Mix Examples - Light – Reflection and Refraction Questions in English

(B) For a spherical mirror,the relationship between the radius of curvature $(R)$ and the focal length $(f)$ is given by $R = 2f$.
The principal focus $(F)$ lies exactly at the midpoint between the pole $(P)$ and the center of curvature $(C)$.
Therefore,the distance between the principal focus $(F)$ and the center of curvature $(C)$ is equal to the focal length $(f)$.

(C) For a concave mirror,a virtual and magnified image is formed only when the object is placed between the pole $(P)$ and the principal focus $(F)$.
This means the object distance $(u)$ must be less than the focal length $(f)$,i.e.,$u < f$.

(A) The magnification $(m)$ of a mirror is defined as the ratio of the image distance $(v)$ to the object distance $(u)$,given by $m = -v/u$.
For a magnified image,the absolute value of magnification must be greater than $1$,i.e.,$|m| > 1$.
This implies $|-v/u| > 1$,which simplifies to $|v| > |u|$.
In the context of a concave mirror,when the object is placed between the focus $(F)$ and the pole $(P)$,or between the focus $(F)$ and the center of curvature $(C)$,the image formed is magnified,meaning the image distance $(v)$ is greater than the object distance $(u)$.

(D) The magnification $m$ is given by the ratio of the height of the image $(h')$ to the height of the object $(h)$,i.e.,$m = h'/h$.
Since the value of $m$ is $+0.5$,the positive sign indicates that the image is virtual and erect.
The magnitude $0.5$ (which is less than $1$) indicates that the image is diminished.
$A$ convex mirror always forms a virtual,erect,and diminished image for all positions of the object.
Therefore,a convex mirror can produce a magnification of $+0.5$.

(B) 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 a spherical mirror,$m = -v/u$.
$1$. $A$ magnification of $+5$ indicates a virtual,erect,and magnified image. This occurs when an object is placed between the pole $(P)$ and the focus $(F)$ of a concave mirror.
$2$. $A$ magnification of $-5$ indicates a real,inverted,and magnified image. This occurs when an object is placed between the focus $(F)$ and the center of curvature $(C)$ of a concave mirror.
Since both cases are possible with a concave mirror,the correct answer is a concave mirror.

(C) In a searchlight or torch,a concave mirror is used as a reflector.
According to the properties of a concave mirror,when a light source is placed at the principal focus $(F)$,the light rays originating from it strike the mirror and are reflected as a parallel beam of light.
These parallel rays do not diverge significantly and can travel over a long distance.
Therefore,the bulb is placed at the principal focus.

(A) The speed of light depends on the refractive index of the medium. The speed of light is given by the formula $v = c/n$,where $c$ is the speed of light in a vacuum and $n$ is the refractive index of the medium.
Since the refractive index of air $(n \approx 1.0003)$ is the lowest among the given options (Water $n \approx 1.33$,Glass $n \approx 1.5$,Diamond $n \approx 2.42$),the speed of light is highest in air.

(C) The refractive index of a medium is a measure of how much the speed of light is reduced when entering that medium.
Air has a refractive index of approximately $1.0003$,which is the lowest among the given options.
Water has a refractive index of approximately $1.33$.
Glass has a refractive index ranging from $1.5$ to $1.9$ depending on the type.
Diamond has a very high refractive index of approximately $2.42$.
Therefore,air has the minimum refractive index.

(B) When a light ray travels from an optically denser medium (glass) to an optically rarer medium (air),it speeds up. According to the laws of refraction,when light enters a rarer medium from a denser medium,it bends away from the normal. Therefore,the correct answer is that it bends away from the normal.

(A) Refraction is the bending of light when it passes from one transparent medium to another.
However,when a light ray is incident normally (perpendicularly) on the surface of a medium,the angle of incidence $(i)$ is $0^{\circ}$.
According to Snell's Law,$n_1 \sin(i) = n_2 \sin(r)$.
Since $\sin(0^{\circ}) = 0$,it follows that $\sin(r) = 0$,which means the angle of refraction $(r)$ is also $0^{\circ}$.
Therefore,the light ray passes undeviated (without bending) through the glass slab when the angle of incidence is $0^{\circ}$.

(C) When a light ray is incident normally (at an angle of $90^{\circ}$ to the surface) on the interface of two media,the angle of incidence $(i)$ is $0^{\circ}$.
According to Snell's Law,$n_1 \sin(i) = n_2 \sin(r)$.
Since $\sin(0^{\circ}) = 0$,it follows that $\sin(r) = 0$,which means the angle of refraction $(r)$ is also $0^{\circ}$.
Therefore,the light ray passes through the glass slab without any deviation or bending.

