Algebraic Expressions

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Data Handling

The word data means information in the form of numerical figures or a set of given facts. E.g. The percentage of marks scored by 10 students.......

The Triangles and its Properties

A closed figure formed by joining three non-collinear points is called a triangle. The three sides and three angles of a triangle are collectively known as elements of the triangle......

Visualising Solid Shapes

A solid is any enclosed three-dimensional shape, i.e., it has 3 dimensions- length, width and height, whereas there are some common (flat) shapes which can be easily drawn on paper. They have only.....

Fractions

Fractions having the same denominator are called like fractions, whereas fractions having different denominator are called unlike fractions......

Perimeter and Area

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The Human Eye and the Colourful World – Summary & Concepts

The Human Eye and the Colourful World

'Eyes are most amazing optical instruments. In fact, as you read this sentence, you are probably unaware of the thousands of pieces of visual information that your eyes are gathering each second. Any defect of eyes can be corrected by using appropriate lenses.'

1.0The human eye

Components and structure

(1) The eye has a nearly spherical shape. The eye ball has a diameter of about one inch (nearly 2.3 cm ). The front portion is more sharply curved and it is covered by a thin, transparent, protective membrane called 'cornea'. This portion is visible from the outside. (2) Behind cornea, there is a liquid called 'aqueous humour' and behind that, there is a crystalline 'lens'. Between the aqueous humour and the lens, there is a muscular diaphragm called 'iris', which has a small hole in it called 'pupil'. (3) The eye lens is composed of a fibrous, jelly like material which is hard in the middle and gradually becomes soft towards the edges. The curvature of the lens is altered by the 'ciliary muscles' to which it is attached. (4) The space between lens and the retina is filled with another liquid called 'vitreous humour'. (5) The light entering the eye forms an image on the 'retina' which is a delicate membrane having enormous number of light-sensitive cells. It contains about 125 million receptors called 'rods' and 'cones' which receive the light rays and about one million optic nerve fibres which transmit the information to the brain. (6) The region on the retina where the 'optic nerve' enters the eye ball is called the 'blind spot'. It is insensitive to the light, that is, any image formed at this spot is not visible. (7) The macula lutea, also called 'yellow spot' is the central part of the retina responsible for sensing fine detail and for looking straight ahead. It has high concentration of nerve endings and it is slightly raised. Its function is to form a very clear image, by sending a large number of electrical signals to the brain. (8) 'Sclera' or 'sclerotic' is the outermost covering of eye and it is made of white fibrous tissue. 'Choroid' is a grey membrane attached to sclera.

The human eye

Rod shapes cells help us to see in dim light.
Cone shape cells help us to see bright light and perceive colours.

Working of eye

(1) When the light enters the eye from air, most of the bending of light occurs at cornea. Some additional bending is done by the lens so as to form an inverted, real image of the object on retina. (2) When the eye is focused on a distant object, the ciliary muscle relaxes allowing ligaments to increase tension on the lens and cause it to flatten i.e., the lens becomes thin or less curved. In this case, the focal length of the eye lens has its maximum value which is equal to its distance from the retina. The parallel rays coming into the eye from the distant object are focused on the retina and we see the object clearly. (3) When the eye is focused on a closer object, the ciliary muscle contracts, allowing the lens, by virtue of its elasticity, to become more curved i.e., the lens becomes thick. In this case, the focal length of the eye lens decreases. The ciliary muscles adjust the focal length in such a way that image is formed on the retina and we see the object clearly. (4) The light-sensitive cells get activated upon illumination and generate electrical signals. These signals are sent to the brain via the optic nerves. The brain interprets these signals, and finally, processes the information so that we perceive objects as they are i.e., the brain makes an inverted image formed on the retina again erect (or upright).

Role of iris

Iris controls the size of the pupil and therefore, helps in regulating the amount of light entering the eye through a variable aperture (the pupil). In low intensity of light, iris expands the pupil to allow more light to enter into the eye. When the light is very bright, iris contracts the pupil and the pupil becomes very small, thus, only a small amount of light enters into the eye.

The iris is that part of the eye which gives it, its distinctive colour. When we say that a person has green or brown eyes, we refer actually to the colour of the iris.

Building Concepts 1 How does the focal length depend on the curvature of eye lens? When is the eye lens thin? When is it thick? Explanation: The change in the curvature of eye lens changes its focal length. When we look at a distant object, the ciliary muscles are relaxed, the lens becomes thin and the focal length of the eye lens is more (about 2.3 cm ). Finally, the image is formed on the retina and we can see the distant object clearly. When we look at objects closer to eye, the ciliary muscles contract, so that the eye lens becomes more rounded in shape i.e., becomes thick. The focal length of the eye lens decreases so as to bring light from the nearby objects into focus on the retina.

Building Concepts 2 You might have experienced that you are not able to see objects clearly for some time when you enter from bright light to a room with dim light. Why? Explanation: The pupil of an eye acts like a variable aperture whose size can be varied with the help of the iris. When the eyes are exposed to bright light, the iris contracts the pupil to allow less light to enter the eye i.e., the size of pupil is reduced in the bright light. When you enter from bright light to a room with dim light, less amount of light enters in your eyes due to reduced size of the pupil. Thus, you are not able to see objects clearly for some time. After some time, iris expands the pupil due to dim light and allows more light to enter the eye and you are able to see things in the room. The pupil opens completely in dark/dim light through the relaxation of the iris.

Active Physics 1

Take a white cardboard and mark a thick cross on its left hand side. Mark a dot on the right side at a distance of 8 cm from the cross. Now, hold the cardboard at arms' length from your eyes. Close your left eye and look continuously on the cross. At this moment, both cross and dot are visible. Move the cardboard slowly towards you but keep your eye on the cross. You will observe that at a certain point, the dot disappears. At this point, the image of dot is formed on the blind spot, and hence, it is not visible.
The similar activity can be done by keeping your right eye closed and looking at the round mark. Move the cardboard slowly towards you keeping your eye on the dot. At some point, the cross disappears. This obviously means that the image of cross is formed on the blind spot.

2.0Active 2 Physics

Stand before a mirror in a darkened room for a few minutes. Then turn on a light in the room and observe your pupils in the mirror as they change size.
You will observe that just after turning the light 'on', your pupil is large in size and it decreases gradually in size. Reason : As you were initially standing in a darkened room, iris expands the pupil so that your eye receive as much light as possible. When you turn on the light, iris gradually contracts the size of pupil so that your eye receives less amount of light. Conclusion : This activity shows that the iris controls the amount of light entering in our eyes by controlling the size of pupil.

3.0Power of accommodation

By contracting or relaxing the ciliary muscles connected to the lens, its shape can be changed such that we can see the nearby as well as the distant objects clearly. This process is called 'power of accommodation' or 'accommodation'.

