Class 10TH Physics Notes Chapter 12 GEOMETRICAL OPTICS

Class 10TH Physics Notes Chapter 12 GEOMETRICAL OPTICS

Class 10TH Physics Notes Chapter 12 GEOMETRICAL OPTICS. Geometric optics, or ray optics, is a model of optics that describes the propagation of light in terms of rays. The ray in geometric optics is a useful abstraction to approximate the paths through which light propagates under certain circumstances.Class 10TH Physics Notes Chapter 12 GEOMETRICAL OPTICS.

Class 10TH Physics Notes Chapter 12 GEOMETRICAL OPTICS. Geometric optics do not take into account certain optical effects such as diffraction and interference. This simplification is useful in practice; it is an excellent approximation when the wavelength is small compared to the size of the structures with which the light interacts. The techniques are particularly useful for describing geometric aspects of imaging, including optical aberrations.Class 10TH Physics Notes Chapter 12 GEOMETRICAL OPTICS.

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Chapter No.12

Major Concepts:

• Reflection of light

• Spherical mirrors and their types

• Important terms related to mirrors

• Sign conventions

• Refraction of light

• Total internal reflection of light

• Refraction through prism

• Ray diagram of lenses

• Application of lenses

• Totally reflecting prism and optical fiber

• Simple microscope

• Compound microscope

• Telescope

• The human eye

• Accommodation. Near point and far point

• Defects of vision (myopia and Hypermetropia.lt)

• Solution of problems

1. What is meant by reflection of light? State and explain laws of Incomprehensive Question reflection with diagram?

Ans. Reflection of light:

The bouncing back of light rays from the surface of a certain medium is called reflection of light.

Explanation:

Consider a place surface “MM- as shown in the figure.

Point of incidence

///// P////////

‘NS s-+ CM

N

Fig: Reflection of light

Let a ray of light coming from the point “C” is allowed to fall on point “P” of the plane surface. This ray “CP” is known as incidence ray. The angle made by the incidence ray is known as angle of incidence (zi).During this process some light rays reflect from the surface at point “P” which is known as ref’ acted ny The angle made by the reflected ray with normal is known as angle of reflection (4 r).

Now from the figure it is clear that; the angle of incidence is equal to the angle of reflection.

• The incident ray, reflected ray, and the normal lies in the same plane.

Law of reflection of light:

History:

The laws of reflection of light were first discovered by a Muslim scientist named Ibn-al-Haitham. There are two laws of reflection which are given below.

a) 1St Law of reflection light:

This law states that “The angle of incidence “ci’ is equal to the angle of reflection “41-” i.e., 4i = LI’

b) 2nd Law of Reflection:

This law states that, The incident ray, reflected ray and normal at the point of incidence all lies in the same plane.”

Q2: What is mirror? Describe types of spherical mirrors?

Mirror:

The smooth and polished surface of a medium which reflect rays of light is called mirror.

Spherical mirror:

A portion of spherical shell (or) sphere whose inner (or) outer surface is shining and deflectable to light is known as spherical mirror. Type of spherical mirror:

i) Concave mirror: A spherical mirror whose inner surface is shining and reflectable to light is known as concave mirror. Silver Coating Protected by Paint

ii) Convex mirror:

A spherical mirror whose outer surface is shining and reflectable to light is known as convex mirror.

Silver Coating Protected by Paint Convex mirror

3. Define the terms related to spherical mirrors?

Ans. Terms related to spherical mirror:

i. Center of curvature:

The Center of sphere of which the mirror is a part is known as center of curvature. It is denoted by “C”.

The radius of sphere of which the mirror is a part is called radius of curvature (R).

iii. Pole of Mirror:

The geometrical Central point of a spherical mirror is called pole. It is denoted by “P”.

iv. Aperture of the spherical mirror:

The diameter of the circular boundary of spherical mirror is called aperture. (OR)

The size of spherical mirror is known as aperture.

V. Principal Axis:

The straight line passing through pole and center of curvature of a spherical mirror is known as principal axis.”

Focal length:

The straight distance between pole and principal focus of a spherical mirror is called focal length. It is denoted by “f”.

Principal Focus (Or) Focal point:

– A point on the principal axis of the spherical mirror at which parallel rays of light after reflection converge (in case of concave mirror) or appear to diverge(in case of convex mirror) is called principal focus.

Q4. Write characteristics of focus of concave and convex mirror.

Ans. Characteristics of concave and convex mirror:

Q 5: Discuss reflection of light rays by spherical mirrors?

Ans: Reflection of light rays from spherical mirror:

The inner (or) outer surface of the mirror is smooth and polished, so the spherical mirror has the ability to reflect light rays. Some special types of reflection of light from the mirror are given below.

Cassel:

If rays of light are parallel to the principal axis. Then after reflection from the spherical mirror these rays of light will passes through the principal focus as shown in figure.

Case (2):

If the incident rays of light passes through the principal focus. Then such rays will become parallel to principal axis after reflection from spherical mirror as shown in fig.

Case (3):

If the incident rays of light passes through the center of curvature, Then such ray of light will reflect back along the same path as shown in figure.

Case 4:

If the incident ray falls on the pole by making an angle “9”with the principal axis. Then such ray of light will make the same angle “0” with principal axis after reflection as shown in figure.

Q6. Discuss the formation of image by concave mirror. Support your answer by drawing figures?

Ans. Image formation by concave mirror:

The position, size and nature of the image of an object placed in front of a concave mirror depend on the location of object with respect to mirror.

1. Object placed beyond “C”:

If an object is placed beyond the center of curvature in front of concave mirror. The image is formed between Centre of curvature and principal focus. The image is real. Inverted and diminished.

1. Object is placed “C”:

If an object is placed on the Centre of curvature. The real, inverted and same size image is formed on the Centre ofcurvature”C”.

1. Object placed between “C”and “F”:

If the object is placed between Centre of curvature and focal point in front of concave mirror. The image is formed beyond “C”. This image is real inverted and magnified.

iv) Object placed at “F”:

If an object is placed at principal focus “F” in front of concave mirror, the real inverted and highly magnified image is formed at infinity.

v) Object placed between “F”and “P”:

If the object is placed in front of a concave mirror between the principal focus “F” and pole “P” the virtual, erect and magnified image is formed ‘behind the mirror.

Q 7: Discuss the formation of image by convex mirror?

