Mix Examples - Light – Reflection and Refraction Questions in English

(A) point source of light placed at the focus of a concave mirror produces a parallel beam of light after reflection.
Similarly,a point source of light placed at the focus of a convex lens produces a parallel beam of light after refraction.
Therefore,both a concave mirror and a convex lens can produce a parallel beam of light from a point source.

(B) Given: Height of object $(h_o)$ = $10\, mm = 1\, cm$. Height of image $(h_i)$ = $-5\, mm = -0.5\, cm$ (since the image is real and inverted,it is formed in front of the mirror). Image distance $(v)$ = $-30\, cm$.
Magnification $(m)$ = $h_i / h_o = -v / u$.
Substituting the values: $-0.5 / 1 = -(-30) / u$.
$-0.5 = 30 / u$.
$u = 30 / -0.5 = -60\, cm$.
Using the mirror formula: $1/f = 1/v + 1/u$.
$1/f = 1/(-30) + 1/(-60)$.
$1/f = (-2 - 1) / 60 = -3 / 60$.
$1/f = -1 / 20$.
Therefore,$f = -20\, cm$.

(B, C) concave mirror can form a magnified (larger) image in two specific cases:
$1$. When the object is placed between the focus $(F)$ and the pole $(P)$ of the mirror (i.e.,at a distance less than its focal length $f$). In this case,the image formed is virtual,erect,and magnified.
$2$. When the object is placed between the focus $(F)$ and the centre of curvature $(C)$. In this case,the image formed is real,inverted,and magnified.
Since both options $B$ and $C$ describe conditions where a magnified image is formed,and typically in multiple-choice questions of this nature,both are valid. However,if only one must be chosen,both represent valid physical scenarios. Given the standard options provided,both $B$ and $C$ result in a magnified image.

(D) According to Snell's law,the refractive index of medium $B$ relative to medium $A$ $(n_{BA})$ is given by the ratio of the sine of the angle of incidence $(i)$ to the sine of the angle of refraction $(r)$:
$n_{BA} = \frac{\sin i}{\sin r}$
From the figure,the light ray travels from medium $A$ to medium $B$. The angle of incidence $i$ is the angle between the incident ray and the normal in medium $A$. Given the angle with the normal is $60^\circ$,so $i = 60^\circ$.
The angle of refraction $r$ is the angle between the refracted ray and the normal in medium $B$. Given the angle with the normal is $45^\circ$,so $r = 45^\circ$.
Substituting these values into Snell's law:
$n_{BA} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \times \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$
Therefore,the refractive index of medium $B$ relative to medium $A$ is $\sqrt{3} / \sqrt{2}$.

(A) From the figure,we can observe the angles of incidence $(i)$ and refraction $(r)$.
By comparing the angles with the normal,we see that the angle of refraction $(r)$ is smaller than the angle of incidence $(i)$,i.e.,$r < i$.
According to Snell's Law,the refractive index of medium $B$ with respect to medium $A$ is given by $n_{BA} = \frac{\sin i}{\sin r}$.
Since $i > r$,it follows that $\sin i > \sin r$,which implies that $\frac{\sin i}{\sin r} > 1$.
Therefore,the refractive index of medium $B$ relative to $A$ is greater than unity.

(B) By observing the figure,we can see that the incident light beams entering through holes $A$ and $B$ emerge through holes $C$ and $D$ respectively,without any change in their direction or convergence/divergence.
This indicates that the light rays have undergone lateral displacement,which is a characteristic property of a rectangular glass slab.
$A$ convex lens would converge the rays,a concave lens would diverge them,and a prism would deviate the rays at an angle.
Therefore,a rectangular glass slab is the correct component inside the box.

(A) In the given figure,a parallel beam of light enters the box from side $A$ and emerges from side $B$ as a diverging beam.
$A$ convex lens is a converging lens,which would cause the parallel rays to converge at a point.
$A$ rectangular glass slab would cause a lateral shift but the rays would remain parallel to each other.
$A$ prism would deviate the rays but would not cause them to diverge in this specific pattern.
$A$ concave lens is a diverging lens,which causes parallel rays of light to spread out (diverge) after passing through it. Therefore,a concave lens is placed inside the box.

