Refraction of Light at Plane Surfaces

(i) The below ray diagram shows the refraction of light from air to glass:

(ii) The below ray diagram shows the refraction of light from glass to air:

Air is a rarer medium while water is denser medium than air. Therefore, when a ray of light travels from air to water (i.e rarer to denser) it bends towards the normal. Below diagram shows reflection and refraction suffered by a light ray at the boundary in passing from air to water:

(a) Below is the completed diagram with the refracted ray B and the reflected ray C labelled:

(b) The relation between angle of incidence i and the angle of refraction r is given by Snell’s law as,

The ratio of the sine of the angle of incidence i to the sine of the angle of refraction r is constant for the pair of given media.

$\dfrac{\text{Sin i}}{\text{sin r}} = \text {constant } _1\mu_2 \\[0.5em]$

The constant ₁μ₂ is called the refractive index of the second medium with respect to the first medium

$\dfrac{\text{sin i}}{\text{sin r}} = \mu w$

(a) (i) The angle of incidence is the angle which the incident ray makes with the normal. Hence, ∠i = 90° – 30° = 60°.

(ii) Angle of refraction is the angle which the refracted ray makes with the normal Hence, ∠r = 90° – 45° = 45°.

(b) We know that according to Snell’s law,

$_{\text{air}}\text{μ}_{\text{liquid}} = \dfrac{\text{sin i}}{\text{sin r}} \\[0.5em] = \dfrac{\text{sin 60} \degree}{\text{sin 45} \degree} \\[0.5em] = \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{\sqrt{2}}} \\[0.5em] = \dfrac{\sqrt{3}}{\sqrt{2}} \\[0.5em] = 1.22$

Hence, the refractive index of liquid with respect to air is 1.22

The perpendicular distance between the incident ray and the emergent ray is defined as lateral displacement. This shift depends upon the angle of incidence, the angle of refraction and the thickness of the medium.

Below diagram shows the lateral displacement of a ray of light when it passes through a parallel-sided glass slab:

(i) Below is the labelled diagram for ray of light at angle of incidence 0°:

(ii) Below is the labelled diagram for ray of light at angle of incidence 45°:

(a) The below diagram shows the path of the ray until it emerges out of the slab:

(b) In diagram, the angle of incidence (i) and the angle of refraction (r) are marked.

Refractive index of glass μ is related to the angle of incidence (i) and angle of reflection (r) by Snell's second law and is represented as —

The ratio of the sine of the angle of incidence i to the sine of the angle of refraction r is constant for the pair of given media

$\dfrac{\text{sin i}}{\text{sin r}} = \text{constant } _1\mu_2 \\[0.5em]$

The constant ₁μ₂ is called the refractive index of the second medium with respect to the first medium.

(c) The angle of emergence (e) is marked in diagram.

As the refraction occurs at two parallel surfaces PQ and RS, therefore angle MOB = angle N₁BO and by the principle of reversibility of the path of a light ray angle of emergence (e) is equal to the angle of incidence (i).

(d) As the refraction occurs at two parallel surfaces PQ and RS, therefore angle MOB = angle N₁BO and by the principle of reversibility of the path of a light ray angle of emergence (e) is equal to the angle of incidence (i).

Hence, the emergent ray BC is parallel to the incident ray AO.

(e) The lateral displacement between the incident ray and the emergent ray is represented by XY.

One of the factors that affects lateral displacement is the the thickness of the medium. More the thickness of the medium more is the lateral displacement.

(a) Refractive index of the liquid is given by Snell's law and is shown as below,

_airμ_liquid = $\dfrac{\text{sin i}}{\text{sin r}}$

_airμ_liquid = $\dfrac{\text{sin 45}\degree}{\text{sin 30}\degree}$

$= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} \\[0.5em] = \dfrac{2}{\sqrt{2}} \\[0.5em] = \sqrt{2} \\[0.5em] = 1.414 \\[0.5em]$

(b) Below labelled diagram shows the path of the ray after it strikes the mirror and re-enters in air:

(c) Below labelled diagram shows the path of the ray when it strikes the plane mirror normally inside the liquid:

The principle of reversibility is used.

A incident ray and B is refracted ray.

Reason — As A is the incoming ray so it is the incident ray and C is the reflected ray. Ray B bends towards the normal and it is the refracted ray.

bends towards the normal

Reason — When a ray of light from air enters a denser medium, it bends towards the normal due to the change in speed. This phenomenon is known as refraction. This bending towards the normal occurs because the speed of light is slower in the denser medium. The change in speed causes the light wavefront to change direction, resulting in the bending of the light ray towards the normal.

0°

Reason — A ray of light incident normally on the surface separating the two media, passes undeviated (i.e., such a ray suffers no bending at the surface). Here, angle of incidence and angle of refraction = 0°

μ = $\dfrac{\text{sin B}}{\text{sin C}}$

Reason — The refractive index of second medium with respect to the first medium is defined as the ratio of the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium.

∴ Refractive index of glass with respect to air (μ) = $\dfrac{\text{sin B}}{\text{sin C}}$

where

B = angle of incidence in air

r = angle of refraction in glass

When a ray of light gets refracted from a denser to a rarer medium, the speed of light decreases.

Reason — When a ray of light gets refracted from a denser to a rarer medium, the speed of light increases due to the change in the refractive index of the medium.

there is no fixed relation

Reason — There is no fixed relation between the incident angle i and refracted angle r because:

1. when a ray of light travels from a rarer medium to a denser medium, it bends towards the normal hence ∠r < ∠i
2. when a ray of light travels from a denser medium to a rarer medium, it bends away from the normal hence ∠r > ∠i
3. when a ray of light is incident normally on the surface separating the two media, it passes undeviated hence ∠r = ∠i.

less than 

Reason — The intensity of the refracted light is always less than the intensity of incident light because a part of the incident light has suffered reflection.

(1) and (4)

Reason — As air is a rarer medium and glass is a denser medium hence, the light ray passes from a rarer to a denser medium hence, the refracted ray will bend in towards normal whereas emergent ray will bend away from normal.

