Geometric Optics

Introduction

__OPTICS__is study of propagation of light. In Geometric optics we are just concerned about, what the path of light is when it gets reflected or refracted. While dealing with Geometric optics we consider light rays to e paraxial i.e. which re very near to each other and thus a lot of assumptions could be encorprarted. In this unit we answer why a coin placed in water appear to be little bit raised above and many such phenomenom.

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**REFLECION**When a ray of light is incident at a point on the surface, the surface throus party or whally the incident energy back into the medium of incidence. This phanomenon is called reflection.

Surface that causes reflection are known as mirrors or reflectors.

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**Important terms related to reflection**__Angle of incidence__: Angle which the incident ray makes with the normal at the point of incidence. (densted i)

__Angle of reflection__: The angle which the reflected ray makes with the plane normal of the point of incidence. (densted by r)

__Glancing Angle__: The angle which the incident ray makes with the plane reflecting surface is called glancin angle. It is generally denoted by g.

g = 90 - i

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**LAWS OF REFLECTION**(i) The incident ray, the reflected ray and the normal to the reflected surface at the point of incidence lie in same plane.

(ii) The angle of incidence is equal to the angle reflection, i.e. i = r. These laws hold for all reflecting surfaces either plane or crowed.

__: Do incident ray and reflected ray differ in terms offrequency, wavelength and speed ?__

**Dumb Question**Ans: None of frequency, wavelength and speed changes due to reflection. However, intensity and have amplitude (I A

^{2}) usually decreases. However in ray optics we do not consider this change.

__: What will happen if incident ray falls along Normal to the surface ?__

**Dumb Question**Ans: In what case i = 0 so r = i = 0

it will retrace its path.

__: How do we determine the 'Normal' in case of spherical surface ?__

**Dumb Question**Ans: Spherical surfaces are a part of circle. So Normal at a point on the surface is the line joining that point with the centre of circle (of which the mirror is a part).

__IMAGE FORMATION__: Image is said to the formed when two rays meet. The image can be real or virtual.

__Real Image__: If the rays actually at a point then the image formed is regarded as real image.

__Virtual image__: If the reflected or refracted rays do not actually meet out only appear to diverge from the point, then it is said that a virtual image is formed at that point.

__: Just as we have real and virtual image, do we have real and virtual objects too ?__

**Dumb Question**Ans: Object is a point from where rays diverge. If the rays actually start from a point then that is called real object. If however the rays appear to be starting from a point than that object is regarded as virtual object.

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**Characteristics of the image formed by o plane mirror**(i) The image formed is atthe same distance behind the reflecting surface as the object is in front of it.

(ii) The size of the image is the same as that of the object.

(iii) The image is virtual and erect which means no light passes through it.

(iv) The image is laterally inverted. i.e. side-wise inverted.

__: What is the meaning of 'side-wise inverted' ?__

**Dumb Question**Ans: The image of right hand appears identical to left hand in a mirror. This is 'side-wise inverted'.

__: Two plane mirrors are indined at angle with each other. A ray strikes one of them. Find its deviation after it has reflected twice-one from each mirror.__

**Illustration**Solution:

__Case I__:

_{1}= Clockwise deviation at A = 180 - 2i

_{1}

_{2}= Anticlockwise deviation at B = 180 - 2i

_{2}

Now from OAB, we have

angle O + angle A + angle B = 180

^{0}

=> i

_{1}- i

_{2}=

as i

_{1}> i

_{2},

_{1}<

_{2}

Hence the net angle clockwise deviation =

_{1}-

_{2}

= (180 - 2i

_{2}) - (180 - 2i

_{1})

= 2(i

_{1}- i

_{2}) = 2

__Case II__:

_{1}= Clockwise deviation at A = 180 - 2i

_{1}

_{2}= clockwise deviation at B = 180 - 2i

_{2}

Now from OAB, we have

angle O + angle A + angle B = 180

^{0}

or + (90 - i

_{1}) + (90 - i

_{2}) = 180

^{0}

=> i

_{1}+ i

_{2}=

Hence, net clockwise deviation =

_{2}+

_{1}

= (180

^{0}- 2i

_{2}) + (180

^{0}- 2i

_{1})

