Mix Examples - Light – Reflection and Refraction Questions in English

(A) When a plane mirror is rotated by an angle $\theta$,the reflected ray rotates by an angle $2\theta$.
However,the size of the image formed by a plane mirror depends only on the size of the object.
Since the rotation of the mirror does not change the size of the object or the distance of the object from the mirror,the magnification remains $1$.
Therefore,the size of the image remains the same.

(B) For a convex mirror,the magnification $m$ is given by $m = -v/u$. Given that the image size is one-third of the object size,$m = 1/3$.
Since $m = -v/u = 1/3$,we have $u = -3v$.
The mirror formula is $1/f = 1/v + 1/u$.
Given $f = +12 \, cm$ (for a convex mirror),we substitute the values: $1/12 = 1/v + 1/(-3v)$.
$1/12 = (3 - 1) / 3v = 2 / 3v$.
$3v = 24$,which gives $v = +8 \, cm$.
$A$ positive value for $v$ indicates that the image is formed $8 \, cm$ behind the mirror.

(C) plane mirror typically forms a virtual image for a real object because the reflected rays diverge.
However,if the incident beam is convergent,the rays are directed towards a point behind the mirror.
In this case,the mirror intercepts these rays before they can meet,reflecting them to a point in front of the mirror.
Since the reflected rays actually meet at this point,a real image is formed.
Therefore,the incident beam must be convergent.

(D) The formula for the number of images $(n)$ formed by two plane mirrors inclined at an angle $(\theta)$ is given by $n = (360^{\circ} / \theta) - 1$.
Given the angle $\theta = 72^{\circ}$.
First, calculate the value of $360^{\circ} / \theta = 360^{\circ} / 72^{\circ} = 5$.
Since the result is an odd integer, the number of images formed is $n = 5 - 1 = 4$.
This formula applies regardless of whether the object is placed symmetrically or asymmetrically when the result of $360^{\circ} / \theta$ is an odd integer.

(A) The magnification $(m)$ of a mirror is defined as the ratio of the height of the image $(h_i)$ to the height of the object $(h_o)$,given by $m = h_i / h_o$.
For a plane mirror,the image formed is virtual,erect,and of the same size as the object.
Since the image is erect,the magnification is positive $(+)$.
Since the size of the image is equal to the size of the object $(h_i = h_o)$,the ratio $h_i / h_o = 1$.
Therefore,the magnification produced by a plane mirror is $+1$.

(C) When an object is placed between the principal focus $(F)$ and the centre of curvature $(C)$ of a concave mirror,the image formed is real,inverted,and magnified.
Since the magnification $(m)$ of a concave mirror depends on the distance of the object from the mirror,different parts of the dice are at different distances from the mirror.
The part of the dice closer to the focus $(F)$ will have a larger magnification compared to the part closer to the centre of curvature $(C)$.
Because the magnification varies along the axis,the edges of the dice parallel to the principal axis will be magnified differently,resulting in a shape that is wider at one end and narrower at the other,resembling a barrel shape.

(D) In this scenario, we have three mirrors arranged mutually perpendicular to each other (two walls and the ceiling).
Let the position of the observer be $(x, y, z)$.
The images are formed by reflections in the $XY$, $YZ$, and $ZX$ planes.
The total number of images formed by three mutually perpendicular mirrors is given by the formula $N = 2^n - 1$, where $n$ is the number of mirrors.
Here, $n = 3$, so $N = 2^3 - 1 = 8 - 1 = 7$.
However, the observer is inside the room, and one of these images is the observer's own position or is obscured by the observer's body.
Specifically, for three mutually perpendicular mirrors, the images are located at $(\pm x, \pm y, \pm z)$ excluding the original position $(x, y, z)$.
There are $2^3 = 8$ total positions including the object.
Subtracting the object itself, we get $8 - 1 = 7$ images.
Since the observer is part of the system, they see $7$ images of themselves.

