Mix Examples - Light – Reflection and Refraction Questions in English

(B) According to the sign convention for spherical mirrors and lenses:
$1$. The focal length of a concave mirror is always negative.
$2$. The focal length of a concave lens is always negative.
Since both the spherical mirror and the thin spherical lens have a focal length of $-15\,cm$ (which is negative),both must be concave in nature.

(C) convex mirror always forms a virtual and erect image of an object,regardless of its distance from the mirror.
Similarly,a plane mirror also always forms a virtual and erect image of an object,regardless of the distance.
Since both plane and convex mirrors produce erect images for all positions of the object,the mirror is likely to be either plane or convex.

(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, 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.

(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.

(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) 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) For lenses,the lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $f$ is the focal length,$v$ is the image distance,and $u$ is the object distance.
For mirrors,the mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,where $f$ is the focal length,$v$ is the image distance,and $u$ is the object distance.

(C) The focal length of a mirror depends only on its radius of curvature $(f = R/2)$ and is independent of the surrounding medium.
Therefore, the focal length of the concave mirror remains unchanged when placed 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 term $(n_g/n_m - 1)$ decreases, causing the focal length $(f)$ of the convex lens to increase.

(A-10, B-20, C-30, D-40) For a concave mirror with focal length $f = 15 \, cm$:
$(1)$ $A$ virtual image is formed when the object is placed between the pole and the focus $(u < 15 \, cm)$. Thus,the position is $10 \, cm$.
$(2)$ $A$ diminished real image is formed when the object is placed beyond the center of curvature $(u > 30 \, cm)$. Thus,the position is $40 \, cm$.
$(3)$ An enlarged real image is formed when the object is placed between the focus and the center of curvature $(15 \, cm < u < 30 \, cm)$. Thus,the position is $20 \, cm$.
$(4)$ An image of the same size is formed when the object is placed at the center of curvature $(u = 2R = 30 \, cm)$. Thus,the position is $30 \, cm$.

(N/A) $(i)$ It is a convex lens. The image is real,inverted,and magnified,which occurs when the object is placed between $F_1$ and $2F_1$.
$(ii)$ It is a concave mirror. The image is virtual,erect,and magnified,which occurs when the object is placed between the focus $(F)$ and the pole $(P)$.

(A) To identify the type of glass,hold the given piece of glass over some printed matter and observe the letters:
$(i)$ 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 as they are,then it is a plane glass plate.

(N/A) ray of light passing through the centre of curvature of a concave mirror strikes the mirror surface at an angle of $90^{\circ}$ (normal incidence). According to the laws of reflection,when a ray falls normally on a surface,it reflects back along the same path.
$(b)$ Convex mirrors are used as rear-view mirrors because they always form an erect,virtual,and diminished image of objects. This provides a wider field of view,allowing the driver to see a much larger area behind the vehicle.
$(c)$ The phenomenon responsible for the apparent displacement of the pencil is the refraction of light.

(N/A) $(a) (i)$ For lenses,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$(ii)$ For mirrors,the mirror formula is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
$(b)$ Since the magnification $m = +4$ is positive,the image is virtual and erect. Since the magnitude of $m$ is $4$ (which is greater than $1$),the image is magnified (enlarged).
$(c)$ $(i)$ When the object is placed between the pole $(P)$ and the focus $(F)$ of a concave mirror,a virtual,erect,and magnified image is formed behind the mirror.
$(ii)$ When the object is placed at the centre of curvature $(C)$ of a concave mirror,a real,inverted image of the same size as the object is formed at the centre of curvature $(C)$.

(D) The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium. Mathematically,$n = c/v$,where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium.
$(b)$ $(i)$ Optical density is directly related to the refractive index. $A$ higher refractive index means higher optical density. Since the angle of refraction is smallest in medium $C$ $(40^{\circ})$,it has the highest refractive index and thus the maximum optical density.
$(ii)$ The speed of light is inversely proportional to the refractive index. Since medium $A$ has the largest angle of refraction $(50^{\circ})$,it has the lowest refractive index,meaning the speed of light will be maximum in medium $A$.
$(iii)$ Light travelling from $A$ to $B$: Since the angle of refraction in $A$ $(50^{\circ})$ is greater than in $B$ $(45^{\circ})$,medium $B$ is optically denser than medium $A$. Therefore,light will bend towards the normal.
$(iv)$ Refractive index of $B$ relative to $C$ $(n_{BC})$ is given by $n_B / n_C$. Since the angle of refraction in $B$ $(45^{\circ})$ is greater than in $C$ $(40^{\circ})$,medium $C$ is optically denser than $B$ $(n_C > n_B)$. Thus,$n_{BC} = n_B / n_C < 1$,which is less than unity.

