Mix Examples - Light – Reflection and Refraction Questions in English

(B) In a compound microscope,the objective lens is placed near the object and must have a short focal length to produce a magnified real image.
The eyepiece acts as a simple magnifier and typically has a longer focal length compared to the objective lens.
Therefore,the lens with a focal length of $5 \, cm$ serves as the objective,and the lens with a focal length of $20 \, cm$ serves as the eyepiece.
Thus,the correct option is $B$.

(C) The speed of light in a vacuum is a fundamental physical constant.
It is denoted by the symbol $c$.
The value of the speed of light in a vacuum is approximately $3 \times 10^{8} \text{ m s}^{-1}$ (or $300,000 \text{ km s}^{-1}$).
Therefore,the correct option is $C$.

(C) The speed of light in a vacuum is approximately $3 \times 10^{8} \ m/s$.
To convert this into $km/s$,we divide the value by $1000$ (since $1 \ km = 1000 \ m$).
Speed in $km/s = (3 \times 10^{8} \ m/s) / 1000 = 3 \times 10^{5} \ km/s$.
Therefore,the correct option is $C$.

(B) The wavelength range of visible light is approximately $400 \; nm$ to $750 \; nm$ (often rounded to $400 \; nm$ to $800 \; nm$).
To convert nanometers $(nm)$ to centimeters $(cm)$:
$1 \; nm = 10^{-9} \; m = 10^{-7} \; cm$.
Therefore,$400 \; nm = 400 \times 10^{-7} \; cm = 4 \times 10^{-5} \; cm$.
Similarly,$800 \; nm = 800 \times 10^{-7} \; cm = 8 \times 10^{-5} \; cm$.
Thus,the range is $4 \times 10^{-5} \; cm$ to $8 \times 10^{-5} \; cm$.

(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) Light waves are electromagnetic waves that do not require a material medium for their propagation.
They are non-mechanical waves,meaning they can travel through a vacuum as well as through various media.
Therefore,the statement that light waves require a medium for their propagation is incorrect.

(A) Light travels in a straight line,a phenomenon known as rectilinear propagation of light. This occurs because the wavelength of visible light is extremely small compared to the size of the objects it interacts with. Due to this very small wavelength,the diffraction effects are negligible in everyday life,causing light to appear to travel in a straight path.

(C) When light rays from an object strike the surface of a plane mirror,they bounce back into the same medium. This phenomenon of light bouncing back from a surface is known as reflection. $A$ plane mirror forms a virtual and erect image of an object due to the regular reflection of light.

(B) lens is a transparent medium bounded by two surfaces,at least one of which is spherical. When light rays pass through a lens,they undergo bending due to the change in the medium (from air to glass/plastic and back to air). This phenomenon of bending of light when it travels from one transparent medium to another is known as refraction. Therefore,the image formation by a lens is based on the principle of refraction.

(A) real image is formed when light rays actually intersect after reflection or refraction.
Because the light rays physically meet at the point of image formation,a real image can be captured or projected onto a screen.
Real images are always inverted relative to the object.
Therefore,the statement 'It can be obtained on a screen' is correct.

(A) For a plane mirror,the distance of the object from the mirror is equal to the distance of the image from the mirror.
Given that the object distance $(u)$ is $10 \ cm$.
According to the properties of a plane mirror,the image distance $(v)$ is equal to the object distance in magnitude.
Therefore,the image will be formed at a distance of $10 \ cm$ behind the mirror.

(D) plane mirror always forms an image that is virtual,erect,and of the same size as the object.
Since the light rays do not actually meet at the image point,the image is virtual.
Therefore,the correct statement is that the image formed is virtual.

(C) According to the laws of reflection,the angle of incidence $(i)$ is always equal to the angle of reflection $(r)$.
Given that the angle of incidence $(i)$ = $30^{\circ}$.
Therefore,the angle of reflection $(r)$ = $30^{\circ}$.

(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.

(C) The centre of the reflecting surface of a spherical mirror is a point called the pole. It lies on the surface of the mirror. It is usually represented by the letter $P$.

(D) When light rays parallel to the principal axis strike a spherical mirror,they reflect and converge at (or appear to diverge from) a specific point on the principal axis.
This point is known as the principal focus,denoted by the letter $F$.
For a concave mirror,the rays actually meet at this point,while for a convex mirror,they appear to diverge from this point.

(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 $F$.

(B) The distance between the pole $(P)$ and the principal focus $(F)$ of a spherical mirror is defined as the focal length $(f)$.
It is half of the radius of curvature $(R)$,represented by the relation $f = R/2$.