The ability or property of the eye to change the shape of lens so as to see the object clearly is called 'accommodation'.

The muscles cannot be strained beyond a limit and thus, an object placed too close to the eye cannot be seen clearly.

In general, young children have a greater power of accommodation because they can clearly see objects that are very close to them. As, the lens in our eye becomes less flexible and its ability to accommodate decreases until we are forced to use corrective eye lens.

Near point

The nearest point for which the image can be formed clearly on the retina, is called the 'near point of the eye'.

Least distance of distinct vision

The minimum distance at which objects can be seen clearly without strain is called 'least distance of distinct vision or clear vision'. In other words, the distance of the near point from the eye is called the 'least distance of distinct vision'.

This distance varies with age; it increases with it because of decreasing effectiveness of the ciliary muscles and the loss of flexibility of the lens. The symbol used for least distance of distinct vision is $D$ . Standard value of $D$ for a young adult with normal vision is 25 cm . For a child of 10 years D is nearly 7 to 8 cm , for an old man of 60 years $D$ is nearly 200 cm .

Active Physics 3

Move this page slowly towards your face. At a certain position of the page you will observe that the letters of this page just begin to blur.
Hold the page at this position and ask your friend to measure the distance between the page and your eye using a scale. This distance from the book to your eye is your near point.

Far point

The farthest point up to which the eye can see objects clearly is called the 'far point' of the eye. For normal eye, far point is at infinity.

Building Concepts 3 Why do we have two eyes for vision and not just one? Explanation: Two eyes provide a wider field of view to us. A single eye has a horizontal region of view of nearly $15 0^{\circ}$ , but with two eyes it is about $18 0^{\circ}$ . Also, the ability to detect faint objects increases with two eyes instead of a single eye. With a single eye, the world looks flat i.e., two-dimensional. The three- dimensional effect can be experienced with two eyes only.

Building Concepts 4 How do we see colours? Explanation: Our retina has a large number of light-sensitive cells having shapes of rods and cones. The rod-shaped cells respond to the intensity of light with different degree of brightness and darkness. In dim light, the rods are sensitive, but the cones are not. But the rods cannot distinguish between various colours. It is the cone-shaped cells which respond to colours. They are active only in bright light and make colour perception possible. The cones are sensitive to red, green and blue light to different extents. When a particular colour like red, falls on the retina, it mainly stimulates the red colour sensitive cones than the other kinds of cones.

The retina of a chick has mostly cones and only a few rods. As the cones are sensitive to bright light only, the chicks wake up with sunrise (dawn) and sleep by the sunset (dusk).

Colour blindness

Some people do not possess some cone cells that respond to certain specific colours only. This is a genetic disorder called 'colour blindness'. The persons who cannot distinguish between certain colours but can see well are called 'colour blind'.

The defect of eye due to which a person cannot distinguish between certain colours is called 'colour blindness'.

Cataract

The crystalline lens of some people in old age becomes hazy or even opaque due to development of membrane over it. This defect is called 'cataract' which leads to decrease or loss of vision of the eye. Cataract can be removed by performing surgery to restore clear vision.

4.0Defects of vision and their correction

Myopia (Near sightedness)

A person suffering from this defect cannot see distant objects clearly. This is because the maximum focal length is less than distance between the lens and the retina. The parallel rays coming from the distant object focus short of the retina. The ciliary muscles are fully relaxed in this case and any strain in it can only further decrease the focal length which is of no help to see distant objects.

Reason This defect arises because the power of eye lens is too great, due to the decrease in focal length of the eye lens. This may arise due to either excessive curvature of the cornea or elongation of the eye ball.

Hypermetropia (far-sightedness)

A person suffering from Hypermetropia cannot see clearly the objects closer to the eye. The least distance of distinct vision is quite larger than 25 cm for that person and the person has to keep the object inconveniently away from the eye. Thus, the image is not formed at the retina if an object is kept at about 25 cm away from the eye. The rays are focused behind the retina (see figure).

Reason This defect arises because either the focal length of the eye lens is too great or eye ball becomes too short. Due to this, light rays from nearby objects cannot be brought to focus on the retina to give a clear image.

In myopic eye far point shift towards the eye.
In hypermetropic eye near point shift away from the eye.

Building Concepts 5 What type of corrective lens would you suggest for the following eye defects and why? (a) Myopia (b) Hypermetropia Explanation: (a) In myopia, a person cannot see the distant object clearly i.e., the far point of a myopic eye is located at some finite distance from the eye. Let this distance be ' $x$ '. Now, the parallel beam from distant object does not form an image on the retina [see figure (a)]. Instead, a diverging beam from the far point forms a clear image on the retina [see figure (b)].

Thus, if we make the parallel beam from the distant object to become a diverging beam using an appropriate lens, a clear image of the distant object can be formed on the retina. We know that a parallel beam can be made diverging easily by using a diverging lens i.e., a concave lens (see figure). Thus, for myopia, a concave lens is used as a corrective lens.

Can be possible using a concave lens Corrective lens for myopia (b) In hypermetropia, a person cannot see the closer object clearly, the near point of a hypermetropic eye is located at a distance greater than 25 cm from the eye. Let this distance be '

y

'. Now, the divergent beam from a point 25 cm away from the eye does not form an image on the retina [see figure(a)]. Instead, a diverging beam from the near point located at a distance greater than 25 cm forms a clear image on the retina [see figure(b)]. Thus, from figure, we can see that if we make the more divergent beam from the object placed at 25 cm to become a less divergent beam as in figure(b) using an appropriate lens, a clear image of the nearby object can be formed on the retina.

(a)

(b) A more divergent beam can be made less divergent easily by using a converging lens i.e., a convex lens (see figure). This is simply like making a more positive number to become less positive by adding a negative number to it. Thus, for hypermetropia, a convex lens is used as a corrective lens.

Corrective lens for hypermetropia

Calculations for corrective lens for myopia

As discussed above, myopia is corrected by using a concave lens (diverging lens) which increases the focal length in order to bring the image of the object back on the retina itself. Let a person can see clearly to a distance ' x ' only (see figure). If we apply lens equation to the lens, then we have $u = - \infty$ and $v = - x$ .

By lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{- x} - \frac{1}{- \infty} = \frac{1}{f}$ or $f = - x$ And, power is given by, $P = \frac{1}{f} = - \frac{1}{x}$

Correction for a myopic eye ('

x

' is always taken

+ v e

, '

x

' and '

f^{'}

' are in metres)

Calculations for corrective lens for hypermetropia

Hypermetropia can be corrected by using a convex lens (converging lens) which decreases the focal length in order to focus the image of a nearby object at the retina. Thus, a person can see the object clearly. Let ' $y$ ' be the minimum distance at which the person can see the object clearly.