Ans: ln case of a convex mirror, for all distances of the object, the image is formed behind the mirror between principal focus “F” and pole “P”. It -is always diminished and erect. Comparing equ (a) and equ (b), we have.

do do— f

di f

By crossing multiplication, we get;

Dof= di (do-f)

Dof=dido-dif

Dividing each term by “dodif’, we get;

Do f dodi dido

Dodif dodif dodif

11 1

— = –

F do di

=

F do di

This is known as spherical mirror formula. It gives the relationship between the image distance “di” object distance “d,” and focal length “f’ of a spherical mirror.

Q10 What is meant by linear magnification?

Ans. Linear magnification:

“The ratio of size of image to the size of an object is known as linear magnification.”

Mathematically;

tm_—

Hi

Ho

Di

Also, M-

-hi = —di

Ho d,

Hi

How does

X ho

Q11 Write sign convention for spherical mirror?

Ans. Sign convention:

The following sign convention is adopted to obtain the correct result in a given problem of image formation by a spherical mirror.

i. All distances are measured from the pole of the mirror.

Ii.Distance of real objects and real images are taken positive.

iii. Distances of virtual objects and virtual images are taken negative.

iv. The focal length of Concave mirror is taken positive and that of a convex mirror is taken negative.

v. For upright image magnification is +ve while magnification is —ve for vertically inverted-image.

Q12. Write uses of spherical mirrors?

Ans. Uses of spherical Mirrors:

There are several important applications of spherical mirrors. A few of them are given below.

i. Convex mirrors are used in vehicles to observe rear view.

ii. Convex mirrors are also used at dangerous road bends for safe drive.

iii. Concave mirrors are used by doctors to examine ear, nose, throat and eyes.

iv. Concave mirrors are used in head—lights of automobiles and search lights.

v. Concave dish is used to collect microwave signals from satellites to a focus.

vi. Concave mirrors are used to concentrate light on the slides of microscope so that the slide can be viewed more clearly.

Q13: “What is meant by refraction of light? State and explain laws of refraction with diagrams?

Ans. Refraction of light:

“The change in speed and direction of light when light enter from one medium to another is called refraction of light.” OR

The phenomena in which change in speed as well as direction of light rays occurs. When light rays enter obliquely from one medium to another medium is known as refraction of light.”

Explanation:

As we know that in a homogeneous transparent medium (having the same density throughout) light travels in a straight line with uniform velocity but when a ray of light enter from one medium to another medium, it undergoes a change in speed as well as in direction.

This change of speed and direction of light rays when enters from one medium to another medium is known as refraction of light.

From rarer to denser medium:

Let’s a ray of Fight -CO” be incident at point Apr “0” of the surface “PC/ .Upon entering the second medium, the ray changes its path. Instead of traveling along “OD”, the ray is refracted and travel along “OE”. The ray “OE” is called refracted ray. Draw a normal “NONE”, the angle LCON=LI is called angle of incidence and angle LEON = z_r is called the angle of refraction. The refracted ray bends towards normal i.e. the angle of refraction is smaller than the angle of incidence.

From denser to rarer medium:

If the light ray travels from denser medium to rarer medium, then the ray is bent Air away from normal,

i.e., the angle of refraction is greater than the angle of incidence as shown in figure.

Law of Refraction:

The refraction of light obeys the following two laws.

i) First law:

This law states that “The incident ray, refracted ray and normal at the point of incident, all lies in the same plane”.

ii) Second Law:

This law states that; “The ratio of sine of angle of incidence to the sine of angle of refraction is constant for a given pair of media.” Mathematically;

Sin LI

Sin LT.

When “n” is constant and is known as refractive index. This law is also called Snell’s law)

Q14. What is mean by index of refraction (refractive index)?

Ans. Refractive index:

“The ratio of sine of angle of incidence to the sine of angle of refraction is known as refractive index.”

SinL!

n=

Sinn-

“OR”

“The ratio of speed of light in vacuum to the speed of light in the given medium is called refractive index.” speed of light in vacuum

n= speed of light in a given medium C

n =

n=

Apparent Depth The refractive index of a substance depends on the nature of the material of the medium and on the wavelength of light used. It is a characteristic property of a substance

Q 15. What is meant by critical angle? Also describe total internal reflection. Derive relation of critical angle and refractor index?

Ans. Critical Angle;

Definition:

The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°, is called critical angle”.

It is denoted by “C”.

Examples:

Critical angle for glass = 42°

Critical angle for diamond = 24.4°

Critical angle for water = 48° etc.

Total internal reflection:

Definition:

When the angle of incidence in the denser medium becomes greater than the critical angle “C”, then there is no refraction, only reflection phenomena take place, such a reflection is known as total internal reflection.

Explanation:

Whenever light enter from an optically denser medium to a rarer medium it bends away from normal. Therefore the angle of refraction “cr” is greater than the angle of incidence “a”. If we increase the angle of incidence gradually, then the angle of refraction”Lr” also increases until for a certain angle of incidence, the angle of refraction become “90°”is known as critical angle. When the angle of incidence in the denser medium becomes greater than the critical angle “C”, there is no refracted beam. All the rays are internally reflected back in the optically denser medium. Such a reflection is called total internal reflection.

Necessary conditions for total internal reflection:

The necessary conditions for total internal reflection are as follows;

a) The incident light must be entering from denser medium to a rarer medium.

b) The angle of incidence “Lr” in denser medium must be greater than critical angle i.e.Li > C. Relationship between refractive we know that; index and critical Angle:

Of angle in denser medium Of angle in rare medium

‘Rare 1 and ndenser = n)

(…Sin 90°=1), township between refractive index and tribe how total internal reflection is rough fiber?

Nrare sine

Denser sln

1 Strict

71 .517190°

Sitt90°

n=

Sin C

1.

n=

sin c Which is the required real critical angle.

Q16.What is optical fiber? Des used in light propagating?

Ans.Optical Fiber:

“A thin hair like rod of glass index called core, which is surr refractive index called cladding.” ( “A thin hair like glass rod I internal reflection.”

Explanation:

Total internal reflection is used in fibre optics which has number of advantages in telecommunication field. Fibre optics consists of hair size threads of glass or plastic through which light can be travelled (see in fig). The inner part of the fibre optics is called core that carries the light and an outer concentric shell is called cladding. The core is made from glass or plastic or reflectively high index of refraction. The cladding is made of glass of plastic, but of relatively low refractive index. Light entering from one end of the core strikes the core — cladding boundary at an angle of incidence greater than critical angle and is reflected back in to the core (see in fig): In this way light travels many kilometers with small loss, of energy. In Pakistan, optical fibre is being used in telephone and advanced telecommunication systems. Now we can listen thousands of phone calls without any disturbance.