(D) convex mirror is a spherical mirror whose reflecting surface is curved outwards.
When an object is placed in front of a convex mirror,the light rays diverge after reflection.
These reflected rays appear to meet behind the mirror when extended backwards.
Therefore,the image formed by a convex mirror is always virtual,erect,and diminished in size,regardless of the position of the object.

(A) virtual image is formed when light rays appear to diverge from a point behind the mirror.
In a plane mirror,the image formed is always virtual,erect,and of the same size as the object.
In a concave mirror,a virtual and erect image is formed only when the object is placed between the pole $(P)$ and the focus $(F)$,but this image is always magnified (larger than the object).
In a convex mirror,the image formed is always virtual,erect,and diminished (smaller than the object) for any position of the object in front of the mirror.
Therefore,a diminished virtual image can be obtained only in a convex mirror.

(B) For a convex mirror,the focal length $f$ is taken as positive $(+f)$.
The object distance $u$ is always taken as negative,so $u = -f$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{v} + \frac{1}{-f}$.
Rearranging the terms: $\frac{1}{v} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$.
Therefore,$v = f / 2$.
The image is formed at a distance of $f / 2$ behind the mirror.

(C) When a light ray is incident normally (perpendicularly) on a plane mirror, the angle of incidence $(i)$ is $0^{\circ}$.
According to the law of reflection, the angle of reflection $(r)$ is also $0^{\circ}$.
The light ray reflects back along the same path.
The deviation $(\delta)$ produced by a mirror is given by the formula $\delta = 180^{\circ} - 2i$.
Substituting $i = 0^{\circ}$, we get $\delta = 180^{\circ} - 2(0^{\circ}) = 180^{\circ}$.
Therefore, the light ray suffers a deviation of $180^{\circ}$.

(D) convex mirror always forms a virtual,erect,and diminished image for any real object placed in front of it.
$1$. The image is always formed behind the mirror between the pole $(P)$ and the focus $(F)$.
$2$. Since the image is formed by the apparent intersection of reflected rays,it is always virtual.
$3$. Therefore,the statement that the image is real is incorrect.

(A) Given: Object distance $u = -40\, cm$,Focal length $f = -20\, cm$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-40}$.
$\frac{1}{v} = \frac{1}{40} - \frac{1}{20} = \frac{1 - 2}{40} = -\frac{1}{40}$.
So,$v = -40\, cm$.
The magnification $m = -\frac{v}{u} = -\frac{-40}{-40} = -1$.
$A$ magnification of $-1$ indicates that the image is real,inverted,and of the same size as the object.

(C) When two plane mirrors are placed at an angle of $90^{\circ}$,the number of images formed is given by the formula $n = (360^{\circ} / \theta) - 1$.
Substituting $\theta = 90^{\circ}$,we get $n = (360^{\circ} / 90^{\circ}) - 1 = 4 - 1 = 3$ images.
Let the man be standing in the first quadrant. The three images are formed at:
$1$. Image $I_1$ in the first mirror (lateral inversion occurs).
$2$. Image $I_2$ in the second mirror (lateral inversion occurs).
$3$. Image $I_3$ which is the image of $I_1$ (or $I_2$) in the other mirror (double lateral inversion occurs).
In $I_1$ and $I_2$,the man appears to use his left hand due to single lateral inversion.
In $I_3$,the image undergoes two lateral inversions,so the left-right orientation is restored to the original state (he appears to use his right hand).
Therefore,in $2$ images,he will be seen using his left hand.

(B) The statement is $False$. The laws of reflection,which state that the angle of incidence equals the angle of reflection $(i = r)$ and that the incident ray,the reflected ray,and the normal all lie in the same plane,are universal. They apply to all reflecting surfaces,including plane mirrors,spherical mirrors (concave and convex),and even irregular surfaces at the point of incidence.

(B) The relationship between the radius of curvature $(R)$ and the focal length $(f)$ of a spherical mirror is given by the formula $R = 2f$.
This means that the radius of curvature is twice the focal length,not half.
Therefore,the given statement is false.

