Mix Examples - The Human Eye and the Colourful World Questions in English

(B) $(i)$ The phenomenon of splitting of white light into its constituent colours (spectrum) when passing through a glass prism is called dispersion of light.
$(ii)$ The deviation of light depends on its refractive index in the medium. $A$ higher deviation implies a higher refractive index,and since refractive index is inversely proportional to the speed of light in that medium $(n = c/v)$,the colour that deviates more $(A)$ travels slower. Therefore,the colour that deviates less $(B)$ has a higher speed in the prism.

(N/A) rainbow is a natural spectrum appearing in the sky after a rainfall. It is caused by the dispersion of sunlight by tiny water droplets present in the atmosphere.
$1$. When sunlight enters a water droplet,it undergoes refraction and dispersion.
$2$. The light then undergoes internal reflection inside the droplet.
$3$. Finally,the light undergoes refraction again as it exits the droplet.
Due to these phenomena (refraction,dispersion,and internal reflection),different colours reach the observer's eye at different angles,resulting in the formation of a rainbow. The observer must always have their back towards the Sun to see a rainbow.

(N/A) The defect is Hypermetropia,also known as long-sightedness. In this condition,the near point of the eye shifts further away than the normal $25 \,cm$.
It can be corrected by using a convex lens of suitable focal length,which converges the light rays from the object at $25 \,cm$ to the eye's actual near point $(N')$,allowing the eye to focus the image on the retina.
$(i)$ Defective eye: Rays from an object placed at the normal near point $(N = 25 \,cm)$ are focused behind the retina because the eye's near point has shifted to $N'$.
$(ii)$ Corrected eye: $A$ convex lens is placed in front of the eye. It forms a virtual image of the object (placed at $N$) at the eye's actual near point $(N')$. The eye then focuses this image onto the retina.

(N/A) The cause of dispersion is that white light is composed of different colours,each having a different wavelength.
When white light enters a glass prism,each colour travels at a different speed within the medium.
Because the refractive index of glass is different for different wavelengths,each colour bends through a different angle with respect to the incident ray.
Specifically,red light has the longest wavelength and bends the least,while violet light has the shortest wavelength and bends the most.
As a result,the rays of each colour emerge along different paths,becoming distinct and forming a band of colours known as a spectrum.
The diagram is as shown below:

(N/A) The person is suffering from myopia (short-sightedness).
$A$ concave lens is used to correct this defect.
The nature of the corrective lens is a concave lens (diverging lens).
We know that the power of a lens $P = 1 / f$ (where $f$ is in meters).
Given $P = -1 \,D$.
Therefore,$f = 1 / P = 1 / (-1) = -1 \,m = -100 \,cm$.
The negative sign indicates that the lens is concave.

(A) $(i)$ The near point of a normal eye is $D = 25 \text{ cm}$. The power of accommodation $P$ is given by $P = 1/f = 100/D$ (where $D$ is in cm).
$P = 100 / 25 = 4 \text{ D}$.
Thus,for a person with normal vision,the power of accommodation is about $4 \text{ Dioptre}$.
$(ii)$ The defect is Hypermetropia (long-sightedness),as the person has difficulty reading nearby objects. She would need a convex lens to correct this defect.
$(iii)$ The pupil regulates and controls the amount of light entering the eye. In bright sunlight,the pupil size is small. When we enter a dark room,it takes some time for the pupil to dilate (expand) to allow more light to enter,which is why we cannot see objects clearly immediately.

(N/A) Iris: It regulates the amount of light entering the eye by dilating the pupil in low-intensity light and contracting the pupil in high-intensity light.
Pupil: It is the aperture that allows light to enter the eye.
Cornea: It is a transparent spherical membrane covering the front of the eyeball. It acts as the primary refractive surface where light is refracted the most upon entering the eye.