Air

Core n=1.53

017. Write the application of optical fibers?

Ans: The light pipe is a bundle of thousands optical fibers bounded together through a flexible pipe jacket. The light pipes are used by doctors and engineers to illuminate those inaccessible places which otherwise are not possible to examine. Light pipes also used to transmit data (image, calls, Movies) from one place to another.

Fibre bundle

Projected image 1–

Lens

Transmitted image

Endoscope:

Endoscope is a medical instrument used for viewing and photographing of internal structure of human body. Due to small size the endoscope is inserted through the mouth and illuminates the internal parts viewed by another bundle of optical fiber. A video camera is fitted outside of this bundle and it makes the interior parts visible to doctors which help them to diagnose easily.

Q18. What is endoscopy? Describe the type of endoscope.

Ans. Endoscope:

A medical procedure in which any type of endoscope is used called endoscopy.

Type of Endoscope:

The endoscope is divided in to different types on the bases of use in medical filed.

i. Gastroscopy:

Gastroscopy is an instrument used to diagnose the stomach.

ii. Cystoscopy:

Cystoscopy is an instrument used to diagnose the bladder.

iii. Bronchoscope:

Bronchoscope is an instrument used to diagnose the bladder and sore throat.

Q19. Describe the behavior of a ray of light after passing through the prism?

Ans.A transparent body having three rectangular and two triangular surfaces is called prism. The angle between two refracting rectangular surfaces opposite to the base is called the angle of prism (LA). The refraction of light through a glass prism is as shown in the figure. The incident ray “EF” makes an angle LEFN = Li with surface “AB”. The incident ray “EF” on entering the glass prism is refracted toward the base “BC” of the prism. When this refracted ray “FG” emerges out of the prism it bends away from the normal.

The inclination of the surface “AC” with “AB” causes the emergent ray “GH” to bend more towards the base as shown in figure. Thus the emergent ray “GH” has been as shown in figure. Thus the emergent ray “GH” has been deviated from its original path EFI. The angle through which the emergent ray has been deviated is called the angle of deviation “D” (L 10H). The value of angle of deviation depends upon the following factors. . + Angle “A” of the prism. The refractive index of the material of the prism. <• The angle of incidence. Experimentally it is verified that when the refracted ray become parallel with the base “BC” of the prism then the angle of deviation become minimum and is called the angle of minimum deviation “Dm”. The refractive index of the material of the prism can be determined by the relation.

Sin (“D n1)

n = A

Sin ()

Q20. Define lens? Describe types of lenses?

Ans. Lens:

A portion of transparent medium like glass which is bounded by at least one curved surface is called lens.

Types of lens:

There are two types of lens which are as follows.

a) Convex lens (or) converging lens:

“A lens which is thicker at the middle and thinner at the edges is called converging lens (or) convex lens.” (OR) A lens which converge parallel rays of light to a single point is called converging lens (or) converging lens.

b) Concave lens (or) Diverging lens:

“A lens which diverge parallel rays of light is called diverging lens (or) concave lens.”

Sub types of convex lens:

There are three types of convex lens which are given below.

i) Double convex lens:

“A convex lens whose both bounding surfaces are convex is called double convex lens.”

ii) Plano convex lens:

“A Convex lens in which one bounding surface is plane and the other is convex is called piano-convex lens.”

iii) Concave —convex Lens:

“A convex lens whose one bounding surface is concave and the other is convex is called concave convex lens.”

Double Convex Plano Convex Concave Convex lens (a) lens (b) lens (c)

Sub types of concave Lens:

There are three types of concave lens which are given below.

i) Double Concave lens:

“A concave lens whose both bounding surfaces are concave is called double concave lens.-

ii) Plano – concave lens:

“A concave lens whose one bounding surfaces is plane and other is concave is called piano— concave lens.”

iii) Convex —concave lens:

-A concave lens whose one bounding surface is convex and other is concave is called Convexo-concave lens.-

Plano-Concave Plano-Convex Convexo-Concave

Inns lens lens

Q 21. Define special terms related with lenses?

Ans. Definitions of special terms:

i) Centre of Curvature:

The Centre of sphere of which the lens is a part is called Centre of curvature. Denoted by “C.’

ii) Principal Axis:-

The straight line joining optical center with center of curvature of lens is known as principal axis.

iv. Optical center:

A single point inside or outside the lens, through which a ray of light passes un-deviated is called optical center. “OR” The geometrical Centre of a lens is called optical Centre.

iv) Aperture:

The geometrical diameter of lens is called aperture. “OR” The size of lens is called aperture”.

v) Principal Focus:

The point where all rays parallel to the principal axis converge after passing through a convex lens (or) appear to diverge after passing through a concave lens-, is called principal focus, denoted by “F”. Concave lenses. Focal Length Focal Length

The radius of sphere of which the lens is a part is called radius of Curvature, denoted by “R”.

vii) Focal length:

The straight distance between optical Centre and principal focus is called focal length. It is denoted by T.

Q 22: Discuss the refraction of light rays by lens?

Ans: Refraction of special Rays:

Ray (1): An incident ray of light passes through optical Centre “0” of lens without bending as show is figure.

Ray (2): An incident ray passes through “F” is refracted parallel to the principal axis as shown in figure.

Ray (3): A ray parallel to principal axis after refraction from a convex lens passes through the principal focus.

In case of a concave lens this ray diverges and appears to be coming from the principal focus “F”.

Q23. Discuss the formation of image by Concave and convex lenses?

Ans. Image formation by Convex Lens:

The position, nature and size of the image of an object that is formed by a convex lens depend on the distance of the object from the lens.

a) Object placed beyond “2F” OR “C”:’

When an object is placed beyond “2F” (Or) “C”in front of a convex lens, the real, inverted and diminished image is formed between F and 2F as shown in figure.

b) Object placed at “2F”:

When an object is placed at “2F” in front of a convex lens. The real. Inverted and same size image is formed at “2F” as shown in fig.

c) Object between “2F” and “F”:

When an object is placed between “2F” and “F”, the real, inverted and magnified image is formed beyond “2F” as shown in fig.

d) Object placed at “F”:

When an object is placed at principal focus “F” of convex lens, the real, inverted and highly magnified image is formed at infinity as shown in figure.

e) Object placed “F” and “O”:

When an object is placed between “F” and “0” the virtual, erect and magnified image is formed on the same side of the lens at which the object is placed as shown in fig.

