Mix Examples - Light – Reflection and Refraction Questions in English

(B) Given that the image is real, inverted, and of the same size as the object, the object must be placed at $2f$ from the convex lens.
Since the image is formed at a distance of $v = 50\, cm$, we have $2f = 50\, cm$.
Therefore, the focal length $f = 25\, cm = 0.25\, m$.
To obtain an image of the same size, the object must be placed at $2f$, which is $50\, cm$ in front of the lens.
The power of the lens $P$ is given by the formula $P = \frac{1}{f(\text{in meters})}$.
Substituting the value, $P = \frac{1}{0.25} = 4\, D$.

(A) Given:
Refractive index of dense flint glass with respect to air,$_{a}\mu_{g} = 1.65$
Refractive index of alcohol with respect to air,$_{a}\mu_{alc} = 1.36$
To find the refractive index of dense flint glass with respect to alcohol $(_{alc}\mu_{g})$,we use the formula:
$_{alc}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{alc}}$
Substituting the given values:
$_{alc}\mu_{g} = \frac{1.65}{1.36} \approx 1.21$
Therefore,the refractive index of dense flint glass with respect to alcohol is $1.21$.

(A) When lenses are placed in contact,the effective power $(P)$ of the combination is the algebraic sum of the individual powers of the lenses.
$P = P_1 + P_2$
Given $P_1 = +3.5 D$ and $P_2 = -2.5 D$.
Therefore,$P = +3.5 D + (-2.5 D) = +1.0 D$.
The focal length $(f)$ of the combination is given by the reciprocal of the total power $(f = 1/P)$.
$f = 1 / 1.0 D = 1.0 m$.
Since the power is positive,the combination behaves as a convex lens with a focal length of $1.0 m$.

(D) Given: Focal length $f = +25\, cm$,Image distance $v = +75\, cm$ (since it is on the opposite side).
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{75} - \frac{1}{u} = \frac{1}{25}$.
Rearranging for $u$: $\frac{1}{u} = \frac{1}{75} - \frac{1}{25} = \frac{1 - 3}{75} = \frac{-2}{75}$.
Thus,$u = -37.5\, cm$.
The negative sign indicates the object is placed $37.5\, cm$ in front of the lens.
Since $v$ is positive and the image is formed on the opposite side,the image is real and inverted. Also,since $|v| > |u|$ $(75 > 37.5)$,the image is magnified.

(A) Given: Object height $O = 5 \, cm$,Object distance $u = -30 \, cm$,Focal length $f = +20 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-30} = \frac{3-2}{60} = \frac{1}{60}$.
Therefore,$v = 60 \, cm$. The positive sign indicates the image is formed on the other side of the lens.
Using magnification formula $m = \frac{I}{O} = \frac{v}{u}$:
$m = \frac{I}{5} = \frac{60}{-30} = -2$.
$I = 5 \times (-2) = -10 \, cm$.
$(i)$ Position: $60 \, cm$ from the lens on the opposite side.
(ii) Nature: Real and inverted (since $v$ is positive and $I$ is negative).
(iii) Size: $10 \, cm$ tall.

(D) Given: Object height $O = 10 \, cm$,object distance $u = -20 \, cm$,focal length $f = +30 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{30} + \frac{1}{-20} = \frac{2 - 3}{60} = -\frac{1}{60}$.
Therefore,the position of the image is $v = -60 \, cm$ (the image is formed on the same side as the object).
Using the magnification formula $m = \frac{I}{O} = \frac{v}{u}$,we have:
$\frac{I}{10} = \frac{-60}{-20} = 3$.
Therefore,the size of the image is $I = 30 \, cm$.
Since $v$ is negative,the image is virtual and erect. Since the height $I$ is positive and greater than $O$,the image is magnified.

(A) Given: Height of object $O = 3 \, cm$,Object distance $u = -20 \, cm$,Focal length $f = +12 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{12} + \frac{1}{-20} = \frac{5 - 3}{60} = \frac{2}{60} = \frac{1}{30}$.
Therefore,$v = +30 \, cm$. The positive sign indicates the image is formed on the other side of the lens.
Using magnification $m = \frac{I}{O} = \frac{v}{u}$,we have:
$\frac{I}{3} = \frac{30}{-20} = -1.5$.
$I = 3 \times (-1.5) = -4.5 \, cm$.
The image is real,inverted,and $4.5 \, cm$ in size.