(A) The power $P$ of a lens is given by the formula $P = 1/f$,where $f$ is the focal length in meters.
For a concave lens,the focal length is negative,and for a convex lens,it is positive.
Given $f = -0.25 \, m$ (for a concave lens),the power $P = 1/(-0.25) = -4 \, D$.
Therefore,a concave lens with a focal length of $-0.25 \, m$ has a power of $-4 \, D$.

(A) rear-view mirror in vehicles is a convex mirror.
Convex mirrors always form virtual,erect,and diminished images of objects placed in front of them.
Since the image is always diminished (smaller than the object),the height of the image $(h')$ is always less than the height of the object $(h)$.
Magnification $(m)$ is defined as the ratio of the height of the image to the height of the object,i.e.,$m = h'/h$.
Because $h' < h$,the magnification $m$ is always less than $1$.

(B) The rays from the Sun are parallel to the principal axis and converge at the focus $(F)$ of the concave mirror. Therefore,the focal length $(f)$ of the mirror is $15 \, cm$.
For a concave mirror,the image size is equal to the object size when the object is placed at the center of curvature $(C)$.
The center of curvature is at a distance equal to twice the focal length $(R = 2f)$.
Thus,$R = 2 \times 15 \, cm = 30 \, cm$.
Therefore,the object should be placed at $30 \, cm$ in front of the mirror.

(C) convex mirror is used to see a full-length image of a distant tall building because it always forms a virtual,erect,and diminished image of objects placed in front of it. Due to its outward-curved surface,it has a wider field of view,allowing it to capture the entire image of a large object within a small mirror area. Plane mirrors would require a mirror of at least half the height of the building to see the full image,which is impractical for a distant tall building.

(D) In torches,searchlights,and vehicle headlights,a concave mirror is used as a reflector.
According to the properties of a concave mirror,when a light source (bulb) is placed at the focus $(F)$,the light rays emanating from it strike the mirror and are reflected as a parallel beam of light.
This parallel beam of light travels over a long distance,which is essential for these devices to illuminate distant objects.
Therefore,the bulb is placed very near to the focus of the reflector.

(A) The laws of reflection state that:
$1$. The angle of incidence is equal to the angle of reflection $(i = r)$.
$2$. The incident ray,the reflected ray,and the normal to the mirror surface at the point of incidence all lie in the same plane.
These laws are universal and apply to all types of reflecting surfaces,including plane mirrors,concave mirrors,and convex mirrors,regardless of their shape or curvature.

(B) When a light ray travels from a rarer medium (air) to a denser medium (glass),it bends towards the normal at the point of incidence.
When the light ray emerges from the denser medium (glass) back into the rarer medium (air),it bends away from the normal.
In figure $B$,the ray bends towards the normal upon entering the glass slab and bends away from the normal upon exiting it,which correctly follows the laws of refraction.
Therefore,the path shown in $B$ is correct.

(C) The bending of light (refraction) depends on the refractive index of the medium. $A$ higher refractive index means the medium is optically denser,causing light to bend more towards the normal when entering from air.
The refractive indices of the given media are approximately:
$1$. Water: $1.33$
$2$. Kerosene: $1.44$
$3$. Mustard oil: $1.46$
$4$. Glycerine: $1.47$
Since glycerine has the highest refractive index among the given options,a ray of light incident obliquely at the same angle will bend the most in glycerine.

(D) According to the rules of reflection for a concave mirror,a ray of light parallel to the principal axis,after reflection,passes through the principal focus $(F)$ of the mirror.
In the given figure,the incident ray is parallel to the principal axis.
Therefore,after reflection,it must pass through the principal focus $(F)$.
Looking at the options,option $D$ correctly shows the reflected ray passing through the principal focus $(F)$.
Thus,the correct ray diagram is $D$.