= 360 - 2(i

_{1}+ i

_{2}) = 360 - 2

=> Net anti-clockwise deviation = 360

^{0}- (360

^{0}- 2) = 2

__: What is deviation and how it is calculated ?__

**Dumb Question**Ans: Deviation is defined as the angle between direction of the incident ray and the reflected ray. It is denoted by .

Here A'OB = = AOA' - AOB

= 180

^{0}- 2i

We know g = 90 - i

= 2g

• If two mirror are kept inclined to each-other at angle with their reflecting surfaces facing each-other, then multiple reflections take place and more than are images are formed.

Number of images formed n = - 1 [if is on even integer]

[if is on odd integer]

•

__Rotation of the reflected ray by a mirror__: Keeping the incident ray fixed, if the plane mirror is rotated through an angle about an axis in the plane of mirror, then the reflected ray is rotatde through an angle 2.

How ??

The glancing angle with M

_{2}= ( + )

Hence, the new angle of deviation (i.e. COP) = 2( + )

The reflected ray has thus been rotated through BOP when the mirror is rotated through an angle and since

BOP = COP - COB or BOP = 2( + ) - 2 = 2

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**SPHERICAL MIRRORS**Some important definition:

• Spherical mirrors are a part of a hollow sphere or spherical surface. They can be concave or convex.

• POLE OR VERTEX: It is the geometrical center of the spherical mirror (P in this case)

• RADIUS OF CURVATURE (R): Radius of curvature is the radius R of the sphere of which the mirror forms a part.

• PRINCIPAL AXIS: The line CP joining the pole and the centre of curvature of the spherical mirror is called the principal axis.

• FOCUS (F): If a parallel beam of rays, parallel to the principal axis and close to it, is incident on a spherical mirror, thereflected rays converge to a point F (in case of a concave mirror) or appear to diverge from a point F (in case of a convex mirror) on the principal axis. The point F is called the focus of the spherical mirror.

• Focal length (f): Focal length is the distance PF between the pole and focus F along the principle axis.

• Aperture: The line joining the end points of a spherical mirror is called the aperture or linear aperture.

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**RELATION BETWEEN f AND R**The magnitude of focal length in spherical mirrors in half the radius of curvature i.e.

f =

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**RULES FOR IMAGE FORMATION**(i) A ray parallel to principal axis passes (or appear to pass) through focus after reflection.

(ii) A ray passing through or directed towards focus after reflection from the spherical mirror becomes parallel to the principal axis.

(iii) A ray passing through or directed towards the centre of curvature, after reflection from the spherical mirror, retroces its path (see dumb question).

__: How do 4 use these rules and where to apply them ?__

**Dumb question**Ans: These rules are to be used for making the ray diagram and find the nature of image. For image formation, we need two rays. So choose one ray to be the one parallel to the principal axis and passing through the object. Apply Rule (ii) for trcking rest of path. Other ray can be choosen as the one which comes from focus .

Image formation by cancave mirror

Position of object | Nature of image | Position |

(i) infinity | real, inverted, highly diminished | at F |

(ii) beyond C | real, inverted and diminished | between F & C |

(iii) at C | real, inverted, same size | at C |

(iv) between F and C | real inveted and enlarged | beyond C |

(v) at F | real, inverted, highly enlarged | at |

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**SIGN CONVENTION**The pole is taken as the origin and rest of the distances are calculated according to it. Thus

• All the distances on the right side are +ve. Thus focus is +ve in case of convex mirror while it is -ve in case of concave mirror.

• Above rule holds, if incident ray is travelling from left to right else directions just reverses.