(A) According to the rules of image formation by spherical mirrors,any light ray travelling parallel to the principal axis of a concave mirror will reflect and pass through its principal focus $(F)$.
This is a fundamental property used to define the focal point of a concave mirror.

(B) convex mirror always forms a virtual,erect,and diminished image of an object placed in front of it,regardless of the distance of the object.
Since the image formed is always smaller than the object,it cannot be used to see a large image of a small object.
Therefore,the given statement is False.

(A) convex mirror is a spherical mirror with its reflecting surface curved outwards.
Due to the nature of its curvature,light rays diverging from an object appear to meet behind the mirror after reflection.
Consequently,for any position of the object in front of a convex mirror,the image formed is always virtual,erect,and diminished in size.
Therefore,the statement is True.

(A) When a small lamp (point source of light) is placed at the principal focus $(F)$ of a concave mirror,the light rays originating from the lamp strike the mirror and are reflected parallel to the principal axis. This property is widely used in torches,searchlights,and vehicle headlights to produce a strong,parallel beam of light.

(B) plane mirror always forms a virtual,erect,and laterally inverted image of an object.
Since the reflected rays never actually meet at a point but only appear to diverge from a point behind the mirror,the image formed is always virtual.
Therefore,a plane mirror cannot form a real image.

(A) concave mirror can form a virtual,magnified,and erect image when the object is placed between the pole $(P)$ and the focus $(F)$ of the mirror. In this position,the light rays appear to diverge from behind the mirror,resulting in a virtual image that is larger than the object and upright.

(B) The statement is False. According to the laws of reflection,when a plane mirror is rotated by an angle $\theta$ while keeping the incident ray fixed,the reflected ray rotates by an angle $2\theta$ in the same direction. Since the original statement claims the reflected ray turns by an angle $\theta$,it is incorrect.

(A) In a plane mirror,the image formed is virtual,erect,and of the same size as the object. One of the fundamental properties of a plane mirror is that the object distance $(u)$ is equal to the image distance $(v)$ from the mirror surface. Therefore,the distance between the image and the mirror is exactly the same as the distance between the object and the mirror.

(B) The number of images $n$ formed by two plane mirrors inclined at an angle $\theta$ is given by the formula $n = (360^{\circ} / \theta) - 1$.
Given $\theta = 45^{\circ}$.
$n = (360^{\circ} / 45^{\circ}) - 1 = 8 - 1 = 7$.
Since the calculated number of images is $7$ and not $9$,the statement is False.

(A) When light reflects off a surface,it remains in the same medium. The properties of light such as frequency,wavelength,and speed depend on the nature of the medium. Since the medium does not change during reflection,these properties remain constant. Therefore,the statement is True.

(B) 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.
It relates the focal length of a spherical mirror to the object distance and the image distance.
The relationship between the height of the image $(h')$ and the height of the object $(h)$ is defined by the magnification formula,which is $m = \frac{h'}{h} = -\frac{v}{u}$.
Therefore,the statement is false.

(A) The mirror formula is given by the expression $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$,where $f$ is the focal length,$v$ is the image distance,and $u$ is the object distance.
This formula is a universal relationship derived from the laws of reflection and geometry.
It applies to both spherical mirrors,i.e.,concave mirrors and convex mirrors,provided that the sign convention is applied correctly during calculation.

(A) The magnification $(m)$ of a spherical mirror is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
It is also related to the object distance $(u)$ and image distance $(v)$ by the formula: $m = \frac{h'}{h} = -\frac{v}{u}$.
Therefore,the magnification equation establishes a direct relationship between the image distance and the object distance.

(A) The magnification $(m)$ produced by a spherical mirror is defined as the ratio of the height of the image $(h')$ to the height of the object $(h)$.
The formula is given by $m = \frac{h'}{h} = -\frac{v}{u}$,where $v$ is the image distance and $u$ is the object distance.
This formula is derived from the geometry of reflection and is universally applicable to all spherical mirrors,whether they are concave or convex,provided that the sign convention is followed correctly.