Image formation by concave Lens:

In case of concave lens, for all positions of the object from concave lens, the virtual, erect and diminished image is formed between “F” and “0” as shown in figure.

Q24. Describe lens formula equation?

Ans. An equation which shows the relationship between focal length of. lens “f’, distance of object from the lens “P” and distance of image from the lens “q” is balled lens formula.

Mathematically,

I = f do DI

Placed within the focal length of a convex lens, the virtual ,erect and magnified image is formed on the same side of object. The image formation of an object by simple microscope is as shown in figure.

Magnifying power (or) Magnification:

The ratio of angle of image seen through a magnifying glass to the angle made by the object when placed at 25cm from an unaided eye is called magnifying power of magnifying glass.

Mathematically;

Nti

cc tanB

M =tan«

A-E’

M= ANB

N

AB

TanB = —N)

And tam (= —AB

M =— ——- (a)

AB

From the figure we see that AABC andaA C are similar, therefore,

AA-SW

_di do A-Er N

—AB =—do di=N= 25 cm)

Putting equation (a)

M = N — (b)

According to lens formula;

=

1 1

f do di

1 1

7 = —do — —N (. d, =-Ni.e, image is virtual)

t .1

— = — — —

F do N

Multiplying each term by “N”

M =1+—f

For normal human eye, N=25cm

Therefore, M=1+-25

Hence, shorter the focal length of convex lens higher will be magnifying power of lens and vice versa.

Q28. What is a compound microscope? Using ray diagram for a compound microscope mention its magnifying power?

Ans. Compound microscope:

Definition;

An optical instrument which helps us to see the magnified images of very small object is called compound microscope.

Construction:

A compound microscope consists of two convex lens which are fitted at each end of two metal tubes whose length is adjustable. The lens near object is called objective and the lens near eye is called eye piece the focal length of objective is less than the focal length eye piece i.e. f„fo

Working:

An object “AB” is placed at a point whose distance “P” is slightly greater than the focal length fc, of objective. A real, inverted and magnified image A”B” of the object is formed after refraction from the objective at distance “r” from objective. The eye piece I adjusted is such a way that the image A’B’ should fall inside its focal length.

The eye piece acts like a magnifying glass and so the final image A”B”of A’B’ is formed at a distance equal to the least distance of distinct vision “d”.

Magnifying power:

The ratio of size of final image seen through compound microscope to the ratio of size of object seen without compound microscope is called magnification (or) magnifying power of compound microscope.

M_stze of final image seen through compound microscope size of object seen without compound microscope

Objective Eyepiece

A’

Pr. D

Astronomical, telescope

NA_ AT- _

AB AB/A’B’

AT- AT’ IV1=— X — AB AB

M=Me x M, (a)

Now RA di

Me =1 + —

Do f

Di

Therefore; M=— (1+—)

FO

As the object “AB” is situated near the principal focus of the objective so; do .f, and ct, =L=the length of microscope

Thus, M=—L (1+N—)

FO FO

This the magnifying power of compound microscope eye.

Q29. What is a telescope? Using a ray diagram explain it’s working and magnifying power?

Ans. Astronomical telescope:

An optical instrument which is used to see heavenly bodies and other distant object clearly is called astronomical telescope.

Contraction:

The astronomical telescope consists of two convex lenses which are fitted at each end of two metals tubes whose length is adjustable. The lens toward the object is called objective and the lens near the eye is called eye piece. The focal length of eye piece is less than the focal length of objective i.e. f„ fo

Working:

The objective form a real, inverted and diminish image KB’ of distant object “AB” . This image “KB- behaves as an object for the eye piece which is lying within focal length of eye piece. The eye piece produces virtual, inverted w.r.t the object and magnified image A”B”.

When the telescope is f6cused at infinity (or) in normal adjustment then the distance between objective and the eye piece is equal to fo + fe i.e. L= fo + fe .

Magnifying power:

The magnifying power of astronomical telescope is given by;

Focal length of objective fo

Magnification M—

Focal length of eye piece =fe

Q30. Define the defect of vision. What is meant by the short sightedness and long sightedness? How these defects can be corrected.

Ans. Defect of vision (sightedness):

The inability of the eye lens to produce sharp images at the retina is called defect of vision, the main defects of vision are as follows.

a. Short sightedness (Myopia)

b. Long sightedness (Hypermetropia.lt)

a) Short sightedness (myopia):

“The defect of eye lens in which a person can see near objects clearly but cannot see distant objects clearly is called shortsightedness

The lens of such an eye is more convergent than that of the normal eye. Due to greater converging power of lens the eye ball becomes too long for the lens system suffering myopia. The parallel rays from the object are focused just in front of the retina as shown is figure Short Sight Incident light.

Removal:

This defect of eye can be removed by using spectacles containing concave lenses.

b) Long sightedness:

“The defect of eye lens in which a person can see distant objects clearly but cannot see near object clearly is called long sightedness.

The lens of such eye is less convergent than that of normal eye. The eye ball becomes too small for the lens system suffering long sightedness. Due to low converging power of lens the parallel rays from the object are focused beyond retina as shown in figure.

Removal:

To remove this defect a convex lens is placed in front of the eye in the form of spectacles. The combination of convex lenses forms the image at retina as shown in figure. Correction Using Convex Lens.

Q.31. How human eye works? How the defects in the eye like short sightedness and long sightedness is corrected by using lenses?

Ans. Human Eye:

In many respects the Human eye is similar to Camera. It has white wall called sclerotic in the front of eye there is a lens

Parts of human eye:

Retina:

The light passing through the lens cross the vitreous humor to form an image on specific wall which is called retina, Retina records the picture and works as film in camera.

Lens:

There is converging lens in the front of eye. Naturally. this lens has ability to change the focal length.

Iris:

The iris is the Coloured part of eye which automatically adjust the size of pupil and controls the amount of light reaching the retina.

Pupil:

Pupil is circular in the center according to intensity of light falling on it. When white light passes, the iris contracts the size of pupil while in dim light pupil is enlarged.

Cornea:

The light enters the eye through transparent membrane called Cornea.

