Spherical Mirror Questions in English

(D) The mirror equation is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Rearranging this into the form of a straight-line equation $y = mx + c$,we get $\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$.
Comparing this with $y = mx + c$,where $y = \frac{1}{v}$ and $x = \frac{1}{u}$,the slope $m = -1$ and the intercept $c = \frac{1}{f}$.
From the graph,the intercept on the $\frac{1}{v}$ axis (when $\frac{1}{u} = 0$) is $-0.10 \ cm^{-1}$.
Therefore,$\frac{1}{f} = -0.10 \ cm^{-1}$.
Calculating the focal length,$f = \frac{1}{-0.10} = -10 \ cm$.

(C) Given: Height of object $h_o = 3 \ cm$,Height of image $h_i = -9 \ cm$ (real image formed on the wall is inverted).
Distance of object from mirror $u = -(x - 300) \ cm$.
Distance of image from mirror $v = -x \ cm$.
Using magnification formula $m = \frac{h_i}{h_o} = -\frac{v}{u}$:
$\frac{-9}{3} = -\frac{-x}{-(x - 300)}$
$-3 = -\frac{x}{x - 300}$
$3 = \frac{x}{x - 300}$
$3(x - 300) = x$
$3x - 900 = x$
$2x = 900$
$x = 450 \ cm$.

(C) For a concave mirror, the magnification $m$ for a real image is negative, so $m = -n$.
By definition of magnification, $m = -\frac{v}{u}$, where $v$ is the image distance and $u$ is the object distance.
Thus, $-n = -\frac{v}{u} \implies v = nu$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Since the mirror is concave, the focal length $f$ is negative, i.e., $-f$. The object distance $u$ is also negative, i.e., $-u$.
Substituting these values: $\frac{1}{-f} = \frac{1}{-nu} + \frac{1}{-u}$.
Multiplying by $-1$: $\frac{1}{f} = \frac{1}{nu} + \frac{1}{u} = \frac{1+n}{nu}$.
Rearranging for $u$: $u = \left(\frac{n+1}{n}\right) f$.

(D) For a spherical mirror,the sign convention states that the direction of incident light is taken as positive.
When an image is formed on the same side as the object,the image distance $v$ is negative,which corresponds to a real image.
When an image is formed on the opposite side of the object (behind the mirror),the image distance $v$ is positive,which corresponds to a virtual image.
Therefore,images formed on the side of the object are real,and images formed on the opposite side are virtual.

(D) 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,they appear to meet behind the mirror. Thus,the image formed is virtual,erect,and magnified.

(D) Given: Magnification $m = -3$ (since the image is real,it must be inverted). Radius of curvature $R = -30 \ cm$. Focal length $f = R/2 = -15 \ cm$.
Using the magnification formula: $m = -v/u \Rightarrow -3 = -v/u \Rightarrow v = 3u$.
Using the mirror formula: $1/f = 1/v + 1/u$.
Substituting the values: $1/(-15) = 1/(3u) + 1/u$.
$1/(-15) = (1 + 3)/(3u) = 4/(3u)$.
$-3u = 60 \Rightarrow u = -20 \ cm$.
The negative sign indicates that the person should stand $20 \ cm$ in front of the concave mirror.

(B) Given: Height of object $h = 10 \text{ cm}$,radius of curvature $R = 15 \text{ cm}$,object distance $u = -10 \text{ cm}$.
Focal length $f = R/2 = 15/2 = 7.5 \text{ cm}$. For a concave mirror,$f = -7.5 \text{ cm}$.
Since the object is placed at $u = -10 \text{ cm}$,and the focal length is $f = -7.5 \text{ cm}$,the object is placed between the focus $(F)$ and the centre of curvature $(C)$ because $f < |u| < R$ (i.e.,$7.5 \text{ cm} < 10 \text{ cm} < 15 \text{ cm}$).
When an object is placed between $F$ and $C$ in front of a concave mirror,the image formed is real,inverted,and magnified.

(C) convex mirror always forms a virtual,erect,and diminished image for any real object placed in front of it,regardless of the object's distance from the mirror.

(D) For a convex mirror,the principal focus $(F)$ is located behind the mirror.
When an object is placed at the principal focus of a convex mirror,the rays of light originating from the object appear to diverge from the focus after reflection.
However,since the focus is behind the mirror,the rays cannot actually meet or appear to meet at any point to form an image in the conventional sense for this specific configuration.
Therefore,no image is formed at the principal focus of a convex mirror.