(A) The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye.
Visible light typically has a wavelength range from approximately $380 \; nm$ to $750 \; nm$.
Converting these values to meters: $380 \; nm = 380 \times 10^{-9} \; m = 3.8 \times 10^{-7} \; m$ and $750 \; nm = 750 \times 10^{-9} \; m = 7.5 \times 10^{-7} \; m$.
Rounding these values,the range is approximately $4 \times 10^{-7} \; m$ to $8 \times 10^{-7} \; m$.

(C) For a spherical mirror of small aperture,the radius of curvature $(R)$ is twice the focal length $(f)$.
This relationship is expressed as $R = 2f$ or $f = R / 2$.
This is because the principal focus lies exactly at the midpoint between the pole and the center of curvature of the mirror.

(B) The surface of a book is rough and uneven at the microscopic level.
When light rays fall on such a rough surface,they are reflected in various directions.
This type of reflection is known as irregular or diffused reflection.
Regular reflection occurs only on smooth,polished surfaces like mirrors.

(B) When a light ray is directed towards the centre of curvature $(C)$ of a concave mirror,it strikes the mirror surface normally (at an angle of $90^{\circ}$ to the tangent at the point of incidence).
According to the laws of reflection,a ray incident normally on a spherical mirror reflects back along the same path.
Therefore,the reflected ray passes back through the centre of curvature $(C)$.

(C) concave mirror produces a virtual and erect image only when the object is placed between the pole $(P)$ and the principal focus $(F)$.
In this position,the light rays appear to diverge from behind the mirror,forming an enlarged,virtual,and erect image.
For all other positions (at $F$,between $F$ and $C$,at $C$,and beyond $C$),the image formed by a concave mirror is real and inverted.

(B) The magnification $(m)$ of a mirror is defined as the ratio of the height of the image $(h_i)$ to the height of the object $(h_o)$,expressed as $m = h_i / h_o$.
In a plane mirror,the image formed is always virtual,erect,and of the same size as the object.
Since the height of the image is equal to the height of the object $(h_i = h_o)$,the magnification is $m = h_i / h_o = 1$.
Additionally,the image is erect,so the magnification is positive,resulting in $m = +1$.

(B) plane mirror can be considered as a spherical mirror with an infinite radius of curvature $(R = \infty)$.
Since the focal length $(f)$ is half of the radius of curvature $(f = R/2)$,it follows that $f = \infty / 2 = \infty$.
Therefore,the focal length of a plane mirror is infinity.

(A) In 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 is at a distance of $2 \, m$ from the mirror.
Therefore,the image is also formed at a distance of $2 \, m$ behind the mirror.
The total distance between the object and its image is the sum of the distance of the object from the mirror and the distance of the image from the mirror.
Total distance = $2 \, m + 2 \, m = 4 \, m$.

(A) In a compound microscope,the objective lens forms a real,inverted,and enlarged image of the object.
This image acts as an object for the eyepiece (ocular lens).
The eyepiece acts as a simple magnifier and forms a virtual,erect (with respect to the intermediate image),and further enlarged image.
Since the intermediate image was inverted,the final image appears inverted with respect to the original object.
Therefore,the final image formed by a compound microscope is virtual,inverted,and enlarged compared to the original object.

(C) For a convex lens,when an object is placed at the centre of curvature $(2F_1)$,the image formed is real,inverted,and of the same size as the object.
This image is formed at the centre of curvature $(2F_2)$ on the other side of the lens.
Therefore,the correct position is at the centre of curvature.

(B) Optical density is closely related to the refractive index of a material. $A$ higher refractive index indicates a higher optical density.
The refractive indices of the given materials are approximately:
$1$. Water: $1.33$
$2$. Glass: $1.50$
$3$. Pearl: $1.53$
$4$. Diamond: $2.42$
Since diamond has the highest refractive index among the given options,it possesses the maximum optical density.