(N/A) Scattering of light is the phenomenon where light rays change their direction upon striking obstacles such as atoms,molecules,dust particles,or water droplets. The colour of the scattered light depends on the size of the scattering particles. Very fine particles scatter mainly blue light of shorter wavelengths,while larger particles scatter light of longer wavelengths.
$(b)$ Two natural phenomena based on the scattering of light are:
$(i)$ The blue colour of the sky.
$(ii)$ The reddish appearance of the Sun at sunrise and sunset.

(N/A) Ciliary muscles help in changing the focal length of the eye lens.
$(b)$ Cataract: The crystalline lens of the eye is made of proteins arranged in a regular pattern,which makes the lens transparent. When a group of these protein molecules clump together in a region,they make the region opaque. Gradually,this membrane grows,and the whole lens becomes opaque.
$(c)$ The colour of scattered light depends on the size of the scattering particles (air molecules,dust particles,and pollutants) in the atmosphere.
$(d)$ From $25 \ cm$ to infinity.
$(e)$ There is no atmosphere in space,hence there are no air molecules to scatter sunlight,making the sky appear dark.

(A) $(i)$ Focal length for distant vision: $f = \frac{1}{P} = \frac{1}{-6} = -0.1667\, m = -16.67\, cm$.
$(ii)$ Focal length for near vision: $f = \frac{1}{P} = \frac{1}{2} = +0.5\, m = +50\, cm$.
$(b)$ $(i)$ Stars twinkle due to atmospheric refraction. The Earth's atmosphere has layers of varying densities and temperatures,causing the refractive index to change continuously. Light from distant stars (point sources) undergoes continuous refraction,causing the apparent position and intensity of the star to fluctuate,which we perceive as twinkling.
$(ii)$ Planets are much closer to Earth than stars and appear as extended sources of light rather than point sources. $A$ planet can be considered as a collection of a large number of point-sized sources. The total variation in the amount of light entering our eye from all these point-sized sources averages out to zero,resulting in a steady light,so they do not twinkle.
$(iii)$ Stars appear raised due to atmospheric refraction. As starlight enters the Earth's atmosphere,it travels from a rarer medium to a denser medium,bending towards the normal at each layer. This continuous bending makes the star appear at a higher position than its actual location.

(N/A) prism $P_{1}$ in the path of white light disperses it into various colours. The red rays are deviated the least and the violet rays are deviated the most. The second prism $P_{2}$ is placed in an inverted position relative to $P_{1}$ such that its refracting edge is at the bottom. The second refracting surface of $P_{1}$ and the first refracting surface of $P_{2}$ are parallel. The dispersion produced by $P_{1}$ is neutralised by $P_{2}$. The colours recombine in prism $P_{2}$ and a beam of white light emerges from it.
$(b)$ Isaac Newton.
$(c)$ Spectrum: It is the band of coloured components of light $(VIBGYOR)$ that is obtained when a beam of white light is passed through a glass prism.

(N/A) The labelled diagram is as shown in the provided image.
Here,the components are as follows:
$(a)$ $PE$ - Incident ray
$(b)$ $FS$ - Emergent ray
$(c)$ $EF$ - Refracted ray
$(d)$ $\angle i$ - Angle of incidence
$(e)$ $\angle D$ - Angle of deviation
$(f)$ $\angle e$ - Angle of emergence

(N/A) Hypermetropia,also known as farsightedness,is a vision defect where a person can see distant objects clearly but cannot see nearby objects distinctly. This happens because the near point of the eye has shifted further away than the normal $25 \ cm$.
Reasons for hypermetropia:
$1$. The eyeball has become too short.
$2$. The focal length of the eye lens has become too large.
Correction: This defect is corrected by using a convex lens of suitable focal length,which converges the light rays before they enter the eye,allowing the image to form on the retina.
$(b)$ The experimental arrangement for observing the Tyndall effect (scattering of light) involves passing a strong beam of light through a colloidal solution (e.g.,sodium thiosulphate solution). The two chemicals used are:
$1$. Sodium thiosulphate $(Na_2S_2O_3)$
$2$. Concentrated sulphuric acid $(H_2SO_4)$