(C) spherical mirror is a part of a sphere. The center of the sphere of which the mirror is a part is called the $Centre of curvature$ $(C)$. The pole $(P)$ is the center of the reflecting surface of the mirror. The principal focus $(F)$ is a point on the principal axis where light rays parallel to the axis converge or appear to diverge. The principal axis is the line passing through the pole and the center of curvature. However,in the context of defining the geometric origin of the mirror,the $Centre of curvature$ is the fundamental point that defines the sphere from which the mirror is cut.

(C) When light rays parallel to the principal axis fall on a spherical mirror,they reflect and converge at (or appear to diverge from) a specific point on the principal axis.
This specific point is known as the Principal focus,denoted by the letter $F$.
For a concave mirror,the rays actually meet at this point,while for a convex mirror,they appear to diverge from this point.

(A) According to the rules of image formation by spherical mirrors,a ray of light passing through the principal focus $(F)$ of a concave mirror becomes parallel to the principal axis after reflection.
This is the converse of the rule where a ray parallel to the principal axis passes through the focus after reflection.

(A) For a concave mirror,when an object is placed beyond the centre of curvature $(C)$,the light rays from the object form a real,inverted,and diminished image between the focus $(F)$ and the centre of curvature $(C)$.
Therefore,the correct position for the object is beyond the centre of curvature.

(B) When an object is placed beyond the center of curvature $(C)$ of a concave mirror,the light rays from the object reflect off the mirror and converge between the center of curvature $(C)$ and the principal focus $(F)$.
Since the reflected rays actually meet at a point,the image formed is real.
Because the image is formed between $C$ and $F$,it is smaller in size than the object,i.e.,diminished.
Therefore,the image is real and diminished.

(A) For a concave mirror,when an object is placed beyond the center of curvature $(C)$,the image formed is real,inverted,and diminished.
$1$. The rays of light coming from the object pass through the mirror and converge between the principal focus $(F)$ and the center of curvature $(C)$.
$2$. Since the rays actually intersect,the image is real.
$3$. The size of the image is smaller than the object,hence it is diminished.

(D) When an object is placed at the center of curvature $(C)$ of a concave mirror,the light rays originating from the object follow the laws of reflection.
One ray passes parallel to the principal axis and reflects through the focus $(F)$.
Another ray passes through the focus $(F)$ and reflects parallel to the principal axis.
These reflected rays intersect exactly at the center of curvature $(C)$.
Therefore,the image formed is real,inverted,and of the same size as the object.

(A) For a concave mirror,when an object is placed at the center of curvature $(C)$,the light rays originating from the object strike the mirror and reflect back through the same path.
As a result,the image is formed at the center of curvature $(C)$ itself.
The image formed is real,inverted,and of the same size as the object.

(B) When an object is placed between the principal focus $(F)$ and the center of curvature $(C)$ of a concave mirror,the light rays from the object follow the rules of reflection:
$1$. $A$ ray parallel to the principal axis passes through the principal focus $(F)$ after reflection.
$2$. $A$ ray passing through the principal focus $(F)$ becomes parallel to the principal axis after reflection.
These reflected rays meet at a point beyond the center of curvature $(C)$.
Therefore,the image formed is real,inverted,and magnified,and it is located beyond the center of curvature $(C)$.

(D) For a concave mirror,when an object is placed between the principal focus $(F)$ and the center of curvature $(C)$:
$1$. The light rays from the object diverge and meet at a point beyond the center of curvature $(C)$.
$2$. The image formed is real because the light rays actually intersect.
$3$. The image is inverted.
$4$. The size of the image is larger than the object (magnified).
Therefore,the image formed is real and magnified.

(B) For a concave mirror,when an object is placed between the principal focus $(F)$ and the center of curvature $(C)$,the image formed is real,inverted,and magnified.
$1$. When the object is beyond $C$,the image is diminished,real,and inverted.
$2$. When the object is at $C$,the image is of the same size,real,and inverted.
$3$. When the object is between $F$ and $C$,the image is formed beyond $C$ and is magnified,real,and inverted.
$4$. When the object is between $F$ and the pole $(P)$,the image is virtual,erect,and magnified.

(B) When an object is placed between the principal focus $(F)$ and the pole $(P)$ of a concave mirror,the light rays diverge after reflection.
By extending these rays backward,they appear to meet behind the mirror.
Therefore,the image formed is virtual,erect,and magnified (larger than the object).
This is the only case where a concave mirror produces a virtual image.

(B) When an object is placed at the centre of curvature $(C)$ of a concave mirror,the light rays originating from the object follow the laws of reflection.
$1$. $A$ ray parallel to the principal axis passes through the principal focus $(F)$ after reflection.
$2$. $A$ ray passing through the principal focus $(F)$ emerges parallel to the principal axis after reflection.
These two reflected rays intersect exactly at the centre of curvature $(C)$.
Therefore,the image formed is real,inverted,and of the same size as the object,located at the centre of curvature $(C)$.