Let the near point of a person having hypermetropia be at $y$ distance (see figure) from the eye ( $y > 25 cm$ ). If we apply lens equation to the lens, then we have $u = - 0.25 m$ and $v = - y$ . $\frac{1}{- y} - \frac{1}{- 0.25} = \frac{1}{f} \frac{1}{f} = \frac{1}{0.25} - \frac{1}{y}$

Correction for a hypermetropic eye

Also, power, $P = \frac{1}{f} = \frac{1}{0.25} - \frac{1}{y} = 4 - \frac{1}{y}$ (' y ' is always taken +ve , ' y ' and ' $f^{'}$ ' are in metres) Sometimes, a person may suffer from both myopia and hypermetropia. Such people often use bifocal lens. Commonly, the upper portion of bifocal lens is a concave lens (used for distant vision) and the lower portion is a convex lens (used for reading purpose).

Presbyopia

Rather than resulting from a change in the shape of the eyeball, the inability to see objects that are close to you can also occur because the lens loses its flexibility. This condition is known as presbyopia. As you age, the lens often loses its flexibility and cannot become round enough to create clear images of near objects.

The defect that arises due to ageing in which a person cannot read comfortably and distinctly without corrective eye glasses is called 'presbyopia'.

Reason The power of accommodation of the eye decreases with ageing. For most of the people, the near point recedes, this means, the least distance of distinct vision increases. This phenomenon arises due to the gradual weakening of ciliary muscles and decreasing flexibility of the crystalline eye lens.

Due to ageing, usually a person can see the distant objects clearly. This is because, the rays from a distant object are less diverging as compared to the rays from a near object. Thus, the lens can still focus rays from distant objects on the retina. To cause the more sharply diverging rays from a near object to be focussed on the retina, the lens has to become quite round to shorten its focal length. Since the lens has lost its flexibility (or elasticity), it cannot become quite round to focus the light rays from the near object. Thus, he/she cannot see the near object clearly.

Correction Presbyopia is literally an "old-age vision" and it is due to a reduction in accommodation ability. The cornea and lens together are not able to bring nearby objects into focus on the retina. The symptoms are the same as with hypermetropia or farsightedness, and the condition can be corrected with converging lens i.e., convex lens. $\to$ These days, it is possible to correct the defects of vision by using 'contact lenses'.

Presbyopia can occur in conjunction with myopia or hypermetropia. If one already wears glasses to correct for myopia, as presbyopia occurs, the bifocal lenses can be used to accommodate both conditions.

Numerical Ability 1

The far point of a myopic person is 80 cm in front of the eye. What is the focal length and power of the lens required to enable him to see very distant objects clearly? Solution: Decode the problem Identify the type of defect Myopia $✓ /$ Hypermetropia Identify the formula $f = - x, P = \frac{1}{f}$ ' $x$ ' is always taken $+ v e$ , ' $x$ ' and ' $f$ ' are in metres Here, $x = 80 cm = 0.8 m$ . Focal length, $f = - 0.8 m$ (A concave lens). Power, $P = \frac{1}{- 0.8}$ $= - 1.25$ dioptres.
The near point of a patient's eye is $50.0 cm$ . (a) What focal length must a corrective lens have to enable the eye to see clearly an object 25.0 cm away? Neglect the eye-lens distance. (b) What is the power of this lens? Solution: Decode the problem Identify the type of defect Myopia/ $✓$ Hypermetropia Identify the formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, P = \frac{1}{f}$ $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{y}$ (' y ' is always taken +ve , ' y ' and ' f are in metres) Here, $u = - 25 cm$ ; v=-50 cm ; f =? (a) By lens equation, $or \frac{1}{f} = \frac{1}{( - 50 )} - \frac{1}{( - 25 )} or \frac{1}{f} = \frac{- 1}{50} + \frac{1}{25} or \frac{1}{f} = \frac{- 1 + 2}{50} = \frac{1}{50} or f = + 50 cm = + 0.5 m$ Alter: Here, $y = 50 cm = 0.5 m$ $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{0.5} or \frac{1}{f} = \frac{100}{25} - \frac{10}{5} = 4 - 2 = + 2 or f = + \frac{1}{2} = + 0.5 m$ (b) Power, $or P = \frac{1}{+ 0.5} = + 2 Dioptre$
Suppose a lens is placed in a device that determines its power as +2.75 diopters. Find (a) the focal length of the lens and (b) the minimum distance at which a patient will be able to focus on an object if the patient's near point is $60.0 cm$ . Neglect the eye-lens distance. Solution: Decode the problem Identify the type of defect Myopia/ $✓$ Hypermetropia Identify the formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $P = \frac{1}{f}$ Here, $P = + 2.75$ dioptre $; f =$ ? (a) $f = \frac{1}{P} = \frac{1}{+ 2.75} = + 0.364 m = + 36.4 cm$ (b) Here, $v = - 60 cm$ ; $u =$ ? By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{( - 60 )} - \frac{1}{u} = \frac{1}{+ 36.4}$ or $\frac{1}{u} = \frac{- 1}{60} - \frac{1}{36.4} = \frac{- 36.4 - 60}{2184} = \frac{- 96.4}{2184}$ or $u = \frac{2184}{- 96.4} = - 22.6 cm$
The near point of a hypermetropic person is $75 cm$ from the eye. What is the focal length and power of the lens required to enable the person to read clearly a book held at $25cm$ from the eye? Solution: Decode the problem Identify the type of defect Myopia/ $✓$ Hypermetropia Identify the formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, P = \frac{1}{f}$ $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (' y ' is always taken +ve , y ' and ' f ' are in metres) Here, $y = 75 cm = 0.75 m$ $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{( 0.75 )} = \frac{100}{25} - \frac{100}{75} = 4 - \frac{4}{3} = \frac{8}{3} = \frac{100}{25} - \frac{100}{75} = 4 - \frac{4}{3} = \frac{8}{3}$ or $f = + \frac{3}{8} = + 0.375 m$ Power, $P = \frac{8}{3} = + 2.67$ dioptres.

5.0Refraction of light through a prism

Prism (triangular prism)

In a triangular prism, there are two triangular base and three rectangular lateral surfaces (refracting faces) [see figure]. These surfaces are inclined to each other at some angle. The angle between its two lateral surfaces is known as the angle of the prism.

Let the incident ray be PQ, QR be the refracted ray and RS

A triangular prism be the emergent ray (see figure). The ray $PQ$ enters from air to glass at the first refracting surface AB.

The light ray after refraction, bends towards the normal (refracted ray QR). At the second surface AC, the ray QR enters from glass to air and bends away from the normal after refraction. The emergent ray RS is not parallel to the incident ray PQ due to the peculiar shape of prism.