(N/A) $(i)$ Scattering of light is the phenomenon in which light rays deviate from their straight path upon striking small particles such as dust,air molecules,or other impurities in the atmosphere.
$(ii)$ The extent of scattering depends primarily on the size of the particles. Very fine particles scatter mainly blue light of shorter wavelengths,while larger particles scatter light of longer wavelengths.
$(iii)$ As sunlight passes through the atmosphere,the fine particles in the air scatter the blue component of white light more strongly than the red component due to its shorter wavelength. This scattered blue light reaches our eyes,making the sky appear blue.
$(iv)$ In space,there is no atmosphere and therefore no particles (like air molecules or dust) to scatter sunlight. Since no light reaches the astronaut's eyes from the sky,it appears dark.

(A-D) Atmospheric refraction is the change in the direction of propagation of light rays as they travel through the atmosphere due to variations in the density of different air layers. The Earth's atmosphere is not uniform; it is optically and molecularly denser near the surface and becomes rarer at higher altitudes,causing the velocity of light to change from layer to layer. Factors such as varying concentrations of gas molecules,dust particles,and temperature differences (hot air is optically rarer than cold air) create layers of different optical densities,leading to refraction.
Stars appear to twinkle because light from distant stars passes through these varying layers of the atmosphere. Fluctuations in the atmosphere cause the amount of starlight reaching the eye to vary,making the star appear alternately bright and dim,which results in the twinkling effect.
Planets are much closer to the Earth and appear as extended sources rather than point sources. The high intensity of light from planets means that slight atmospheric refraction does not cause significant changes in their perceived brightness,so they do not twinkle.
$(b)$ Two effects of atmospheric refraction on the Sun as observed from the Earth are:
$1$. Advance sunrise: The Sun is visible about $2$ minutes before the actual sunrise.
$2$. Delayed sunset: The Sun remains visible for about $2$ minutes after the actual sunset.

(N/A) Cornea: $A$ thin transparent membrane that allows light to enter and refracts it to focus on the crystalline lens.
$(b)$ Crystalline lens: $A$ double convex lens that changes its focal length with the help of ciliary muscles and suspensory ligaments,enabling light rays to focus on the retina.
$(c)$ Iris: The coloured portion of the eye that regulates the size of the pupil and controls the amount of light entering the eye.
$(d)$ Retina: Consists of light-sensitive cells,known as rod and cone cells,which enable us to see in dim and bright light.

(N/A) The two main causes of myopia are:
$1$. Excessive curvature of the eye lens.
$2$. Elongation of the eyeball.
To correct this, a concave lens of suitable focal length is used. The lens diverges the incoming parallel rays from a distant object so that they appear to come from the far point of the myopic eye, allowing the image to be formed on the retina.
$(b)$ Given:
Far point of the myopic eye $(v)$ = $-150 \ cm = -1.5 \ m$.
For a distant object, the object distance $(u)$ = $\infty$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{f} = \frac{1}{-150} - \frac{1}{\infty} = \frac{1}{-150} - 0 = -\frac{1}{150}$.
Thus, the focal length $(f)$ = $-150 \ cm = -1.5 \ m$.
Power $(P)$ = $\frac{1}{f(in \ meters)} = \frac{1}{-1.5} = -0.667 \ D \approx -0.67 \ D$.

(N/A) $(i)$ Myopia (Near-sightedness): $A$ defect where a person can see nearby objects clearly but cannot see distant objects distinctly. It is corrected using a concave lens.
$(ii)$ Astigmatism: $A$ defect where the cornea or lens has an irregular curvature,causing blurred vision at all distances. It is corrected using cylindrical lenses.
$(iii)$ Bifocal lenses: Lenses consisting of both concave and convex parts,used to correct Presbyopia,where a person has both Myopia and Hypermetropia.
$(iv)$ Far-sightedness (Hypermetropia): $A$ defect where a person can see distant objects clearly but cannot see nearby objects distinctly. It is corrected using a convex lens.
$(b)$ In Myopia,the image of a distant object forms in front of the retina. $A$ concave lens of suitable focal length is placed in front of the eye to diverge the incoming light rays,shifting the image back onto the retina.