(A) convex mirror always forms a virtual,erect,and diminished image for any position of the object in front of it.
When an object is placed at any finite distance (including $2f$) in front of a convex mirror,the light rays diverge after reflection.
These rays appear to originate from a point behind the mirror.
The image is always formed between the pole $(P)$ and the principal focus $(F)$ behind the mirror.
Therefore,the correct option is $A$.

(B) For a concave mirror,magnification $m = -v/u$. Given $m = -2$.
Substituting this,$-2 = -v/u$,which implies $v = 2u$.
According to the mirror formula,$1/f = 1/v + 1/u$.
Substituting $v = 2u$,we get $1/f = 1/(2u) + 1/u = 3/(2u)$.
This gives $u = 1.5f$.
Since the focal length $f$ is between the pole and the center of curvature $(2f)$,an object placed at $1.5f$ lies between the principal focus $(F)$ and the center of curvature $(C)$.
Thus,the correct position is between the principal focus and the center of curvature.

(A) The magnification $m$ of a spherical mirror is given by $m = -v/u = h'/h$.
$1.$ Magnification $m = -1$: The negative sign indicates a real and inverted image. The magnitude $1$ indicates the image size is equal to the object size. Thus, $(1-b)$.
$2.$ Magnification $m = +5$: The positive sign indicates a virtual and erect image. The magnitude $5$ $( > 1)$ indicates an enlarged image. Thus, $(2-d)$.
$3.$ Magnification $m = -5$: The negative sign indicates a real and inverted image. The magnitude $5$ $( > 1)$ indicates an enlarged image. Thus, $(3-a)$.
$4.$ Magnification $m = -0.5$: The negative sign indicates a real and inverted image. The magnitude $0.5$ $( < 1)$ indicates a diminished image. Thus, $(4-c)$.
Therefore, the correct matching is $(1-b), (2-d), (3-a), (4-c)$.

(B) For a concave mirror,a real image is always inverted.
According to the sign convention,the height of an inverted image is taken as negative $(h_i < 0)$ and the height of the object is taken as positive $(h_o > 0)$.
Magnification $(m)$ is defined as the ratio of the height of the image to the height of the object: $m = h_i / h_o$.
Since $h_i$ is negative and $h_o$ is positive,the magnification $m$ must be negative $(m < 0)$.

(B) According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Given that the object is at infinity,$u = \infty$.
Substituting this into the formula: $\frac{1}{v} + \frac{1}{\infty} = \frac{1}{f}$.
Since $\frac{1}{\infty} = 0$,we get $\frac{1}{v} + 0 = \frac{1}{f}$.
Therefore,$v = f$.
When an object is placed at infinity in front of a concave mirror,the rays of light are parallel to the principal axis,and the image is formed at the principal focus $(F)$.

(B) The magnification $(m)$ of a mirror is given by the ratio of the height of the image $(h')$ to the height of the object $(h)$,which is also equal to $(-\frac{v}{u})$.
Given $m = -\frac{1}{3}$.
$1$. The negative sign $(-)$ indicates that the image is real and inverted.
$2$. The magnitude of magnification is $\frac{1}{3}$,which is less than $1$,indicating that the image is diminished.
$3$. $A$ plane mirror always produces a virtual,erect image with a magnification of $+1$.
$4$. $A$ convex mirror always produces a virtual,erect,and diminished image with a magnification between $0$ and $+1$.
$5$. $A$ concave mirror can produce real,inverted,and diminished images when the object is placed beyond the center of curvature $(C)$.
Therefore,the mirror must be a concave mirror.

(D) In the context of spherical mirrors,the magnification $m$ is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$,i.e.,$m = h'/h$.
For real images,the image is formed below the principal axis,making the height of the image $(h')$ negative,while the object height $(h)$ is taken as positive.
Therefore,a negative magnification $(m < 0)$ indicates that the image is real and inverted.
Since all real images formed by spherical mirrors are inverted,the correct nature of the image is inverted.

(D) concave mirror can form various types of images depending on the position of the object:
$1$. When the object is placed between the pole $(P)$ and the focus $(F)$,the image formed is virtual,erect,and enlarged.
$2$. When the object is placed at the focus $(F)$,the image is real,inverted,and highly enlarged.
$3$. When the object is placed between $F$ and $C$,the image is real,inverted,and enlarged.
$4$. When the object is placed at $C$,the image is real,inverted,and of the same size.
$5$. When the object is placed beyond $C$,the image is real,inverted,and diminished.
$6$. $A$ concave mirror can never form a virtual and diminished image. $A$ virtual and diminished image is only formed by a convex mirror.