0°

Reason — A ray of light incident normally on the surface separating the two media, passes undeviated (i.e., such a ray suffers no bending at the surface). Here, angle of incidence and angle of refraction = 0°

remain the same

Reason — If the refractive indices of two media are the same, it means that light travels at the same speed in both media.

increases

Reason — The speed of light differs in different mediums due to their varying optical densities. In denser mediums, light travels slower, while in rarer mediums, it travels faster. According to the equation c = λ × f, where c is the speed of light, λ is the wavelength, and f is the frequency, if the speed of light decreases (as it does when entering a denser medium), and f remains constant, then λ must increase to compensate. Conversely, when light from a denser medium enters a rarer medium, its speed increases, and hence, its wavelength increases as well.

all of the above

Reason — When a ray of light passes from a rarer medium to a denser medium, the speed decreases due to the change in the refractive index of the medium. Frequency remains constant due to the nature of light itself, whereas, wavelength decreases as a consequence of the change in speed. Intensity also decreases as part of the light ray suffers reflection.

9/8

Reason — Given,

_aμ_w = $\dfrac{4}{3}$

_aμ_g = $\dfrac{3}{2}$

_wμ_g = $\dfrac{\text{Refractive index of glass }}{\text {Refractive index of water}}$

Substituting we get,

_wμ_g = $\dfrac{\text{3/2 }}{\text {4/3}}$ = $\dfrac{\text{3 x 3 }}{\text {2 x 4}}$ = $\dfrac{9}{8}$

Ray 2

Reason — Here water is rarer medium and turpentine is denser medium and when light travels from denser to rarer medium it bends away from the normal and vice versa.

Ray 1 : As it enters from turpentine to water then it should bend away from the normal but here it is bending towards the normal.

Ray 2 : It follows laws of refraction perfectly since it is bending towards normal while entering turpentine and away from it while entering water.

Ray 3 : It shows no deflection while entering any medium which doesn't happen in refraction.

Ray 4 : It is bending away from the normal while entering turpentine whereas it should bend towards the normal.

Angle of incidence is always greater than the angle of refraction

Reason — As light enters the denser medium, it bends towards the normal, resulting in a smaller angle of refraction compared to the angle of incidence.

As we know,

$μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in glass (V)}} \\[0.5em]$

Given,

Refractive index of glass, μ = 1.5

Speed of light in air, C = 3 × 10⁸ m / s

Substituting the values in the formula above we get,

$1.5 = \dfrac{3 \times 10^8 }{V} \\[0.5em] V = \dfrac{3 \times 10^8}{1.5} \\[0.5em] \Rightarrow V = 2 \times 10^8$

Hence, Speed of light in glass = 2 × 10⁸ m s^-1

As we know,

$μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in diamond (V)}} \\[0.5em]$

Given,

Speed of light in diamond = 125,000 km s^-1 = 125 x 10⁶ m/s

Speed of light in air, c = 3 × 10⁸ ms^-1

Substituting the values in the formula above we get,

$μ = \dfrac{3 \times 10^8 }{125 \times 10^6} \\[0.5em] μ = \dfrac{3 \times 10^2}{125} \\[0.5em] μ = \dfrac{300}{125} \\[0.5em] \Rightarrow μ = 2.4 \\[0.5em]$

Hence, refractive index = 2.4

As it is given that,

_aμ_w = $\dfrac{4}{3}$

And we know that —

_wμ_a = $\dfrac{1}{\dfrac{4}{3}}$

_wμ_a = $\dfrac{3}{4}$

_wμ_a = 0.75

Hence, the refractive index of air with respect to water is 0.75

As we know that,

$<em>{\text{a}}\text{μ}</em>{\text{g}} = \dfrac{\text {wavelength of light in air}}{\text {wavelength of light in glass}} \\[0.5em]$

Given,

Refractive index of glass with respect to air is given by _aμ_g = $\dfrac{3}{2}$

Wavelength of light in air = 6600 Å

Substituting the values in the formula above we get,

$\dfrac{3}{2} = \dfrac{\text {wavelength of light in air}}{\text{wavelength of light in glass}} \\[0.5em] \dfrac{3}{2} = \dfrac{6600}{\text {wavelength of light in glass}} \\[0.5em] \text {wavelength of light in glass} = \dfrac{6600}{\dfrac{3}{2}} \\[0.5em] \text {wavelength of light in glass} = 6600 \times \dfrac{2}{3} \\[0.5em] \text {wavelength of light in glass} = 4400 \\[0.5em]$

Hence, wavelength of light in glass = 4400 Å

The change in the direction of the path of light, when it passes from one transparent medium to another transparent medium, is called refraction of light. The refraction of light is essentially a surface phenomenon.

When a ray of light passes from one medium to another medium, its direction (or path) changes because of the change in speed of light in going from one medium to another.

In passing from one medium to other, if light slows down, it bends towards the normal and if light speeds up, it bends away from the normal.

If the speed of light remains same in passing from one medium to another, the ray of light does not bend then it passes undeviated.

Hence, we can say that the cause of refraction of light is that its speed is different in different mediums.

The Snell’s laws of refraction of light are:

(i) The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.

(ii) The ratio of the sine of the angle of incidence i to the sine of the angle of refraction r is constant for the pair of given media.

$\dfrac{\text{sin i}}{\text{sin r}} = \text{constant } _1\mu_2 \\[0.5em]$

The constant ₁μ₂ is called the refractive index of the second medium with respect to the first medium.

The refractive index of the second medium with respect to the first medium is defined as the ratio of the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium.

$μ = \dfrac{\text{Speed of light in vacuum or air(c)}}{\text{Speed of light in that medium (V)}}$

The refractive index of a transparent medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum ( i.e. V < c ) and μ = 1 for air or vacuum.

(a) As we know that light is an electromagnetic wave and it travels with same speed for all wavelength, which is equal to 3 x 10⁸.

Hence, the ratio of speeds of light of wavelength 4000Å (i.e. violet light) and 8000Å (i.e. red light) in vacuum = 1:1.

(b) The 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.

