Plane Mirror Questions in English

(D) The number of images $n$ formed by two plane mirrors inclined at an angle $\theta$ is given by the formula $n = \left( \frac{360^{\circ}}{\theta} - 1 \right)$ if $\frac{360^{\circ}}{\theta}$ is an even integer.
Given that the number of images $n = 3$, we substitute this into the formula:
$3 = \frac{360^{\circ}}{\theta} - 1$
$4 = \frac{360^{\circ}}{\theta}$
$\theta = \frac{360^{\circ}}{4} = 90^{\circ}$
Thus, the angle between the two plane mirrors should be $90^{\circ}$.

(A) The angle of incidence is defined as the angle between the incident ray and the normal to the surface at the point of incidence.
However,the angle between the incident ray and the reflecting surface itself is known as the glancing angle.
In the provided diagram,the angle $\theta$ represents the angle between the incident ray $I$ and the surface $XY$,which is the glancing angle.
Therefore,the correct option is $A$.

(B) When two plane mirrors are placed at an angle $\theta$,the total deviation $\delta$ produced by two successive reflections is given by $\delta = 360^{\circ} - 2\theta$.
Given that the mirrors are perpendicular,$\theta = 90^{\circ}$.
Substituting the value of $\theta$ in the formula:
$\delta = 360^{\circ} - 2(90^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}$.
$A$ deviation of $180^{\circ}$ means the final ray is directed exactly opposite to the incident ray,which implies the emergent ray is parallel to the incident ray.

(D) The focal length $f$ of a spherical mirror is related to its radius of curvature $R$ by the formula $f = R/2$. $A$ plane mirror can be considered as a spherical mirror with an infinite radius of curvature $(R = \infty)$. Therefore,the focal length of a plane mirror is $f = \infty/2 = \infty$. Thus,the correct option is $D$.

(B) The focal length of a plane mirror is $f = \infty$.
The power $P$ of a mirror is defined as the reciprocal of its focal length in meters,given by $P = \frac{1}{f}$.
Substituting the value of $f$:
$P = \frac{1}{\infty} = 0 \text{ dioptre}$.
Therefore,the power of a plane mirror is $0$.

(C) Given,the angle of incidence,$i = 60^{\circ}$.
According to the law of reflection,the angle of reflection $r$ is equal to the angle of incidence $i$,so $r = 60^{\circ}$.
The angle of deviation $\delta$ produced by a plane mirror is given by the formula:
$\delta = 180^{\circ} - (i + r)$
Since $i = r = 60^{\circ}$,we have:
$\delta = 180^{\circ} - (60^{\circ} + 60^{\circ})$
$\delta = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

(A) When a ray of light is incident normally on a plane mirror,it means the ray travels along the normal to the mirror surface.
Therefore,the angle of incidence,$i$,which is the angle between the incident ray and the normal,is $0^{\circ}$.
According to the laws of reflection,the angle of incidence is equal to the angle of reflection $(i = r)$.
Thus,the angle of reflection,$r = 0^{\circ}$.

(C) Let the girl's eye be at point $O$ at a height of $140 \,cm$ from the ground. The mirror extends from $85 \,cm$ to $160 \,cm$ $(85 + 75 = 160 \,cm)$ above the ground.
$1$. The girl can see the portion of her body below her eye level by looking at the lower part of the mirror. The distance from her eye to the lower edge of the mirror is $140 - 85 = 55 \,cm$. By the property of reflection,the lowest point she can see on her body is $55 \,cm$ below the level of the lower edge of the mirror,which is $85 - 55 = 30 \,cm$ from the ground.
$2$. The girl can see the portion of her body above her eye level by looking at the upper part of the mirror. The distance from her eye to the top edge of the mirror is $160 - 140 = 20 \,cm$. By the property of reflection,the highest point she can see on her body is $20 \,cm$ above the level of the top edge of the mirror,which is $160 + 20 = 180 \,cm$. However,the girl's height is only $150 \,cm$. Thus,she can see up to the top of her head $(150 \,cm)$.
$3$. The total height of the image visible to the girl is the distance from the lowest point she can see $(30 \,cm)$ to the highest point she can see $(150 \,cm)$.
Height visible = $150 \,cm - 30 \,cm = 120 \,cm$.
Therefore,the correct option is $(c)$.

