Plane Mirror Questions in English

(C) For a plane mirror,a real object always forms a virtual image.
Conversely,a virtual object forms a real image.
$A$ virtual object is formed when incident light rays are converging towards a point behind the mirror.
Since the problem states that a real image is formed by a plane mirror,the incident rays must be converging.

(B) Let the angle between the two mirrors be $\theta.$
According to the problem,the incident ray is parallel to the mirror $(M_2).$ Therefore,the angle of incidence on the first mirror $(M_1)$ is $\theta.$
By the law of reflection,the angle of reflection from $(M_1)$ is also $\theta.$
This reflected ray strikes the second mirror $(M_2).$
In the triangle formed by the two mirrors and the light ray,the angle at the second mirror is $\theta$ (since the reflected ray is parallel to the first mirror $(M_1)$).
Thus,the sum of angles in the triangle is $\theta + \theta + \theta = 180^o.$
$3\theta = 180^o$
$\theta = 60^o.$

(C) Let the mirror be placed along the $y$-axis from $y = -d/2$ to $y = d/2$. The source $S$ is at $(L, 0)$. The image $S'$ of the source $S$ is formed at $(-L, 0)$.
The man walks along the line $x = 2L$. The rays from the image $S'$ that reach the man must pass through the edges of the mirror.
The rays from $S'$ passing through the top edge $(0, d/2)$ and bottom edge $(0, -d/2)$ of the mirror define the field of view.
Using similar triangles,the height $h$ of the field of view at a distance $x = 2L$ from the mirror is given by the ratio of distances.
The distance from the image $S'$ to the mirror is $L$,and the distance from the image $S'$ to the man is $L + 2L = 3L$.
By similar triangles,the width of the field of view $h$ is given by $\frac{h}{d} = \frac{3L}{L} = 3$.
Therefore,$h = 3d$.

(D) According to the law of reflection,the angle of incidence is equal to the angle of reflection.
At point $P$,the angle of incidence is $30^o$,so the angle of reflection is $30^o$.
Since the walls are perpendicular,the angle of incidence at point $Q$ is $90^o - 30^o = 60^o$. Thus,the angle of reflection at $Q$ is $60^o$.
At point $R$,the angle of incidence is $90^o - 60^o = 30^o$. Thus,the angle of reflection at $R$ is $30^o$.
At point $S$,the angle of incidence is $90^o - 30^o = 60^o$. Thus,the angle of reflection at $S$ is $60^o$.
At point $T$,the angle of incidence is $90^o - 60^o = 30^o$. Thus,the angle of reflection at $T$ is $30^o$.
Therefore,the angles of reflection at $Q, R, S$ and $T$ are $60^o, 30^o, 60^o, 30^o$ respectively.

(B) The object is located on the $y$-axis at a distance of $10 \text{ cm}$ from the origin $c$. Its coordinates are $(0, 10)$.
The mirror makes an angle of $30^{\circ}$ with the negative $x$-axis,which means it makes an angle of $150^{\circ}$ with the positive $x$-axis.
The angle of the normal to the mirror with the $y$-axis is $30^{\circ}$.
The image of an object at distance $d$ from the mirror along a line making angle $\theta$ with the mirror surface will be at distance $d$ on the other side.
Using the geometry of reflection,the image $I$ is at a distance of $10 \text{ cm}$ from the origin $c$.
The angle the line joining the origin to the image makes with the positive $x$-axis is $-60^{\circ}$ (or $300^{\circ}$).
Thus,the coordinates of the image are $(10 \cos(-60^{\circ}), 10 \sin(-60^{\circ})) = (10 \times 0.5, 10 \times -\frac{\sqrt{3}}{2}) = (5, -5\sqrt{3})$.
Therefore,the correct option is $B$.

(B) Given: Velocity of object $\vec{v}_o = 10 \cos(60^\circ) \hat{i} + 10 \sin(60^\circ) \hat{j} = 5 \hat{i} + 5\sqrt{3} \hat{j} \text{ cm/s}$.
Velocity of mirror $\vec{v}_m = 2 \hat{i} \text{ cm/s}$.
The velocity of the image $\vec{v}_i$ is given by the formula: $\vec{v}_{i,x} = 2\vec{v}_{m,x} - \vec{v}_{o,x}$ and $\vec{v}_{i,y} = \vec{v}_{o,y}$.
Calculating the x-component: $\vec{v}_{i,x} = 2(2) - 5 = 4 - 5 = -1 \text{ cm/s}$.
Calculating the y-component: $\vec{v}_{i,y} = 5\sqrt{3} \text{ cm/s}$.
Thus,the velocity of the image with respect to the ground is $\vec{v}_i = -1 \hat{i} + 5\sqrt{3} \hat{j} \text{ cm/s}$.
The velocity of the image with respect to the mirror is $\vec{v}_{im} = \vec{v}_i - \vec{v}_m = (-1 \hat{i} + 5\sqrt{3} \hat{j}) - (2 \hat{i}) = -3 \hat{i} + 5\sqrt{3} \hat{j} \text{ cm/s}$.