“IMac/a \ – X

Bite

HOMO

Lent

Flurel

Formation of image:

The image formation in human eye is shown in fig. Human eye acts like a camera. In place of the film, the retina records the picture. The eye has a refracting system containing a converging lens. The lens forms an image on the retina which is a light sensitive layer at the back of the eye, the lens changes focal length. Light enters the eye through a transparent membrane called is the cornea. The iris is the coloured portion of the eye and controls the amount of light reaching the retina. It has an opening at its Centre called the pupil. The iris controls the size of the pupil. In bright light, iris contracts the size of the pupil while in dim light pupil enlarged. The lens of the eye is flexible and accommodates objects over a wide range of distances.

Accommodation:

The variation of the focal length of eye lens is called accommodation. OR “The sharpness of image formed in retina for objects at different distances from the eye is controlled by an alternation an alternation in the focal length of eye lens. This is called accommodation.”

For Object:

If the object is far away from eye,

• The biliary muscles relax

• The curvature of lens is decreased.

• The focal length decreased.

• The divergent rays from mean object are thus bent more and come to focus on retina. ‘

Nearer Object:

If an object close to eye,

• The ciliary muscles contracts.

• The curvature of lens increased.

• The focal length decreased.

• The divergent rays from near object are thus bent more and come to focus on retina.

Accommodation is larger of healthy eye in young people while it goes on decreasing with increasing the age. Defects in accommodation be corrected by using different types of lens in eyeglasses.

Example12.1:

The concave side of a spoon has a focal length of 5.00cm Find the image distance for this ‘mirror’ when the object distance is (a) 12.0 cm, (b)10.0cm, (c)7.50cm,(d)5.00cm and (e) 2.00cm..

Given:

Focal length ‘f = 5.00cm.

(a) Object Distance ‘do = 12.0cm,

(b) Object Distance ‘do’ = 12:0cm,

(c) Object Distance ‘do’ = 12.0cm,

(d) Object Distance’s,’ = 12.0cm,

(e) Object Distance ‘do = 12.0cm, Required:

(a) Image Distance =?

(b) Image Distance =?

(c) Image Distance ‘di’ =?

(d) Image Distance’s,’ =?

(e) Image Distance’s,’ =? Solution:

By mirror equation

1 1, 1 1 1 1

— = — + —or— —

F do di di f does

Taking LCM

Do— f f do

— = order=—– (1)

Di Rio does—fa

Putting the values in equation1

F does

d =

5.00cm x 12.00 cm

12.00cm-5.00cm 60.0cM2

Di=

7.00cyn

Therefore,

Di= 10.00cm-5.00cm

50.0cm2

D,

Therefore,

Di= 8.6cm

b) Putting value in equation 1

!do

5.00 arts

2.00cm-5.00cm 10,0t-m2

Di-

-3.009K

Therefore,

c) Putting value in equation 1

Di= FD°

Do-to

5.00cm x 7.50cm

Di-

– 7.50cm-5.00cm 37.5erft2

Di-

2.50p? F

Therefore,

Cli= 15.0cm

d) Putting value in equation 1

Di= f d°

Do- FO

5.00cm x 5.00cm

Di

– 5.00cm-5.00cm 2 5 .0.ern 2

Di-

Therefore,

Di- 00

e) Putting value in equation 1

Di- FD°

Do- h,

5.00cm x 2.00cm

Therefore,

Example 12.2:

An external rearview car mirror is convex with a radius of curvature of16.0m (fig). Determine (a) the focal length of the mirror, (b) location of the image and (c) its magnification for an object 10.0m from the mirror.

Given:

Object distance 1(101= 10m Radius of curvature =R=16.0m

Required:

(a)focal length 1=?

(b) Image distance ‘clo’=?

(c) Magnification ‘m’=?

Solution:

(a) Focal length 11=? We know that:

Putting values

Therefore,

(b) By. Mirror equation

Or

Taking LCM

f= R

2

– 1 16m

2

f= -8.0m

1= 1

F d does

1

= – — —

D, f d0

I do-f

D, FD,

d. = FD°

1 do-f,

Putting the values

8- .0m x10.0m

Di- 10m-(-8.0m)

ao.om2

18.0m

Therefore, cli= – 4.4m

Putting values d,

m= —

Do

– 4.4m

Life

Example 12.3:

In water having index of refraction as 1.33?

Given:

Speed of light in vacuum ‘C’= 3.00x108m/s Index of refraction for water ‘n’=1.33

Required:

Speed of light in water ‘y’=?

Solution:

We know that;

n =—V or v = —Cn 3.00 x108m/s

Putting values V-

1.33

Conclusion:

Therefore, the speed of light in water is 2.2610°m/5

Example 12.4:

Light travels from crown glass (n9= 1.52) in to water (nw=1.33).The angle of incidence in crown glass is 40.0°. What is the angle of refraction in water?

Given:

Index of refraction for crown glass ‘n9’=1.52

Index of refraction for water ‘n,’=1.33

Angle of incidence 09=40.0°

Required:

Angle of refraction/V-9

Solution:

By Snell’s law

ni sin el = n2 sin 02

Or ng sin By = n, sin Ow

sin Ow = 7-19-

nw

sin Ow = —1113532 sin40°

Or sin Ow = 0.735 and Ow. sin-1 0.735

Example 12.5:

Find the critical angle for light traveling from glass (n=1.502) to (a) air (n=1.002) and (b) water (n=1.332)

Given:

Index of refraction for ordinary glass ‘ng’=1.502

Index of refraction for air na=1.002

Index of refraction for water nw=1.002

Required:

Critical Angle Q=?

Solution:

The critical angle is

sin 0, = –‘Ora,

n,

(a) When light goes from glass to air I

0, =sin-viL

nn ni

he critical angle

Putting the value,

Oci = sin_11.00

1.50

Oc =41.8°

(b)When light goes from glass to water the critical angle is

0, =sin-1741

ns

Putting the value,

0, =

1.50

O =62.5°

Example12.6:

A converging lens of focal length 10.0cm forms images of an object situated at various distances,.(a) if the object is placed 30.0cm from the lens, locate the image, state whether it’s real or virtual, and find its magnification. Repeat the problem (b) when the object is at 20.0 cm and (c) when the object is 15.0cm from the lens (d) when the object is 10.0cm from the lens and (e) when the objects 5.00cm from lens

Given:

Focal length T=10.00cm.