(N/A) $(i)$ Yes,the lens can produce an image of the complete object. Although the intensity of the image will be reduced because less light passes through the lens,the image will still be formed at the same position.
$(ii)$ The ray diagram shows the object placed beyond $2F_1$. The image is formed between $F_2$ and $2F_2$ on the other side of the lens.
$(iii)$ Given: Height of object $h = 4\, cm$,focal length $f = +20\, cm$,object distance $u = -15\, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-15} = \frac{3 - 4}{60} = \frac{-1}{60}$
Therefore,$v = -60\, cm$.
The negative sign indicates that the image is formed on the same side as the object.
Nature: Virtual and erect.
Size of the image: $m = \frac{v}{u} = \frac{-60}{-15} = +4$.
$h' = m \times h = 4 \times 4\, cm = +16\, cm$. The image is magnified and erect.

(N/A) The mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,where $f$ is the focal length,$v$ is the image distance,and $u$ is the object distance.
$(b)$ The ray diagram shows the object placed between the pole $(P)$ and the focus $(F)$ of a concave mirror,resulting in a virtual,erect,and magnified image behind the mirror.
$(c)$ Given:
Focal length of concave mirror,$f = -15\, cm$
Object distance,$u = -10\, cm$
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-10} = -\frac{1}{15} + \frac{1}{10} = \frac{-2 + 3}{30} = \frac{1}{30}$
Therefore,$v = +30\, cm$.
Since $v$ is positive,the image is formed $30\, cm$ behind the mirror.
The image is virtual,erect,and magnified (as magnification $m = -\frac{v}{u} = -\frac{30}{-10} = +3$).

(A-D) Concave mirror is used in a solar furnace because it converges parallel rays of sunlight to a single point (focus),producing high heat.
$(b)$ Convex mirror is used as a side/rear-view mirror in vehicles because it provides a wider field of view and always forms a virtual,erect,and diminished image.
Concave mirror can form a magnified and virtual image when the object is placed between the pole $(P)$ and the focus $(F)$ of the mirror.
The ray diagram for the magnified and virtual image is as follows: When an object is placed between $P$ and $F$,the reflected rays appear to diverge from a point behind the mirror,forming a virtual,erect,and magnified image.

(N/A) For the ray diagrams of a convex lens:
$(i)$ When the object is between the optical centre and the principal focus $(F)$,the image formed is virtual,erect,and magnified,and it is formed on the same side of the lens as the object.
$(ii)$ When the object is between $F$ and $2F$,the image formed is real,inverted,and magnified,and it is formed beyond $2F$ on the other side of the lens.
$(iii)$ When the object is at $2F$,the image formed is real,inverted,and of the same size as the object,and it is formed at $2F$ on the other side of the lens.
$(b)$ If the convex lens is replaced by a concave lens,the nature and position of the image will change as follows:
In both cases $(i)$ and $(ii)$,the concave lens will always form a virtual,erect,and diminished image. The image will always be formed between the optical centre and the principal focus $(F)$ on the same side as the object,regardless of the object's position.

(C) The principal focus is a point on the principal axis where rays parallel to the principal axis converge (in a concave mirror) or appear to diverge from (in a convex mirror) after reflection.
$(b)$ $A$ concave mirror forms a real,inverted,and diminished image when the object is placed beyond the center of curvature $(C)$.
$(c)$ Given: Object height $h = 4 \,cm$,object distance $u = -6 \,cm$,focal length $f = -12 \,cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\frac{1}{-12} = \frac{1}{v} + \frac{1}{-6}$
$\frac{1}{v} = \frac{1}{6} - \frac{1}{12} = \frac{2-1}{12} = \frac{1}{12}$
$v = 12 \,cm$.
The image is formed at a distance of $12 \,cm$ behind the mirror (virtual image).

(N/A) The optical centre is the central point of a lens through which a ray of light passes without undergoing any deviation.
$(b)$ To obtain a real,inverted,and highly enlarged image with a convex lens,the object must be placed at the principal focus $(F_1)$. The image will be formed at infinity.
$(c)$ Given: Focal length of concave lens $f = -20 \, cm$,Image distance $v = -15 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{-20} = \frac{1}{-15} - \frac{1}{u}$
$\frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20}$
$\frac{1}{u} = \frac{-4 + 3}{60} = -\frac{1}{60}$
$u = -60 \, cm$.
Thus,the object should be placed at a distance of $60 \, cm$ in front of the lens.

(N/A) convex lens can be used as a magnifying glass.
$(a)$ When the object is placed between the optical centre $O$ and the principal focus $F_{1}$,the convex lens forms a virtual,erect,and magnified image on the same side as the object.
$(b)$ When the object is placed at $2F_{1}$,the convex lens forms a real,inverted image of the same size as the object at $2F_{2}$ on the other side of the lens.
[Refer to the provided ray diagrams for the visual representation of these cases.]