Refraction of light through a prism

The angle between the incident ray and the emergent ray is called 'angle of deviation ( $δ$ )'.

6.0Dispersion of white light by a prism

If a beam of white light is made to fall on one face of a prism, the light emerging out from the other face of prism consists of seven colours namely Violet (V), Indigo (I), Blue (B), Green $(G)$ , Yellow (Y), Orange ( O ) and Red (R). The deviation suffered by the red light is minimum and for the violet light, it is maximum (see figure).

The phenomenon of splitting up of white light into its constituent colours is called 'dispersion of light'.

Dispersion of white light through a prism The beautiful, sparkling colours produced by ice crystals on a small branch of a tree in winter, the vibrant colours of a rainbow, and the brilliant flashes of colour you see when light passes through diamonds, all these are examples of the phenomenon known as 'dispersion'.

Reason The wavelengths of different colours of light are different and the refractive index of glass is different for different wavelengths. Higher the wavelength, lower will be the refractive index and thus, lower will be the deviation and vice-versa. Thus, deviation of red light is low as its wavelength is large. Similarly, deviation of violet light is high as its wavelength is small. Initially, it was a matter of debate, whether the prism itself creates colours in some way or it only separates the colours already present in white light. Sir Isaac Newton settled the issue by performing a simple experiment. He put another similar prism, but in an inverted position, and let the emergent, separated beam fall on the second prism. The resulting emergent beam was found to be white light (see figure). The first prism separated the white light into its component colours, which were recombined by the inverted prism to give white light. Thus, white light itself consists of colours which can be separated by the prism.

Refractive index $\propto \frac{1}{Wavelength}$ or $μ \propto \frac{1}{λ}$

As $λ_{V} < λ_{Y} < λ_{R}$ As $μ_{V} > μ_{Y} > μ_{R}$

Speed of violet light inside the prism is slowest and that of red is highest.
Deviation of violet is maximum and that of red is minimum.

7.0The Rainbow

A rainbow is a natural spectrum appearing in the sky after the rain shower. It is caused by dispersion of sunlight by tiny water droplets, present in the atmosphere. A rainbow always formed in a direction opposite to that of the sun. The water droplets act like small prisms. They refract and disperse the incident sunlight, then reflect it internally, and finally refract it again when it comes out the raindrop. Due to dispersion of light and internal reflection, different colour reach the observer's eye.

You can also see a rainbow on a sunny day when you look at the sky through a waterfall or through a water fountain, with the sun behind you.

Phenomena involved in the formation of a rainbow are refraction, dispersion and internal reflection.

8.0Atmospheric refraction

The refraction of light caused by the earth's atmosphere due to variable refractive index of air at different zones is called 'atmospheric refraction'.

Twinkling of stars

Twinkling of stars can be seen on a clear night. This is due to atmospheric refraction of light coming from the stars (star light). As the star light enters into the earth's atmosphere, atmospheric refraction takes place due to gradually changing refractive index of the air. Since the physical conditions of the refracting medium (earth's atmosphere) are not stationary, star light flux (luminous flux) entering the eye of an observer continuously fluctuates. This means luminous energy reaching our eyes per second from the star increases and decreases with time. Thus, the star sometimes appears brighter and at some other times fainter, causing the 'twinkling of stars'. Star appears slightly higher than its actual position As the star light enters the earth's atmosphere, it undergoes refraction continuously before it reaches the earth's surface. This refraction occurs due to the gradually changing refractive index of air. As the star light enters from rarer medium to comparatively denser medium, it bends more and more towards the normal as it is reaching the earth's surface (see figure). Thus, apparent position of star is slightly higher than its actual position when it is viewed from the earth's surface. Also, this apparent position is not stationary but it changes with time because of variable physical conditions of the refracting medium (earth's atmosphere).

Apparent star position due to atmospheric refraction

Building Concepts 6 Why do planets not twinkle? Explanation: The apparent size of stars is very small as compared to apparent size of planets. Thus, the star may be considered as a 'point sized' source of light and the planet as an 'extended source' of light. So, the planet can be considered as a collection of large number of 'point sized' sources of light, such that the dimming effect of some 'point sources' is nullified by the brighter effect of the other 'point sources'. The variable atmospheric conditions are unable to create variations in light flux from planet entering our eye and thus, planets do not twinkle. Delayed sunset and advanced (early) sunrise The sun is visible before actual sunrise and after actual sunset because of atmospheric refraction. Actual sunrise means the 'sunrise on actual crossing of the horizon by the sun'. The refractive index of air with respect to free space or vacuum is 1.00029 ( $\sim 1.0003$ ). Due to this, sunlight bends towards the surface of Earth because of refraction. Thus, the sun appears to be raised above the horizon when it is slightly below the horizon (see figure). The apparent shift in the direction of sun is about $1/ 2^{\circ}$ and corresponding time difference between actual and apparent sunrise (or sunset) is about 2 minutes.

Earth atmosphere Effect of atmospheric refraction at the sunrise or sunset Thus, sun is visible to us about 2 minutes before actual sunrise and about 2 minutes after actual sunset.

9.0Scattering of light

The process in which the light rays are deflected by the particles of the medium through which they pass is called scattering of light. Tyndall effect When rays of light fall on the fine particles of a colloidal solution, the path of the beam is visible due to the scattering of light by the colloidal particles. This phenomenon is called 'Tyndall effect'.

The scattering of light as it passes through a medium containing small particles is called 'Tyndall effect'. The colour of the scattered light depends on the size of the scattering particles. Very fine particles scatter shorter wavelengths (colours at violet end) by greater amount as compared to the light having longer wavelengths (colours at red end). If the size of particles is quite large, then, the scattered light is almost white.

Tyndall effect is observed when a fine beam of sunlight enters a room through a small hole and the light get scattered from the smoke or dust particles present in the room ; when sunlight passes through a canopy of a dense forest, tiny water droplets present in the air scatter the light. (T) The 'danger' signals are made of red colour because red colour is scattered least by the air particles, smoke or fog due to its longer wavelength. Hence, red colour can be seen in the same shade even at the large distance.

Very small particles scatter light of shorter wavelength better than longer wavelength.
The scattering of longer wavelength of light increases as the size of the particles increases.
Larger particles scatter light of all wavelengths equally well.