(D) Given: Object height $O = 4 \ cm$,object distance $u = -27 \ cm$,focal length $f = +18 \ cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{18} + \frac{1}{-27} = \frac{3-2}{54} = \frac{1}{54}$.
Therefore,the image position $v = +54 \ cm$.
Using magnification formula $m = \frac{I}{O} = \frac{v}{u}$,we have:
$\frac{I}{4} = \frac{54}{-27} = -2$.
Therefore,the image size $I = 4 \times (-2) = -8 \ cm$.
The positive value of $v$ indicates that the image is formed on the other side of the lens,and the negative value of $I$ indicates that the image is real and inverted.

(A) Given: Object height $h = 3 \, cm$,Object distance $u = -9 \, cm$,Focal length $f = -18 \, cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-18} = \frac{1}{v} + \frac{1}{-9}$.
$\frac{1}{v} = \frac{1}{9} - \frac{1}{18} = \frac{2-1}{18} = \frac{1}{18}$.
Therefore,$v = +18 \, cm$. Since $v$ is positive,the image is formed $18 \, cm$ behind the mirror,which means it is virtual and erect.
Now,magnification $m = \frac{h'}{h} = -\frac{v}{u}$.
$h' = -\frac{v \times h}{u} = -\frac{18 \times 3}{-9} = 6 \, cm$.
Thus,the image is virtual,erect,$18 \, cm$ behind the mirror,and $6 \, cm$ in size.

(1.18) The refractive index of a medium $2$ with respect to medium $1$ is given by the formula: ${}_{1}n_{2} = \frac{{}_{a}n_{2}}{{}_{a}n_{1}}$.
Here,medium $1$ is ice $(i)$ and medium $2$ is rock salt $(r)$.
Given: Refractive index of ice with respect to air $({}_{a}n_{i})$ = $1.31$.
Refractive index of rock salt with respect to air $({}_{a}n_{r})$ = $1.54$.
Therefore,the refractive index of rock salt with respect to ice $({}_{i}n_{r})$ = $\frac{1.54}{1.31} \approx 1.18$.

(A) Given:
Speed of light in air $(c)$ = $3.0 \times 10^{8} \text{ m s}^{-1}$
Refractive index of diamond $(n)$ = $2.42$
Speed of light in diamond $(v)$ = ?
Formula:
The refractive index $(n)$ is defined as the ratio of the speed of light in air $(c)$ to the speed of light in the medium $(v)$:
$n = \frac{c}{v}$
Calculation:
Rearranging the formula to solve for $v$:
$v = \frac{c}{n}$
$v = \frac{3.0 \times 10^{8}}{2.42}$
$v \approx 1.24 \times 10^{8} \text{ m s}^{-1}$
Therefore,the speed of light in diamond is $1.24 \times 10^{8} \text{ m s}^{-1}$.

(A) Given: Object height $O = 3.0 \text{ cm}$,focal length $f = -7.5 \text{ cm}$,image distance $v = -5.0 \text{ cm}$.
$(i)$ Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-5.0} - \frac{1}{-7.5} = -0.2 + 0.1333 = -0.0667 \text{ cm}^{-1}$.
Thus,$u = -15 \text{ cm}$. The object is placed $15 \text{ cm}$ in front of the concave lens.
(ii) Magnification $m = \frac{I}{O} = \frac{v}{u}$.
$I = \frac{v}{u} \times O = \frac{-5.0}{-15.0} \times 3.0 = 1.0 \text{ cm}$.
Since the magnification is positive $(m = +0.33)$,the image is virtual,erect,and $1.0 \text{ cm}$ in size.

(A) Given: Object distance $u = -12 \ cm$. Magnification $m = -4$ (since the image is real and inverted).
Using the magnification formula for a spherical mirror: $m = -\frac{v}{u}$.
Substituting the values: $-4 = -\frac{v}{-12}$.
Solving for $v$: $-4 = \frac{v}{12} \implies v = -48 \ cm$.
The negative sign indicates that the image is formed in front of the mirror,which is consistent with a real image. Thus,the distance of the image from the mirror is $48 \ cm$.

(A) Given: Object distance $u = -12 \, cm$,Image distance $v = 18 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{18} - \frac{1}{-12} = \frac{1}{18} + \frac{1}{12}$.
Taking the $LCM$ of $18$ and $12$ as $36$: $\frac{1}{f} = \frac{2+3}{36} = \frac{5}{36}$.
Therefore,$f = \frac{36}{5} = 7.2 \, cm$.
The magnification $m$ is given by $m = \frac{v}{u} = \frac{18}{-12} = -1.5$.
Since the magnitude of magnification $|m| = 1.5 > 1$,the image is magnified.