(C) Using the mirror formula,we have $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Here,the object distance $u = -20 \ cm$ and the focal length $f = -10 \ cm$ (for a concave mirror).
Substituting these values into the formula:
$\frac{1}{-10} = \frac{1}{v} + \frac{1}{-20}$.
Rearranging to solve for $v$:
$\frac{1}{v} = \frac{1}{20} - \frac{1}{10}$.
$\frac{1}{v} = \frac{1 - 2}{20} = \frac{-1}{20}$.
Therefore,$v = -20 \ cm$.

(A) Given: Object distance $u = -20 \ cm$ (by sign convention).
Magnification $m = -3$ (since the image is real and magnified).
Using the magnification formula $m = -\frac{v}{u}$,we have:
$-3 = -\frac{v}{-20} \Rightarrow v = -60 \ cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,we get:
$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = \frac{-4}{60} = -\frac{1}{15}$.
Therefore,the focal length $f = -15 \ cm$.
The magnitude of the focal length is $15 \ cm$.

(B) For a convex mirror, the focal length $f = +30 \,cm$.
Magnification $m = \frac{h_i}{h_o} = +\frac{1}{4}$ (since the image formed by a convex mirror is always virtual and erect).
Using the magnification formula $m = -\frac{v}{u}$, we get $\frac{1}{4} = -\frac{v}{u}$, which implies $v = -\frac{u}{4}$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{30} = \frac{1}{-u/4} + \frac{1}{u}$
$\frac{1}{30} = -\frac{4}{u} + \frac{1}{u}$
$\frac{1}{30} = -\frac{3}{u}$
$u = -30 \times 3 = -90 \,cm$.
The distance of the object from the mirror is $90 \,cm$.

(B) For a convex mirror,the magnification $m$ is given by $m = \frac{h_i}{h_o} = -\frac{v}{u}$.
Given that the size of the image is $1/n$ times the size of the object,we have $m = 1/n$.
Thus,$1/n = -v/u$,which implies $v = -u/n$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,where $f$ is the focal length of the convex mirror (positive).
Substituting $v = -u/n$ into the mirror formula:
$\frac{1}{-u/n} + \frac{1}{u} = \frac{1}{f}$
$-\frac{n}{u} + \frac{1}{u} = \frac{1}{f}$
$\frac{1-n}{u} = \frac{1}{f}$
$u = f(1-n)$.
Since the object distance $u$ is conventionally negative,the magnitude is $|u| = f(n-1)$.

(C) Given: Object distance $u = -18 \ cm$ (in front of the mirror).
Image distance $v = +4 \ cm$ (on the other side of the mirror).
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{4} + \frac{1}{-18} = \frac{9 - 2}{36} = \frac{7}{36}$.
Therefore,$f = \frac{36}{7} \ cm \approx 5.14 \ cm$.
Since the focal length $f$ is positive,the mirror is a convex mirror.
Since the image is formed on the other side of the mirror,it is a virtual image.

(D) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Using sign convention,$u$ is replaced by $-u$ and $f$ by $-f$.
So,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{f}$.
Rearranging for $v$,we get $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f-u}{uf}$.
Thus,$v = \frac{uf}{f-u}$.
The magnitude of the image position is $|v| = \left| \frac{uf}{f-u} \right| = \frac{uf}{u-f}$.
The rod extends from $u$ to $\infty$. The image of the nearer end is at $v_1 = \frac{uf}{u-f}$.
The image of the farther end (at $\infty$) is at the focus $v_2 = f$.
The length of the image is $L = |v_1 - v_2| = \left| \frac{uf}{u-f} - f \right|$.
$L = \left| \frac{uf - f(u-f)}{u-f} \right| = \left| \frac{uf - uf + f^2}{u-f} \right| = \frac{f^2}{u-f}$.

(A) When an object is placed between the focal point $(F)$ and the center of curvature $(C)$ of a concave mirror,the light rays from the object reflect off the mirror and converge at a point beyond the center of curvature $(C)$.
This results in an image that is real,inverted,and magnified (larger than the object).
Therefore,the image is formed farther than the center of curvature from the mirror.

(B) Given,focal length of the concave mirror $f = 20 \ cm$.
The radius of curvature $R = 2f = 2 \times 20 \ cm = 40 \ cm$.
The object distance $u = 40 \ cm$.
Since the object is placed at the centre of curvature $(u = R)$,the image formed by a concave mirror is real,inverted,and of the same size as the object,and it is formed at the centre of curvature itself.