(C) The total deviation $\delta$ produced by two plane mirrors inclined at an angle $\theta$ is given by $\delta = 360^o - 2\theta$.
Given $\delta = 270^o$,we have $270^o = 360^o - 2\theta$.
$2\theta = 360^o - 270^o = 90^o$,so $\theta = 45^o$.
The number of images $n$ formed by two mirrors inclined at an angle $\theta$ is given by $n = \frac{360^o}{\theta} - 1$.
Substituting $\theta = 45^o$,we get $n = \frac{360^o}{45^o} - 1 = 8 - 1 = 7$.
Therefore,the number of observable images is $7$.

(C) The distance between the two mirrors is $d = 6\,cm + 15\,cm = 21\,cm$. Let $x_1 = 6\,cm$ (distance from $M_1$) and $x_2 = 15\,cm$ (distance from $M_2$).
Images formed by $M_1$ are at distances:
$I_1 = x_1 = 6\,cm$
$I_3 = x_1 + 2d = 6 + 42 = 48\,cm$
$I_5 = x_1 + 4d = 6 + 84 = 90\,cm$
Images formed by $M_2$ are at distances from $M_2$:
$I_2 = x_2 = 15\,cm$
$I_4 = x_2 + 2d = 15 + 42 = 57\,cm$
To find the position of the $4^{th}$ image formed by $M_1$,we look at the sequence of images. The images formed by $M_1$ are at $6\,cm, 48\,cm, 90\,cm, \dots$ from $M_1$. The images formed by $M_2$ are at $15\,cm, 57\,cm, 99\,cm, \dots$ from $M_2$.
Converting the $M_2$ image distances to be relative to $M_1$:
$I_2$ is at $15 + 21 = 36\,cm$ from $M_1$.
$I_4$ is at $57 + 21 = 78\,cm$ from $M_1$.
Ordering the images by distance from $M_1$: $6\,cm (M_1), 36\,cm (M_2), 48\,cm (M_1), 78\,cm (M_2), 90\,cm (M_1), \dots$
The $4^{th}$ image is at $78\,cm$ from $M_1$.

(B) When a plane mirror rotates by an angle $\theta$, the reflected ray rotates by an angle $2\theta$.
Given angular acceleration $\alpha = 4\pi \, rad/s^2$ and time $t = \frac{1}{4} \, s$.
The angle of rotation of the mirror is $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
Since the mirror starts from rest, $\omega_0 = 0$.
$\theta = \frac{1}{2} \times (4\pi) \times (\frac{1}{4})^2 = 2\pi \times \frac{1}{16} = \frac{\pi}{8} \, radians$.
Converting radians to degrees: $\theta = \frac{\pi}{8} \times \frac{180^\circ}{\pi} = 22.5^\circ$.
The reflected ray rotates by $2\theta = 2 \times 22.5^\circ = 45^\circ$.

(C) The boy's height is $H = 1.5\,m$ and his eye level is $h_e = 1.4\,m$. The distance from his eyes to the top of his head is $1.5 - 1.4 = 0.1\,m$. The distance from his eyes to his feet is $1.4\,m$.
For a person to see their full image,the mirror must cover the range from half the distance to the head to half the distance to the feet relative to the eye level.
Required mirror height above eye level: $0.1 / 2 = 0.05\,m$.
Required mirror height below eye level: $1.4 / 2 = 0.7\,m$.
The mirror's lower edge is at $0.8\,m$ from the floor. Since the eye level is at $1.4\,m$,the lower edge of the mirror is $1.4 - 0.8 = 0.6\,m$ below the eye level.
Since the required distance below the eye level is $0.7\,m$ and the mirror only extends $0.6\,m$ below the eye level,the boy cannot see his feet.

(B) The object $O$ is placed between two parallel mirrors $AB$ and $CD$.
Each mirror forms an image of the object $O$ behind it.
Since the mirrors are not facing each other (they are arranged along parallel lines but are offset),light rays from the object $O$ can reflect off mirror $AB$ to form an image $I_1$,and light rays can reflect off mirror $CD$ to form an image $I_2$.
Because the mirrors do not overlap in the field of view of the reflected rays from the other mirror,no further multiple reflections occur.
Therefore,only two images are formed,one by each mirror.

(A) In the given figure,the incident rays $A$ and $B$ are parallel to each other. After reflection from the mirror,the reflected rays $A'$ and $B'$ also remain parallel to each other.
For a spherical mirror (concave or convex),parallel incident rays will either converge to a focus or appear to diverge from a focus,meaning they will not remain parallel after reflection.
Only a plane mirror can reflect a set of parallel incident rays as a set of parallel reflected rays,regardless of the angle of incidence,provided the surface is flat.
Therefore,the mirror must be a plane mirror.