(D) $(i)$ For power $P = +2 \text{ D}$,the lens is a convex lens. Focal length $f = \frac{1}{P} = \frac{1}{+2} = +0.5 \text{ m} = +50 \text{ cm}$.
$(ii)$ For power $P = -4 \text{ D}$,the lens is a concave lens. Focal length $f = \frac{1}{P} = \frac{1}{-4} = -0.25 \text{ m} = -25 \text{ cm}$.
$(b)$ Given: Power $P = -4 \text{ D}$,object distance $u = -100 \text{ cm}$.
Focal length $f = \frac{1}{P} = -25 \text{ cm}$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-25} + \frac{1}{-100} = \frac{-4 - 1}{100} = \frac{-5}{100} = -\frac{1}{20}$.
Therefore,$v = -20 \text{ cm}$. The image is formed at a distance of $20 \text{ cm}$ on the same side as the object.

(N/A) Given: Magnification $m = -3$ (for a real image),Object distance $u = -10\, cm$. We need to find the focal length $f$.
Using the magnification formula: $m = -v/u$
$-3 = -v / (-10)$
$-3 = v / 10$
$v = -30\, cm$
Using the mirror formula: $1/f = 1/v + 1/u$
$1/f = 1/(-30) + 1/(-10)$
$1/f = (-1 - 3) / 30 = -4 / 30$
$f = -30 / 4 = -7.5\, cm$
Thus,the focal length of the concave mirror is $-7.5\, cm$.
$(b)$ The ray diagram for an object placed $6\, cm$ in front of a convex mirror is shown below.

(A) Given: Object height $O = 1 \, cm$,Object distance $u = -15 \, cm$,Focal length $f = -10 \, cm$.
Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = \frac{-3 + 2}{30} = \frac{-1}{30}$.
Thus,$v = -30 \, cm$. The negative sign indicates the image is formed in front of the mirror.
Magnification $m = \frac{I}{O} = -\frac{v}{u}$.
$I = -\frac{v}{u} \times O = -\frac{-30}{-15} \times 1 = -2 \, cm$.
The image is $30 \, cm$ in front of the mirror,$2 \, cm$ tall,and is real and inverted.

(A) Given: Object height $O = 2.0 \text{ cm}$,Focal length $f = +10 \text{ cm}$,Object distance $u = -15 \text{ cm}$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-15} = \frac{3 - 2}{30} = \frac{1}{30}$.
Therefore,$v = +30 \text{ cm}$. The positive sign indicates the image is formed on the other side of the lens.
Magnification $m = \frac{v}{u} = \frac{I}{O}$.
$I = \frac{v}{u} \times O = \frac{30}{-15} \times 2.0 = -4 \text{ cm}$.
The image is real,inverted,and $4 \text{ cm}$ tall,formed at a distance of $30 \text{ cm}$ from the lens.

(N/A) The beam of light must be passing through the optical centre of the lens.
When a light ray passes through the optical centre of a lens,it does not undergo any refraction or deviation.
This is because the central part of the lens acts like a very thin glass slab,allowing the light to pass straight through without changing its path.

(D) When a light ray passes through a rectangular glass slab,it undergoes refraction at two parallel surfaces ($AB$ and $CD$).
According to the laws of refraction,when a light ray travels from a rarer medium (air) to a denser medium (glass),it bends towards the normal.
When it emerges from the denser medium (glass) back into the rarer medium (air),it bends away from the normal.
Because the opposite surfaces of the rectangular glass slab are parallel to each other,the emergent ray is parallel to the incident ray.
In the given diagram,the ray $S$ is parallel to the incident ray $IM$,which is the characteristic property of refraction through a rectangular glass slab.

(N/A) $(i)$ $A$ convex lens can produce a magnified and virtual image when the object is placed between the optical center and the principal focus.
$(ii)$ $A$ concave lens always produces a diminished and virtual image for any position of the object.