(A) Given: Height of object $h_o = 3.6 \ cm$,Radius of curvature $R = 30 \ cm$.
Focal length $f = R/2 = 30/2 = 15 \ cm$.
Since it is a concave mirror,$f = -15 \ cm$.
The object is at a distance of $10 \ cm$ from the principal focus.
Position of object $u = -(f + 10) = -(15 + 10) = -25 \ cm$.
Using mirror formula: $1/v + 1/u = 1/f$.
$1/v = 1/f - 1/u = 1/(-15) - 1/(-25) = -1/15 + 1/25 = (-5 + 3)/75 = -2/75$.
$v = -75/2 = -37.5 \ cm$.
Magnification $m = -v/u = h_i/h_o$.
$m = -(-37.5) / (-25) = -37.5 / 25 = -1.5$.
Height of image $h_i = m \times h_o = -1.5 \times 3.6 = -5.4 \ cm$.
The magnitude of the height of the real image is $5.4 \ cm$.

(B) For a concave mirror,the distances of the object $(x_1)$ and the real image $(x_2)$ from the principal focus are related to the focal length $(f)$ by Newton's formula: $f^2 = x_1 \cdot x_2$.
Given,$x_1 = 16 \ cm$ and $x_2 = 9 \ cm$.
Substituting these values into the formula:
$f^2 = 16 \ cm \times 9 \ cm = 144 \ cm^2$.
Taking the square root on both sides:
$f = \sqrt{144} \ cm = 12 \ cm$.
Therefore,the focal length of the mirror is $12 \ cm$.

(D) Given: Focal length $f = -9 \ cm$ (concave mirror). The rod is placed along the principal axis. The end closer to the mirror is at $u_1 = -15 \ cm$. The length of the rod is $6 \ cm$,so the farther end is at $u_2 = -15 - 6 = -21 \ cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we find the image positions $v_1$ and $v_2$.
For $u_1 = -15 \ cm$: $\frac{1}{v_1} = \frac{1}{-9} - \frac{1}{-15} = \frac{-5 + 3}{45} = \frac{-2}{45} \implies v_1 = -22.5 \ cm$.
For $u_2 = -21 \ cm$: $\frac{1}{v_2} = \frac{1}{-9} - \frac{1}{-21} = \frac{-7 + 3}{63} = \frac{-4}{63} \implies v_2 = -15.75 \ cm$.
The length of the image is $|v_1 - v_2| = |-22.5 - (-15.75)| = |-6.75| = 6.75 \ cm$.

(C) The given situation is shown in the figure.
Let $AB$ be the object and $A^{\prime}B^{\prime}$ be the image.
Here,$AB = l$,$A^{\prime}B^{\prime} = l^{\prime}$,and the object distance $u = -X$.
Using the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,where $f = -F$ for a concave mirror:
$\frac{1}{-F} = \frac{1}{v} + \frac{1}{-X}$
$\frac{1}{v} = \frac{1}{X} - \frac{1}{F} = \frac{F-X}{FX}$
$v = \frac{FX}{F-X}$
For a small object placed along the principal axis,the longitudinal magnification $M$ is given by:
$M = \frac{l^{\prime}}{l} = -\frac{dv}{du} = -\frac{d}{du} \left( \frac{fu}{u-f} \right) = -\frac{f^2}{(u-f)^2}$
Substituting $u = -X$ and $f = -F$:
$M = -\frac{(-F)^2}{(-X - (-F))^2} = -\frac{F^2}{(F-X)^2}$
The magnitude of the longitudinal magnification is:
$\left| \frac{l^{\prime}}{l} \right| = \left( \frac{F}{F-X} \right)^2$

(C) Given, distance between object and mirror, $u = -7 \,cm$.
Radius of curvature, $R = -12 \,cm$.
Therefore, focal length, $f = \frac{R}{2} = \frac{-12}{2} = -6 \,cm$.
Using the mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the given values:
$\frac{1}{v} + \frac{1}{-7} = \frac{1}{-6}$
$\frac{1}{v} = \frac{1}{7} - \frac{1}{6} = \frac{6 - 7}{42} = -\frac{1}{42}$
$v = -42 \,cm$.
The negative sign indicates that the image is formed at a distance of $42 \,cm$ to the left of the mirror.