(A) According to the law of reflection,the angle of incidence $(i)$ is equal to the angle of reflection $(r)$.
For a plane mirror,if two incident rays strike the same point,their corresponding reflected rays must maintain the same angular relationship with the normal.
Specifically,if an incident ray makes an angle $i$ with the normal,its reflected ray must make an angle $r = i$ with the normal on the opposite side.
In diagram $(I)$,the reflected rays maintain the same relative angular separation from the normal as the incident rays,satisfying the law of reflection for both rays.
Diagrams $(II)$ and $(III)$ do not correctly show the symmetry required by the law of reflection relative to the normal for both rays simultaneously.
Therefore,only diagram $(I)$ correctly represents the reflection of two rays from the same point on a plane mirror.

(B) Let the mirror $M$ be inclined at an angle $\theta$ with the horizontal $H$. The incident ray makes an angle $\theta$ with the mirror surface. According to the law of reflection,the angle of reflection is equal to the angle of incidence. Therefore,the reflected ray also makes an angle $\theta$ with the mirror surface. From the geometry of the setup,the angle between the reflected ray and the horizontal is the sum of the angle the reflected ray makes with the mirror and the angle the mirror makes with the horizontal. Thus,the angle with the horizontal is $\theta + \theta = 2\theta$.

(B) When two plane mirrors are placed at an angle $\theta$ to each other,the images formed by an object placed between them lie on a circle.
The center of this circle is the point of intersection of the two mirrors.
For $\theta = 90^o$,the number of images formed is $n = (360^o / 90^o) - 1 = 4 - 1 = 3$.
These $3$ images are located at the vertices of a square (with the object and the intersection point forming the other vertices),all lying on a circle centered at the intersection point of the mirrors.

(C) For a person of height $H$ to see their complete image in a plane mirror,the minimum length of the mirror required is $L = H/2$.
Given $H = 170 \, cm$,the minimum length of the mirror is $L = 170/2 = 85 \, cm$. Thus,statement $(b)$ is correct.
The bottom edge of the mirror should be placed at a height equal to half the distance from the ground to the eyes. Let $h$ be the height of the eyes from the ground.
Given $h = 160 \, cm$,the height of the bottom of the mirror from the ground is $h/2 = 160/2 = 80 \, cm$. Thus,statement $(c)$ is correct.
Therefore,the correct statements are $(b)$ and $(c)$.

(C) Let the angle of incidence on mirror $M_1$ be $i_1 = 30^o$. The angle of reflection is also $30^o$. The angle the reflected ray makes with the surface of $M_1$ is $90^o - 30^o = 60^o$.
In the triangle formed by the two mirrors and the ray,the angle at the vertex is $40^o$. The angle at the base on $M_1$ is $60^o$. Thus,the angle at the base on $M_2$ is $180^o - (40^o + 60^o) = 80^o$.
The angle of incidence on $M_2$ is $90^o - 80^o = 10^o$. Since $10^o < 90^o$,the ray reflects from $M_2$.
The angle of reflection from $M_2$ is $10^o$. The angle this ray makes with the surface of $M_2$ is $90^o - 10^o = 80^o$.
Now,consider the triangle formed by the ray and the mirrors. The angle at the vertex is $40^o$. The angle at the base on $M_2$ is $80^o$. The angle at the base on $M_1$ is $180^o - (40^o + 80^o) = 60^o$.
Since the angle with the surface of $M_1$ is $60^o$,the angle of incidence is $90^o - 60^o = 30^o$. This is the same as the initial angle of incidence,meaning the ray will continue to reflect between the mirrors.
However,the question asks for the number of reflections. Based on the geometry,the ray will strike $M_1$ again,then $M_2$,and so on. Given the mirrors are long,the ray will undergo a finite number of reflections until it exits the system. For an angle of $40^o$,the total number of reflections is calculated as $N = \frac{180^o}{\theta} = \frac{180^o}{40^o} = 4.5$. The ray will undergo $4$ reflections.

(D) Let the width of the rear window be $W = 120 \, cm$ and the height be $H = 45 \, cm$. The distance from the driver to the rear window is $d_1 = 2 \, m$. The mirror is placed at a distance $d_2 = 0.5 \, m$ in front of the driver. The image of the rear window is formed at a distance $d_2 = 0.5 \, m$ behind the mirror. Thus,the total distance from the driver to the image of the rear window is $D = d_1 + d_2 + d_2 = 2 + 0.5 + 0.5 = 3 \, m$.
Using similar triangles,the ratio of the mirror size $(w, h)$ to the window size $(W, H)$ is equal to the ratio of the distance of the mirror from the driver to the total distance of the image from the driver:
$\frac{w}{W} = \frac{d_2}{D} \Rightarrow w = W \times \frac{0.5}{3} = 120 \times \frac{1}{6} = 20 \, cm$
$\frac{h}{H} = \frac{d_2}{D} \Rightarrow h = H \times \frac{0.5}{3} = 45 \times \frac{1}{6} = 7.5 \, cm$
Therefore,the minimum size of the mirror is $(20 \, cm \times 7.5 \, cm)$.