(C) Snell's law describes the relationship between the angles of incidence and refraction when light passes through the boundary between two different isotropic media.
It states that the product of the refractive index of the first medium $(n_{1})$ and the sine of the angle of incidence $(\sin \theta_{1})$ is equal to the product of the refractive index of the second medium $(n_{2})$ and the sine of the angle of refraction $(\sin \theta_{2})$.
Mathematically,this is expressed as $n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}$.

(D) 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,$n = \frac{c}{v}$.
Since both $c$ and $v$ are speeds measured in the same units (e.g.,$m/s$),their ratio is a dimensionless quantity.
Therefore,the refractive index has no unit.

(C) The refractive index of a medium is a measure of how much the speed of light is reduced inside the medium. It is commonly denoted by the symbol $n$ or $\mu$. In the context of the $NCERT$ physics curriculum,both symbols are used interchangeably to represent the refractive index. Since both $n$ and $\mu$ are standard,and given the options,$n$ is the most frequently used symbol in modern textbooks.

(B) 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 the medium $(v)$: $n = c / v$.
Given that the refractive index $n = 1.5 = 3/2$.
We need to find the ratio of the velocity of light in the medium $(v)$ to the velocity of light in a vacuum $(c)$,which is $v / c$.
From the formula $n = c / v$,we have $v / c = 1 / n$.
Substituting the value of $n$: $v / c = 1 / (3/2) = 2/3$.
Therefore,the velocity of light in the medium is $2/3$ times the velocity of light in a vacuum.

(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,$n = c/v$.
The refractive index of water is approximately $4/3$.
Therefore,$4/3 = c/v$,which implies $c = (4/3)v$.
This means the velocity of light in a vacuum is $4/3$ times the velocity of light in water.

(A) The refractive index of medium $(2)$ with respect to medium $(1)$ is defined as the ratio of the speed of light in medium $(1)$ to the speed of light in medium $(2)$.
Mathematically,this is expressed as $\eta_{21} = \frac{v_{1}}{v_{2}}$.
Also,since the refractive index is inversely proportional to the velocity of light in the medium,it can be expressed in terms of refractive indices as $\eta_{21} = \frac{\eta_{2}}{\eta_{1}}$.
Given the options provided,the correct formula representing the ratio of velocities is $\frac{v_{1}}{v_{2}}$.

(A) The refractive index $(n)$ of a medium is defined as the ratio of the speed of light in a vacuum or air $(c)$ to the speed of light in that medium $(v)$.
Mathematically,$n = \frac{c}{v}$.
Given:
Speed of light in air $(c)$ = $3 \times 10^{8} \, m \, s^{-1}$.
Refractive index of water $(n)$ = $\frac{4}{3}$.
We need to find the velocity of light in water $(v)$.
Rearranging the formula: $v = \frac{c}{n}$.
Substituting the values: $v = \frac{3 \times 10^{8}}{\frac{4}{3}} = 3 \times 10^{8} \times \frac{3}{4} = \frac{9}{4} \times 10^{8} = 2.25 \times 10^{8} \, m \, s^{-1}$.
Therefore,the velocity of light in water is $2.25 \times 10^{8} \, m \, s^{-1}$.

(C) The lateral shift $(d)$ produced by a rectangular glass slab is given by the formula: $d = \frac{t \cdot \sin(i - r)}{\cos(r)}$,where $t$ is the thickness of the glass slab,$i$ is the angle of incidence,and $r$ is the angle of refraction.
From this formula,it is evident that the lateral shift is directly proportional to the thickness $(t)$ of the glass slab.
Therefore,the lateral shift depends on the thickness of the glass slab.

(D) According to Snell's Law of refraction,the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive index of the second medium to the refractive index of the first medium.
Mathematically,this is expressed as: $n_1 \sin \theta_1 = n_2 \sin \theta_2$.
Rearranging the terms to find the ratio $\frac{\sin \theta_1}{\sin \theta_2}$,we get:
$\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}$.

(C) The absolute refractive index 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 $(v)$.
For common glass (crown glass),the refractive index is approximately $1.5$.
This value indicates that light travels $1.5$ times slower in glass than in a vacuum.