Active Physics 4

Place a strong source ( S ) of white light at the focus of a converging lens ( $L_{1}$ ). This lens provides a parallel beam of light. Allow the light beam to pass through a transparent glass tank (T) containing clear water. Allow the beam of light to pass through a circular hole (C) made in a cardboard. Obtain a sharp image of the circular hole on a screen (M) using a second converging lens (L2), as shown in figure. Dissolve about 200 g of sodium thiosulphate (hypo) in about 2 L of clean water taken in the tank. Add about 1 to 2 mL of concentrated sulphuric acid to the water.
You will find fine microscopic sulphur particles precipitating in about 2 to 3 minutes. As the sulphur particles begin to form, you can observe the blue light from the three sides of the glass tank. This is due to scattering of short wavelengths by minute colloidal sulphur particles. Observe the colour of the transmitted light from the fourth side of the glass tank facing the circular hole. It is interesting to observe, at first, the orange red colour and then bright crimson red colour on the screen.
This activity demonstrates the scattering of light that helps you to understand the bluish colour of the sky and the reddish appearance of the Sun at the sunrise or the sunset. We conclude that very fine particles scatter mainly shorter wavelength light (blue light) and longer wavelength light (red light) gets transmitted.

Building Concepts 7 Why sky appears blue? Explanation: The molecules of air and other fine particles in the atmosphere have size quite smaller as compared to the wavelength of the visible light. These particles scatter the light having shorter wavelengths at the violet or blue end more than the light having longer wavelengths at the red end. Thus, when the sunlight passes through the earth's atmosphere, the fine particles present in air scatter the blue colour more strongly in comparison to the red colour. The scattered blue light falls on our eyes. Thus, the sky appears blue.

Colour of the sun at sunrise and sunset At the sunrise or sunset, the sun is near the horizon. Light reaching the eye of an observer travels a larger distance through the atmosphere. Thus, most of the blue light and shorter wavelength rays are scattered away by the air particles. Hence, the light that reaches the eye of an observer is of longer wavelengths (of red end). Thus, the sun appears reddish at sunrise or sunset (see figure).

Sun looks white at noon At the noon, the light has to travel a comparatively shorter distance through the atmosphere before reaching the eye of an observer. Thus, only a very little amount of blue or violet light is scattered away. Hence, the sun at noon looks almost white (see figure).

Clouds are generally white The clouds are generally white because the larger particles like dust and water drops scatter light of all colours, almost equally and all the colours reach our eyes equally and combine to form white light.

10.0Concept Map

11.0Some Basic Terms

Diameter : The distance from the center to any point on the circle is called the radius of the circle. Two radii form the diameter.
Diaphragm : In optics, a diaphragm is a thin opaque structure with an opening at its centre, the role of diaphragm is to stop the passage of light, except for the light passing through the aperture.
Membrane : A membrane is a thin, flexible layer of tissue that covers, lines, separate or connect cells or parts of an organism.
Elasticity: It is the ability of a deformed material body to return to its original shape and size when the forces causing the deformation are removed.
Ligaments : It is a ring like fibrous membrane that connects the ciliary muscles and the lens of the eye, holding the lens in place.
Illumination : The quantity of light emitted by a lighting source.
Two dimensional : Two dimensional means having only two dimensional such as length and width. Two dimensional things are flat and have no depth. Examples : Geometrical shapes like squares, circles, polygons etc.
Three dimensional : Three dimensional shapes are solid that have three dimensions such as length, width and height. Examples : Cone-shaped ice cream, cubical box, ball, sphere etc.
Hazy vision : Hazy vision or cloudy vision means that you can't see things clearly.
Opaque: This means that no light is able to pass through them.
Elongation : It is the ability of a material to stretch when under strain.
Bifocal lens : A bifocal lens is an eyeglass lens with two different optical powers two correct vision at both long and short distances. The lens is divided into two segments with top of the lens is concave lens and the bottom of the lens is convex lens.
Spectrum : Spectrum is a band of colours that is produced when white light passes through a prism and splits into its constituent colour.
Deviation : A light that slightly diverges from its path because a change in the medium is known as a deviation of light.
Colloidal solution : The colloidal solution can be define as a mixture of particles of substances. These particles are microscopically dispersed and soluble/insoluble which are suspended in a fluid regularly.
Colloidal particles : Colloidal particles are small solid particles that are suspended in fluid.
Molecules: It is a group of two or more atoms that are held together by a chemical bond.
Monochromatic light : Light that has a single wavelength or colour.
Choroid : A molecule is the smallest unit of matter that has distinct chemical and physical properties.
Range of vision : Range of vision for a normal human eye is the range of distance for which human eye can see an object clearly. It ranges from infinity to $25 cm$ .

12.0SOLVED EXAMPLES

The near point of a patient's eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 25.0 cm away? Neglect the eye-lens distance. (b) What is the power of this lens? Solution: Here, $u = - 25 cm$ ; $v = - 50 cm$ ; $f =$ ? (a) By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{f} = \frac{1}{( - 50 )} - \frac{1}{( - 25 )}$ or $\frac{1}{f} = \frac{- 1}{50} + \frac{1}{25}$ or $\frac{1}{f} = \frac{- 1 + 2}{50} = \frac{1}{50}$ or $f = + 50 cm = + 0.5 m$ Alter: Here, $y = 50 cm = 0.5 m$ $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{y}$ or $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{0.5}$ or $\frac{1}{f} = \frac{100}{25} - \frac{10}{5} = 4 - 2 = + 2$ or $f = + (1/2) = + 0.5 m$ (b) Power, $P = \frac{1}{f}$ or $= + 2$ Dioptre
Suppose a lens is placed in a device that determines its power as +2.75 diopters. Find (a) the focal length of the lens and (b) the minimum distance at which a patient will be able to focus on an object if the patient's near point is 60.0 cm . Neglect the eye-lens distance. Solution: Here, $P = + 2.75$ dioptre ; $f =$ ? (a) $f = \frac{1}{P} = \frac{1}{+ 2.75} = + 0.364 m$ $= + 36.4 cm$ (b) Here, $v = - 60 cm; u =$ ? By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{( - 60 )} - \frac{1}{u} = \frac{1}{+ 36.4}$ or $\frac{1}{u} = \frac{- 1}{60} - \frac{1}{36.4}$ or $\frac{1}{u} = \frac{- 1}{60} - \frac{1}{36.4} = \frac{- 36.4 - 60}{2184} = \frac{- 96.4}{2184}$ or $u = \frac{2184}{- 96.4} = - 22.6 cm$
A particular near-sighted patient can't see objects clearly when they are beyond 25 cm (the far point of the eye). (a) What focal length should the prescribed contact lens have to correct this problem? (b) Find the power of the lens, in diopters. Neglect the distance between the eye and the corrective lens. Solution: Here, $u = - \infty; v = - 25 cm; f =$ ? (a) By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{f} = \frac{1}{( - 25 )} - \frac{1}{( - \infty )} = \frac{- 1}{25}$ or $f = - 25 cm = - 0.25 m$ Alter : Here, $x = 25 cm$ $f = - x = - 25 cm = - 0.25 m$ (b) Power, $P = \frac{1}{f}$ or $P = \frac{1}{- 0.25} = - 4$ Dioptre
A retired bank officer can easily read the fine print of the financial page when the newspaper is held no closer than arm's length, 60.0 cm from the eye. What should be the focal length of an eyeglass lens that will allow her to read at the more comfortable distance of 24.0 cm ? What is the power of this lens? Solution: Here, $v = - 60 cm; u = - 24 cm; f =$ ? By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{f} = \frac{1}{( - 60 )} - \frac{1}{( - 24 )}$ or $\frac{1}{f} = \frac{- 1}{60} + \frac{1}{24}$ or $\frac{1}{f} = \frac{- 2 + 5}{120} = \frac{3}{120} = \frac{1}{40}$ or $f = + 40 cm = + 0.4 m$ Power, $P = \frac{1}{f}$ or $P = \frac{1}{+ 0.4}$ = + 2.5 Dioptre
A boy uses spectacles of focal length -60 cm . Name the defect of vision he is suffering from. Which lens is used for the correction of this defect? Compute the power of this lens. Solution: Given, $f = - 60 cm$ $= - 0.60 m$ The negative sign of focal length indicates that the lens is concave. Thus the boy suffers from myopia (short sightedness) vision. Power of the lens $= \frac{1}{f} = \frac{1}{( - 0.60 )}$ $= - 1.67 D$