(N/A) Long-sightedness,also known as $Hypermetropia$,is a vision defect in which a person can see distant objects clearly but cannot see nearby objects distinctly.
Two causes for the development of $Hypermetropia$ are:
$1$. The focal length of the eye lens is too long.
$2$. The eyeball has become too small.
Correction:
This defect is corrected by using a convex lens of appropriate power. The convex lens converges the light rays coming from a nearby object before they enter the eye,allowing the image to be formed on the retina.
[Ray Diagram Description: $A$ diagram showing the image forming behind the retina in a defective eye,and a second diagram showing a convex lens placed in front of the eye,which converges the rays so that the image is formed exactly on the retina.]

(N/A) Dispersion: The splitting of white light into its constituent seven colours is called dispersion of light.
Cause: Ordinary white light is a superposition of waves of wavelengths extending throughout the visible spectrum. The speed of light in a vacuum is the same for all wavelengths,but the speed in a material substance is different for different wavelengths. As a result,different colours are deviated by different angles when they pass through a prism. This is called dispersion of light.
$(b)$ In a prism,the refraction of light takes place at the two slant surfaces. The dispersion of white light occurs at the first surface of the prism where its constituent colours are deviated through different angles. At the second surface,these split colours undergo further refraction and get separated more. However,in a rectangular glass slab,the refraction of light takes place at two parallel surfaces. At the first surface,although the white light splits into its constituent colours on refraction,these split colours,upon undergoing refraction at the second parallel surface,emerge as a parallel beam,which gives the impression of white light.

(A) The defect of vision is Hypermetropia (farsightedness).
$(b)$ Two reasons for this defect are:
$(i)$ The eyeball has become too short,so that light rays from a nearby object are focused at a point behind the retina.
$(ii)$ The focal length of the eye lens is too long because the ciliary muscles are unable to make the lens sufficiently convex.
$(c)$ Given: Object distance $u = -25\, cm$,Image distance $v = -50\, cm$ (the near point of the defective eye).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} = \frac{-1 + 2}{50} = \frac{1}{50}$
$f = +50\, cm = +0.5\, m$
Power $P = \frac{1}{f(m)} = \frac{1}{0.5} = +2.0\, D$.
The nature of the lens is a convex lens.
$(d)$ The ray diagrams are as shown in the image.

(N/A) The eye defect is Myopia (nearsightedness).
$(b)$ The two reasons for this defect are:
$(i)$ Excessive curvature of the eye lens.
$(ii)$ Elongation of the eyeball.
$(c)$ To see distant objects clearly,the person needs a lens that forms a virtual image of an object at infinity at his far point $(2\, m)$.
Given: Object distance $u = \infty$,Image distance $v = -2\, m$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-2} - \frac{1}{\infty} = -0.5\, m^{-1}$.
Power $P = \frac{1}{f(m)} = -0.5\, D$.
The negative sign indicates that the lens is a concave lens (diverging lens).
$(d)$ The ray diagrams are as shown in the provided image.

(N/A) The Earth's atmosphere contains air molecules and other fine particles. These particles are smaller in size than the wavelength of visible light. According to Rayleigh scattering, the amount of scattering is inversely proportional to the fourth power of the wavelength $(I \propto 1/\lambda^4)$. Since blue light has a shorter wavelength compared to red light, it is scattered much more strongly by the atmospheric particles in all directions. When this scattered blue light enters our eyes, the sky appears blue.
$(b)$ At the time of sunrise or sunset, the Sun is near the horizon. Sunlight has to travel through a much thicker layer of the Earth's atmosphere to reach the observer. During this long journey, most of the blue and shorter wavelength light is scattered away by atmospheric particles. Only the light with the longest wavelength, which is red light, is able to reach the observer's eyes without being scattered significantly. Therefore, the Sun appears reddish at sunrise and sunset.