(N/A) converging mirror is a concave mirror. The focal length $f$ is $20 \, cm$.
Initially,the object is at $25 \, cm$,which is between the center of curvature $(C = 40 \, cm)$ and the focus $(F = 20 \, cm)$. In this position,the image formed is real,inverted,and magnified,located beyond the center of curvature.
As the object is moved from $25 \, cm$ to $20 \, cm$ (the focus),the image shifts further away from the mirror and its size increases.
When the object is at the focus ($15 \, cm$ is between the focus and the pole),the image is formed at infinity.
When the object is moved from $20 \, cm$ to $15 \, cm$,it is now placed between the focus and the pole.
In this new position,the image becomes virtual,erect,and magnified,and it is formed behind the mirror.

(N/A) We prefer a convex mirror for observing the traffic behind us because its field of view is much larger than that of a plane mirror.
Although it gives an erroneous idea about the speed of the vehicles behind us,we still prefer it due to its wider field of view.
The view shown by a plane mirror and a convex mirror is as shown in the figure. It is clear from the figure that the view produced by a convex mirror is wider than that produced by a plane mirror.

(B) The reason for this warning is that a convex mirror forms a virtual, erect, and diminished image of the objects behind the vehicle.
Because the image is diminished in size, the driver may perceive the object as being further away than it actually is.
This optical illusion can lead a driver to believe that the vehicle behind is at a safe distance when it is actually closer, potentially causing an accident if the driver changes lanes without caution.

(B) As the object is moved away from a convex mirror,the distance of the virtual image,formed behind the mirror,increases (the image remains between the pole $P$ and the focus $F$).
Specifically,the image shifts from the pole $P$ towards the focus $F$ as the object distance increases.
Simultaneously,the size of the image gradually decreases.
When the object is at infinity,the image is formed at the focus $F$ and is highly diminished in size.

(N/A) $1$. When the object is at infinity,the image is formed at the focus $(F)$,and it is highly diminished and inverted.
$2$. As the object moves from infinity towards the centre of curvature $(C)$,the image moves from the focus $(F)$ towards the centre of curvature $(C)$,and its size increases.
$3$. When the object is at the centre of curvature $(C)$,the image is formed at the centre of curvature $(C)$,and its size is equal to the size of the object.
$4$. As the object moves from the centre of curvature $(C)$ towards the focus $(F)$,the image moves away from the centre of curvature $(C)$ towards infinity,and its size becomes magnified.
$5$. When the object is at the focus $(F)$,the image is formed at infinity.
$6$. When the object is placed between the focus $(F)$ and the pole $(P)$,the image is formed behind the mirror,and it becomes virtual,erect,and magnified.

$(55^{\circ})$ According to the law of reflection, the angle of incidence $(i)$ is equal to the angle of reflection $(r)$.
Given that the total angle between the incident ray and the reflected ray is $70^{\circ}$, we have $i + r = 70^{\circ}$.
Since $i = r$, we can write $2i = 70^{\circ}$, which gives $i = 35^{\circ}$.
The normal to the mirror surface makes an angle of $90^{\circ}$ with the mirror.
Therefore, the angle $\theta$ between the incident ray and the mirror surface is $\theta = 90^{\circ} - i$.
Substituting the value of $i$, we get $\theta = 90^{\circ} - 35^{\circ} = 55^{\circ}$.

(C) The focal length of a mirror depends only on its radius of curvature $(f = R/2)$ and is independent of the refractive index of the surrounding medium. Therefore, the focal length of the concave mirror remains unchanged in water.
However, the focal length of a lens is determined by the Lens Maker's Formula: $1/f = (n_g/n_m - 1)(1/R_1 - 1/R_2)$, where $n_g$ is the refractive index of the lens material and $n_m$ is the refractive index of the medium. Since the refractive index of water $(n_m \approx 1.33)$ is greater than that of air $(n_m \approx 1.0)$, the relative refractive index $(n_g/n_m - 1)$ decreases, causing the focal length of the convex lens to increase.

(A) Yes,it is possible. The focal length $f$ of a lens in a medium is given by the Lens Maker's Formula: $\frac{1}{f} = (\frac{n_l}{n_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
If the refractive index of the surrounding medium $(n_m)$ is greater than the refractive index of the lens material $(n_l)$,the term $(\frac{n_l}{n_m} - 1)$ becomes negative.
This causes the lens to change its nature: a convex lens (convergent) will behave as a concave lens (divergent),and vice versa.

(B) Hold the given piece of glass over some printed matter.
$(i)$ If the letters appear magnified,the given piece is a convex lens.
$(ii)$ If the letters appear diminished,the given piece is a concave lens.
$(iii)$ If the letters appear to be of the same size,then it is a plane glass plate.