• Distances measured above principal axis are taken to be positive while distances measured below principal axis are taken to be negative.

notations used:-

u: Distance of the object from pole of spherical mirror.

v: distance of the image from the pole of the spherical mirror.

f: focal length of the spherical mirror.

R: Radius of curvature of the spherical mirror.

• All the symbols used are assigned values with proper sign convention. i.e. for concave mirror f is always -ve and so on.

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**MIRROR FORMULA**All values have to placed in the equation with proper sign.

__: Is the above formula valid for simple mirror also ?__

**Dumb Question**Ans: This is a general formula for all spherical mirror. For flat mirrors R = , as it can be assumed to be a part of sphere with infinite radius, then this equation can be used. It yeilds

= 0 => v = - u

which correctly describes image formation in a flat mirror.

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**LINER MAGNIFICATION**For linear objects, the ratio of the image size (I) to the object size (O) is called linear magnification or transverse magnification or lateral magnification. If is generally denoted by m.

m =

m =

__: A plane mirror is placed 22.5 cm in front of a concave mirror of focal length 10 cm. Find where an object can be placed between the two mirrors, so that the first image in both the mirrors coincides.__

**Illustration**Solution:

As shown in figure, if the object is placed at a distance x from the concave mirror, its distance from the place mirror will be (212.5 - x). So, plane mirror will form equal and erect emage of object at a distance (22.5 - x) behind the mirror.

Now as according to given problem the image formed by concave mirror coicides with the image formed by concave mirror, therefore for cancave mirror

v = - [22.5 + (22.5 - x)] = - (45 - x) and u = - x

So,

i.e. x

^{2}- 45x + 450 = 0 or (x - 30)(x - 15) = 0

i.e. x = 30 cm or x = 15 cm

But as the distance between two mirrors is 22.5 cm, x = 30 cm is not admissible. So the object must be at a distance of 15 cm from concave mirror.

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**REFRACTION**Whenever a ray of light from one transparent medium to another, it gets deviated from its original path while crossing the interface of the two media (except in case of normal incidence). This phenmenon of devition or bending of light rays from their original path while passing from one medium toanother is called refraction.

AB incident ray BC refracted ray NBN' Normal

i angle of incidence r angle f refraction KL interface

• If the refracted ray bends towards the normal with respect to the incident ray, then the second medium is said to be optically denser as compared to the first medium.

• If the refracted ray bends away from the normal, then the second medium is said to be (optically) as compared to the first medium.

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**LAWS OF REFRACTION**(i) The incident ray, there fracted ray and the normal to the refracting surface at the point of incidence, all lie in the same plane.

(ii)The ratio of sine of the angle of incidence to the sine of the angle of refraction is constant for any two given media

i.e. = constant = 1µ

_{2}= Refractive index.

This is called Snell's law.

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**Refractive index**• 1µ

_{2}= then 1µ

_{2}(or

_{1}n

_{2}) is a constant called refractive index of the second medium w.r.t. first medium (containing incident ray)

• 1µ

_{2}= where µ

_{2}is absolute refractive index of second medium and µ

_{1}is of first.

• Refractive index is a relative quantity. If the first medium is air (vaccum to be precise) then = µ or n called absolute refractive index.

=> µ

_{1}sin i = µ

_{2}sin r = constant.

product of absolute R % of medium and sine of rays with normal in that medium is constant.

__: What actually occurs in the process of refraction and what is true significances of µ ?__

**Dumb Question**Ans: During refraction, the velocity of light changes at the interface and thus light follows an alternative path.

µ is actually ratio of light in vaccum with the speed of light in given medium

µ 1µ

_{2}=

__: What parameter of wave changes during refraction.__

**Dumb Question**Ans: During the process of refraction only frequency of light remains constant, while wavelength and velocity change

__: What is the meaning of 'rare' and 'dense' when associated with medium ?__

**Dumb Question**Ans: 'rare' and 'dense' are also relative terms. When compared the medium with higher value of refractive index is called 'denser medium' and the other one is called rarer medium.