(B) The absolute 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 $(v)$: $n = c/v$.
Since the speed of light in a vacuum is the maximum possible speed of light in any medium, $c$ is always greater than $v$ $(c > v)$.
Therefore, the ratio $c/v$ is always greater than $1$ for any medium other than a vacuum (where it is exactly $1$).
Thus, the absolute refractive index of any medium is always greater than $1$ $(n > 1)$.

(D) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters,given by the formula $P = \frac{1}{f(m)}$.
Since $P \propto \frac{1}{f}$,a lens with a smaller focal length will have a higher power.
Comparing the given focal lengths: $10 \, cm < 20 \, cm < 25 \, cm < 50 \, cm$.
Therefore,the lens with the focal length of $10 \, cm$ will have the maximum power $(P = \frac{1}{0.1} = 10 \, D)$.

(D) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters,given by the formula: $P = \frac{1}{f(m)}$.
Given,$P = +5.0 \,D$.
Substituting the value in the formula: $5.0 = \frac{1}{f}$.
Therefore,$f = \frac{1}{5.0} \,m = 0.2 \,m$.
To convert the focal length into centimeters,multiply by $100$: $f = 0.2 \times 100 \,cm = +20 \,cm$.
Since the power is positive,the lens is a convex lens,and its focal length is positive.

(A) The relative refractive index of medium $2$ with respect to medium $1$ is given by the formula: $n_{21} = n_2 / n_1$, where $n_2$ is the refractive index of the second medium and $n_1$ is the refractive index of the first medium.
Given refractive indices: $n_{\text{water}} = 1.33$, $n_{\text{benzene}} = 1.50$, $n_{\text{sapphire}} = 1.77$.
Calculating the values for each option:
$A$: Sapphire relative to water = $1.77 / 1.33 \approx 1.3308$
$B$: Sapphire relative to benzene = $1.77 / 1.50 = 1.18$
$C$: Benzene relative to water = $1.50 / 1.33 \approx 1.1278$
$D$: Water relative to benzene = $1.33 / 1.50 \approx 0.8867$
Comparing these values, the relative refractive index of sapphire with respect to water $(1.3308)$ is the maximum.

(C) plane mirror always forms a virtual and erect image.
In a plane mirror,the image formed is of the same size as the object and is laterally inverted.
Since the light rays do not actually meet at the image point but appear to diverge from it,the image is virtual.
Therefore,the correct option is $C$.

(C) 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 $(v)$,given by $n = \frac{c}{v}$.
From this,the velocity of light in a medium is $v = \frac{c}{n}$.
Let $n_w = \frac{4}{3}$ be the refractive index of water and $n_g = \frac{3}{2}$ be the refractive index of glass.
The velocity of light in water is $v_w = \frac{c}{n_w}$ and in glass is $v_g = \frac{c}{n_g}$.
The ratio of the velocity of light in water to that in glass is $\frac{v_w}{v_g} = \frac{c/n_w}{c/n_g} = \frac{n_g}{n_w}$.
Substituting the given values: $\frac{v_w}{v_g} = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$.
Therefore,the ratio is $\frac{9}{8}$.

(C) The optical density of a medium is directly related to its refractive index. $A$ higher refractive index indicates a higher optical density.
Given refractive indices are:
Water: $n = 1.33$
Glass: $n = 1.50$
Diamond: $n = 2.42$
Comparing these values,$2.42 > 1.50 > 1.33$. Since diamond has the highest refractive index,it is the most optically dense medium among the three.

(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) When a light ray is incident normally (perpendicularly) on the surface separating two media,the angle of incidence $(i)$ is $0^{\circ}$.
According to Snell's Law,$n_1 \sin(i) = n_2 \sin(r)$.
Since $i = 0^{\circ}$,then $\sin(i) = 0$.
Therefore,$n_1 \times 0 = n_2 \sin(r)$,which implies $\sin(r) = 0$.
Thus,the angle of refraction $(r)$ is $0^{\circ}$.
This means the light ray passes undeviated through the interface.