(A) Using the mirror equation,we have:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
Given that for a concave mirror,the focal length $f = -10 \,cm$ and the object distance $u = -25 \,cm$.
Substituting these values into the equation:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-10} - \frac{1}{-25}$
$\frac{1}{v} = -\frac{1}{10} + \frac{1}{25} = \frac{-5 + 2}{50} = -\frac{3}{50}$
$v = -\frac{50}{3} \approx -16.7 \,cm$
The magnification $m$ is given by:
$m = -\frac{v}{u} = -\frac{-16.7}{-25} = -0.67$
Thus,the image is formed at $-16.7 \,cm$ and the magnification is $-0.67$.

(A) Given: Distance of light source $(u) = -1.5 \ m = -\frac{3}{2} \ m$. Radius of curvature of the convex mirror $(R) = +1 \ m$. Focal length $(f) = \frac{R}{2} = +0.5 \ m$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{0.5} = \frac{1}{v} + \frac{1}{-1.5} \Rightarrow 2 = \frac{1}{v} - \frac{2}{3}$.
Solving for $v$: $\frac{1}{v} = 2 + \frac{2}{3} = \frac{6+2}{3} = \frac{8}{3}$.
Therefore,$v = \frac{3}{8} = 0.375 \ m$.
Since $v$ is positive,the image is formed behind the mirror,making it virtual and upright.
Magnification $(m) = -\frac{v}{u} = -\frac{0.375}{-1.5} = \frac{0.375}{1.5} = 0.25$.

(C) The equation of the reflecting surface is $x^{2}+y^{2}=R^{2}$.
Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.
This $\frac{dy}{dx}$ represents the slope of the tangent at point $M$. The normal at point $M$ makes an angle of $45^{\circ}$ with the $x$-axis because the incident ray is horizontal and the reflected ray is vertical,making the angle of incidence equal to the angle of reflection ($22.5^{\circ}$ each relative to the normal).
Alternatively,for the ray to turn $90^{\circ}$ (from $+x$ to $+y$),the normal at $M$ must make an angle of $45^{\circ}$ with the $x$-axis.
The slope of the normal is $\tan(45^{\circ}) = 1$. Since the normal is perpendicular to the tangent,the slope of the tangent is $-1$.
Thus,$-\frac{x}{y} = -1$,which means $x = y$.
Substituting $x = y$ into the equation $x^{2}+y^{2}=R^{2}$,we get $2x^{2} = R^{2}$,so $x = \pm \frac{R}{\sqrt{2}}$.
Based on the geometry of the reflection (ray coming from left,reflecting to $+y$),point $M$ must be in the second quadrant,where $x < 0$ and $y > 0$.
Therefore,the coordinates are $M = \left(-\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right)$.

(D) The angle $\theta$ subtended by the sun at the mirror is given by the ratio of the diameter of the sun $(D)$ to the distance between the earth and the sun $(d_{SE})$.
Since the sun is at a very large distance,its image is formed at the focus of the concave mirror.
The angular size $\theta$ is given by $\theta = \frac{D}{d_{SE}} = 0.009 \ rad$.
The diameter of the image $(d)$ formed at the focal plane is given by $d = f \times \theta$,where $f$ is the focal length of the mirror.
The focal length $f$ is half of the radius of curvature $R = 0.4 \ m$,so $f = \frac{R}{2} = \frac{0.4}{2} = 0.2 \ m$.
Substituting the values,we get $d = 0.2 \ m \times 0.009 = 0.0018 \ m$.
Therefore,$d = 1.8 \times 10^{-3} \ m$.

(D) For a concave mirror, the magnification $m$ for a real image is negative, so $m = -3$ and $m = -2$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and magnification $m = -\frac{v}{u}$, we have $v = -mu$.
Case $1$: $u_1 = -x$, $m_1 = -3$, so $v_1 = -(-3)(-x) = -3x$.
$\frac{1}{-3x} + \frac{1}{-x} = \frac{1}{f} \implies \frac{-1-3}{3x} = \frac{1}{f} \implies \frac{1}{f} = \frac{-4}{3x} \quad (i)$
Case $2$: $u_2 = -(x+5)$, $m_2 = -2$, so $v_2 = -(-2)(-(x+5)) = -2(x+5)$.
$\frac{1}{-2(x+5)} + \frac{1}{-(x+5)} = \frac{1}{f} \implies \frac{-1-2}{2(x+5)} = \frac{1}{f} \implies \frac{1}{f} = \frac{-3}{2(x+5)} \quad (ii)$
Equating $(i)$ and $(ii)$:
$\frac{-4}{3x} = \frac{-3}{2(x+5)} \implies 8(x+5) = 9x \implies 8x + 40 = 9x \implies x = 40 \ cm$.
Substituting $x = 40$ into $(i)$:
$\frac{1}{f} = \frac{-4}{3(40)} = \frac{-4}{120} = \frac{-1}{30} \implies f = -30 \ cm$.
The magnitude of the focal length is $30 \ cm$.