Dense medium is the one in which light travels slower than in rarer medium.

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**LATERAL DEVIATION (through a glass slas)**If there exists a region of index through which ray has to pass; such that there exists a medium of same refractive index on the entry and exit of the ray then the ray gets laterally shifted by distance d

d = where t length of region

Why ??

In wright angled OBK, we have BOK = i - r

sin(i - r) = or d = OB sin(i - r) ................................. (i)

In right angled triangle ON'B, we have

cos r = or OB =

Substituting the above of OB in equation (i), we get

d = sin(i - r)

Ans:

d = where t length of region

Why ??

In wright angled OBK, we have BOK = i - r

sin(i - r) = or d = OB sin(i - r) ................................. (i)

In right angled triangle ON'B, we have

cos r = or OB =

Substituting the above of OB in equation (i), we get

d = sin(i - r)

__: In the adjacant system what is the angle between incident ray and emergent ray ?__**Dumd Question**Ans:

Applying Snell's law at first interface

sin i = sin r

_{1}............................................. (i)

Applying Snell's law at 2, 3 & 4

We get

µ

_{1}sin r

_{1}= µ

_{2}sin r

_{2}.................................... (ii)

µ

_{2}sin r

_{2}= µ

_{3}sin r

_{3}.................................... (iii)

µ

_{3}sin r

_{3}= sin e ......................................... (iv)

Combining all the form equation we get

sin i = µ

_{1}sin r

_{1}= µ

_{2}sin r

_{2}= µ

_{3}sin r

_{3}= sin e

or sin i = sin e

=> i = e

emergent ray is parallel to incident ray. This result is independent of no. of slabs used and holds as long as incident ray and emergent ray are in same medium.

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**Apparent depth**When the object is in different medium and observer is in different medium, then the object seems to be

(i) nearer to the observer (if object is in denser o medium).

(ii) farther to the observer (if object is in rarer medium).

Then the actual depth

Apparent depth =

where µ is refrective index of object medium w.r.t. that of observer's medium.

for case (i) the apparent shift is given by t -

Why ??

In the above figure

sin i tan i = II'ly sin r tan r =

(the approxiamation holds because i & r are small angles)

from Snell's law

µ = => µ =

=> y =

Similarly the other case can be proved (when object medium is rarer)

__: Whar will happen if a glass slab of thickness 't' is placed between object and a mirror.__

**Dumb Question**Ans:

In that case, the point of intersection of ray shifts by t(1 - 1/µ) in the direction in which the ray is moving.

(where µ is refractive index of glass slab) And calculation needs to be made according to this new position.

__: A fish rising vertically to the surface of water in a lake uniformly at the rate of 3 m/s observes 9 m/s vertically above it. If the refractive index of water is 4/3 find the actual velocity of the dive of the bird.__

**Illustration**Solution: If at any instant the fish is at the depth 'x' below fish water surface while the bird at a height y above the surface, then the apparent height of the bird from the surface as seen by the fish will be given by

µ = or Apparent height = µy

So, the total apparent distance of the bird as seen by the fish in water will be

h = x + µy

or or 9 = 3 + µ()

or

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**CRITICAL ANGLE & TOTAL INTERNAL REFLECTION**• When a ray of light goes from adenser medium to a rarer medium, the angle of refraction is greater than the angle of incidence. If the angle of incidence is increased, the angle of refraction may enentually become 90

^{0}.

• This incidence angle is called critical angle (C).

• If the angle is increased furthe, then the ray comes back in the medium of incidence. This phenomenon is called total INTERNAL REFLECTION.

• Crtical angle is sin C = where µ =

__: Is Snell's law applicable for total internal reflection ?__

**Dumb Question**Ans: Snell's law no more holds for total internal reflection, however we can apply laws of Reflection to totally internally reflected ray.