$μ =\dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in that medium (V)}} \\[0.5em]$

Denser medium has a higher refractive index. Hence, the speed of light in such medium is lower in comparison to the speed of light in a medium which has a lower refractive index.

As we know that, when a ray of light gets refracted from a rarer to a denser medium, the speed of light decreases, while if it is refracted from a denser to a rarer medium, the speed of light increases.

(a) (i) When a ray of light passes from water to air, its speed increases as it is moving from a denser to a rarer medium.

(b) (i) When a ray of light passes from water to glass, its speed decreases as it is moving from a rarer to a denser medium.

(a) (ii) The refractive index of the medium (air) is less than that of water as the speed of refracted ray increases.

(b) (ii) The refractive index of the medium (glass) is more than that of water as the speed of refracted ray decreases.

This statement indicates that white light travels in air 1.5 times faster than in glass or speed of light in glass is $\dfrac{1}{1.5}$ times the speed of light in air.

(a) The relation between the speed of light in air c and in glass V is given by —

$μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in that medium (V)}} \\[0.5em] 1.5 = \dfrac{c}{V} \\[0.5em] \Rightarrow c = 1.5 V \\[0.5em]$

(b) Let wavelength of light in glass be λ₁ so we get,

$μ = \dfrac{λ}{λ_1} \\[0.5em] 1.5 = \dfrac{λ}{λ_1} \\[0.5em] λ_1 = \dfrac{λ}{1.5} \\[0.5em]$

The speed of blue light in glass is less that of red light.

i.e. c_b < c_r

We know that,

$μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in glass (V)}} \\[0.5em]$

Hence, the refractive index of blue light is greater than that of red light.

i.e. μ_b > μ_r.

Hence, the refractive index of red and blue light in glass will be different.

The factors on which the refractive index of a medium depends are as follows —

(i) Nature of the medium (on the basis of speed of light) — Less the speed of light in the medium as compared to that in air, more is the refractive index of the medium.

We know that,

$μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in glass (V)}} \\[0.5em]$

V_glass = 2 x 10 ⁸ ms^-1, μ_glass = 1.5

V_water = 2.25 x 10 ⁸ ms^-1, μ_water = 1.33

(ii) Physical condition such as temperature — With an increase in temperature, the speed of light in medium increases, so the refractive index of the medium decreases.

(i) The glass piece is not seen when the refractive index of the liquid becomes equal to the refractive index of glass.

(ii) The light of a single colour is used as the refractive index of a medium (glass or liquid) is different for the light of different colours.

As shown in the figure, if an illuminated object is held in front of a thick plane glass mirror, and is viewed obliquely, a number of images are seen and out of these the second image is the brightest.

The reason for this is that when a ray of the light falls on the surface of the mirror from the source, a small part of light (nearly 4%) is reflected forming a faint virtual image, while a large part of light (nearly 96%) is refracted inside the glass.

Now, we can see that this ray is strongly reflected back by the silvered surface inside the glass. This ray is then partially refracted in air and this refracted ray forms another virtual image.

This image is the brightest image because it is due to the light suffering a strong first reflection at the silvered surface.

(i) The ray of light incident normally on a plane glass slab passes undeviated, ( i.e. such a ray suffers no bending at the surface). Thus, if angle of incidence ∠i = 0°, then the angle of refraction ∠r = 0°.

(ii) In the case when ray of light is incident normally on a plane glass slab, it passes undeviated. Hence, the angle of deviation is 0°.

We observe that, when a ray of light passing from one medium to other is obliquely incident it bends at the surface due to change in speed in going from one medium to another.

However, for normal incidence from one medium to another, angle of incidence ∠i = 0° then the angle of refraction ∠r = 0°. Hence the speed of light changes but the direction of light does not change.

Speed, intensity and wavelength are the quantities which will differ for a refracted ray in comparison to those of a incident ray.

The frequency of light depends on the source of light, so it does not change on refraction.

When a ray of light gets refracted from a rarer to a denser medium, the speed of light decreases; while if it is refracted from a denser to a rarer medium, the speed of light increases.

When light passes from a rarer to a denser medium, the wavelength decreases and when it passes from denser to rarer medium, its wavelength increases.

As we know,

Refractive index of air, μ_a = 1.0003

Refractive index of water, μ_w = 1.33

Refractive index of ordinary glass, μ_g = 1.5

With the help of the above values, we can see that μ_g > μ_w > μ_a

When a ray of light gets refracted from a rarer to a denser medium, the speed of light decreases, while if it is refracted from a denser to a rarer medium, the speed of light increases.

So on the basis of above information, we can say that —

(i) When a ray of light passes from water to air, its speed increases as it is moving from a denser to a rarer medium.

(ii) When a ray of light passes from water to glass, its speed decreases as it is moving from a rarer to a denser medium.

(a) (i) The speed of red light is maximum. Therefore, the refractive index of a transparent medium is least for red colour.

(ii) The speed of violet light is least. Therefore, the refractive index of a transparent medium is most for violet colour.

(b) The wavelength of visible light increases from violet to red colour. It is highest for red colour. Hence, red colour travels fastest in any medium except air.

In a given medium, the speed of red light is maximum. Therefore, the refractive index of that medium is maximum for violet light and least for red light. (i.e. μ_violet > μ_red).

The wavelength of visible light increases from violet to red end, so the refractive index of a medium decreases with increase in wavelength of light used.

Hence, the refractive index of a medium for violet light which has the least wavelength is greater than that for red light which has the greatest wavelength.

With increase in temperature, the speed of light in a medium increases, so the refractive index of the medium decreases.

The refractive index of water with respect to air is _aμ_w.

$_{\text{a}}\text{μ}_{\text{w}} = \dfrac{\text{Speed of light in air}}{\text {Speed of light in water}} \\[0.5em]$

The refractive index of glass with respect to air is _aμ_g.