(D) concave mirror can form both real and virtual images.
$1$. Real images: These can be enlarged (when the object is between $F$ and $C$),diminished (when the object is beyond $C$),or of the same size (when the object is at $C$).
$2$. Virtual images: $A$ concave mirror forms a virtual image only when the object is placed between the pole $(P)$ and the focus $(F)$. In this case,the image formed is always virtual,erect,and enlarged.
$3$. Therefore,a concave mirror can never form a virtual and diminished image. $A$ virtual and diminished image is only formed by a convex mirror.

(A) convex mirror always forms a virtual,erect,and diminished image for all positions of the object placed in front of it.
For an erect image,the height of the image $(h')$ and the height of the object $(h)$ are both taken as positive.
Magnification $(m)$ is defined as the ratio of the height of the image to the height of the object: $m = h'/h$.
Since both $h'$ and $h$ are positive for an erect image,the magnification $(m)$ is always positive.
Additionally,for a convex mirror,the image is always diminished,meaning the size of the image is smaller than the size of the object $(h' < h)$,so the magnification is always less than $1$ $(m < 1)$.

(A) The mirror formula expresses the relationship between the object distance $(u)$,image distance $(v)$,and the focal length $(f)$ of a spherical mirror.
It is given by the equation: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Here,$u$ is the distance of the object from the pole,$v$ is the distance of the image from the pole,and $f$ is the focal length of the mirror.

(B) For a spherical mirror,the relationship between the focal length $(f)$ and the radius of curvature $(R)$ is given by the formula: $R = 2f$.
Given that the focal length $f = 20 \ cm$.
Substituting the value into the formula: $R = 2 \times 20 \ cm = 40 \ cm$.
Therefore,the radius of curvature is $40 \ cm$.

(A) The relationship between the focal length $(f)$ and the radius of curvature $(R)$ of a spherical mirror is given by the formula: $f = \frac{R}{2}$.
Given that the radius of curvature $R = 30 \ cm$.
Substituting the value of $R$ into the formula: $f = \frac{30 \ cm}{2} = 15 \ cm$.
Therefore,the focal length of the spherical mirror is $15 \ cm$.

(A) Magnification $(m)$ produced by a spherical mirror or lens is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
Mathematically,it is expressed as: $m = \frac{h'}{h}$.
Therefore,magnification is the ratio of the height of the image to the height of the object.

(D) Given: Object height $(h_o)$ = $4 \, cm$,Object distance $(u)$ = $-12 \, cm$ (by sign convention).
Since the image is formed behind the mirror,it is a virtual image. Thus,Image distance $(v)$ = $+24 \, cm$.
The magnification $(m)$ of a mirror is given by the formula: $m = \frac{h_i}{h_o} = -\frac{v}{u}$.
Substituting the values: $\frac{h_i}{4} = -\frac{24}{-12}$.
$\frac{h_i}{4} = 2$.
$h_i = 2 \times 4 = 8 \, cm$.
Therefore,the height of the image is $8 \, cm$.

(B) Given: Focal length $(f)$ = $-12 \ cm$ (for a concave mirror),Object distance $(u)$ = $-18 \ cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-12} = \frac{1}{v} + \frac{1}{-18}$.
Rearranging: $\frac{1}{v} = \frac{1}{18} - \frac{1}{12}$.
Calculating the common denominator $(36)$: $\frac{1}{v} = \frac{2 - 3}{36} = -\frac{1}{36}$.
Therefore,$v = -36 \ cm$.
The center of curvature $(C)$ is at $2f = 2 \times 12 = 24 \ cm$.
Since the image distance $(36 \ cm)$ is greater than the center of curvature $(24 \ cm)$,the image is formed beyond the center of curvature.

(D) Given: Focal length $f = -10 \text{ cm}$ (for a concave mirror),Object distance $u = -20 \text{ cm}$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-10} = \frac{1}{v} + \frac{1}{-20}$.
Rearranging: $\frac{1}{v} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = -\frac{1}{20}$.
Therefore,$v = -20 \text{ cm}$.
Since the object is placed at $2f$ $(2 \times 10 \text{ cm} = 20 \text{ cm})$,the image is also formed at $2f$,which is the center of curvature.

(C) Given: Focal length $(f)$ = $-20 \text{ cm}$ (for a concave mirror),Object distance $(u)$ = $-10 \text{ cm}$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-10}$.
Rearranging: $\frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{2-1}{20} = \frac{1}{20}$.
So,$v = +20 \text{ cm}$.
Since the image distance $(v)$ is positive,the image is formed behind the mirror,which means it is virtual.
Magnification $(m)$ = $-\frac{v}{u} = -(\frac{20}{-10}) = +2$.
Since $m$ is positive and greater than $1$,the image is virtual and magnified.