__: A ray of light from a denser medium strikes a rarer medium at an angle of incidence i If the reflected and the refracted rays are mutually perpendicular to each other, what is the value of the critical angle ?__

**Illustration**Solution: From Snell's law, we have

or µ = ........................................ (i)

According to the given problem i + r 90 = 180

or r = 90 - i

Substituting the above value of r in equation (i), we get

µ =

or µ = cot i ................................................ (ii)

By definition

C = sin

^{-1}() or C = sin

^{-1}

Using equation (ii)

or C = sin

^{-1}(tan i)

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**REFRACTION THROUGH PRISM**If A is the angle of prism, deviation caused due to it, i, angle of incidence and e angle of emergence than

+ A = i + e

Why ??

In the quadrilateral AQNR, two of the angle (at the verticles Q and R) are right angle. Therefore, the sum of the other angles of the quadrilateral is 180

^{0}.

A + QNR = 180

^{0}from the triangles QNR, r

_{1}+ r

_{2}+ QNR = 180

^{0}

Comparing these two equations, we get r

_{1}+ r

_{2}= A

The total deviation is the sum of deviations at the two faces:

= (i - r

_{1}) + (e - r

_{2}) = (i + e) - (r

_{1}+ r

_{2})

=> = (i + e) - A

=> + A = i + e

__: For same prism do e have some deviation for all rays ?__

**Dumb Question**Ans: At the two interface, applying Snell's law we have

= µ and =

deviation depends upon the refraction index, which further is depend upon wavelength ray. Thus deviation is different for rays of different wavelengths.

__: Can we have same deviation for more than one angle of incidence ?__

**Dumb Question**Ans: It can be easily seen that if we reverse the emergent ray, it goes back along the same path. The angles of incidence and emergence get interchanged but the angle of deviation remains the same.

Thus the same angle of deviation is possible for two different angles of incidence: i and e such that i + e = A +

__: A ray of light is incident on one face of a prism (µ = 1.5) at an angle of 60__

**Illustration**^{0}. The refracting angle of the prism is also 60

^{0}. Find the angle of emergence and the angle of deviation. Is there any other angle of incidence which will produce the same deviation ?

Solution:

Angle of incidence = i = 60

^{0}

at point P,

=> sin r

_{1}=

or r

_{1}35

^{0}6' using r

_{1}+ r

_{2}= A

At point Q, => sin e = 1.5 sin 24

^{0}44

sin e = 0.63

=> e = 39

^{0}

Deviation = = (i + e) - A = (60 + 39) - (60) = 39

^{0}

If i and e are interchanged, deviation remains the same. These same deviation is obtained for angles of incidence 60

^{0}and 39

^{0}.

__: What will happen if the angle of the prism is greater than twice the critical angle for the material of the prism ?__

**Dumb Question**Ans: We know r

_{1}+ r

_{2}= A > 2C or r

_{1}+ r

_{2}> 2C

these occurs two cases:

(i) r

_{2}> C the ray not go out of prism. rather will be reflected back inside. This is physically not possible as in that case incident ray.

(ii) r

_{1}> C this is physically not possible as in that case incident ray couldn't have entered the prism.

These occurs third case too r

_{1}> C & r

_{2}> C here also result of (ii) will hold.

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**MINIMUM DEVIATION**Adjacant figure shows the experimental plot of Vs i graph for a prism.

for this case i = e and r

_{1}= r

_{2}

The equation modifies as

+ A = i + e = 2i or i =

II'ly r

_{1}+ r

_{2}= A or r

_{1}= A/2

Applying Snell's law for first surface, the equation yields

µ = [Refraction index of prisom]

for prism of small angle (<>0) we have approximations for sin e

µ = => µ =

or

_{m}= (µ - 1)A

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**FEW IMPORTANT TERMS**• Grazing Incident: When i = 90

^{0}, in this angle of refraction becomes equal to critical angle.

• Grazing Emergence: When e = 90

^{0}, here r

_{2}= Critical angle.

• Maximum Deviation: We know = (i + e) - A

deviation will be maximum when the angle of incidence i is maximum i.e. i = 90

^{0}.