(N/A) The angle of incidence is the angle between the incident ray and the normal at the point of incidence.
$A$ line joining the centre of curvature $(C)$ and the point of incidence $(Y)$ acts as a normal to the spherical mirror at that point.
Therefore,the two points are $C$ and $Y$.
After reflection,a ray of light incident parallel to the principal axis of a concave mirror passes through its principal focus $(F)$. The course of the reflected ray is shown in the figure.

(N/A) The incident ray $XY$,which is parallel to the principal axis,after reflection from the concave mirror will pass through the principal focus $F$.
To mark the normal,we draw a line from the center of curvature $C$ to the point of incidence $Y$.
The angle between the incident ray $XY$ and the normal $CY$ is the angle of incidence $(i)$.
The angle between the reflected ray and the normal $CY$ is the angle of reflection $(r)$.
According to the laws of reflection,the angle of incidence is equal to the angle of reflection $(i = r)$.

(B) convex mirror always produces a virtual and erect image,regardless of the position of the object in front of the mirror.
In contrast,a concave mirror produces a virtual and erect image only when the object is placed between the pole $(P)$ and the focus $(F)$.
Since the image remains virtual and erect for any position of the object,the spherical mirror must be convex.

(N/A) $(i)$ When $n_{1} > n_{2}$,the lens acts as a denser medium compared to the surroundings. Therefore,the concave lens behaves as a diverging lens,and the parallel rays diverge after passing through it.
$(ii)$ When $n_{1} = n_{2}$,the lens and the surrounding medium have the same refractive index. There is no change in the optical density,so the light rays pass through the lens without any deviation or refraction.
$(iii)$ When $n_{1} < n_{2}$,the lens acts as a rarer medium compared to the surroundings. In this case,the nature of the lens reverses,and the concave lens behaves as a converging lens,causing the parallel rays to converge after passing through it.

(N/A) $(i)$ The refracted ray $BD$ is shown in the figure.
$(ii)$ The point through which the refracted ray passes on the principal axis is known as the principal focus (denoted as $F_2$ in the figure) of the lens.

(N/A) When an object is placed at the principal focus $(F)$ of a convex lens,the refracted rays become parallel to each other and meet at infinity. This results in the formation of a real,inverted,and highly enlarged image at infinity.
Therefore,the object should be placed at a distance of $10\, cm$ from the optical centre of the lens.
The ray diagram is as shown below.

(N/A) $(i)$ The magnification of the image formed is unity (or $1$).
$(ii)$ The image formed is $(1)$ real and $(2)$ inverted,and it is formed at the same position as the object.
$(iii)$ The distance between the object and the optical centre of the lens is called the focal length of the lens.

(N/A) $(i)$ Light speeds up when it travels from an optically denser to an optically rarer medium. In other words, it travels from a medium of high refractive index to a medium of low refractive index. Therefore, one such pair is from crown glass $(1.52)$ into water $(1.33)$.
$(ii)$ Light slows down when it travels from an optically rarer to an optically denser medium. In other words, it travels from a medium of low refractive index to a medium of high refractive index. One such pair is from water $(1.33)$ into diamond $(2.42)$.

(WATER) The refractive index $(n)$ of a medium is defined by the formula $n = \frac{c}{v}$,where $c$ is the speed of light in a vacuum and $v$ is the speed of light in the given medium.
Since $c$ is a constant,the relationship between speed and refractive index is $v = \frac{c}{n}$,which implies $v \propto \frac{1}{n}$.
This means that the speed of light is inversely proportional to the refractive index of the medium.
Comparing the given refractive indices: Water $(1.33)$ < Kerosene $(1.44)$ < Turpentine $(1.47)$.
Since water has the lowest refractive index among the three,light travels the fastest through water.

(N/A) The lateral displacement depends upon:
$(i)$ The angle of incidence of the incident ray $PQ$ on the face $DC$ of the given glass slab.
$(ii)$ The thickness of the glass slab.
The lateral displacement is defined as the perpendicular distance between the emergent ray and the original path of the incident ray $PQ$ when produced forward.

(N/A) The object is placed between the center of curvature $(C)$ and the principal focus $(F)$ of a concave mirror.
To find the image,we consider two rays:
$1$. $A$ ray parallel to the principal axis,after reflection,passes through the principal focus $(F)$.
$2$. $A$ ray passing through the principal focus $(F)$,after reflection,becomes parallel to the principal axis.
The intersection of these two reflected rays gives the position of the image.
The image formed is real,inverted,and magnified,and it is located beyond the center of curvature $(C)$.