(A) This person suffers from hypermetropia (farsightedness).
For the corrective lens, the object distance $u = -30 \ cm$ and the image must be formed at the person's near point $v = -150 \ cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{f} = \frac{1}{-150} - \frac{1}{-30} = \frac{-1 + 5}{150} = \frac{4}{150}$.
$f = \frac{150}{4} = +37.5 \ cm = +0.375 \ m$.
The positive sign indicates a convex lens.
Power $P = \frac{1}{f(\text{in } m)} = \frac{1}{0.375} = +2.67 \ D$.

(A) For a myopic eye, the correcting lens is a concave lens.
The far point of the myopic person is $x = 80 \, cm = 0.8 \, m$.
To see distant objects (at infinity) clearly, the lens must form a virtual image of an object at infinity at the person's far point $(80 \, cm)$.
Thus, the focal length $f$ of the corrective lens is equal to the negative of the far point distance:
$f = -x = -80 \, cm = -0.8 \, m$.
The power $P$ of the lens is given by the formula $P = \frac{1}{f(in \, meters)}$.
$P = \frac{1}{-0.8 \, m} = -1.25 \, D$.
Therefore, a concave lens of power $-1.25 \, D$ is required.

(A) myopic person suffers from myopia (nearsightedness),which is corrected using a concave lens.
For a myopic person,the far point is at $40 \, cm$. To see distant objects (at infinity) clearly,the lens must form a virtual image of the object at the person's far point.
Given: Object distance $u = \infty$,Image distance $v = -40 \, cm = -0.4 \, m$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-0.4} - \frac{1}{\infty} = -2.5 \, m^{-1}$.
Thus,the focal length $f = -0.4 \, m = -40 \, cm$.
The power of the lens $P = \frac{1}{f \text{ (in meters)}} = \frac{1}{-0.4} = -2.5 \, D$.
Therefore,a concave lens of focal length $40 \, cm$ and power $-2.5 \, D$ is required.

(B) Initially,the person suffered from myopia (nearsightedness),which was corrected by a concave lens of power $-1.00\, D$. This lens allowed him to see distant objects clearly.
As the person aged,he developed presbyopia,a condition where the eye loses its ability to focus on near objects due to the weakening of ciliary muscles and reduced flexibility of the eye lens.
To correct this,he requires a convex lens for reading. The power of $+2.00\, D$ indicates that his near point has shifted further away from the normal $25\, cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $f = +0.5\, m$ $(+50\, cm)$ and $u = -25\, cm$ (the standard near point),we find the image distance $v$ that his eye can currently accommodate:
$\frac{1}{v} = \frac{1}{50} + \frac{1}{-25} = \frac{1-2}{50} = -\frac{1}{50}$.
Thus,$v = -50\, cm$. This means his new near point has shifted to $50\, cm$ from his eyes,necessitating the use of reading glasses.

(D) Given: The object distance $u = \infty$ (for distant objects) and the image distance $v = -150 \ cm = -1.5 \ m$ (since the image must be formed at the person's far point).
Using the lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-1.5} - \frac{1}{\infty}$.
Since $\frac{1}{\infty} = 0$, we get $\frac{1}{f} = -\frac{1}{1.5}$.
Therefore, the focal length $f = -1.5 \ m$.
The power of the lens $P$ is given by $P = \frac{1}{f(\text{in meters})}$.
$P = \frac{1}{-1.5} = -0.67 \ D$.
Thus, the required focal length is $-1.5 \ m$ and the power is $-0.67 \ D$.

