Refraction of Light Questions in English

(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.

(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.

(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 $C$.

(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.

(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) 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) 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.

(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.

(C) Total internal reflection $(TIR)$ occurs when light travels from a denser medium to a rarer medium.
For $TIR$ to take place,two conditions must be satisfied:
$1$. Light must travel from an optically denser medium to an optically rarer medium.
$2$. The angle of incidence $(i)$ must be greater than the critical angle $(c)$ for the pair of media $(i > c)$.

(B) When light travels from an optically denser medium to an optically rarer medium, it bends away from the normal.
As the angle of incidence increases, the angle of refraction also increases.
The specific angle of incidence at which the angle of refraction becomes $90^{\circ}$ (i.e., the refracted ray grazes the interface of the two media) is known as the $\text{Critical angle}$.