(A) The man can see the image of the source $S$ in the mirror as long as he remains within the field of view of the image.
Let the mirror be $AB$ with length $L = 20\,cm$. The source $S$ is at a distance $d_1 = 10\,cm$ from the mirror.
The man walks along a line at a distance $d_2 = 20\,cm$ from the mirror. The total distance of the man from the source $S$ is $d_1 + d_2 = 10\,cm + 20\,cm = 30\,cm$.
By using similar triangles,the width of the field of view $PQ$ at a distance $d_1 + d_2$ from the mirror is given by:
$\frac{PQ}{L} = \frac{d_1 + d_2}{d_1} = \frac{30\,cm}{10\,cm} = 3$
$PQ = 3 \times L = 3 \times 20\,cm = 60\,cm$.
The man walks at a speed $v = 10\,cm/s$.
The time $t$ for which he can see the image is $t = \frac{PQ}{v} = \frac{60\,cm}{10\,cm/s} = 6\,s$.

(A) Let the two mirrors be $M_1$ and $M_2$ separated by distance $L$. The object $O$ is at distance $x = L/3$ from $M_1$ and $y = 2L/3$ from $M_2$.
Images formed by $M_1$ are at distances $x, 2L+x, 4L+x, \dots$ from $M_1$,i.e.,$L/3, 7L/3, 13L/3, \dots$
Images formed by $M_2$ are at distances $y, 2L+y, 4L+y, \dots$ from $M_2$,i.e.,$2L/3, 8L/3, 14L/3, \dots$
Converting all positions relative to $M_1$:
Positions from $M_1$: $L/3, 7L/3, 13L/3, \dots$ and $(L - 2L/3) = L/3$ (reflected),$(L + 2L/3) = 5L/3, (3L + 2L/3) = 11L/3, \dots$
Sorted positions from $M_1$: $L/3, 5L/3, 7L/3, 11L/3, 13L/3, 17L/3, \dots$
The gaps between consecutive images are: $4L/3, 2L/3, 4L/3, 2L/3, 4L/3, \dots$
Since the possible distances between any two images are multiples of $2L/3$ (specifically $2L/3, 4L/3, 6L/3=2L, 8L/3, \dots$),the value $\frac{3L}{2}$ is not possible.

(C) Let $d = 10 \, cm = 0.1 \, m$ be the distance between the mirrors and $L = 20 \, m$ be the length of the mirrors.
After each reflection,the ray covers a horizontal distance $x$ along the mirror.
From the geometry of the path,we have $\tan 53^{\circ} = \frac{x}{d}$.
Given $\tan 53^{\circ} = \frac{4}{3}$,we get $x = d \tan 53^{\circ} = 0.1 \times \frac{4}{3} \, m = \frac{0.4}{3} \, m$.
The total number of reflections $n$ is given by the ratio of the total length $L$ to the horizontal distance covered per reflection $x$.
$n = \frac{L}{x} = \frac{20}{0.4 / 3} = \frac{20 \times 3}{0.4} = \frac{60}{0.4} = 150$.
Thus,the light undergoes $150$ reflections.

(C) Let the incident ray make an angle of $10^o$ with the horizontal. The reflected ray is vertical, meaning it makes an angle of $90^o$ with the horizontal.
The angle between the incident ray and the reflected ray is the total deviation $\delta$.
Since the incident ray is $10^o$ below the horizontal and the reflected ray is $90^o$ above the horizontal, the angle between them is $\delta = 90^o + 10^o = 100^o$.
The law of reflection states that the angle of deviation $\delta$ is related to the angle of incidence $i$ by $\delta = 180^o - 2i$.
Therefore, $100^o = 180^o - 2i$, which gives $2i = 80^o$, or $i = 40^o$.
The angle of incidence $i$ is the angle between the incident ray and the normal to the mirror.
Let $\theta$ be the angle the mirror makes with the horizontal. The angle between the incident ray and the mirror surface is $i = 90^o - (\text{angle between incident ray and mirror surface})$.
From the geometry, the angle between the incident ray and the mirror surface is $\theta + 10^o$.
Thus, the angle of incidence $i = 90^o - (\theta + 10^o)$.
Substituting $i = 40^o$, we get $40^o = 90^o - \theta - 10^o$.
$40^o = 80^o - \theta$, which implies $\theta = 40^o$.

(C) To see the full-length image of an object of height $H$ in a plane mirror,the minimum length of the mirror required is $H/2$.
This is because the light rays from the top of the head and the feet must reflect off the mirror to reach the eyes.
Given the height of the man $H = 160\,cm$.
The minimum length of the mirror $L = H/2 = 160/2 = 80\,cm$.
The height of the eyes does not affect the minimum length of the mirror required to see the full image,provided the mirror is positioned correctly.