(a) Object Distance td,;=30.0cm,

(b) Object Distance ‘clo1=20.0cm,

(c) Object Distance’s, =15.0cm,

(d) Object Distance Idoi=10.00cm,

(e) Object Distance’do1=5.00cm,

Required:

(a) Image Distance’s,’=?

Magnification m=?

(b) Image Distance’s,’=?

Magnification m=?

(c) Image Distance’s,’=?

Magnification m=?

(d) Image Distance ‘d1’=?

Magnification m=?

(e) Image Distance ‘d11=? Magnification m=?

1= 1 I

f di do

1 1 -1

Or — = – — —

d, f do

Taking LCM

1 do- f

— =

di fd0

rdo

(1)

dc,-/

1. To get image distance putting values in equation

Putting value

10.0cm x30.0cm d,= 30.0cm-10.0cm

3 00 pre

d=

213.efli

The magnification formula is

di

m=—

do

Therefore 15cm

m –

30cm

m= -0.500

(b) To get image distance putting values in equation 1

10.0cm x 20.0cm- 10.0cm 200enfr

The magnification formula is

et;20cm

m=— m-

do 20cm

Therefore, -1.00

(c) To get image distance putting values in equation 1

Moon x15.0crn

15.0cm-10.0cm 150c2/12-

d, –

501

Therefore = d,=30 cm

The magnification formula is

di

m=— =

do 15cm

Therefore m= -2.00

Therefore, when in an object distance is less than twice the focal length but Greater than the focal length (2f)

(d) To get image distance putting values in equation 1

dI= io.ocm x to.ocm

io.ocm-io.ocm

100eif2

cl-

ort

di- cc

Lues in equation 1

x5.00cm

d,=

s- .00cm-io.ocm

SocA2 d, = _sem

The magnification formula is

di

m=— — m= do —10cm

Sum

Therefore, m= +2

Example12.7:

A biologist with a near-point distance of N=26cm. examines an insect wing through a magnifying glass whose focal length is 4.3cm. Find the angular magnification when the image produced by the magnifier is (a) at the near point and (b) at infinity.

Given:

Focal length of magnifying glass T=4.3cm.

Near -point distance of N=26cm

Object Distance ‘do’= 1.1cm

Required:

Angular magnification ‘m81=?

Solution:

The magnification when the image is at near point for magnifying glass is

Me— +1

Putting the values

m =26cm. +,

0 —

4.3cm

Me=7

When the image is at infinity for magnifying glass is

Me

t

The magnification putting the values

26cm me 4 2cm me= 6

Example 12.8:

In biology class, a student with a near. Point distance of N=25cm uses a microscope to view an amoeba. If the objective has a focal length of 1.0cm, the eyepiece has a focal length of 2.5cm, and amoeba is 1.1 cm from the objective, what is the magnification produced by the microscope?

Given:

Focal length of eye piece ‘fe’=2.5cm.

Focal length of objective lo’=1.0cm.

Near- point distance of NI= 25cm

Object distance ‘do’=1.1cm

Required:

Total magnification ‘m’=?

1

= — — f di do

1 1

Or — = — —

di

fo do

Taking LCM

1= do-fo

Di f dd

Rod,

Di-

Do-fo

Putting values

Di= ticm x1.Ocm

Locm-ticm

Licm2

D,-

0.1cm,

D, = 11cm

The magnification formula for compo and microscope is

rn= — X —

fo fe

11cm 25cm

M=- X

1.0cm 2.5cm

Therefore, m= -110

Putting values

Example 12.9:

The largest optical located at the Yerkes observatory ir focal length of 19m, and the eye p calculate the total magnifying pow ?.r of this telescope. (b) Estimate the focal length of the telescope.

Given:

Refracting telescope in the world is Wisconsin. The objective lens has a

Focal length of objective ‘f;=1 iece has a focal length of 10cm. (a)

Focal length of eye piece ‘fe.’=11m. Required:

(A) Angular magnification me=?

(B) Approximate length ‘L’=?

Solution:

(a)The magnification of telescope is

jo

me=—

fe

Putting values,

19m

Me =

0.10m

Therefore,

b) For a relaxed eye, the image is at the focal point of both the eye piece and the objective lenses. The distance between the two lenses is thus

L z-; fo+ fe –=-; 19m +0.10m

Therefore, L 19m

Answer which is essentially of the length of telescope.

90 Assignment 12.1:

A dentist uses a concave mirror with focal length 2.0cm to examine some teeth. If the tooth under examination is 1.2 cm high and mirror is placed at 0.9cm. Calculate the distance of image formed the height of the image and magnification.

Solution:

Given:

Focal length =f =2cm

Object (tooth) Height = ho = 1.1cm

Distance of object (tooth) =d0=0.9cm

Required:

(a) Distance of image = di =?

(b) Height of image = hi =?

(c) Magnification = in =?

Calculation:

(a) Distance of image = di:

We know that;

1, 1

Fdo di

1 1 1

(1)

Di f do

Putting value in eq (1), we get

11 di= 2 0.9

1 09-2

Di (2) (0.9)

1 —1.1

— =

Di 1.8

.8

Di = -11.1= —1.64 (b) Height of image:

We know that;

Hi— at

Ho do

DiXho

0.9

hi = —2.0 cm

The —ve sign shows that the image is virtual.

hi

M = —

ho

—2

= —

1.1

The –ye sign shows that the image is vertically inverted.

Assignment 12.3:

A convex security mirror in a warehouse has a -0.50 m focal length. A 2.0 m tall forklift is 5.0 m from the mirror. What is the image’ position and image height?

Solution:

Given:

Focal length of convex mirror=f=-0.5m

Height of object = ho = 2m

Distance of object from the mirror =5m

Required:

(a) Image position = di =7 (b) Height of image = it; =?

Calculation:

(a) Image position (c11):

We know that

1= +

1 1

f di do

1 1 1

— = — — —

di f do

1 1 1

di —0.5 5

1 1 1

di 0.5 5

1 —5—(0.5)

=

di 2.5

1= -5.5 di 2.5

di = —15

5.5

di = —1*-555

di = —0.45m

The -ve sign show that the image is virtual.

(b) Height of image (hi):

We know that;

hi di

di z_

= – X no

d0

Putting value

hi =,S 5×2

2

—0.9

The ve sign show that the image is vertically inverted.