The Human Eye and the Colourful World

1.0The human eye

Components and structure

The human eye

Rod shapes cells help us to see in dim light.
Cone shape cells help us to see bright light and perceive colours.

Working of eye

Role of iris

The iris is that part of the eye which gives it, its distinctive colour. When we say that a person has green or brown eyes, we refer actually to the colour of the iris.

Active Physics 1

Take a white cardboard and mark a thick cross on its left hand side. Mark a dot on the right side at a distance of 8 cm from the cross. Now, hold the cardboard at arms' length from your eyes. Close your left eye and look continuously on the cross. At this moment, both cross and dot are visible. Move the cardboard slowly towards you but keep your eye on the cross. You will observe that at a certain point, the dot disappears. At this point, the image of dot is formed on the blind spot, and hence, it is not visible.
The similar activity can be done by keeping your right eye closed and looking at the round mark. Move the cardboard slowly towards you keeping your eye on the dot. At some point, the cross disappears. This obviously means that the image of cross is formed on the blind spot.

2.0Active 2 Physics

Stand before a mirror in a darkened room for a few minutes. Then turn on a light in the room and observe your pupils in the mirror as they change size.
You will observe that just after turning the light 'on', your pupil is large in size and it decreases gradually in size. Reason : As you were initially standing in a darkened room, iris expands the pupil so that your eye receive as much light as possible. When you turn on the light, iris gradually contracts the size of pupil so that your eye receives less amount of light. Conclusion : This activity shows that the iris controls the amount of light entering in our eyes by controlling the size of pupil.

3.0Power of accommodation

The ability or property of the eye to change the shape of lens so as to see the object clearly is called 'accommodation'.

The muscles cannot be strained beyond a limit and thus, an object placed too close to the eye cannot be seen clearly.

In general, young children have a greater power of accommodation because they can clearly see objects that are very close to them. As, the lens in our eye becomes less flexible and its ability to accommodate decreases until we are forced to use corrective eye lens.

Near point

The nearest point for which the image can be formed clearly on the retina, is called the 'near point of the eye'.

Least distance of distinct vision

This distance varies with age; it increases with it because of decreasing effectiveness of the ciliary muscles and the loss of flexibility of the lens. The symbol used for least distance of distinct vision is $D$ . Standard value of $D$ for a young adult with normal vision is 25 cm . For a child of 10 years D is nearly 7 to 8 cm , for an old man of 60 years $D$ is nearly 200 cm .

Active Physics 3

Move this page slowly towards your face. At a certain position of the page you will observe that the letters of this page just begin to blur.
Hold the page at this position and ask your friend to measure the distance between the page and your eye using a scale. This distance from the book to your eye is your near point.

Far point

The farthest point up to which the eye can see objects clearly is called the 'far point' of the eye. For normal eye, far point is at infinity.

The retina of a chick has mostly cones and only a few rods. As the cones are sensitive to bright light only, the chicks wake up with sunrise (dawn) and sleep by the sunset (dusk).

Colour blindness

The defect of eye due to which a person cannot distinguish between certain colours is called 'colour blindness'.

Cataract

4.0Defects of vision and their correction

Myopia (Near sightedness)

Hypermetropia (far-sightedness)

In myopic eye far point shift towards the eye.
In hypermetropic eye near point shift away from the eye.

y

(a)

Corrective lens for hypermetropia

Calculations for corrective lens for myopia

By lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{- x} - \frac{1}{- \infty} = \frac{1}{f}$ or $f = - x$ And, power is given by, $P = \frac{1}{f} = - \frac{1}{x}$

Correction for a myopic eye ('

x

' is always taken

+ v e

, '

x

' and '

f^{'}

' are in metres)

Calculations for corrective lens for hypermetropia

Correction for a hypermetropic eye

Presbyopia

The defect that arises due to ageing in which a person cannot read comfortably and distinctly without corrective eye glasses is called 'presbyopia'.

Presbyopia can occur in conjunction with myopia or hypermetropia. If one already wears glasses to correct for myopia, as presbyopia occurs, the bifocal lenses can be used to accommodate both conditions.