$_{\text{a}}\text{μ}_{\text{g}} = \dfrac{\text{Speed of light in air}}{\text {Speed of light in glass}} \\[0.5em]$

So, the refractive index of glass with respect to water is given by

$_{\text{w}}\text{μ}_{\text{g}} = \dfrac{_a\mu_g}{_a\mu_w} \\[1em]$

(a) When light travels from a rarer to a denser medium, its speed decreases

(b) When light travels from a denser to a rarer medium, its speed increases.

A prism is defined as a transparent medium bounded by five plane surfaces with a triangular cross section. The below diagram shows the principal section of a prism with its refracting surfaces, refracting angle and base labelled:

(i) The below diagram shows the refraction of a ray of light of single colour through a prism with the incident, refracted, emergent rays and the angle of deviation properly labelled:

The below diagram shows the refraction of a ray of light of single colour through a parallel sided glass slab with the incident, refracted, emergent rays and the angle of deviation properly labelled:

(ii) In a prism, the refraction of light occurs at two inclined faces, so the emergent ray is not parallel to the incident ray but it is deviated towards the base of the prism. On the other hand, in a parallel sided glass slab, the refraction of light occurs at two parallel faces, so the emergent ray is parallel to the incident ray with a lateral displacement.

In a prism, the ray of light suffers refraction at two inclined faces. In each refraction, the ray bends towards the base of the prism. The below diagram shows the refraction of a ray of light of single colour through a prism:

At the first face AB, the ray suffers a deviation equal to δ₁. Then, at the second face AC, the ray PQ suffers a deviation δ₂. Thus, the prism has produced a deviation which is the angle between the direction of incident ray and the emergent ray. This is called the angle of deviation produced by a prism.

The angle of deviation is caused as the ray passing through a prism suffers refraction at two inclined planes.

The three factors on which the angle of deviation depends are as follows:

1. The angle of incidence (i)
2. The material of prism (i.e on refractive index μ)
3. The angle of prism (A)

(a) It is experimentally observed that as the angle of incidence increases, the angle of deviation first decreases, reaches to a minimum value for a certain angle of incidence and then on further increasing the angle of incidence, the angle of deviation begins to increase. The below curve shows the variation in the angle of deviation with the angle of incidence at a prism surface:

(b) For a given prism and given colour of light, 𝛿_min (angle of minimum deviation) is unique since only one horizontal line can be drawn parallel to i – axis at the lowest point of i – 𝛿 curve i.e only for one value of angle of incidence i, the refracted ray inside the prism is parallel to its base.

Below diagram shows the refraction of a monochromatic light ray through a prism when it suffers minimum deviation:

In an equilateral prism, when the prism is in minimum deviation, the angle of incidence i₁ is equal to the angle of emergence i₂.

Hence,

Angle of emergence (i₂) = Angle of incidence (i₁).

Below ray diagram shows the formation of the image of an object when seen by an observer through a glass prism with its vertex pointing upwards:

(a) The angle of incidence on the first face of the prism when a ray of light is normally incident is 0°.

(b) The angle of refraction from the first face of the prism r₁ = 0° as the angle of refraction of a normally incident ray is always 0°.

(c) As the prism is equilateral therefore A = 60°.

Given, r₁ = 0°.

Hence, taking into consideration the two conditions we get,

the angle of incidence at the second face of the prism, equal to 60°.

(d) No, the light ray will not suffer minimum deviation as the angle of incidence is 0°.

The below diagram shows the complete path of the ray as it enters prism A and emerges out of prism B:

bends at both the surfaces of the prism towards its base

Reason — This phenomenon occurs because light undergoes refraction twice as it enters and exits the prism. At each surface, the light bends towards the normal due to the change in medium, causing it to deviate from its original path. Since the prism is thicker at the base and thinner at the apex, the net effect of refraction at both surfaces is to bend the light towards the base of the prism.

size of prism

Reason — The deviation produced by the prism depends on :

1. the angle of incidence
2. the material of prism
3. the angle of prism
4. the colour or wavelength of light used.

It does not depend on the size of prism.

line AD is the refracting edge

Reason — ABED and ACFD are the refracting surfaces, angle a is the angle of prism and BCFE is the base of the prism.

r₁ + r₂

Reason — The angle of a prism is typically represented by the sum of its refracting angles, hence, r₁ + r₂.

i₁ = i₂

Reason — The angle of deviation becomes minimum when angle of incidence (i₁) becomes equal to the angle of emergence (i₂). This phenomenon occurs when the light ray inside the prism is incident at the minimum deviation angle and is known as the angle of minimum deviation.

parallel

Reason — In position of minimum deviation, the refracted ray inside the prism is parallel to its base.

In a prism as well as a parallel sided glass slab the emergent ray is parallel to the incident ray with a lateral displacement .

Reason — In a parallel-sided glass slab, the emergent ray is parallel to the incident ray without any lateral displacement. However, in a prism, there is a lateral displacement of the emergent ray due to the angle of the prism.

0°

Reason — When a ray of light is incident normally (perpendicularly) on a surface, it means the angle of incidence is 0°.

Both A and R are true and R is the correct explanation of A.

Explanation

Assertion (A) is true. If the angles of the base of a prism are equal, then at minimum deviation condition for prism the angle of incidence is equal to the angle of emergence and the refracted ray inside the prism is parallel to the base of the prism

Reason (R) is true. For the minimum deviation to occur, the angle of incidence must be equal to the angle of emergence. This ensures that the deviation caused by the prism is minimized.

As we know,

$δ_ {min} = 2i - A$

Given,

i = 48°

A = 60°

So, substituting the values in the formula above we get,

$δ_ {min} = (2 \times 48°) - 60° \\[0.5em] δ_ {min} = 96° - 60° \\[0.5em] \Rightarrow δ_ {min} = 36° \\[0.5em]$

Hence, the angle of minimum deviation is equal to 36°

As we know,

$δ_{min} = 2i - A$

Given,

$δ_{min} = 36°$

A = 60°

So, substituting the values in the formula above we get,

$36° = (2 \times i) - 60° \\[0.5em] \Rightarrow i = \dfrac{(36° + 60°)}{2} \\[0.5em] \Rightarrow i = \dfrac{96}{2} \\[0.5em] \Rightarrow i = 48° \\[0.5em]$

Hence, the angle of incidence is equal to 48°.