Assignment 12.3:

If the ‘speed of light is kerosene oil is 2.08 x 108m/sec. Calculate the index of refraction?

Solution:

Given:

Speed of light in kerosene oil = v = 2.08 x 108m/sec Speed of light in vacuum= .0 = 3 x 108m/sec

Required:

Index of refraction = n =?

Calculation:

We know that;

C

n =

3 x108m/sec

n=

2.08 x108m/sec

n = 1.44

Conclusion: Hence;

Index of refraction of kerosene oil =1.44

Assignment 12.3:

Find the index of refraction for medium -2, if medium -1 is air with index of refraction na = 1.00, the incident angle is (30.0)° and the angle of refraction is (22.0)°. Compare the result with the table and identify the nature of medium -2

Given:

Refractive index of medium -1(air) = n1 = 1.00

Angle of incident = 191.00.00

Angle of refraction = 02=. (22.0)0

Required:

Refractive index of medium —2 = n2 =?

Calculation:

We know that;

n2 — sine1

In sinO2

=

112

sin02

Sin02 X 111

N sin30a 100

2 xs

sin270 0.5k 1.00

0.375

n2 =1.33

Comparison:

If we compare the refractive index of medium -2 i.e. n2 = 1.33 with the table, this value match with the refractive index of water. This medium -2 is water.

Assignment 12.5:

What is the critical angle for light traveling in polystyrene (a type of plastic with index of refraction for polystyrene as 1.49) pipe surrounded by air take index o refraction for air to be 1.00?

Solution:

Given:

Refractive index of polystyrene =) 1, = 1.49 Refractive index of air = n2 = 1.00

Required:

Critical angle = 8, =?

Calculate:

We know that,

Sine, =

Ni

Sine, = 1179

Sine, = 0.671

9, = sin-‘(0.671) =42.2°

Assignment 12.6:

An object is placed 30.0cm in front p a converging lens and then 12.5 cm in front of a diverging lens. Both lenses have a focal, length of 10.0cm.for both cases: find the image distance and the magnification. Describe the images.

Solution:

(a) Lens = converging lens = convex lens:

Given:

Object distance = do = 30cm

Focal length = f = +10cm (convex lens ).

Required:

Image distance = d, =?

Magnification = in =?

(b) Lens= Diverging lens= concave lens:

Given:

Object Distance = do = 12.5cm

Focal length = f = —10cm (concave lens )

Required:

Image distance = d, =?

Magnification = M =?

Calculation:

(a) Lens= converging lens = convex lens di =?

We know that; 1

= —— f do d,

1 1 1

(1)

d, f

Putting value in eq(1) , we get,

1

= — —

di 10 30

1 3-1

— = —

di 30

1 2

— = —

di 30

di = —30

2

di = +15 cm Magnification = m =

15

M = — —

30

111 = —0.5

(1) The image is real (Distance positive

(2) The image is vertically inverted (magnification —v3) •

(3) The image is smaller than the object (magnification is less than”1″) (b) lens= Diverging lens= concave lens di =?

We know that;

1 1 1

1= 1 1

f do di

— = — — —

Putting value in eq (1), we get

di f do

1

= — —

di —10 12.5

1 1 1

di 10 12.5

1 —12.5-10

di (12.5)(10)

1 —22.5

— =

di 125

125

22.5 (1)

di = —5.56 cm

Magnification= m = —

a

(-5.56)

m=

12.5

+5.56

=

12.5

M = +0.445

The image is virtual. (Distance negative)

The image is upright. (Magnification +ve)

The image is smaller than the object (magnification is less than) Conclusion: Hence;

Image Distance (in case of convex lens) = di + 1Scm

Image Distance (In case of concave lens) = d, = -5.56cm Magnification (in case of convex lens) = m = -0.5

Magnification (in case of concave lens)= m = +0.445

Assignment.12.7:

An 8 cm focal length converging lens is used as “Jeweler’s loupe” which is a magnifying glass. Estimate the magnification (a) when the eye is relaxed and (b) if the eye is focused its near point.

Solution:

Given:

Lens = converging lens= convex lens.

Focal length of converging lens=f=+8 cm

Near —point distance =N=25cm (least Distance of Distinct vision) Required:

(a) Magnification (when eye is relaxed)= m =?

(b) Magnificent (when eye is focused at near point)= m

Calculation:

Magnification (when eye is relaxed = m =?

We know that;

m=-

2s

m

m = 3.12

Magnification (when eye is focused at near point)= m =?

We know that,

Putting value, • m = 1 + 7

25

M = 1 –

8

M = 1 1- 3.12

Conclusion:

Hence;

(A) 3 time’s magnified image is formed when the eye.is relaxed.

(B) 4 times magnified image is formed when the eye is focused at its near point.

Assignment 12.8:

If the focal length of eye piece is increased, does the magnitude of the magnification increase (or) decrease? Check your response by calculating the magnification when the focal length of the .eye piece is 3.6 cm

Solution:

Given:

Focal length of objective = to = 1.0cm

Near -point Distance = N = 25cm

Image distance = cl, = 11cm

Focal of eye —piece = fe = 3.5

Required:

(a) Effect-on magnification=?

(b) Magnification=?

Calculation:

Magnification of compound microscope and eye piece focal length: The magnification of compound microscope is given by;

N

771 = — Fe

X (1)

FO

From eq(1) it is clear that the magnification of compound microscope is inversely proportional to the focal length of eye — piece (fe) i.e. increase in focal length of eye — piece causes decrease in magnification of compound microscope and vice versa.

(b) Magnification when fe=3.5cm:

We know that;

Di N

771 = — — X —

FO Fe

Putting value

11 25

Trt = — — X —

1.0 3.5

M = —78.6 = —79

m = —79

The magnification of compound microscope is decreased when eye piece focal length is increased

Problems:1

A1, 50 cm high object placed 20.0 cm from a concave mirror with radius of curvature 30.0 cm. determine (a) the position of the image, and (b) is size, also draw the ray diagrams.

Solution:

Given:

Mirror = Concave Mirror

Height of object = ho = 1.50 cm

Position of object = do = 20.0 cm

Radius of Curvature = R = 30.0 cm

Focal length of concave mirror = f = R = — = +15 cm (concave mirror)

Required:

(a) Position of image = d, =?