(A) According to the rules of refraction for a convex lens,a ray of light passing through the principal focus $(F)$ of a convex lens becomes parallel to the principal axis after refraction.
In the given figure,the incident ray is passing through the principal focus $(F)$ before hitting the lens.
Therefore,after refraction through the convex lens,this ray must emerge parallel to the principal axis.
Comparing this with the given options,diagram $A$ correctly depicts this behavior where the ray becomes parallel to the principal axis after passing through the lens.

(B) $1$. $A$ concave mirror produces a magnified image when the object is placed close to it. Since the head appears bigger, the top part of the mirror is concave.
$2$. $A$ plane mirror produces an image of the same size as the object. Since the middle portion appears of the same size, the middle part of the mirror is plane.
$3$. $A$ convex mirror always produces a diminished (smaller) image. Since the legs appear smaller, the bottom part of the mirror is convex.
$4$. Therefore, the order of the mirrors from top to bottom is concave, plane, and convex.

(C) When an object is placed at infinity,the rays of light coming from it are parallel to the principal axis.
In a concave mirror,these rays converge at the focus $(F)$,forming a highly diminished,point-sized,real,and inverted image.
In a convex mirror,these rays appear to diverge from the focus $(F)$ behind the mirror,forming a highly diminished,point-sized,virtual,and erect image.
In a concave lens,these rays appear to diverge from the focus $(F)$ on the same side,forming a highly diminished,point-sized,virtual,and erect image.
In a convex lens,these rays converge at the focus $(F)$ on the other side,forming a highly diminished,point-sized,real,and inverted image.
Therefore,all four optical elements produce a highly diminished and point-sized image for an object at infinity.

(A-D) Concave mirror: When an object is placed between the pole and the focus of a concave mirror,a virtual,erect,and enlarged image is formed behind the mirror.
$(b)$ Convex lens: When an object is placed between the optical centre and the focus of a convex lens,a virtual,erect,and enlarged image is formed on the same side as the object.
$(c)$ Concave lens: $A$ concave lens always forms a virtual,erect,and diminished image between the optical centre and the focus,regardless of the object's position (between infinity and the lens).
$(d)$ Convex mirror: $A$ convex mirror always forms a virtual,erect,and diminished image between the pole and the focus,behind the mirror,for any object position between infinity and the pole.

(N/A) When a light ray enters a rectangular glass slab,it undergoes refraction at two parallel interfaces.
$1$. At the first interface (air-glass),the light ray bends towards the normal because it travels from a rarer to a denser medium.
$2$. At the second interface (glass-air),the light ray bends away from the normal as it travels from a denser to a rarer medium.
$3$. According to Snell's Law,the angle of incidence at the first surface is equal to the angle of emergence at the second surface because the opposite sides of the rectangular slab are parallel.
$4$. Consequently,the emergent ray is parallel to the incident ray,but it undergoes a lateral displacement.

(B) No,the pencil will not appear to be bent to the same extent. The extent of bending of the pencil depends on the refractive index of the liquid used. Since different liquids like kerosene and turpentine have different refractive indices compared to water,the angle of refraction will vary. The bending of light at the interface of two media is determined by the relative refractive index,which dictates the change in the velocity of light as it passes from one medium to another.

(N/A) The refractive index $(n)$ of a medium is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in that medium $(v)$. Mathematically,it is expressed as: $n = \frac{c}{v}$.
To find the refractive index of medium $2$ with respect to medium $1$ $(n_{21})$,we take the ratio of the speed of light in medium $1$ $(v_1)$ to the speed of light in medium $2$ $(v_2)$. The expression is: $n_{21} = \frac{v_1}{v_2}$.

(D) The refractive index of diamond with respect to glass is given by $n_{dg} = \frac{n_d}{n_g} = 1.6$.
Here,$n_d$ is the absolute refractive index of diamond and $n_g$ is the absolute refractive index of glass.
Given that $n_g = 1.5$.
Substituting the values,we get $\frac{n_d}{1.5} = 1.6$.
Therefore,$n_d = 1.6 \times 1.5 = 2.4$.