Angle of deviation is the angle between the direction of incident ray and the emergent ray. It is represented by the greek alphabet δ (delta).

(i) For a given angle of incidence, the prism with a higher refractive index produces greater deviation than the prism which has a lower refractive index.

For example — A flint glass prism produces more deviation than a crown glass prism for same refracting angle since μ_flint > μ_crown

(ii) The refractive index of a given transparent medium is different for the light of different colours. It decreases with the increase in the wavelength of light.

Thus, the refractive index of the material of a prism for visible light is maximum for the violet colour and minimum for the red colour.

Consequently, a given prism deviates violet light the most and red light the least.

(i.e. δ _violet > δ _red since μ_violet > μ_red )

(i) When a ray of light incident at an angle of incidence i₁ passes through an equilateral glass prism such that the refracted ray inside the prism is parallel to its base and emerges at an angle of emergence i₂ then angle of incidence i₁ is equal to the angle of emergence i₂.

Hence,

i₁ = i₂

(ii) When a ray of light incident at an angle of incidence i₁ passes through an equilateral glass prism such that the refracted ray inside the prism is parallel to its base and emerges at an angle of emergence i₂ then angle of deviation is minimum.

(i) As we know that in an equilateral glass prism the ray suffers minimum deviation.

So, angle of incidence i₁ is equal to angle of emergence i₂

Given, angle of incidence i₁ = 48°

Applying the rule given above, we get,

    i₁ = i₂
⇒ i₂ = 48°

Hence, angle of emergence i₂ = 48°

(ii) (a) As the minimum angle of deviation is 36° for yellow light, at an incident angle of 48° so for any angle of incidence other than 48° the angle of deviation will be more than 36°.

Hence, when the angle of incidence is changed to 30°, the angle of deviation will be more than 36°.

(b) As the minimum angle of deviation is 36° for yellow light, at an incident angle of 48° so for any angle of incidence other than 48° the angle of deviation will be more than 36°.

Hence, when the angle of incidence is changed to 60°, the angle of deviation will be more than 36°.

The angle of deviation (𝛿) increases with increase in the angle of prism (A).

Angle of deviation is the angle which the emergent ray makes with the direction of incident ray.

The statement given above is False.

Reason — The deviation produced by a prism is not independent of the angle of incidence and the deviation is different for different colors of light due to the phenomenon of dispersion, where different wavelengths of light are refracted by different amounts, resulting in the separation of colors. So, the deviation produced by a prism is not the same for all colours of light

(i) The angle of minimum deviation produced by a prism decreases with increase in the wavelength of incident light.

(ii) The angle of minimum deviation produced by a prism increases with increase in the refracting angle of the prism.

The relation between the angle of incident (i₁), angle of emergence (i₂), angle of prism (A) and angle of deviation (𝛿) for a ray of light passing through an equilateral prism is given by —

𝛿 = (i₁ + i₂) - A

(i) On passing through a prism, violet colour will deviate the most.

(ii) On passing through a prism, red colour will deviate the least

For a given angle of incidence, the prism with a higher refractive index produces greater deviation than the prism which has a lower refractive index.

As refractive index of flint glass prism (B) is more than a crown glass prism (A) (i.e. μ_flint > μ_crown)

Hence, a flint glass prism (B) produces more deviation than a crown glass prism (A).

Consider a point object O kept at the bottom of a transparent medium ( such as water or glass ) separated from air by the surface PQ.

A ray of light OA, starting from the object O, is incident on the surface PQ normally, so it passes undeviated along the path AA'. Another ray OB, starting along the object O, strikes the boundary surface PQ at B and suffers refraction.

Since, the ray travels from denser to rarer medium so it bends away from normal N'BN drawn at the point of incidence B on the surface PQ and travels along BC in air.

When viewed by the eye , the ray BC appears to be coming from point I which is the virtual image of O, obtained on producing A'A and CB backwards.

Thus, any point seen from air will appear to be at I, which is lesser depth = AI than its actual depth AO.

Angle of incidence = ∠OBN'

Angle of refraction = ∠CBN.

Since, AO and BN’ are parallel and OB is transversal line, so

∠AOB = ∠OBN' = i

Similarly, IA’ and BN are parallel and IC is the transversal line, so

∠BIA’ = ∠CBN = r

In right-angle triangle BAO,

$\text {sin i} = \dfrac{BA}{OB}$

and,

In right-angle triangle IAB,

$\text {sin r} = \dfrac{BA}{IB}$

For refraction from medium to air, by Snell’s law

$_mμ _a = \dfrac{\text {sin i}}{\text {sin r}} = \dfrac {\dfrac {BA}{OB}}{\dfrac {BA}{IB}} \\[0.5em] \Rightarrow _mμ _a = \dfrac{IB}{OB} \\[0.5em]$ Hence, refractive index of medium with respect to air is,

$_aμ _m = \dfrac{1}{_mμ_a} = \dfrac{OB}{IB} \\[0.5em]$

The object is viewed from a point vertically above the object O, since point B is very close to the point A. ∴ IB = IA and OB = OA

Hence,

$_aμ _m = \dfrac{1}{_mμ_a} = \dfrac{OA}{IA} \\[0.5em] \Rightarrow \dfrac{\text {Real depth}}{\text {Apparent depth}} \\[0.5em]$

(a) When a tank is viewed normally from above from a rarer medium, the apparent depth of the tank is always less than its real depth.

As _aμ_w = 4 / 3, therefore, the depth of the pond appears three fourth of its real depth on seeing it from air in a nearly vertical direction.

(b) Below is the labelled ray diagram of a tank of water viewed normally from above:

(a) Water in a pond appears to be only three-quarters of its actual depth due to the refraction of light while travelling from one medium to another.

The light rays suffer refraction while travelling from denser medium (water or glass) to rarer medium (air), so it bends away from the normal. Below is the ray diagram for this:

(b) Let us say an object (B) is at the bottom of a pond. Consider a ray of light BC from the object that moves from water to air.