(A) The focal length $f$ of a lens is given by the formula $f = 1 / P$,where $P$ is the power of the lens in dioptres $(D)$ and $f$ is in metres $(m)$.
$(i)$ For distant vision:
Given power $P = -0.5 \text{ D}$.
$f = 1 / (-0.5) = -2 \text{ m}$.
$(ii)$ For near vision:
Given power $P = +1.5 \text{ D}$.
$f = 1 / 1.5 = 0.67 \text{ m}$ (approximately).

(N/A) The person is suffering from both myopia (short-sightedness) and presbyopia.
$1$. The near point of a normal eye is $25 \text{ cm}$. Since the person's near point is $50 \text{ cm}$, he cannot see objects closer than $50 \text{ cm}$, which indicates presbyopia (or hypermetropia).
$2$. The far point of a normal eye is infinity $(\infty)$. Since the person's far point is limited to $300 \text{ cm}$, he cannot see distant objects clearly, which indicates myopia (short-sightedness).

(N/A) $(i)$ The near point for a person with normal vision is $25 \, cm$ and the far point is infinity.
$(ii)$ The defect is Hypermetropia (farsightedness). The person needs a convex lens to correct this defect.
$(iii)$ In bright sunlight,the pupil of the eye is small to allow less light to enter. When we enter a dark room,the pupil takes some time to dilate (expand) to allow more light to enter,which is why we cannot see objects clearly immediately.

(N/A) As a result of dispersion produced by the prism,the white light is split up into seven colours. These colours are represented by the acronym '$VIBGYOR$' in the order of increasing deviation: $1$ (Red),$2$ (Orange),$3$ (Yellow),$4$ (Green),$5$ (Blue),$6$ (Indigo),$7$ (Violet). Thus,the colour at position $3$ is yellow and at position $5$ is blue. The student's statement is incorrect because the sky is blue and gold is yellow,which corresponds to positions $5$ and $3$ respectively,not $3$ and $5$.
$(b)$ The positions of the colours of the given objects are represented by:
$(i)$ Brinjal (Violet) $- 7$
$(ii)$ Danger signal (Red) $- 1$
$(iii)$ Neel (Indigo) $- 6$
$(iv)$ Orange (Orange) $- 2$

(N/A) Take a thick sheet of cardboard and make a small hole or narrow slit in its middle. Allow sunlight to fall on the narrow slit. This gives a narrow beam of white light. Now,take a glass prism and allow the light from the slit to fall on one of its faces as shown in the figure.
Turn the prism slowly until the light that comes out of it appears on a nearby screen. We will observe that sunlight has split into seven colours. This shows that sunlight is made up of seven colours.
$(b)$ Red colour bends least while violet colour bends the most. The different colours bend at different angles with respect to the incident ray due to refraction. This happens because the speed of light of different colours is different in the glass medium,leading to different refractive indices for different wavelengths.

(N/A) The diagram is as shown in the reference image.
$(i)$ The phenomenon is called the dispersion of light. It occurs because different wavelengths of light travel with different speeds in the glass prism,causing them to bend at different angles.
$(ii)$ This phenomenon is observed in nature in the formation of a rainbow.
$(iii)$ White light consists of seven different colors (wavelengths),which are violet,indigo,blue,green,yellow,orange,and red.

(N/A) In bright light,the iris contracts the pupil of the eye to allow less light to enter. When we enter a dim-lit room,the iris takes time to expand the pupil to allow more light to enter the eye. Thus,for a short duration,one is unable to see clearly in a dim-lit room.

(N/A) For a normal eye,the near point is at $25\, cm$ and the far point is at infinity. The given person cannot see objects clearly closer than $50\, cm$ (near point defect) or farther than $300\, cm$ (far point defect). Therefore,he is suffering from both myopia (nearsightedness) and hypermetropia (farsightedness),a condition often referred to as presbyopia.
$(b)$ To correct these defects,a bifocal lens is required. $A$ concave lens of suitable focal length is used in the upper part to correct myopia (to see distant objects up to infinity),and a convex lens of suitable focal length is used in the lower part to correct hypermetropia (to see near objects up to $25\, cm$).