(N/A) Yes,the statement is correct.
$1$. To obtain a magnified virtual image,the object must be placed between the optical center $(O)$ and the principal focus $(F_1)$,i.e.,at a distance less than $20 \, cm$ from the lens.
$2$. To obtain a magnified real image,the object must be placed between the principal focus $(F_1)$ and the center of curvature $(2F_1)$,i.e.,at a distance greater than $20 \, cm$ but less than $40 \, cm$ from the lens.

(A) When the object is at a large distance (like a building outside),the rays of light coming from it are considered parallel to the principal axis.
According to the lens formula,for an object at infinity,the image is formed at the principal focus $(F)$.
Since the image of the nearby window pane was formed at $15\, cm$,the building (which is much further away) will form an image closer to the focal point of the lens.
Therefore,Sudha should move the screen towards the lens to obtain a sharp image of the building.
The approximate focal length of the lens is equal to the image distance of the distant object,which is $15\, cm$.

(A) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters,expressed as $P = \frac{1}{f(m)}$.
This indicates that power is inversely proportional to the focal length $(P \propto \frac{1}{f})$.
$A$ lens with a shorter focal length has a higher power and thus a greater ability to converge light rays.
Comparing the two lenses,the lens with a focal length of $20\, cm$ has a higher power than the lens with a focal length of $40\, cm$.
Therefore,the lens with a focal length of $20\, cm$ will provide more convergent light.

(N/A) When two plane mirrors are placed at an angle of $90^{\circ}$ to each other,the incident ray and the final reflected ray will always be parallel to each other,regardless of the angle of incidence.
Explanation:
$1$. Let the two mirrors be $M_1$ and $M_2$ placed at $90^{\circ}$.
$2$. The incident ray strikes $M_1$ at an angle of incidence $i$. According to the law of reflection,the angle of reflection $r = i$.
$3$. The reflected ray from $M_1$ acts as the incident ray for $M_2$. By geometry,the angle of incidence at $M_2$ will be $(90^{\circ} - i)$.
$4$. The final angle of reflection at $M_2$ will also be $(90^{\circ} - i)$.
$5$. By calculating the total deviation or the angles with the normal,it can be shown that the final reflected ray is parallel to the initial incident ray.

(N/A) When light travels from a rarer medium (air) to a denser medium (water),it bends towards the normal. In this case,the angle of refraction $(r)$ is less than the angle of incidence $(i)$.
When light travels from a denser medium (water) to a rarer medium (air),it bends away from the normal. In this case,the angle of refraction $(r)$ is greater than the angle of incidence $(i)$.
The ray diagrams are provided in the image.

(N/A) To draw the ray diagrams for a concave mirror,follow these steps for each case:
$(a)$ Between $P$ and $F$: The image is virtual,erect,and magnified,formed behind the mirror.
$(b)$ Between $F$ and $C$: The image is real,inverted,and magnified,formed beyond $C$.
$(c)$ At $C$: The image is real,inverted,and of the same size as the object,formed at $C$.
$(d)$ Beyond $C$: The image is real,inverted,and diminished,formed between $F$ and $C$.
$(e)$ At infinity: The image is real,inverted,and highly diminished (point-sized),formed at the focus $F$.

(N/A) To draw the ray diagrams for a convex lens,follow these steps for each case:
$(a)$ When the object is between the optical centre $(O)$ and focus $(F_1)$: The image is virtual,erect,and magnified,formed on the same side as the object.
$(b)$ When the object is between $F_1$ and $2F_1$: The image is real,inverted,and magnified,formed beyond $2F_2$ on the other side.
$(c)$ When the object is at $2F_1$: The image is real,inverted,and of the same size as the object,formed at $2F_2$.
$(d)$ When the object is at infinity: The image is real,inverted,and highly diminished (point-sized),formed at the focus $F_2$.
$(e)$ When the object is at the focus $F_1$: The image is real,inverted,and highly magnified,formed at infinity.

(N/A) The laws of refraction are as follows:
$1$. The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence all lie in the same plane.
$2$. Snell's Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for the light of a given color and for a given pair of media, expressed as $\frac{\sin i}{\sin r} = \text{constant} = n_{21}$.
When a light ray passes through a rectangular glass slab:
- The ray enters from air to glass (denser medium) and bends towards the normal.
- It travels through the glass and exits from glass to air (rarer medium), bending away from the normal.
- The emergent ray is parallel to the incident ray but laterally displaced. This perpendicular distance between the original path of the incident ray and the emergent ray is called lateral displacement.