The ray now moves away from the normal N along the path CD, after refraction from the water surface.

When we produce CD it appears from the point B’.

The virtual image of the object B appears at B’.

Hence, we get,

$\text {Refractive index of water} = \dfrac {\text {Real depth}}{\text {Apparent depth}} \\[0.5em]$

The above figure shows a stick placed obliquely in water. The portion OP of the stick under water when seen from air appears to be shortened and raised up as OP'. This is due to refraction of light from water to air.

The rays of light coming from tip P of the stick when pass from water to air, bend away from the normal and appear to be coming from point P' which is the virtual image of the point P.

Thus, the part PO of the stick appears to be P'O.

Hence, the immersed part of the stick appears to be raised and therefore, bent at the point O at the surface of water and the stick XOP appears as XOP'.

When a fish is looking at a 1.0 m high plant at the edge of the pond, it appears to be taller than its actual height due to the refraction of the light rays travelling from one medium to another.

When the fish is looking at the plant, as air is a rarer medium in comparison to water, the ray OP will bend away from the normal while emerging out from the water to air.

But, when we extend the ray it appears to be coming from a higher point.

Hence, the plant will look taller than its actual height.

(i) When a student puts his pencil into an empty trough, and later pours water in the same trough, the part of the pencil which is immersed in water will look short and raised up.

(ii) The above observation is based on the phenomenon of refraction of light.

(iii) Below is the diagram showing how the student’s eye sees the pencil through water:

μ = real depth/apparent depth

Reason — Refractive index of a liquid can be found using μ = real depth/apparent depth

refraction of light

Reason — When you look at a small air bubble in a glass block from above, the light rays coming from the air bubble bend as they pass from the denser medium (glass) to the less dense medium (air). This bending of light causes the apparent position of the air bubble to be shifted, making it appear raised or displaced from its actual position. Therefore, the apparent displacement of the air bubble is due to the refraction of light.

violet light

Reason — When light travels from a rarer medium to a denser one, it bends towards the normal. The amount of bending (and consequently the apparent shift in the position of the object) is inversely proportional to the wavelength of light, which is characterized by its color.

This means that shorter wavelengths (such as violet light) will experience a greater degree of refraction compared to longer wavelengths (such as red light).

all of the above

Reason — The shift by which an object appears raised depends on:

1. The refractive index of the medium — higher the refractive index of the medium, more is the shift.
2. The thickness of the denser medium — thicker the medium, more is the shift
3. Colour of the incident light — the shift decreases with the increase in the wavelength of light used.

greater 

Reason — When an object is placed in a rarer medium (like air) and viewed from a denser medium (like water), the light rays coming from object when passing from air to water, bend towards the normal and appear to be coming from a new point such that the distance appears greater than the actual distance.

violet

Reason — The shift increases with the decrease in the wavelength of light used. Since, wavelength of violet light is least, hence, the shift is maximum and it appears to be the most raised.

Both A and R are true and R is the correct explanation of A.

Explanation

Assertion (A) is true. Planets do not twinkle because, unlike stars, they are seen as extended sources of light. When observing a planet, which appears as a disk rather than a point of light, the combined effect of a large number of point-sized sources averages out the total variation in the amount of light entering the observer's eye. This averaging effect nullifies the twinkling phenomenon, making planets appear steady in brightness.

Reason (R) is true. Planets are much closer to us than stars, so they are not point-sized objects to our eye. Hence, the fluctuations have a negligible effect and they don't seem to twinkle.

assertion is true but reason is false.

Explanation

Assertion (A) is true. An empty test tube placed in water in a beaker with its mouth outside the water surface appears silvery when viewed from a suitable direction because of the phenomenon of total internal reflection at the air-water interface, which can create a silvery appearance.

Reason (R) is false. Substance in water appears silvery due to the refraction of light, is incorrect. The silvery appearance typically arises due to total internal reflection, not refraction.

assertion is false but reason is true.

Explanation

Assertion (A) is false. When a ray of light travels from a rarer medium to a denser medium, it slows down and bends towards the normal. This is due to the change in speed and direction of light when it crosses the boundary between two different mediums, which is known as refraction.

Reason (R) is true. The speed of light is higher in a rarer medium (e.g., air) compared to a denser medium (e.g., glass or water). This difference in speed is what causes the refraction of light when it passes from one medium to another.

We know that,

Real depth = Apparent depth x Refractive index of water

Given,

Apparent depth = 3 m

Refractive index of the water = $\dfrac{4}{3}$

So, substituting the values in the formula we get,

Real depth = $3\times \dfrac{4}{3} = 4 \text{ m}$

Hence, actual depth = 4 m

As we know,

$\text {Shift} = \text {Real depth} \times (1 - \dfrac{1}{_aμ_m})$

Given,

$\text {Refractive index of the water} = \dfrac{4}{3} \\[0.5em]$

and

Real depth = 16 cm

So, substituting the values in the formula we get,

$\text {Shift} = 16 \times (1 - \dfrac{3}{4}) \\[0.5em] \text {Shift} = 16 \times \dfrac{1}{4} \\[0.5em] \text {Shift} = \dfrac{16}{4} \\[0.5em] \Rightarrow \text {Shift} = 4 \text{ cm} \\[0.5em]$

Hence, the coin appears to be raised by a height of 4 cm when seen from vertically above.

As we know,

$\text {Shift} = \text {Real depth} \times (1 – \dfrac{1}{_aμ_m})$

Given,

Shift in the image = 7 mm or 0.7 cm

Refractive index of the glass block = 1.5

So, substituting the values in the formula we get,

$0.7 = \text {Real depth} \times (1 – \dfrac{1}{1.5}) \\[0.5em] 0.7 = \text {Real depth} \times \dfrac{0.5}{1.5} \\[0.5em] \text {Real depth} = \dfrac{0.7 \times 1.5}{0.5} \\[0.5em] \Rightarrow \text {Real depth} = 2.1 \\[0.5em]$

Hence, the thickness of glass slab = 2.1 cm.
(Thickness of glass slab is same as the real depth of the postage stamp).