(N/A) $(i)$ When the student sits at the front desk,the object is too close. The light rays from the blackboard focus behind the retina,resulting in a blurred image on the retina.
$(ii)$ When the student sits at the last desk,the object is at a sufficient distance. The light rays from the blackboard focus exactly on the retina,resulting in a clear image.
$(b)$ The student is suffering from hypermetropia,also known as long-sightedness.
$(c)$ This defect can be corrected by using a convex lens of suitable focal length. The convex lens converges the light rays before they enter the eye,allowing the image to be formed on the retina.

(N/A) $(i)$ When the student is seated at the last desk,the image of the distant object forms in front of the retina because the eye lens is unable to focus light rays from distant objects onto the retina.
$(ii)$ When the student is seated at the front desk,the object is closer,and the eye lens can focus the light rays onto the retina,resulting in a clear image.
$(b)$ The student is suffering from myopia (near-sightedness).
$(c)$ The defect can be corrected by using a concave lens of suitable focal length. The concave lens diverges the incoming light rays from the distant object,making them appear to come from the student's far point,thus allowing the eye lens to focus them onto the retina.

(N/A) In a prism,the refraction of light takes place at two slant surfaces that are not parallel to each other. The dispersion of white light occurs at the first surface of the prism,where its constituent colours are deviated through different angles. At the second surface,these split colours undergo further refraction and separation,resulting in a spectrum.
In a rectangular glass block,the refraction of light takes place at two parallel surfaces. At the first surface,white light splits into its constituent colours upon refraction. However,because the second surface is parallel to the first,these split colours undergo refraction at the second surface in such a way that they emerge as a parallel beam,which recombines to give the impression of white light.

(N/A) The colours at positions $3$ and $5$ are yellow and blue respectively. The student has identified them as yellow (i.e.,colour of the core of a hard-boiled egg) and blue (i.e.,colour of the sky) respectively. Hence,the statement is correct.
$(b)$ $(i)$ Position $7$ is the position of violet colour,which corresponds to the colour of a solution of potassium permanganate.
$(ii)$ Position $1$ is the position of red colour,which corresponds to the colour of danger or stop signal lights.

(N/A) To represent the refraction of a light ray through a prism,we follow these steps:
$1$. Draw a triangular prism $ABC$.
$2$. Draw an incident ray $PQ$ striking the face $AB$.
$3$. Draw a normal $N$ at the point of incidence $Q$ on face $AB$.
$4$. The angle between the incident ray $PQ$ and the normal $N$ is the angle of incidence,denoted by $i$.
$5$. The refracted ray $QR$ travels inside the prism. The angle between the refracted ray $QR$ and the normal $N$ at face $AB$ is the angle of refraction,denoted by $r_1$.
$6$. At face $AC$,the ray emerges as $RS$. Draw a normal $N'$ at the point of emergence $R$.
$7$. The angle between the emergent ray $RS$ and the normal $N'$ is the angle of emergence,denoted by $e$.
$8$. Extend the incident ray $PQ$ forward and the emergent ray $RS$ backward to meet at a point $D$. The angle between these two lines is the angle of deviation,denoted by $\delta$.

(N/A) In the minimum deviation position, the refracted ray inside the prism is parallel to the base $BC$. The emergent ray is shown in the provided solution image.
The angle of deviation $\delta$ depends upon the following factors:
$(i)$ The angle of the prism $(A)$.
$(ii)$ The refractive index of the material of the prism $\mu$.
$(iii)$ The angle of incidence $(i)$.
$(iv)$ The wavelength of the light used $\lambda$.