(A) virtual object is formed when incident rays are converging towards a point behind the mirror. In this case,the point $P$ behind the mirror acts as a virtual object.
When these rays strike the plane mirror,they are reflected and actually meet at a point $Q$ in front of the mirror.
Since the reflected rays actually meet at point $Q$,the image formed at $Q$ is a real image.
Therefore,the image of a virtual object due to a plane mirror is indeed real.
Both the Assertion and the Reason are correct,and the Reason correctly explains why the image is real.

(A) The number of images $n$ formed by two plane mirrors inclined at an angle $\theta$ is given by $n = \frac{360^o}{\theta} - 1$ if $\frac{360^o}{\theta}$ is an even integer.
If $\frac{360^o}{\theta}$ is an odd integer,then $n = \frac{360^o}{\theta} - 1$ if the object is placed symmetrically,and $n = \frac{360^o}{\theta}$ if the object is placed asymmetrically.
Given $\theta = 72^o$,so $\frac{360^o}{72^o} = 5$,which is an odd integer.
Since the object is placed asymmetrically,the number of images formed is $n = 5$.
Thus,the Assertion is correct.
The Reason states the general rule for the number of images formed by two inclined plane mirrors,which is correct.
Since the Assertion is a direct application of the rule stated in the Reason,the Reason is the correct explanation of the Assertion.

(B) plane mirror forms a real image when the object is virtual. $A$ virtual object is formed when converging light rays are incident on the mirror. The reflected rays then converge at a point in front of the mirror,forming a real image.
Conversely,if the object is real (diverging rays from a point source),the plane mirror forms a virtual image behind the mirror.
Therefore,the Assertion is correct because a plane mirror can form a real image if the incident rays are converging (virtual object).
The Reason is also correct because a plane mirror forms a virtual image for a real object.
However,the Reason does not explain why the Assertion is true; it describes a different case (real object vs. virtual object). Thus,both are correct,but the Reason is not the correct explanation of the Assertion.

(D) Given:
Angle of deflection of the mirror,$\theta = 3.5^{\circ}$.
Distance of the screen from the mirror,$D = 1.5\; m$.
When a plane mirror is rotated by an angle $\theta$,the reflected ray rotates by an angle $2\theta$.
Therefore,the angle of deflection of the reflected ray is $2\theta = 2 \times 3.5^{\circ} = 7.0^{\circ}$.
The displacement $(d)$ of the reflected spot of light on the screen is given by the relation:
$\tan(2\theta) = \frac{d}{D}$
Substituting the values:
$d = D \times \tan(7.0^{\circ})$
$d = 1.5\; m \times \tan(7.0^{\circ})$
Using $\tan(7.0^{\circ}) \approx 0.12278$
$d = 1.5 \times 0.12278 = 0.18417\; m$
Converting the displacement into centimeters:
$d = 0.18417 \times 100\; cm = 18.417\; cm \approx 18.4\; cm$.
Thus,the displacement of the reflected spot of light is $18.4\; cm$.

Reflection of light is the phenomenon of bouncing back of light rays into the same medium when they strike a polished surface.
Laws of reflection:
$(i)$ The incident ray,the reflected ray,and the normal to the surface at the point of incidence all lie in the same plane.
$(ii)$ The angle of incidence $(i)$ is equal to the angle of reflection $(r)$,i.e.,$\angle i = \angle r$.

(A) Mirrors are mainly classified into two types:
$(1)$ Plane mirror
$(2)$ Spherical mirror
Spherical mirrors are further divided into two types:
$(1)$ Concave mirror: If the inner (curved) surface of a spherical mirror is made reflecting,it is called a concave mirror.
$(2)$ Convex mirror: If the outer (bulging) surface of a spherical mirror is made reflecting,it is called a convex mirror.

(N/A) Wavelength of incident light,$\lambda = 5000 \,\mathring{A} = 5000 \times 10^{-10} \,m$.
Speed of light,$c = 3 \times 10^{8} \,m/s$.
Frequency of incident light is given by the relation,$v = \frac{c}{\lambda}$.
$v = \frac{3 \times 10^{8}}{5000 \times 10^{-10}} = 6 \times 10^{14} \,Hz$.
The wavelength and frequency of reflected light remain the same as that of the incident light.
Hence,the wavelength of reflected light is $5000 \,\mathring{A}$ and its frequency is $6 \times 10^{14} \,Hz$.
When the reflected ray is normal to the incident ray,the sum of the angle of incidence,$\angle i$,and the angle of reflection,$\angle r$,is $90^{\circ}$.
According to the law of reflection,the angle of incidence is always equal to the angle of reflection,i.e.,$\angle i = \angle r$.
Therefore,$\angle i + \angle i = 90^{\circ} \implies 2\angle i = 90^{\circ} \implies \angle i = 45^{\circ}$.
The angle of incidence for the given condition is $45^{\circ}$.