(A) Given: Object height $h_o = 2.0 \,cm$,Object distance $u = -15 \,cm$,Focal length $f = -10 \,cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} + \frac{1}{-15} = \frac{1}{-10} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} = -\frac{1}{30}$.
So,$v = -30 \,cm$.
The magnification $m = -\frac{v}{u} = -\frac{-30}{-15} = -2$.
Image height $h_i = m \times h_o = -2 \times 2.0 = -4.0 \,cm$.
The negative sign indicates that the image is real and inverted. The size of the image is $4.0 \,cm$.

(D) For a real image,magnification $m = -3$. Since $m = -v/u$,we have $-3 = -v/u$,which implies $v = 3u$.
Given that the distance between the object and the image is $40 \ cm$,and for a real image formed by a concave mirror,both object and image are on the same side,the distance is $|v - u| = 40 \ cm$.
Substituting $v = 3u$,we get $|3u - u| = 40$,so $|2u| = 40$,which gives $u = -20 \ cm$ (using sign convention).
Then $v = 3(-20) = -60 \ cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = \frac{-4}{60} = \frac{-1}{15}$.
Therefore,$f = -15 \ cm$.

(B) In the parallax method,we need to form a real image of the object to observe the parallax between the image and the needle.
For a concave mirror,a real image is formed only when the object is placed beyond the focus $(F)$.
If the object is placed between the pole $(P)$ and the focus $(F)$,the image formed is virtual,erect,and magnified,which cannot be used for the parallax method as it cannot be captured on a screen or observed as a real point of coincidence.
Therefore,the object must be placed at any point beyond the focus $(F)$.

(A) For a concave mirror,the focal length $f = -15 \text{ cm}$ and the object distance $u = -10 \text{ cm}$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-10} = -\frac{1}{15} + \frac{1}{10} = \frac{-2 + 3}{30} = \frac{1}{30}$.
Thus,$v = +30 \text{ cm}$.
Since $v$ is positive,the image is formed behind the mirror,which means it is virtual and erect.
The magnification $m = -\frac{v}{u} = -\frac{30}{-10} = +3$.
Since the magnification $m$ is positive and $|m| > 1$,the image is virtual,erect,and magnified.

(B) For a concave mirror,the focal length $f = -10 \text{ cm}$.
The rod is placed from $u_1 = -20 \text{ cm}$ to $u_2 = -30 \text{ cm}$ (since the length is $10 \text{ cm}$ and the near end is at $20 \text{ cm}$ from the pole).
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
For the near end at $u_1 = -20 \text{ cm}$:
$\frac{1}{v_1} - \frac{1}{20} = -\frac{1}{10} \implies \frac{1}{v_1} = -\frac{1}{10} + \frac{1}{20} = -\frac{1}{20} \implies v_1 = -20 \text{ cm}$.
For the far end at $u_2 = -30 \text{ cm}$:
$\frac{1}{v_2} - \frac{1}{30} = -\frac{1}{10} \implies \frac{1}{v_2} = -\frac{1}{10} + \frac{1}{30} = -\frac{2}{30} = -\frac{1}{15} \implies v_2 = -15 \text{ cm}$.
The length of the image is the distance between the images of the two ends:
$\text{Length of image} = |v_1 - v_2| = |-20 - (-15)| = |-5| = 5 \text{ cm}$.

(B) For a concave mirror,the magnification $m$ can be $\pm 2$ because the image can be real (inverted) or virtual (erect).
The focal length $f = -10$ cm.
$1$) For real image,$m = -2$:
Using $m = \frac{f}{f-u}$,we have $-2 = \frac{-10}{-10-u} \Rightarrow -2 = \frac{10}{10+u} \Rightarrow -20 - 2u = 10 \Rightarrow 2u = -30 \Rightarrow u_1 = -15$ cm.
$2$) For virtual image,$m = +2$:
Using $m = \frac{f}{f-u}$,we have $2 = \frac{-10}{-10-u} \Rightarrow 2 = \frac{10}{10+u} \Rightarrow 20 + 2u = 10 \Rightarrow 2u = -10 \Rightarrow u_2 = -5$ cm.
The distance between the two positions is $|u_1 - u_2| = |-15 - (-5)| = |-10| = 10$ cm.