(N/A) We have the angle of incidence at the face $AB = 0^{\circ}$ as the ray $PQ$ falls normally on this face.
The normally incident ray $PQ$ passes through the prism undeviated and strikes the face $AC$ at the point $R$.
We draw the normal $MN$ to the face $AC$ at the point $R$.
The angle of incidence at the point $R$ is the angle $NRQ$.
From the geometry of the equilateral triangle $ABC$,the angle at vertex $A$ is $60^{\circ}$. In the triangle formed by the ray $PQ$ and the sides of the prism,the angle at $Q$ is $90^{\circ}$ and the angle at $A$ is $60^{\circ}$,so the angle at $R$ inside the triangle is $180^{\circ} - (90^{\circ} + 60^{\circ}) = 30^{\circ}$.
Since $MN$ is normal to $AC$,the angle of incidence $i = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
Hence,the angle of incidence at face $AB = 0^{\circ}$ and the angle of incidence at face $AC = 60^{\circ}$.

$(i)$ No, the images of letters of different colours will be raised by slightly different heights.
$(ii)$ The letter $V$, corresponding to the violet colour, will be raised to the maximum. This is because the apparent depth is given by the formula: $\text{Apparent depth} = \frac{\text{Real depth}}{n}$. Since the refractive index $n$ is maximum for the violet colour, its apparent depth will be the least. Consequently, the shift (raised height) is maximum for the violet colour.

(A) The ability of the eye lens to adjust its focal length to focus objects at varying distances is known as $accommodation$.
This process is facilitated by the ciliary muscles,which change the curvature and thickness of the lens.
When the ciliary muscles relax,the lens becomes thin,increasing the focal length to see distant objects.
When the ciliary muscles contract,the lens becomes thick,decreasing the focal length to see nearby objects.

(B) Cinematography relies on the phenomenon known as the $persistence \ of \ vision$.
This is the ability of the human eye to retain an image on the retina for approximately $1/16$ of a second after the object has been removed from the field of view.
In cinematography,a series of still images are projected in rapid succession (usually $24$ frames per second).
Because the brain perceives these images as continuous motion due to the $persistence \ of \ vision$,the viewer sees a smooth,moving picture rather than individual static frames.

(C) The human eye functions like a camera. Light enters the eye through the cornea and pupil,passes through the crystalline lens,and is focused onto the light-sensitive screen located at the back of the eye,known as the retina. The retina contains photoreceptor cells (rods and cones) that convert light energy into electrical signals,which are then transmitted to the brain via the optic nerve. Therefore,the correct answer is the retina.

(D) The ability of the eye lens to adjust its focal length to focus objects at varying distances is known as accommodation.
This process is controlled by the ciliary muscles.
When the ciliary muscles contract,the curvature of the lens increases,making it thicker and decreasing its focal length,which allows the eye to focus on nearby objects.
When the ciliary muscles relax,the lens becomes thinner,increasing its focal length,which allows the eye to focus on distant objects.

(A) When white light passes through a prism,it splits into its component colours,a phenomenon known as dispersion.
According to Cauchy's dispersion formula,the refractive index of a material depends on the wavelength of the light.
The refractive index is inversely proportional to the wavelength of light.
Red light has the longest wavelength in the visible spectrum,which results in the lowest refractive index for the prism material.
Since the angle of deviation is directly related to the refractive index,the colour with the lowest refractive index undergoes the least deviation.
Therefore,red light is deviated the least,while violet light,having the shortest wavelength,is deviated the most.

(B) The visible spectrum of light follows the order of $VIBGYOR$ (Violet, Indigo, Blue, Green, Yellow, Orange, Red).
In this spectrum, the wavelength $\lambda$ increases from violet to red.
Comparing the given colors: Blue, Green, and Yellow.
According to the $VIBGYOR$ sequence, the order of increasing wavelength is: $\lambda_{\text{blue}} < \lambda_{\text{green}} < \lambda_{\text{yellow}}$.
Therefore, the correct order of decreasing wavelength is $\lambda_{\text{yellow}} > \lambda_{\text{green}} > \lambda_{\text{blue}}$.