(C) According to the law of reflection,the angle of incidence equals the angle of reflection,and the incident ray,reflected ray,and normal lie in the same plane.
Let the normal vector be $\overrightarrow{ c }$. The component of the incident ray $\overrightarrow{ a }$ along the normal is $(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$.
The component of $\overrightarrow{ a }$ perpendicular to the normal is $\overrightarrow{ a } - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$.
Since the reflected ray $\overrightarrow{ b }$ has the same component perpendicular to the normal but the opposite component along the normal,we have:
$\overrightarrow{ b } = (\overrightarrow{ a } - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }) - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$
$\overrightarrow{ b } = \overrightarrow{ a } - 2(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$

(C) Let the mirror be placed along the $y$-axis with its centre at the origin $(0,0)$. The mirror extends from $y = -25 \,cm$ to $y = +25 \,cm$.
The light source $S$ is at $(60 \,cm, 0)$.
The image $S'$ of the source $S$ in the plane mirror is formed at $(-60 \,cm, 0)$.
The man walks along the line $x = 120 \,cm - 60 \,cm = 60 \,cm$ relative to the mirror plane,or more simply,the man is at a distance of $1.2 \,m = 120 \,cm$ from the mirror.
The rays from the image $S'$ that define the field of view pass through the edges of the mirror at $y = 25 \,cm$ and $y = -25 \,cm$.
Using similar triangles,the ratio of the height to the distance from the image $S'$ is constant:
$\frac{y_{edge}}{distance_{S'}} = \frac{x}{distance_{S'} + distance_{man}}$
Here,the distance of the image $S'$ from the mirror is $60 \,cm$. The distance of the man from the mirror is $120 \,cm$.
So,the total distance of the man from the image $S'$ is $60 \,cm + 120 \,cm = 180 \,cm$.
Using the property of similar triangles:
$\frac{25 \,cm}{60 \,cm} = \frac{x}{180 \,cm}$
$x = \frac{25 \times 180}{60} = 25 \times 3 = 75 \,cm$.
This $x$ is the distance from the central axis to the extreme point on one side.
The total distance between the two extreme points is $2x = 2 \times 75 \,cm = 150 \,cm$.

(B) Let the corner where the mirrors meet be the origin $(0,0)$. The position of the point source $P$ is $(a, 2a)$.
The image formed by mirror ${M}_{1}$ (at $x=0$) is ${I}_{1} = (-a, 2a)$.
The image formed by mirror ${M}_{2}$ (at $y=0$) is ${I}_{2} = (a, -2a)$.
The image formed by both mirrors (the intersection of the reflections) is ${I}_{3} = (-a, -2a)$.
We need to find the distances between all pairs of images:
$1$. Distance between ${I}_{1}$ and ${I}_{2}$: $\sqrt{(a - (-a))^2 + (-2a - 2a)^2} = \sqrt{(2a)^2 + (-4a)^2} = \sqrt{4a^2 + 16a^2} = \sqrt{20a^2} = 2\sqrt{5}a = 2 \times 2.3a = 4.6a$.
$2$. Distance between ${I}_{1}$ and ${I}_{3}$: $\sqrt{(-a - (-a))^2 + (-2a - 2a)^2} = \sqrt{0 + (-4a)^2} = 4a$.
$3$. Distance between ${I}_{2}$ and ${I}_{3}$: $\sqrt{(-a - a)^2 + (-2a - (-2a))^2} = \sqrt{(-2a)^2 + 0} = 2a$.
The longest distance is $4.6a$. Thus,the correct option is $B$.

(C) The ripples act as plane mirrors reflecting light to form an image with an angular extent $\delta$.
Given that $O$ and $E$ are at the same height,$O E B C$ forms a trapezium with $O E \parallel B C$.
Let $\beta$ be the angle of incidence at point $B$ and $\gamma$ be the angle of incidence at point $C$.
From the geometry of the reflection,the angle of the reflected ray with the horizontal is affected by the tilt $\alpha$ of the mirror surface.
Using the law of reflection and the geometric properties of the angles shown in the diagram:
$\angle 1 = 90^{\circ} - \beta - \alpha$
$\angle 2 = 90^{\circ} - \gamma - \alpha$
Since $O$ and $E$ are at the same height,the symmetry implies $\beta = \gamma$.
The angular extent $\delta$ is the angle subtended by the reflected rays at the observer $E$.
From the triangle geometry,the total deviation caused by the tilted mirrors results in $\delta = 2 \alpha$.

(C) For the light beam to exit through the same hole after reflecting off every other wall exactly once,the path must be symmetric with respect to the diagonal of the square.
Let the angle of incidence be $\theta_i$. The angle of reflection at the first wall will also be $\theta_i$.
By tracing the path through the square,the geometry requires that the angle of incidence at the subsequent walls leads to a path that returns to the opening at an angle of exit $e = \theta_i$.
For the beam to exit through the same hole after hitting all three other walls,the total deviation must be such that the beam is parallel to its original path but reversed,or follows a symmetric path.
In a square enclosure,for the light to hit all three walls and return to the opening,the geometry dictates that the angle $\theta_i$ must satisfy the condition $2\theta_i = 90^{\circ}$,which gives $\theta_i = 45^{\circ}$.
Thus,the beam will come out only at $\theta_i = 45^{\circ}$.