(N/A) For a concave lens,the image formed is always virtual,erect,and diminished,regardless of the object's position.
$(a)$ When the object is at the focus $(F_1)$,the rays appear to diverge from a point between the optical center $(O)$ and the focus $(F_1)$. The image is virtual,erect,and highly diminished.
$(b)$ When the object is between the focus $(F_1)$ and twice the focal length $(2F_1)$,the image is formed between the optical center $(O)$ and the focus $(F_1)$. The image is virtual,erect,and diminished.
$(c)$ When the object is beyond twice the focal length $(2F_1)$,the image is formed between the optical center $(O)$ and the focus $(F_1)$. The image is virtual,erect,and diminished.

(N/A) For a convex mirror:
$(a)$ When the object is at infinity,the incident rays are parallel to the principal axis. After reflection,they appear to diverge from the principal focus $(F)$ behind the mirror. The image formed is virtual,erect,highly diminished,and located at the principal focus $(F)$ behind the mirror.
$(b)$ When the object is at a finite distance from the mirror,the image formed is always virtual,erect,and diminished. It is located between the pole $(P)$ and the principal focus $(F)$ behind the mirror,regardless of the object's distance.

(A) Given that the image is formed on a screen,it must be a real image.
For a real image,the magnification $m$ is negative. Given $m = -3$.
We know that $m = \frac{v}{u}$,where $v$ is the image distance and $u$ is the object distance.
Since the image is on the other side of the lens,$v = +80 \, cm$.
Using the magnification formula: $-3 = \frac{80}{u}$.
Therefore,$u = -\frac{80}{3} \, cm \approx -26.67 \, cm$.
The negative sign indicates that the object is placed in front of the lens at a distance of $26.67 \, cm$.
The nature of the image is real and inverted,and the lens used is a convex lens.

(D) Given: Focal length $f = -20\, cm$ (for a concave mirror),Magnification $m = -1/3$ (for a real and inverted image).
Using the magnification formula $m = -v/u$,we get $-1/3 = -v/u$,which implies $v = u/3$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,substitute $v = u/3$:
$\frac{1}{u/3} + \frac{1}{u} = \frac{1}{-20}$
$\frac{3}{u} + \frac{1}{u} = -\frac{1}{20}$
$\frac{4}{u} = -\frac{1}{20}$
$u = -80\, cm$.
The object is placed at a distance of $80\, cm$ from the mirror.
The image is real and inverted,and the mirror is concave.

(A) The power of a lens is defined as the reciprocal of its focal length in meters,given by the formula $P = \frac{1}{f(m)}$.
The $SI$ unit of power of a lens is Dioptre $(D)$.
For the first student,the focal length $f = 50\, cm = 0.5\, m$. Since the focal length is positive,the lens is a convex lens. Its power $P = \frac{1}{0.5} = +2\, D$.
For the second student,the focal length $f = -50\, cm = -0.5\, m$. Since the focal length is negative,the lens is a concave lens. Its power $P = \frac{1}{-0.5} = -2\, D$.

(A) $(i)$ The object distance $u = 50.0 - 12.0 = 38.0 \, cm$. The image distance $v = 88.0 - 50.0 = 38.0 \, cm$. Since $u = v$,the object is placed at $2f$. Thus,$2f = 38.0 \, cm$,which gives the focal length $f = 19.0 \, cm$.
$(ii)$ If the candle is shifted to $31.0 \, cm$,the new object distance $u' = 50.0 - 31.0 = 19.0 \, cm$. Since the object is now placed at the focus $(u' = f = 19.0 \, cm)$,the image will be formed at infinity.