Numerical Ability 1

The far point of a myopic person is 80 cm in front of the eye. What is the focal length and power of the lens required to enable him to see very distant objects clearly? Solution: Decode the problem Identify the type of defect Myopia $✓ /$ Hypermetropia Identify the formula $f = - x, P = \frac{1}{f}$ ' $x$ ' is always taken $+ v e$ , ' $x$ ' and ' $f$ ' are in metres Here, $x = 80 cm = 0.8 m$ . Focal length, $f = - 0.8 m$ (A concave lens). Power, $P = \frac{1}{- 0.8}$ $= - 1.25$ dioptres.
The near point of a patient's eye is $50.0 cm$ . (a) What focal length must a corrective lens have to enable the eye to see clearly an object 25.0 cm away? Neglect the eye-lens distance. (b) What is the power of this lens? Solution: Decode the problem Identify the type of defect Myopia/ $✓$ Hypermetropia Identify the formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, P = \frac{1}{f}$ $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{y}$ (' y ' is always taken +ve , ' y ' and ' f are in metres) Here, $u = - 25 cm$ ; v=-50 cm ; f =? (a) By lens equation, $or \frac{1}{f} = \frac{1}{( - 50 )} - \frac{1}{( - 25 )} or \frac{1}{f} = \frac{- 1}{50} + \frac{1}{25} or \frac{1}{f} = \frac{- 1 + 2}{50} = \frac{1}{50} or f = + 50 cm = + 0.5 m$ Alter: Here, $y = 50 cm = 0.5 m$ $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{0.5} or \frac{1}{f} = \frac{100}{25} - \frac{10}{5} = 4 - 2 = + 2 or f = + \frac{1}{2} = + 0.5 m$ (b) Power, $or P = \frac{1}{+ 0.5} = + 2 Dioptre$
Suppose a lens is placed in a device that determines its power as +2.75 diopters. Find (a) the focal length of the lens and (b) the minimum distance at which a patient will be able to focus on an object if the patient's near point is $60.0 cm$ . Neglect the eye-lens distance. Solution: Decode the problem Identify the type of defect Myopia/ $✓$ Hypermetropia Identify the formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $P = \frac{1}{f}$ Here, $P = + 2.75$ dioptre $; f =$ ? (a) $f = \frac{1}{P} = \frac{1}{+ 2.75} = + 0.364 m = + 36.4 cm$ (b) Here, $v = - 60 cm$ ; $u =$ ? By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{( - 60 )} - \frac{1}{u} = \frac{1}{+ 36.4}$ or $\frac{1}{u} = \frac{- 1}{60} - \frac{1}{36.4} = \frac{- 36.4 - 60}{2184} = \frac{- 96.4}{2184}$ or $u = \frac{2184}{- 96.4} = - 22.6 cm$
The near point of a hypermetropic person is $75 cm$ from the eye. What is the focal length and power of the lens required to enable the person to read clearly a book held at $25cm$ from the eye? Solution: Decode the problem Identify the type of defect Myopia/ $✓$ Hypermetropia Identify the formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, P = \frac{1}{f}$ $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (' y ' is always taken +ve , y ' and ' f ' are in metres) Here, $y = 75 cm = 0.75 m$ $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{( 0.75 )} = \frac{100}{25} - \frac{100}{75} = 4 - \frac{4}{3} = \frac{8}{3} = \frac{100}{25} - \frac{100}{75} = 4 - \frac{4}{3} = \frac{8}{3}$ or $f = + \frac{3}{8} = + 0.375 m$ Power, $P = \frac{8}{3} = + 2.67$ dioptres.

5.0Refraction of light through a prism

Prism (triangular prism)

Let the incident ray be PQ, QR be the refracted ray and RS

A triangular prism be the emergent ray (see figure). The ray $PQ$ enters from air to glass at the first refracting surface AB.

Refraction of light through a prism

The angle between the incident ray and the emergent ray is called 'angle of deviation ( $δ$ )'.

6.0Dispersion of white light by a prism

The phenomenon of splitting up of white light into its constituent colours is called 'dispersion of light'.

Refractive index $\propto \frac{1}{Wavelength}$ or $μ \propto \frac{1}{λ}$

As $λ_{V} < λ_{Y} < λ_{R}$ As $μ_{V} > μ_{Y} > μ_{R}$

Speed of violet light inside the prism is slowest and that of red is highest.
Deviation of violet is maximum and that of red is minimum.

7.0The Rainbow

You can also see a rainbow on a sunny day when you look at the sky through a waterfall or through a water fountain, with the sun behind you.

Phenomena involved in the formation of a rainbow are refraction, dispersion and internal reflection.

8.0Atmospheric refraction

The refraction of light caused by the earth's atmosphere due to variable refractive index of air at different zones is called 'atmospheric refraction'.

Twinkling of stars

Apparent star position due to atmospheric refraction

Earth atmosphere Effect of atmospheric refraction at the sunrise or sunset Thus, sun is visible to us about 2 minutes before actual sunrise and about 2 minutes after actual sunset.

9.0Scattering of light

The scattering of light as it passes through a medium containing small particles is called 'Tyndall effect'. The colour of the scattered light depends on the size of the scattering particles. Very fine particles scatter shorter wavelengths (colours at violet end) by greater amount as compared to the light having longer wavelengths (colours at red end). If the size of particles is quite large, then, the scattered light is almost white.

Very small particles scatter light of shorter wavelength better than longer wavelength.
The scattering of longer wavelength of light increases as the size of the particles increases.
Larger particles scatter light of all wavelengths equally well.

Active Physics 4

Place a strong source ( S ) of white light at the focus of a converging lens ( $L_{1}$ ). This lens provides a parallel beam of light. Allow the light beam to pass through a transparent glass tank (T) containing clear water. Allow the beam of light to pass through a circular hole (C) made in a cardboard. Obtain a sharp image of the circular hole on a screen (M) using a second converging lens (L2), as shown in figure. Dissolve about 200 g of sodium thiosulphate (hypo) in about 2 L of clean water taken in the tank. Add about 1 to 2 mL of concentrated sulphuric acid to the water.
You will find fine microscopic sulphur particles precipitating in about 2 to 3 minutes. As the sulphur particles begin to form, you can observe the blue light from the three sides of the glass tank. This is due to scattering of short wavelengths by minute colloidal sulphur particles. Observe the colour of the transmitted light from the fourth side of the glass tank facing the circular hole. It is interesting to observe, at first, the orange red colour and then bright crimson red colour on the screen.
This activity demonstrates the scattering of light that helps you to understand the bluish colour of the sky and the reddish appearance of the Sun at the sunrise or the sunset. We conclude that very fine particles scatter mainly shorter wavelength light (blue light) and longer wavelength light (red light) gets transmitted.

10.0Concept Map

11.0Some Basic Terms

Diameter : The distance from the center to any point on the circle is called the radius of the circle. Two radii form the diameter.
Diaphragm : In optics, a diaphragm is a thin opaque structure with an opening at its centre, the role of diaphragm is to stop the passage of light, except for the light passing through the aperture.
Membrane : A membrane is a thin, flexible layer of tissue that covers, lines, separate or connect cells or parts of an organism.
Elasticity: It is the ability of a deformed material body to return to its original shape and size when the forces causing the deformation are removed.
Ligaments : It is a ring like fibrous membrane that connects the ciliary muscles and the lens of the eye, holding the lens in place.
Illumination : The quantity of light emitted by a lighting source.
Two dimensional : Two dimensional means having only two dimensional such as length and width. Two dimensional things are flat and have no depth. Examples : Geometrical shapes like squares, circles, polygons etc.
Three dimensional : Three dimensional shapes are solid that have three dimensions such as length, width and height. Examples : Cone-shaped ice cream, cubical box, ball, sphere etc.
Hazy vision : Hazy vision or cloudy vision means that you can't see things clearly.
Opaque: This means that no light is able to pass through them.
Elongation : It is the ability of a material to stretch when under strain.
Bifocal lens : A bifocal lens is an eyeglass lens with two different optical powers two correct vision at both long and short distances. The lens is divided into two segments with top of the lens is concave lens and the bottom of the lens is convex lens.
Spectrum : Spectrum is a band of colours that is produced when white light passes through a prism and splits into its constituent colour.
Deviation : A light that slightly diverges from its path because a change in the medium is known as a deviation of light.
Colloidal solution : The colloidal solution can be define as a mixture of particles of substances. These particles are microscopically dispersed and soluble/insoluble which are suspended in a fluid regularly.
Colloidal particles : Colloidal particles are small solid particles that are suspended in fluid.
Molecules: It is a group of two or more atoms that are held together by a chemical bond.
Monochromatic light : Light that has a single wavelength or colour.
Choroid : A molecule is the smallest unit of matter that has distinct chemical and physical properties.
Range of vision : Range of vision for a normal human eye is the range of distance for which human eye can see an object clearly. It ranges from infinity to $25 cm$ .