(D) When a light ray undergoes two successive reflections from two plane mirrors inclined at an angle $\theta$,the total deviation $\delta$ produced is given by the formula $\delta = 360^{\circ} - 2\theta$.
In this problem,the emergent ray is parallel to the incident ray,which means the total deviation $\delta$ is $180^{\circ}$.
Substituting this value into the formula:
$180^{\circ} = 360^{\circ} - 2\theta$
Rearranging the terms to solve for $\theta$:
$2\theta = 360^{\circ} - 180^{\circ}$
$2\theta = 180^{\circ}$
$\theta = 90^{\circ}$
Therefore,the angle between the mirrors must be $90^{\circ}$ for the reflected ray to be parallel to the incident ray.

(C) The incident ray travels in the negative $y$-direction,so its direction vector is $-\hat{j}$.
The mirror makes an angle of $30^{\circ}$ with the $Y$-axis. The normal to the mirror makes an angle of $30^{\circ}$ with the $X$-axis (or $60^{\circ}$ with the $Y$-axis).
According to the law of reflection,the angle of incidence equals the angle of reflection.
The incident ray makes an angle of $30^{\circ}$ with the mirror surface. Thus,the reflected ray also makes an angle of $30^{\circ}$ with the mirror surface.
Looking at the geometry,the reflected ray makes an angle of $30^{\circ}$ with the $X$-axis in the fourth quadrant.
The direction vector of the reflected ray can be written as $\vec{v} = v_x \hat{i} + v_y \hat{j}$.
Since it is in the fourth quadrant,$v_x > 0$ and $v_y < 0$.
The angle with the positive $X$-axis is $-30^{\circ}$.
Thus,the direction is proportional to $(\cos(-30^{\circ})\hat{i} + \sin(-30^{\circ})\hat{j}) = (\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j})$.
Multiplying by $2$,we get the vector $\sqrt{3}\hat{i} - \hat{j}$.
Therefore,the correct option is $(c)$.

(C) From the geometry of the figure,the shadow length on the screen is $CD = H$.
From similar triangles $\triangle BGF$ and $\triangle DEF$,we have:
$\frac{DE}{BG} = \frac{FE}{GF} \Rightarrow \frac{h'}{h} = \frac{d-x}{x} \Rightarrow \frac{d}{x} = \frac{h' + h}{h} \quad \dots(i)$
Now,from similar triangles $\triangle ABG$ and $\triangle ACE$,we have:
$\frac{CE}{AE} = \frac{BG}{AG} \Rightarrow \frac{H + h'}{d + x} = \frac{h}{x} \Rightarrow \frac{d+x}{x} = \frac{H + h'}{h} \Rightarrow \frac{d}{x} + 1 = \frac{H + h'}{h} \Rightarrow \frac{d}{x} = \frac{H + h' - h}{h} \quad \dots(ii)$
Equating equations $(i)$ and $(ii)$:
$\frac{h' + h}{h} = \frac{H + h' - h}{h}$
$h' + h = H + h' - h$
$H = 2h$
Thus,the height of the shadow on the wall is $2h$.

(A) When a plane mirror is rotated by an angle $\delta$ about an axis in its plane,the normal to the mirror also rotates by the same angle $\delta$.
Let the angle of incidence be $i$. The angle of reflection is also $i$.
When the mirror rotates by $\delta$,the new angle of incidence becomes $i \pm \delta$ (depending on the direction of rotation).
The new angle of reflection will also be $i \pm \delta$.
The change in the direction of the reflected ray is the difference between the new angle of reflection and the original angle of reflection relative to the incident ray.
Mathematically,it is a standard result in optics that if a plane mirror is rotated by an angle $\delta$,the reflected ray rotates by an angle $2 \delta$ in the same direction.

(A) Digital movie projectors,specifically those using Digital Light Processing $(DLP)$ technology,utilize a Digital Micromirror Device $(DMD)$.
This device consists of millions of microscopic mirrors arranged on a semiconductor chip.
Each mirror corresponds to a single pixel in the projected image.
By tilting these micromirrors,the projector reflects light either towards the lens (to create a bright pixel) or away from it (to create a dark pixel).
Therefore,the fundamental principle is the reflection of light from micromirrors.

(B) The day and night setting in a rearview mirror uses a thick wedge-shaped mirror.
In the 'day' position,the front surface of the glass reflects the light,providing a bright image.
In the 'night' position,the mirror is tilted so that the light reflects from the silvered back surface of the wedge,which is much dimmer,thereby reducing glare from the headlights of vehicles behind.