(N/A) First,calculate the focal length $(f)$ of the lens:
Object distance $u = 12.0 - 50.0 = -38.0\, cm$
Image distance $v = 88.0 - 50.0 = +38.0\, cm$
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{38} - \frac{1}{-38} = \frac{1}{f} \implies \frac{2}{38} = \frac{1}{f} \implies f = 19.0\, cm$
$(i)$ New object position $= 31.0\, cm$. New $u = 31.0 - 50.0 = -19.0\, cm$. Since the object is at the focus $(u = -f)$,the image will be formed at infinity.
$(ii)$ If the candle is shifted further towards the lens (i.e.,$u < f$),the object lies between the optical center and the focus. The image formed will be virtual,erect,and magnified.
$(iii)$ The ray diagram shows the object placed between $F$ and $O$,resulting in a virtual image formed on the same side as the object.

(A) prism is a transparent optical element with flat,polished surfaces that refract light.
It is typically a glass object with two triangular bases and three rectangular lateral surfaces inclined at an angle to each other.

(N/A) The phenomenon of bouncing back of light rays when they strike a smooth or polished surface is known as reflection of light. When light hits a surface,it changes its direction and returns to the same medium.

(N/A) $(i)$ Angle of incidence: It is the angle between the incident ray and the normal at the point of incidence.
$(ii)$ Angle of reflection: It is the angle between the reflected ray and the normal at the point of incidence.
$(iii)$ Plane of incidence: It is the plane that contains the incident ray,the reflected ray,and the normal at the point of incidence.

(N/A) $(i)$ The angle of incidence $(i)$ is equal to the angle of reflection $(r)$,i.e.,$i = r$.
$(ii)$ The incident ray,the normal to the mirror at the point of incidence,and the reflected ray all lie in the same plane.

(B) The image formed by a plane mirror undergoes lateral inversion,which means the left side of the object appears as the right side in the image.
When a driver looks into their rear-view mirror,they see the word $AMBULANCE$ written in reverse on the vehicle behind them.
Due to the lateral inversion caused by the rear-view mirror,the word appears correctly oriented to the driver.
This allows the driver to quickly identify the vehicle as an $AMBULANCE$ and give way.

(B) plane mirror always forms a virtual and erect image.
The size of the image is equal to the size of the object (unmagnified).
The image formed is laterally inverted,meaning it exhibits right-left reversal.
Therefore,the image is virtual,erect,and of the same size as the object.

(A) Yes,a plane mirror can be considered a spherical mirror with an infinite radius of curvature $(R = \infty)$.
Since the focal length $(f)$ is half of the radius of curvature $(f = R/2)$,the focal length of a plane mirror is also infinite $(f = \infty)$.

(B) plane mirror can be considered as a part of a spherical mirror with an infinitely large radius of curvature. Since the surface of a plane mirror is perfectly flat,it does not curve at any point. Mathematically,the radius of curvature $(R)$ is related to the focal length $(f)$ by the formula $R = 2f$. For a plane mirror,the focal length is considered to be infinity,therefore the radius of curvature is also infinity.

(C) When two plane mirrors are placed parallel to each other,the light rays reflect back and forth between the two mirrors repeatedly. Each reflection creates a new image of the previous image. Since the mirrors are parallel,the angle between them is $0^{\circ}$. The formula for the number of images formed is $n = (360^{\circ} / \theta) - 1$. Substituting $\theta = 0^{\circ}$,we get $n = (360^{\circ} / 0^{\circ}) - 1$,which approaches infinity. Therefore,an infinite number of images are formed.

(A) When a ray of light is incident normally (perpendicularly) on a plane mirror,it strikes the surface along the normal.
By definition,the angle of incidence $(i)$ is the angle between the incident ray and the normal at the point of incidence.
Since the ray is incident along the normal,the angle between them is $0^{\circ}$,so $i = 0^{\circ}$.
According to the laws of reflection,the angle of incidence is equal to the angle of reflection $(i = r)$.
Therefore,the angle of reflection $(r)$ is also $0^{\circ}$.

(N/A) $(i)$ The image is of the same size as the object (unmagnified).
(ii) The image is virtual,meaning it cannot be caught on a screen.
(iii) The image is erect (upright).
(iv) The image exhibits lateral inversion (right-left reversal).