(B) The absolute refractive index 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,$n = c/v$.
For water,the speed of light is approximately $2.25 \times 10^8 \text{ m/s}$,and the speed of light in a vacuum is $3 \times 10^8 \text{ m/s}$.
Therefore,the refractive index of water $n = (3 \times 10^8) / (2.25 \times 10^8) = 3 / 2.25 = 300 / 225 = 4/3 \approx 1.33$.

(B) convex lens is a converging lens. When light rays traveling parallel to the principal axis strike the surface of a convex lens,they undergo refraction. After passing through the lens,these rays converge at a specific point on the principal axis. This specific point is known as the principal focus $(F)$ of the lens.

(C) For a convex lens,when an object is placed between the principal focus $(F)$ and the center of curvature $(2F)$,the image formed is real,inverted,and magnified.
This image is formed beyond $2F$ on the other side of the lens.
Therefore,the correct position of the object is between $F$ and $2F$.

(B) When an object is placed at $2F_1$ in front of a convex lens,the light rays originating from the object pass through the lens and converge at $2F_2$ on the other side.
According to the rules of image formation by a convex lens,when the object distance $u = 2F$,the image distance $v = 2F$.
The image formed is real,inverted,and of the same size as the object.

(C) When an object is placed between the principal focus $(F)$ and the optical center $(O)$ of a convex lens,the light rays diverge after refraction.
By extending these rays backward,they appear to meet behind the object on the same side of the lens.
Therefore,the image formed is virtual,erect,and magnified (larger than the object).
This property is used in simple microscopes or magnifying glasses.

(D) When an object is placed at the principal focus $(F)$ of a convex lens,the light rays coming from the object pass through the lens and become parallel to the principal axis.
These parallel rays do not meet on the same side of the lens and appear to meet at infinity.
Therefore,the image is formed at infinity,is real,inverted,and highly enlarged.

(B) For a convex lens,when an object is placed at $2F$ (twice the focal length),the light rays pass through the lens and converge at $2F$ on the other side.
This results in an image that is real,inverted,and exactly the same size as the object.
Therefore,the correct position is at $2F$.

(C) ray of light passing through the optical center $(O)$ of a thin lens goes straight without any deviation or refraction.
This happens because the optical center is the central point of the lens where the incident ray strikes the surface at an angle of $90^{\circ}$ (normal incidence),resulting in no bending of the light path.

(A) For a convex lens,when an object is placed between the optical center $(O)$ and the principal focus $(F)$,the light rays diverge after refraction.
When these rays are traced backward,they appear to meet at a point on the same side of the lens as the object.
This results in the formation of a virtual,erect,and magnified image on the same side as the object.
Therefore,the correct position is between the principal focus and the optical center.

(D) convex lens can form real images (when the object is placed beyond the focus) which can be either magnified,diminished,or the same size as the object.
It can also form a virtual and magnified image when the object is placed between the optical center and the focus.
However,a convex lens cannot form a virtual and diminished image. $A$ virtual and diminished image is only formed by a concave lens or a convex mirror.

(B) concave lens is a diverging lens.
For any position of an object placed in front of a concave lens,the light rays diverge after refraction.
When these rays are extended backward,they appear to meet at a point on the same side of the lens as the object.
Therefore,the image formed is always virtual,erect,and diminished in size,regardless of the object's distance from the lens.

(A) concave lens is a diverging lens. Regardless of the position of the object (placed anywhere between the optical center and infinity),the light rays diverge after refraction. When these rays are extended backward,they appear to meet between the optical center $(O)$ and the principal focus $(F_1)$ on the same side as the object. The image formed is always virtual,erect,and diminished.

(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.

(B) lens is typically formed by the intersection of two spherical surfaces.
Each of these spherical surfaces forms a part of a sphere.
The centre of each such sphere is called the centre of curvature of the lens.
Since a lens has two spherical surfaces,it has two centres of curvature,usually represented as $C_1$ and $C_2$.

(C) convex mirror always forms a virtual,erect,and diminished image for any position of the object in front of it.
Similarly,a concave lens always forms a virtual,erect,and diminished image for any position of the object in front of it.
Therefore,both convex mirrors and concave lenses are characterized by their ability to always produce virtual images regardless of the object's distance.

(B) The lens formula relates the focal length $(f)$,the object distance $(u)$,and the image distance $(v)$ for a spherical lens.
It is given by the expression: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Here,$v$ is the distance of the image from the optical center,$u$ is the distance of the object from the optical center,and $f$ is the focal length of the lens.

(B) lens is a transparent optical medium bounded by two surfaces,at least one of which is spherical.
When light rays pass through a lens,they change their direction due to the change in the optical medium (from air to glass or vice versa).
This bending of light as it passes from one transparent medium to another is known as refraction.
Therefore,the formation of images by lenses is based on the principle of refraction.