(CONCAVE LENS) Since the incident ray of light parallel to the principal axis is diverged by the lens,the lens used is a concave lens.
To complete the ray diagram:
$1$. Draw a concave lens at the position indicated.
$2$. Extend the diverged ray backwards until it appears to pass through the focus $F_2$.
$3$. Draw a second ray from the top of the object $B$ passing through the optical center of the lens. This ray will pass undeviated.
$4$. The point where these two rays appear to intersect gives the position of the image $A'B'$.
The completed figure is as shown below.

(N/A) $(i)$ The lens formed by the combination is a concave lens.
$(ii)$ The lens is a diverging lens.
$(iii)$ $A$ concave lens causes parallel rays of light to diverge after refraction. The rays appear to originate from a single point on the principal axis,known as the principal focus. This is illustrated in the provided diagram.

(N/A) The object $AB$ is placed between the optical center $O$ and the principal focus $F_{1}$ of the convex lens.
$(i)$ $A$ ray of light starting from $B$ and passing through the optical center $O$ goes straight without any deviation.
$(ii)$ $A$ ray of light starting from $B$ and moving parallel to the principal axis passes through the focus $F_{2}$ after refraction by the lens.
$(iii)$ The two refracted rays appear to diverge from a point $B'$. By dropping a perpendicular from $B'$ to the principal axis,we get the point $A'$. Thus,$A'B'$ is the virtual image of the object $AB$.
$(iv)$ The image formed is virtual and erect.

(N/A) $(i)$ The distance $OS$ is called the focal length of the lens.
$(ii)$ The rays will still return to the same point $S$. When the mirror $M$ is placed in contact with the convex lens $L$,the rays from the point source $S$ (placed at the focus) pass through the lens and become parallel to the principal axis. These parallel rays strike the plane mirror $M$ normally and reflect back along the same path. After passing through the lens again,they converge at the point source $S$ itself.

(N/A) The lens used is a concave lens because it forms an erect,virtual,and diminished image for an object placed anywhere in front of it.
$(i)$ The outline of the concave lens is drawn at the position $L$.
(ii) $A$ ray from $B$ passing through the optical center $O$ goes undeviated.
(iii) $A$ ray from $B$ parallel to the principal axis,after refraction,appears to diverge from the focus $F_2$.
(iv) The final image $A'B'$ is formed between the optical center $O$ and the focus $F_2$ on the same side as the object.

(N/A) The lens is a convex lens.
Since the image is virtual,erect,and magnified,the object must be placed between the optical center and the focus of the convex lens.
The lens is situated between the object $O$ and the image $I.$
To locate the lens,draw a line from the top of the object to the top of the image and extend it to intersect the principal axis; this point is the optical center of the lens.
Draw a vertical line through this point to represent the lens.
To locate the focus $F,$ draw a ray from the top of the object parallel to the principal axis,which will pass through the focus after refraction. Extend this refracted ray backward to meet the image point. The point where the parallel ray meets the principal axis after refraction is the focus $F.$

(N/A) Power is related to the focal length as $P = \frac{1}{f}$.
An increase in power means a decrease in focal length.
Since the sunlight comes from infinity,the image is formed at the focal point of the lens.
If the power is doubled $(P' = 2P)$,the new focal length becomes half of the original focal length $(f' = \frac{f}{2})$.
Therefore,to focus the sunlight on the card,she will have to move the lens closer to the card,specifically at a distance equal to half of the previous focal length.

(C) thick plane mirror consists of a silvered back surface and a front glass surface.
When light falls on the mirror,a small portion of it is reflected from the front surface (first reflection),while the majority of the light enters the glass,reflects from the silvered back surface,and exits through the front surface (second reflection).
Because the reflection from the silvered back surface involves a much larger amount of light compared to the partial reflection from the front glass surface,the second image is the brightest of all the images formed.

(C) plane mirror produces a lateral inversion of the object.
When an object is placed in front of a mirror facing north,the mirror reflects the image such that the east-west component is reversed while the north-south component remains the same.
An arrow pointing north-east has a northward component and an eastward component.
In the mirror image,the northward component remains pointing north,but the eastward component is reversed to point west.
Therefore,the arrow in the image will point north-west.