(b) Height of image =h, =? Calculation:

(a) Position of image (ci,):

We know that;

=

1 1 1

— = – —

Di .f do

1 1 1

— = — —

Di 15 20

1 20-15

Di (15)(20)

1 5

— = —

D, 300

= 300

(b) Height of image (hi):

We know that; hi = di ho dc(1:1

Hi =—who

Do

Putting values. We get:

= hi = —60×1.50

20

Conclusion:

Hence:

a) Position of image=d, =60cm

b) Height of image =4.5cm

Problems:2

A candle of height 8.0 cm is located at a distance of 300 mm from a concave mirror; its virtual image is formed behind the mirror at a distance of 3.0 cm from the pole (or vertex). Find the focal length of the mirror and height of the mirror formed.

Solution:

Given:

Mirror =Convex mirror

Height of object =h0=8.0 cm

Position of object = do=300.00tnm = 30.0cm Position of image = d,=-3.0cm

Required:

a) Focal length 4=2

b) Height of image=hi=?. Calculations:

a) Focal length (f):

We know that:

1= 1 I

I do di

1 1

_ =

30 -3

1= 1 , 1

– — –

f 30-t- 3

1 1-10

– =

30

I -9

– = —

V 30

= -30

9

The ve sign show that the mirror is convex.

b) Height of image:

We know that:

Hi + di

How does

Di

Hi + –x ho

Do

Putting values, we get:

Hi + x 8

30

The —ve sign show that the image is virtual Conclusion:

a) Focal length =f= —3.33cm

b) Height of image=hi= —0.8cm

Problems:3

Calculate the speed of light in zircon with index of refraction n=1.923, a material is used in jewelry to replicate diamond?

Solution:

Given:

Speed of light in vacuum = c= 3x108m/sec Refractive index of zircon = n =1.923

Required:

Speed of light in zircon=v=?

Calculation:

We know that:

n = – v = –

Putting the values

3x1on

v = (—) m/sec

1.923

V = 1.56 x 108m/sec

Conclusion:

Hence:

The speed of light in zircon (v) is 1.56x108m/sec

Problems:4

A light ray strike in air / water surface at an angle of 26° with respect to the normal. The refractive index of water is 1.33. Find the angle of refraction when the direction of the rays is (a) from air to water (b) from water to air.

Solution:

Given:

Angle of incidence = 0, = 46″

Refractive index of air = n.„,. = 1

Refractive index of water= n„. = 1.33

Required:

(a) Angle of refraction (from air to water) = =’

(b) Angle of refraction (from water to air) = Or_ =7

(a) From air to water:

According to smell’s law:

(Lair sinner, Water sin0 RI Putting value we get. Sin°

Nor

X sin0i n water

1 x sin46 (1

sin0r, =

1.33

Sirlerl

= ixsin0.72

Siilert = 0.54

Ort = sin-` (0.54)

Fb) from water to Air:

According to smell’s law; nwater siner2

Sine;

Siner2 = nwater X sine]

Nair

Putting value we get; n

sin0r2 =

Water X sine

Flair

Sine

1.33 xsin46°

r,

1.33 x0.72

sin0 r2 =

Siner2 = 0.96

Ere = sin-1(0.96)

Conclusion:

Hence;

(a) Angle of refraction (from air to water) =Ori = 33o

(b) Angle of refraction (from water to air) =Or, = 74°

Problems:5

An optical fiber is made from flint glass with index of refraction 1.66 and is surrounded by a cladding made of crown glass with index of refraction 1.52. What is the critical angle?

Solution:

Given:

Required:

Calculation:

We know that

Putting value, we get;

Refractive index of flint glass=critical Angle= Qc=?

Refractive Index of crown glass ri

RF

C

SinQ, =

ni

QC =

n,

nZ

QC=sin…1 1.52

QC= 66.3°1.66

tt =1.66

= nc.G =1.66

Conclusion:

Hence;

Critical Angle=Q,= 66.3

Problems:6

Suppose the book page is held 7.50 cm from a concave lens of focal length 10.0 cm. What magnification is produced in each case i

Solution:

Given:

(a) Lens=convex lens

Focal length= f T 10.0cm

Position of object = do = 7.50cm

(b) Lens= concave lens

Focal length = f = —10.0

Position of object = do = 7.50cm

Calculation:

(a) Magnification for convex lens (M):

1 1 1

_…._

di do

1- = 1

– —

di f do

Putting value we get:

1, 1 1

– =

di 10 7.5

1 7.5-10

di (10)(75)

1 -2.5

di = 75

-75

o di =

2.5

– 750

di =

25

di = —30cm

– , M =d re

M = (-30)

Required:

(a) Magnification for convex lens = M =?

(b) Magnification for concave lens = M at?

7.5

M = +4—— Answer the +ve sign shows that the image is upright

(b) Magnification for Concave Lens:

1.

f do d,

1 = 1 1

d, f

Putting values, we get;

1 1 1

di —10 7.5

I I

10 7.5

d, ( I )(7.5)

1_ -17.5

d; 75

75

di= :1-

17.5

d, = .56 cm

The —ve sign shows that the image is virtual, Now;

d

d

M = –

Putting value we get;

M= C-5.56

7.5

M= (-5.56)

7.5

M = 5.56

7.5

The +ye sign shows that the image is upright.

Problems:7

Gualala is viewing a flea using a magnifier with f =3.0 cm if her near point is at N=25 cm then calculate the maximum magnification she can get.

Solution:

Given:

Focal length of magnifier= = 3cm

Near point distance = N = 25cm

Required:

Maximum magnification = M„,„ =?

Calculation:

We know that

Maw, = + 1 Putting values, we get;

An 25

‘Max = 3

+

Mmax = 8-33 + 1

Mmax = 9.33

Conclusion:

Hence;

Maximum magnification=Mmax = 9.33

Problems:8

A telescope has a magnification of 40.0 and a length of 1230 mm. What are the focal lengths of the objective and eyepiece?

Solution:

Given:

Magnification of telescope = M = 40 Length of telescope = L = 1230mm

Required:

(a) Focal length of objective= Ai =?

(b) Focal length of eyepiece = re =?

Calculation:

We know that,

M=L

Fe To = M

Also;

L. = FO + f,

Putting equation (1) and Equation (2) we get;

L=Mfe+le

L = (M + 1) f,

=

m+1

Putting value we get;

1230

Fe = 40+1

1

Fe =230

41

Putting eq (3) in eq (1)

FO = (M)(30)

Fu = (40)(30)