The relation of refractive index μ with real and apparent depth is given as below,

$_a\mu_m = \dfrac{\text{Real Depth}}{\text{Apparent Depth}}$

The shift by which the object appears to be raised depends on —

(i) The refractive index of the medium — Higher the refractive index of the medium, more is the shift.

(ii) The thickness of the denser medium — Thicker the medium, more is the shift.

(iii) The colour or wavelength of incident light — the shift decreases with the increase in the wavelength of light used.

When a ray of light passes from a denser to a rarer medium, at a certain angle of incidence i = C, the angle of refraction becomes 90°.

This angle C is called the critical angle.

Thus, critical angle is defined as the angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90°.

(a) When a ray of light travelling in a denser medium, is incident at the surface of a rarer medium at the angle of incidence greater than the critical angle for the pair of media, the ray is totally reflected back into the denser medium. This phenomenon is known as total internal reflection.

(b) The two necessary conditions for total internal reflection are —

1. The light must travel from a denser medium to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the pair of media.

(c) Below diagram illustrate total internal reflection:

Total Internal Reflection Diagram:

Completed ray diagram with refracted rays labelled is shown below:

Ray A is incident normally at water air boundary so its angle of incidence is 0°. Hence, the refracted ray shown as AA' in the diagram passes undeviated.

For Ray B, the angle of incidence is equal to the critical angle for water air boundary so it is partly reflected as BC' and partly refracted as BD'.

Two necessary conditions for the total internal reflection are —

1. The light must travel from a denser to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the pair of medium.

(a) Below diagram shows the path of the rays after striking the water surface:

(b) Rays A and B show the phenomenon of ‘refraction of light’ and ray C shows the phenomenon of ‘total internal reflection’.

(a) The angle of incidence ∠PQN is known as the critical angle.

(b) For the refracted ray QS, the angle of refraction is 90°.

(c) When the angle of incidence ∠PQN is increased, then the phenomenon that occurs is total internal reflection.

(d) For the ray PR, the angle of incidence is less than ∠PQN ( i.e the critical angle ). So according to the laws of reflection, ray PR suffers partial reflection and refraction.

(e) Below diagram shows the refracted ray and the position of image of the object P when seen vertically from above:

(a) Diagram with rays PA, PB and PC is shown below:

(b) Given,

μ_g = 1.5

As we know,

$\text {sin i}_c = \dfrac {1}{μ} \\[0.5em]$

Substituting the values in the formula we get,

$\text {sin i}_c = \dfrac{1}{1.5} \\[0.5em] \Rightarrow \text {sin i}_c = 0.667 \\[0.5em] \Rightarrow i_c = 41.8$

Hence, we can round off i_c = 42°

Applying,

$\dfrac{\text {sin r}}{\text {sin i}} = _a\mu_g \\[0.5em] \text {sin r} = _a\mu_g \times \text {sin i} \\[0.5em] \text {sin r} = _a\mu_g × \text {sin 42°} \\[0.5em]$

Given,

$\text {sin 42°} = \dfrac{2}{3}$

and

$_aμ_g = \dfrac{3}{2}$

Substituting the values in the formula we get,

$\text{sin r} = \dfrac{3}{2} \times \dfrac{2}{3} \\[0.5em] \Rightarrow \text {sin r} = 1 \\[0.5em]$

As sin r = 1 , so angle of refraction = 90°.

Completed diagram showing the path of the ray till it emerges out from the slab is shown below:

A prism having an angle of 90° between its two refracting surfaces and the other two angles each equal to 45° is called a total reflecting prism. In such a prism, the light incident normally on any of its faces suffers total internal reflection inside the prism.

Due to this property, a total reflecting prism is used for the following three purposes —

1. To deviate a ray of light through 90°.
2. To deviate a ray of light through 180°.
3. To erect the inverted image without producing deviation in its path.

Below diagram shows the deviation of a ray of light through 90° by a total reflecting prism:

Below diagram shows the deviation of a ray of light through 90° by a total reflecting prism:

ABC is a total reflecting prism. A beam of light is incident normally at the face AB. It passes undeviated into the prism and strikes at the face AC making an angle of incidence equal to 45°.

For glass air interface, the critical angle is equal to 42°, therefore the beam of light suffers total internal reflection at the face AC because the angle of incidence is greater than the critical angle.

The reflected beam inside the prism then strikes the face BC, where it is incident normally and therefore passes undeviated. Hence, the incident beam gets deviated through 90°.

This action of prism is used in a periscope where a total reflecting prism is preferred over a plane mirror.

(a) The angle through which the incident ray deviates and emerges out of the prism is the angle of deviation.

The angle of deviation = the angle which the emergent ray makes with the incident ray XY = 90°.

(b) The instrument where this action of prism is put into use is the periscope.

(c) The surface AB of the prism behaves as a mirror.

Below diagram shows the usage of a right angled isosceles prism which is used to make an inverted image erect:

This action of prism is used in a slide.

(a) Below diagram shows the path of the ray PQ till it emerges from the prism with all the angles labelled:

(b) The angle of deviation of the ray PQ is 180°.

(c) This action of the prism is used in Binoculars.

(a) The angle of incidence at the faces AB, i = 0°.
The angle of incidence at the faces AC, i = 60°.

(b) The phenomenon which the ray of light suffers at:
face AB — refraction
face AC — total internal reflection
face BC — refraction

Below is the labelled ray diagram showing total internal reflection of a ray of light incident normally on one face of a 30°, 90°, 60° prism

Below is the completed diagram showing the path of the light ray till it emerges out of the prism with all angles marked:

A total reflecting prism is used to turn a ray of light through 180°.

This device is used in a binocular and camera to invert the image without the loss of intensity.

