = I_{direct} + I_{reflected} = \frac{L}{r^2} + \frac{L}{r^2} = \frac{2L}{r^2}$.
$6$. Therefore,the ratio of illuminances is $\frac{I_2}{I_1} = \frac{2L/r^2}{L/r^2} = 2:1$.

(C) The mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
For a concave mirror,using sign convention,$u$ and $v$ are negative. Let their magnitudes be $u$ and $v$. Then $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,which can be written as $v = \frac{uf}{u-f}$.
As $u \to f$,$v \to \infty$.
As $u \to \infty$,$v \to f$.
This represents a rectangular hyperbola where $v$ decreases as $u$ increases,which corresponds to the curve shown in graph $C$.

(A) For a concave mirror,the mirror formula is given by: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Using sign convention,$f = -|f|$. Thus,$\frac{1}{v} = -\frac{1}{|f|} - \frac{1}{u}$.
$1$. When $u = 0$,$\frac{1}{v} = -\infty$,so $v = 0$.
$2$. When $u = |f|$,$\frac{1}{v} = -\frac{1}{|f|} + \frac{1}{|f|} = 0$,so $v = \infty$.
$3$. When $u = \infty$,$\frac{1}{v} = -\frac{1}{|f|}$,so $v = -|f|$.
As $u$ increases from $0$ to $|f|$,$v$ changes from $0$ to $-\infty$. As $u$ increases from $|f|$ to $\infty$,$v$ changes from $+\infty$ to $-|f|$.
Comparing these characteristics with the given graphs,option $A$ correctly represents this behavior.

(C) For a spherical mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
At point $P$,the graph intersects the line $v = u$,which implies $u = v$. Substituting this into the mirror formula: $\frac{1}{u} + \frac{1}{u} = \frac{1}{f} \implies \frac{2}{u} = \frac{1}{f} \implies u = 2f$. Thus,at point $P$,$u = v = 2f$.
For points on the curve above $P$,the value of $v$ is greater than the value of $v$ at $P$. Since $v_P = 2f$,any point above $P$ on the curve corresponds to $v > 2f$.

(B) For a convex mirror,the mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Since the focal length $f$ is positive for a convex mirror,we have $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$.
Using sign convention,$u$ is negative,so let $u = -x$ where $x > 0$. Then $\frac{1}{v} = \frac{1}{f} + \frac{1}{x} = \frac{x+f}{xf}$.
Thus,$v = \frac{xf}{x+f}$.
As $u$ varies from $0$ to $-\infty$,$x$ varies from $0$ to $+\infty$.
When $x = 0$,$v = 0$.
When $x \to \infty$,$v \to f$.
As $x$ increases,$v$ increases from $0$ to $f$. The graph shows $v$ on the $y$-axis and $u$ on the $x$-axis (where $u$ is negative). The correct representation is Graph $B$.

(D) For a concave mirror,the magnification $m$ is given by $m = \frac{f}{f - u}$.
For a real image,the object distance $u$ is greater than the focal length $f$ $(u > f)$. Let $x$ be the distance of the object from the focus,so $x = u - f$.
Substituting $u = f + x$ into the magnification formula:
$m = \frac{f}{f - (f + x)} = \frac{f}{-x} = -\frac{f}{x}$.
Taking the magnitude of magnification $|m| = \frac{f}{x}$,we see that $|m| \propto \frac{1}{x}$.
This relationship represents a rectangular hyperbola,which corresponds to the graph shown in option $D$.

(B) According to Newton's formula for mirrors, the relationship between the distance of the object from the focus $(x)$ and the distance of the image from the focus $(y)$ is given by $xy = f^2$, where $f$ is the focal length of the mirror.
Since $f$ is constant for a given mirror, $xy = \text{constant}$.
This equation represents a rectangular hyperbola.
Among the given options, the graph that represents a rectangular hyperbola is Graph $B$.

(D) For a spherical mirror,the magnification $m$ is given by $m = \frac{f}{f - u}$.
In the first case,the object distance $u_1 = -25 \; cm$. Thus,$m_1 = \frac{f}{f - (-25)} = \frac{f}{f + 25}$.
In the second case,the object is moved $15 \; cm$ further away,so the new object distance $u_2 = -(25 + 15) = -40 \; cm$. Thus,$m_2 = \frac{f}{f - (-40)} = \frac{f}{f + 40}$.
Given the ratio $\frac{m_1}{m_2} = 4$,we have:
$\frac{f / (f + 25)}{f / (f + 40)} = 4$
$\frac{f + 40}{f + 25} = 4$
$f + 40 = 4(f + 25)$
$f + 40 = 4f + 100$
$3f = -60$
$f = -20 \; cm$.

(A) The focal length of a spherical mirror depends only on its radius of curvature $(R)$ and is given by the formula $f = R/2$.
Since the radius of curvature of a mirror is a geometric property and does not depend on the refractive index of the surrounding medium, the focal length remains unchanged when the mirror is immersed in water.
Therefore, the focal length in water remains $f$.

(C) Given: Height of object $h_1 = 5 \, cm$,Object distance $u = -100 \, cm$ $(1 \, m = 100 \, cm)$,Radius of curvature $R = -20 \, cm$.
Focal length $f = R/2 = -20/2 = -10 \, cm$.
Using the magnification formula for a mirror: $m = \frac{h_2}{h_1} = -\frac{v}{u}$.
Also,$m = \frac{f}{f - u}$.
Substituting the values: $\frac{h_2}{5} = \frac{-10}{-10 - (-100)} = \frac{-10}{-10 + 100} = \frac{-10}{90} = -\frac{1}{9}$.
$h_2 = 5 \times (-\frac{1}{9}) = -\frac{5}{9} \approx -0.55 \, cm$.
The magnitude of the height of the image is $0.55 \, cm$.

(C) Transverse magnification is given by $m = -\frac{v}{u}$.
Longitudinal magnification $(m_L)$ is defined as the ratio of the change in image position to the change in object position: $m_L = \frac{dv}{du}$.
From the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Differentiating both sides with respect to $u$:
$0 = -\frac{1}{v^2} \frac{dv}{du} - \frac{1}{u^2}$.
Rearranging the terms:
$\frac{dv}{du} = -\frac{v^2}{u^2}$.
Since $m = -\frac{v}{u}$,it follows that $m^2 = \frac{v^2}{u^2}$.
Therefore,$m_L = -m^2$. In terms of magnitude,the longitudinal magnification is $m^2$.

(C) Using the mirror equation,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Rearranging for image distance,$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$.
For a concave mirror,the focal length $f$ is negative $(f < 0)$.
If the object is virtual,the object distance $u$ is positive $(u > 0)$. Substituting these into the equation,$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ results in a negative value for $\frac{1}{v}$,meaning $v < 0$. Since the image distance is negative,the image is formed by the actual intersection of reflected rays,making it a real image.
If the object is real,the object distance $u$ is negative $(u < 0)$. In this case,the image distance $v$ can be positive or negative depending on whether the object is placed between the pole and the focus or beyond the focus.