12.0SOLVED EXAMPLES

The near point of a patient's eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 25.0 cm away? Neglect the eye-lens distance. (b) What is the power of this lens? Solution: Here, $u = - 25 cm$ ; $v = - 50 cm$ ; $f =$ ? (a) By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{f} = \frac{1}{( - 50 )} - \frac{1}{( - 25 )}$ or $\frac{1}{f} = \frac{- 1}{50} + \frac{1}{25}$ or $\frac{1}{f} = \frac{- 1 + 2}{50} = \frac{1}{50}$ or $f = + 50 cm = + 0.5 m$ Alter: Here, $y = 50 cm = 0.5 m$ $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{y}$ or $\frac{1}{f} = \frac{1}{0.25} - \frac{1}{0.5}$ or $\frac{1}{f} = \frac{100}{25} - \frac{10}{5} = 4 - 2 = + 2$ or $f = + (1/2) = + 0.5 m$ (b) Power, $P = \frac{1}{f}$ or $= + 2$ Dioptre
Suppose a lens is placed in a device that determines its power as +2.75 diopters. Find (a) the focal length of the lens and (b) the minimum distance at which a patient will be able to focus on an object if the patient's near point is 60.0 cm . Neglect the eye-lens distance. Solution: Here, $P = + 2.75$ dioptre ; $f =$ ? (a) $f = \frac{1}{P} = \frac{1}{+ 2.75} = + 0.364 m$ $= + 36.4 cm$ (b) Here, $v = - 60 cm; u =$ ? By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{( - 60 )} - \frac{1}{u} = \frac{1}{+ 36.4}$ or $\frac{1}{u} = \frac{- 1}{60} - \frac{1}{36.4}$ or $\frac{1}{u} = \frac{- 1}{60} - \frac{1}{36.4} = \frac{- 36.4 - 60}{2184} = \frac{- 96.4}{2184}$ or $u = \frac{2184}{- 96.4} = - 22.6 cm$
A particular near-sighted patient can't see objects clearly when they are beyond 25 cm (the far point of the eye). (a) What focal length should the prescribed contact lens have to correct this problem? (b) Find the power of the lens, in diopters. Neglect the distance between the eye and the corrective lens. Solution: Here, $u = - \infty; v = - 25 cm; f =$ ? (a) By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{f} = \frac{1}{( - 25 )} - \frac{1}{( - \infty )} = \frac{- 1}{25}$ or $f = - 25 cm = - 0.25 m$ Alter : Here, $x = 25 cm$ $f = - x = - 25 cm = - 0.25 m$ (b) Power, $P = \frac{1}{f}$ or $P = \frac{1}{- 0.25} = - 4$ Dioptre
A retired bank officer can easily read the fine print of the financial page when the newspaper is held no closer than arm's length, 60.0 cm from the eye. What should be the focal length of an eyeglass lens that will allow her to read at the more comfortable distance of 24.0 cm ? What is the power of this lens? Solution: Here, $v = - 60 cm; u = - 24 cm; f =$ ? By lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{f} = \frac{1}{( - 60 )} - \frac{1}{( - 24 )}$ or $\frac{1}{f} = \frac{- 1}{60} + \frac{1}{24}$ or $\frac{1}{f} = \frac{- 2 + 5}{120} = \frac{3}{120} = \frac{1}{40}$ or $f = + 40 cm = + 0.4 m$ Power, $P = \frac{1}{f}$ or $P = \frac{1}{+ 0.4}$ = + 2.5 Dioptre
A boy uses spectacles of focal length -60 cm . Name the defect of vision he is suffering from. Which lens is used for the correction of this defect? Compute the power of this lens. Solution: Given, $f = - 60 cm$ $= - 0.60 m$ The negative sign of focal length indicates that the lens is concave. Thus the boy suffers from myopia (short sightedness) vision. Power of the lens $= \frac{1}{f} = \frac{1}{( - 0.60 )}$ $= - 1.67 D$

Related Article:-

Algebraic Expressions

Data Handling

The Triangles and its Properties

Visualising Solid Shapes

Fractions

Perimeter and Area

Related Article:-

Algebraic Expressions

Data Handling

The Triangles and its Properties

Visualising Solid Shapes

Fractions

Perimeter and Area

Join ALLEN!

The Human Eye and the Colourful World

1.0The human eye

Components and structure

Working of eye

Role of iris

2.0Active 2 Physics

3.0Power of accommodation

Near point

Least distance of distinct vision

Far point

Colour blindness

Cataract

4.0Defects of vision and their correction

Myopia (Near sightedness)

Hypermetropia (far-sightedness)

Calculations for corrective lens for myopia

Calculations for corrective lens for hypermetropia

Presbyopia

5.0Refraction of light through a prism

Prism (triangular prism)

6.0Dispersion of white light by a prism

7.0The Rainbow

8.0Atmospheric refraction

Twinkling of stars

9.0Scattering of light

10.0Concept Map

11.0Some Basic Terms

12.0SOLVED EXAMPLES

On this page

Join ALLEN!

The Human Eye and the Colourful World

1.0The human eye

Components and structure

Working of eye

Role of iris

2.0Active 2 Physics

3.0Power of accommodation

Near point

Least distance of distinct vision

Far point

Colour blindness

Cataract

4.0Defects of vision and their correction

Myopia (Near sightedness)

Hypermetropia (far-sightedness)

Calculations for corrective lens for myopia

Calculations for corrective lens for hypermetropia

Presbyopia

5.0Refraction of light through a prism

Prism (triangular prism)

6.0Dispersion of white light by a prism

7.0The Rainbow

8.0Atmospheric refraction

Twinkling of stars

9.0Scattering of light

10.0Concept Map

11.0Some Basic Terms

12.0SOLVED EXAMPLES

On this page