(A) dioptre meter (also known as a lensmeter or focimeter) is an ophthalmic instrument used to measure the refractive power of a lens.
Since the power of a lens $(P)$ is defined as the reciprocal of its focal length $(f)$ in meters $(P = 1/f)$,the instrument directly measures the dioptric power of spectacle lenses.

(A) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters.
Formula: $P = \frac{1}{f(m)}$.
Given: Focal length $(f)$ = $50 \ cm = 0.5 \ m$.
Since it is a convex lens,the focal length is positive $(f = +0.5 \ m)$.
Calculation: $P = \frac{1}{+0.5 \ m} = +2.0 \ D$.
Therefore,the power of the convex lens is $+2.0 \ D$.

(A) The power $(P)$ of a lens is defined as the reciprocal of its focal length $(f)$ in meters.
Formula: $P = \frac{1}{f(m)}$.
Given: Power $(P)$ = $+2.0 \ D$.
Substituting the value in the formula: $2.0 = \frac{1}{f}$.
Therefore,$f = \frac{1}{2.0} = 0.5 \ m$.
Since the power is positive,the lens is a convex lens,and the focal length is positive $(+0.5 \ m)$.

(C) The focal length $(f)$ of the convex lens is given as $20 \text{ cm}$.
Since the focal length must be in meters for the power calculation,we convert it: $f = 20 \text{ cm} = 0.2 \text{ m}$.
The formula for the power $(P)$ of a lens is $P = 1/f$ (where $f$ is in meters).
Substituting the value: $P = 1 / 0.2 = 5 \text{ D}$.
Since it is a convex lens,the power is positive.
Therefore,the power is $+5.0 \text{ D}$.

(A) The power of a lens is defined as the reciprocal of its focal length in meters $(P = 1/f)$.
The $SI$ unit of focal length is the meter $(m)$.
Therefore,the $SI$ unit of power is $m^{-1}$,which is commonly known as the Diopter $(D)$.

(B) In an astronomical telescope,the objective lens is designed to collect as much light as possible from distant objects,which requires a large aperture and a long focal length $(f_o)$.
Conversely,the eyepiece is designed to magnify the image formed by the objective,which requires a short focal length $(f_e)$.
Therefore,the focal length of the objective is much greater than the focal length of the eyepiece $(f_o > f_e)$.

(C) An astronomical telescope consists of two lenses: an objective lens and an eyepiece.
The objective lens forms a real,inverted,and diminished image of a distant object at its focal plane.
This image acts as an object for the eyepiece,which is adjusted such that the image lies within its focal length.
The eyepiece then acts as a simple magnifier,producing a final image that is virtual,inverted,and magnified relative to the original object.
Therefore,the final image is virtual,inverted,and enlarged.

(B) compound microscope consists of two main lenses: the objective lens and the eyepiece.
$1$. The objective lens is placed near the object being observed.
$2$. The eyepiece (or ocular lens) is the lens placed near the eye of the observer.
Therefore,the lens placed near the eye is called the eyepiece.

(A) simple microscope is essentially a convex lens used to view small objects.
To obtain a magnified,virtual,and erect image,the object must be placed between the optical center $(O)$ and the principal focus $(F)$ of the convex lens.
When the object is placed in this position,the light rays diverge after refraction,and when traced backward,they form a virtual image on the same side as the object.
Therefore,the correct position is between the principal focus and the optical center.

(B) In a compound microscope,the objective lens is a convex lens with a short focal length.
To obtain a real,inverted,and magnified image,the object is placed at a distance slightly greater than the focal length $(f_o)$ of the objective lens.
This image acts as an object for the eyepiece,which then produces a final virtual and highly magnified image.

(C) In a compound microscope,the objective lens is a convex lens with a short focal length. To obtain a real,inverted,and magnified image,the object is placed slightly beyond the focal length $(f_o)$ of the objective lens. This image acts as an object for the eyepiece,which then produces a final virtual and magnified image.

(A) In an astronomical telescope,the objective lens is designed to collect as much light as possible from distant objects,which requires a large aperture and a large focal length $(f_o)$.
Conversely,the eyepiece is used as a magnifying glass to view the image formed by the objective lens,which requires a short focal length $(f_e)$.
Therefore,for high magnification $(M = -f_o / f_e)$,the focal length of the objective lens is much larger than the focal length of the eyepiece $(f_o > f_e)$.