Below diagram shows the path of the light ray entering the prism P till it emerges out of the prism Q:

Below diagram shows the path of ray PQ through the glass prism ABC till it emerges out of the prism:

Completed diagram showing the path of the ray PQ till it emerges out of the block is shown below:

90°

Reason — When light travels from a denser medium to a rarer medium and the angle of incidence is equal to the critical angle, the angle of refraction becomes 90° as shown in the below fig:

light ray passes from a denser to rarer medium

Reason — The two necessary conditions for total internal reflection are —

1. The light must travel from a denser medium to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the pair of media.

Option (d)

Reason — As a ray travels from denser medium to rarer medium and if it's angle of incidence is greater than the critical angle then total internal reflection occurs so scientist must increase angle X till it beocmes greater than the critical angle to achieve total internal reflection.

Option (d)

Reason — There are two necessary conditions for total internal reflection :

1. The light must travel from an optically denser to a rarer medium.

2. The angle of incidence must be greater than the critical angle for the pair of media.

60°

Reason — An equilateral prism can be used to deviate a ray of light through 60°.

Reason — Below is the completed diagram showing the deviation of ray of light A through 180°:

suffer total internal reflection

Reason — We can see from the below diagram that the ray is travelling from a denser to a rarer medium and the angle of incidence is greater than the critical angle. In such a case no refracted ray is obtained and the incident ray is reflected back into the same medium. This phenomenon is known as total internal reflection.

100%

Reason — In the process of total internal reflection, all of the light is reflected back into the original medium, and none of the energy (or intensity) of the light is lost. Therefore, 100% of the energy of the light is reflected back.

violet

Reason — Critical angle is inversely proportional to the refractive index, μ_Violet > μ_Red, hence critical angle would be least for violet light for a pair of media.

The statement 'the critical angle for diamond is 24°' implies that at an incident angle of 24°, the angle of refraction in the air will be 90° within the diamond.

If the incident angle is more than the critical angle then the ray will suffer total internal reflection without any refraction.

The critical angle depends on two factors:

(a) The colour or wavelength of light — The refractive index of a transparent medium decreases with the increase of wavelength of light. (It is most for violet light and least for red light), therefore the critical angle for a pair of media is least for the violet light and most for the red light, i.e. the critical angle increases with the increase in wavelength of light.

(b) The temperature — On increasing the temperature of medium, its refractive index decreases. So, the critical angle for the pair of medium increases with increase in temperature.

(a) The expression for _aμ_g = $\dfrac {\text {sin r}}{\text {sin i}}$

(b) When refractive angle r = 90° then, the corresponding angle of incidence i will be equal to critical angle.

(c) The phenomenon of total internal reflection occurs when the angle of incidence exceeds the value of i obtained in part (b).

(a) Diamond has lowest critical angle and water has highest critical angle.

(b) If the temperature of all the substance is raised by 30^o C then the critical angle of the all the media will increase.

Reason :

(a) As the relation between critical angle and refractive index of a medium is given by,

$\text{sin C} = \dfrac {1}{_a\text{μ}_{medium}}$

where $_a\text{μ}_{medium}$ is the refractive index of the given medium with respect to air

From above relation it is clear that critical angle (C) has inverse relation with refractive index ($_a\text{μ}_{medium}$) so lowest C is the one with highest value of $_a\text{μ}_{medium}$ and similarly highest C value corresponds to lowest $_a\text{μ}_{medium}$.

(b) On increasing the temperature of a medium, it's refractive index decreases, therefore the critical angle increases with the increase in temperature.

The advantage of using a total reflecting prism as a reflector in place of a plane mirror, is that the image is much brighter and the brightness remains unchanged even after the long use of the total reflecting device.

In the case of a plane mirror, the image is less bright and the brightness gradually decreases as the silvering on mirror becomes old and rough.

Knowing the refractive index of the denser medium with respect to the rarer medium, we can calculate the critical angle C for that pair of medium with the help of the relation

$\text{sin C} = \dfrac{1}{ _a\mu_g}$

(a) As we know,

$_aμ_g = \dfrac{1}{\text {sin C}} \\[0.5em] = \text {cosec C}$

Refractive index is

$_aμ_g = \dfrac{3}{2}$

Substituting the values in the formula we get,

$\dfrac{3}{2} = \dfrac{1}{\text {sin C}} \\[1em] \text {sin C} = \dfrac{1}{\dfrac{3}{2}} \\[1em] \text {sin C} = \dfrac{2}{3}$

As,

$\text {sin 42°} = \dfrac {2}{3}$

Hence, critical angle for glass air surface is 42°.

(b) As we know,

$_aμ_w = \dfrac{1}{\text {sin C}} = \text {cosec C} \\[0.5em]$

Refractive index is

$_aμ_w = \dfrac {4}{3} \\[0.5em]$

Substituting the values in the formula we get,

$\dfrac{4}{3} = \dfrac{1}{\text {sin C}} \\[1em] \text {sin C} = \dfrac{1}{\dfrac{4}{3}} \\[1em] \text {sin C} = \dfrac{3}{4} \\[1em] \text {sin 49°} = \dfrac{3}{4} \\[0.5em]$

Hence, critical angle for water air surface is 49°.

The angle of refraction becomes equal to 90° when a light ray is incident from a denser medium on the boundary separating it from a rarer medium at an angle of incidence equal to the critical angle. Below diagram shows the path of the ray when angle of incidence is equal to the critical angle:

As the wavelength of light decreases (or increases) refractive index becomes more (or less) and critical angle becomes less (or more).

(i) As the wavelength for red light is more than that for yellow colour, the refractive index becomes less and the critical angle will be more than 45°.

(ii) As the wavelength for blue light is less than that for yellow colour, the refractive index becomes more and the critical angle will be less than 45°.

As we know, the relation between critical angle and refractive index is

$\text {sin c} = \dfrac{1}{μ}$

where, c is critical angle and μ is refractive index.

As μ for red light is less as compared to that for green light and μ and c are inversely related.

Hence, the critical angle of red light is higher than that of green light.

(a) Total internal reflection occurs only when a ray of light passes from a denser medium to a rarer medium.

(b) Critical angle is the angle of incidence in denser medium for which the angle of refraction in rarer medium is 90°.

True

Reason — When the angle of incidence is greater than the critical angle, light undergoes total internal reflection and effectively remain within the denser medium. Hence, no refracted ray is obtained.