(C) The object is at the bottom of the vessel,which is filled with water to a height of $10 \,cm$. The plane mirror is placed $7 \,cm$ above the water surface.
Due to refraction at the water-air interface,the object at the bottom appears to be at a shallower depth.
The apparent depth $d'$ is given by the formula:
$d' = \frac{\text{Real depth}}{n} = \frac{10 \,cm}{1.33} \approx 7.52 \,cm$.
The total distance of the apparent position of the object from the mirror is the sum of the distance of the mirror from the water surface and the apparent depth of the object:
$D = 7 \,cm + 7.52 \,cm = 14.52 \,cm$.
Since a plane mirror forms an image at the same distance behind it as the object is in front of it,the distance of the image from the mirror is $14.52 \,cm$,which is approximately $14.5 \,cm$.

(A) Let $\hat{t}$ be the unit vector along the normal to the surface.
Let $\hat{n}_1$ be the unit vector along the incident ray and $\hat{n}_2$ be the unit vector along the reflected ray.
From the geometry of reflection,the angle of incidence equals the angle of reflection,$\theta$.
We can resolve $\hat{n}_1$ and $\hat{n}_2$ into components parallel and perpendicular to the normal $\hat{t}$.
$\hat{n}_1 = -\cos \theta \hat{t} + \sin \theta \hat{u}$,where $\hat{u}$ is a unit vector parallel to the surface.
$\hat{n}_2 = \cos \theta \hat{t} + \sin \theta \hat{u}$.
Subtracting the two equations: $\hat{n}_2 - \hat{n}_1 = 2 \cos \theta \hat{t}$.
Since $\hat{n}_1 \cdot \hat{t} = \cos(180^\circ - \theta) = -\cos \theta$,we have $\cos \theta = -(\hat{n}_1 \cdot \hat{t})$.
Substituting this into the subtraction equation: $\hat{n}_2 - \hat{n}_1 = -2(\hat{n}_1 \cdot \hat{t}) \hat{t}$.
Therefore,$\hat{n}_2 = \hat{n}_1 - 2(\hat{n}_1 \cdot \hat{t}) \hat{t}$.

(B) The incident ray vector is $\hat{e}_0$ and the reflected ray vector is $\hat{e}$. The normal vector is $\hat{n}$.
From the law of reflection,the angle of incidence equals the angle of reflection,and all three vectors lie in the same plane.
Let $\theta$ be the angle between the incident ray and the normal. Since $\hat{e}_0$ is directed towards the mirror,the angle between $\hat{e}_0$ and $\hat{n}$ is $(180^\circ - \theta)$.
Thus,$\hat{e}_0 \cdot \hat{n} = |\hat{e}_0| |\hat{n}| \cos(180^\circ - \theta) = -\cos \theta$.
We can resolve the incident ray $\hat{e}_0$ into components parallel and perpendicular to the normal $\hat{n}$:
$\hat{e}_0 = \hat{e}_{0\perp} + \hat{e}_{0\parallel} = (\hat{e}_0 \cdot \hat{n}) \hat{n} + (\hat{e}_0 - (\hat{e}_0 \cdot \hat{n}) \hat{n})$.
The reflected ray $\hat{e}$ has the same component parallel to the mirror surface but the component along the normal is reversed:
$\hat{e} = -(\hat{e}_0 \cdot \hat{n}) \hat{n} + (\hat{e}_0 - (\hat{e}_0 \cdot \hat{n}) \hat{n})$.
$\hat{e} = \hat{e}_0 - 2(\hat{e}_0 \cdot \hat{n}) \hat{n}$.
Therefore,the correct option is $(B)$.

(A) Let the source be at $S(0,1)$ and the point on the mirror be $M$. The reflected ray passes through $P(3,3)$.
Since the mirror is along the $X$-axis,the image of the source $S(0,1)$ formed by the mirror is $I(0,-1)$.
The path length of the reflected ray is $SM + MP$.
By the law of reflection,$SM = IM$.
Therefore,the total path length is $IM + MP = IP$.
The distance $IP$ is the distance between points $I(0,-1)$ and $P(3,3)$.
Using the distance formula: $IP = \sqrt{(3-0)^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(C) The number of images $n$ formed by two plane mirrors inclined at an angle $\theta$ is given by $n = \frac{360^{\circ}}{\theta} - 1$ if $\frac{360^{\circ}}{\theta}$ is an even integer,or if it is an odd integer and the object is placed symmetrically.
Given $n = 5$,we have $5 = \frac{360^{\circ}}{\theta} - 1$,which implies $\frac{360^{\circ}}{\theta} = 6$.
Thus,$\theta = \frac{360^{\circ}}{6} = 60^{\circ}$.
When the angle is decreased by $30^{\circ}$,the new angle $\theta' = 60^{\circ} - 30^{\circ} = 30^{\circ}$.
The new number of images $n'$ is given by $n' = \frac{360^{\circ}}{30^{\circ}} - 1 = 12 - 1 = 11$.