(A) real image is formed when light rays actually intersect after reflection.
For a plane mirror,a real image is formed only when the incident beam of light is converging towards a point behind the mirror.
In this case,the point behind the mirror acts as a virtual object for the plane mirror.
The reflected rays then converge at a point in front of the mirror,forming a real image.
Therefore,the incident beam must be converging.

(C) Let the velocity of the ball be $\vec{v}_b = u \cos \theta \hat{i} + u \sin \theta \hat{j}$.
Assuming the mirror is placed vertically in the $yz$-plane,the velocity of the image of the ball $\vec{v}_i$ will be $\vec{v}_i = -u \cos \theta \hat{i} + u \sin \theta \hat{j}$.
The velocity of the image with respect to the ball is given by $\vec{v}_{ib} = \vec{v}_i - \vec{v}_b$.
Substituting the values: $\vec{v}_{ib} = (-u \cos \theta \hat{i} + u \sin \theta \hat{j}) - (u \cos \theta \hat{i} + u \sin \theta \hat{j})$.
$\vec{v}_{ib} = -2u \cos \theta \hat{i}$.
Since the velocity is constant and directed along the negative $x$-axis,the motion is a straight line and horizontal.

(A) plane mirror always forms an image that is virtual,erect,of the same size as the object,and laterally inverted.
Since the image is virtual,it is not real.
Since the image is erect,it satisfies condition $B$.
Since the image is of the same size,it does not satisfy condition $C$.
Since the image is laterally inverted,it satisfies condition $D$.
Therefore,the correct characteristics are $B$ (Erect) and $D$ (Laterally inverted).

(B) Let the distance between the mirrors be $d = 10\,cm$. The object $O$ is at $x_A = 2\,cm$ from mirror $A$ and $x_B = 8\,cm$ from mirror $B$.
$1$. The first image $I_1$ formed by mirror $A$ is at a distance of $2\,cm$ behind mirror $A$.
$2$. The first image $I_2$ formed by mirror $B$ is at a distance of $8\,cm$ behind mirror $B$.
$3$. The second image formed by mirror $A$ $(I_3)$ is the image of $I_2$ formed by mirror $A$. The distance of $I_2$ from mirror $A$ is $10\,cm + 8\,cm = 18\,cm$. Therefore,$I_3$ is formed at $18\,cm$ behind mirror $A$.
The images behind mirror $A$ are at distances $2\,cm$ $(I_1)$ and $18\,cm$ $(I_3)$. The second nearest image behind mirror $A$ is at a distance of $18\,cm$ from mirror $A$.

(B) Initially,the object is at $x = -12\,cm$ and the mirror is at $x = 0$. The image is formed at $x = +12\,cm$.
When the mirror is moved $4\,cm$ towards the object,the new position of the mirror is $x = -4\,cm$.
The distance of the object from the new mirror position is $u = |-12 - (-4)| = 8\,cm$.
Since the image is formed at the same distance behind the mirror,the new image position $x_i$ satisfies $|x_i - (-4)| = 8\,cm$,so $x_i = 4\,cm$.
The initial position of the image was $12\,cm$ and the final position is $4\,cm$.
Therefore,the shift in the position of the image is $12\,cm - 4\,cm = 8\,cm$ towards the mirror.

(B) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$. Given $r = 30^{\circ}$,therefore $i = 30^{\circ}$.
The angle of deviation $\delta$ for a plane mirror is given by the formula $\delta = 180^{\circ} - (i + r)$.
Substituting the values: $\delta = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,the angle of deviation is $120^{\circ}$.

(C) Let the incident unit vector be $\vec{a} = \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}$ and the reflected unit vector be $\vec{b} = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}$.
The angle $\theta$ between these two vectors is given by the dot product formula: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
Since these are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.
$\vec{a} \cdot \vec{b} = (\frac{1}{2})(\frac{1}{2}) + (\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2}) = \frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2}$.
Therefore,$\cos \theta = -\frac{1}{2}$,which implies $\theta = 120^{\circ}$.
The angle between the incident ray and the reflected ray is $120^{\circ}$.
The angle of deviation $\delta = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
The angle of deviation is also given by $\delta = 180^{\circ} - 2i$,where $i$ is the angle of incidence.
$60^{\circ} = 180^{\circ} - 2i \implies 2i = 120^{\circ} \implies i = 60^{\circ}$.
Alternatively,looking at the vectors,the angle with the $y$-axis (normal) is $\cos^{-1}(\frac{\sqrt{3}}{2}) = 30^{\circ}$ if the mirror is along the $x$-axis,but the vectors indicate the normal is the $y$-axis. The angle with the normal is $60^{\circ}$.

(B) The magnification $(m)$ of a plane mirror is $+1$.
This is because a plane mirror creates an image that is the same size as the object $(h_i = h_o)$.
The image formed by a plane mirror is virtual and erect,which means the magnification must be positive.
Since the height of the image is equal to the height of the object,the ratio is $1$.
Therefore,the magnification is calculated as:
$m = \frac{h_i}{h_o} = 1$.