(N/A) $(i)$ Activity: Hold a concave mirror in your hand and direct it towards a distant object (like a window or a tree). Direct the light reflected by the mirror onto a sheet of paper held close to the mirror. Move the sheet back and forth slowly until a bright,sharp spot of light is seen on the paper. The distance of this image from the position of the mirror gives the approximate value of the focal length of the mirror.
$(ii)$ As an object is moved gradually away from a convex mirror,the size of the image becomes smaller and it shifts towards the focus.
$(iii)$ When a light ray passes through a rectangular glass slab,it undergoes refraction at two parallel surfaces,$PQ$ (air-glass interface) and $SR$ (glass-air interface). The extent of bending of the ray of light at the first surface is equal and opposite to the bending at the second surface. Thus,the emergent ray is parallel to the incident ray,but it is laterally displaced. This is shown in the diagram below.

(N/A) $(i)$ Principal focus: $A$ number of rays parallel to the principal axis,after reflection from a concave mirror,meet at a point on the principal axis which is called the principal focus.
$(ii)$ $A$ convex mirror always forms an erect and diminished image. It provides a wider field of view because it is curved outwards,allowing the driver to see a much larger area behind the vehicle.
$(iii)$ To obtain an erect image with a convex lens,the object must be placed between the optical center $(O)$ and the principal focus $(F_1)$. The image formed is virtual,erect,and magnified,as shown in the diagram.

(N/A) This means that the ratio of the speed of light in air/vacuum to the speed of light in diamond is $2.42$.
$(b)$ The two figures are shown below,completing the path of the light rays.
$(c)$ $A$ virtual image produced by a concave mirror is magnified,whereas a virtual image produced by a convex mirror is always diminished. $A$ virtual image produced by a plane mirror is of the same size as the object.
$(d)$ The negative sign in the magnification value indicates that the image formed is real and inverted.

(N/A) The ray diagrams are as shown in the provided image.
$(b)$ Magnification $(m)$ is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$,which is also equal to $-\frac{v}{u}$,where $v$ is the image distance and $u$ is the object distance.
$m = \frac{h'}{h} = -\frac{v}{u}$
$m = -1$ signifies:
$(i)$ The size of the image is equal to the size of the object.
$(ii)$ The negative sign indicates that the image is real and inverted.
This magnification is produced by a concave mirror when the object is placed at the centre of curvature $(C)$.

(N/A) $(i)$ At the centre of curvature $(C)$ of a concave mirror: The image is formed at $C$,is real,inverted,and of the same size as the object.
$(ii)$ Within the focal length $(f)$ of a convex lens: The image is formed on the same side as the object,is virtual,erect,and magnified.
$(iii)$ Between the pole $(P)$ and focus $(F)$ of a concave mirror: The image is formed behind the mirror,is virtual,erect,and magnified.
$(iv)$ In front of a convex mirror: The image is formed behind the mirror between the pole and focus,is virtual,erect,and diminished.
$(v)$ In front of a concave lens: The image is formed between the optical centre and focus on the same side as the object,is virtual,erect,and diminished.

(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.

(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.

(B) $1$. A Convex mirror always forms a virtual, erect, and diminished image for any position of the object in front of it.
$2$. A Concave lens always forms a virtual, erect, and diminished image for any position of the object in front of it.
$3$. In contrast, Concave mirrors and Convex lenses can form both real and virtual images depending on the position of the object.
$4$. Therefore, the correct combination is Convex mirror and Concave lens.

(D) Light waves are electromagnetic waves because they consist of oscillating electric and magnetic fields.
Unlike mechanical waves (such as sound waves),light waves do not require a material medium for their propagation and can travel through a vacuum.
Therefore,the statement that light waves are electromagnetic waves is correct.

(D) plane mirror always forms a virtual and erect image of an object,regardless of its distance from the mirror.
$A$ convex mirror also always forms a virtual and erect image of an object,regardless of its distance from the mirror.
$A$ concave mirror can form both real and virtual images depending on the object's position.
Therefore,both plane and convex mirrors always produce an erect image for any object distance.

A virtual image is an image that cannot be projected onto a screen and is formed when light rays appear to diverge from a point.
$1$. A Plane mirror always forms a virtual, erect, and same-sized image.
$2$. A Convex mirror always forms a virtual, erect, and diminished image.
$3$. A Concave mirror forms a virtual, erect, and magnified image when the object is placed between the pole $(P)$ and the focus $(F)$.
Therefore, all three types of mirrors can produce a virtual image under specific conditions.

(B) According to the rules of reflection for a concave mirror,any light ray that is parallel to the principal axis will pass through the principal focus $(F)$ after reflection from the mirror.
In the given figure,the incident ray is parallel to the principal axis.
Therefore,after reflection,this ray must pass through the principal focus,which is marked as point $F$.

(A) According to the sign convention for spherical mirrors,the focal length of a concave mirror is negative,while that of a convex mirror is positive.
According to the sign convention for spherical lenses,the focal length of a concave lens is negative,while that of a convex lens is positive.
Since the focal length is given as $-15 \; cm$ for both,the mirror must be concave and the lens must also be concave.
Therefore,both the mirror and the lens are concave.