Here,

()

_{max}= (90

^{0}+ e) - A expression for maximum deviation.

__: Find the manimum and maximum angle of deviation for a prism with angle A = 60__

**Illustration**^{0}and µ = 1.5.

Solution:

__Minimum deviation__:

The angle of minimum deviation occurs when i = e and r

_{1}= r

_{2}and is given by

µ = =>

_{m}= 2 sin

^{-1}(µ sin A/2) - A

Substituting µ = 1.5 and A = 60

^{0}, we get

_{m}= 2 sin

^{-1}(0.75) - 60

^{0}= 37

^{0}

__Maximum Deviation__:

The deviation is maximum when i = 90

^{0}or e = 90

^{0}that is at grazing incidence or grazing emergence.

Let i = 90

^{0}

=> r

_{1}= C = sin

^{-1}() => r

_{2}= sin

^{-1}() = 42

^{0}

=> r

_{2}= A - r

_{1}= 60

^{0}- 42

^{0}= 18

^{0}

Using , we have

=> sin e = µ sin r

_{2}= 1.5 x sin 18

^{0}= 4.63

=> e = 28

^{0}

Deviation =

_{max}= (i + e) - A = (90

^{0}+ 28

^{0}) - 60

^{0}= 58

^{0}

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**DISPERSION OF LIGHT THROUGH PRISM**When a ray of white light is passed through a prism, it spits up into its constituent colours. This phenomenon of splitting up of white light into constituent colours is called DISPERSION.

• Cauchy's expression forrefractive index µ = a + a,b -> Constant of material.

Since each colour has different wavelength hence refractive index will be different for different colours.

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**ANGULAR DISPERSION**_{V}-

_{R}= (µ

_{V}- µ

_{R})A

it is the angle between the extreme rays of dispassed colour band, which are violet and Red.

__: The ratio of (angular) dispersion to the deviation of the mean ray (yellow) is called the dispersive power of the prism. It is denoted by W.__

**DISPERSIVE POWER (W)**W = , where is the deviation of the mean ray

In term of refractive index

W = , where µ =

__: Calculate the diapersive power of crown and flint glass-prism from the following data: for crown glass for crown glass__

**Illustration**µ

_{V}= 1.522; µ

_{R}= 1.514 µ

_{V}= 1.662; µ

_{R}= 1.644

Solution: For crown glass µ

_{V}= 1.522; µ

_{R}= 1.514

µ

_{y}= = 1.518

Hence, the dispersive power of crown glass

W = = 0.01544

for flint glass µ

_{V}= 1.662; µ

_{R}= 1.644

µ = = 1.6

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**DISPERSION WITHOUT DEVIATION**Let two prisms of angle A & A' be placed as shown in figure. If the system produces net deviation as zero

then where µ & µ' are the refractive index of mean wavelength.

__: What is the meaning of 'dispersion without deviation' ?__

**Dumb Question**Ans: The rays gets dispersed, however, the mean ray remains parallel to the incident ray, i.e. angle between incident and mean emergent ray is zero.

Condition for it os + ' = 0 or (µ - 1)A + (µ' - 1)A' = 0

this yields the required result.

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**DEVIATION WITHOUT DISPERSION (ARCHROMATIC COMBINATION OF PRISM)**The emergent ray is undispersed. for this

this is the condition for achromatism or no dispersion.

Why ??

Let µ'

_{V}, µ'

_{R}and µ be the corresponding values for the material of the second prism, while µ

_{V}, µ

_{R}& µ be of the first prism.

Angular deviation by first prism is

(

_{V}-

_{R}) = (µ

_{V}- µ

_{R})A

Angular dispersion produced by second prism

('

_{V}- '

_{R}) = (µ'

_{V}- µ'

_{R})A'

Since combination does not produce any dispersion

(

_{V}-

_{R}) + ('

_{V}- '

_{R}) = 0

or A(µ

_{V}- µ

_{R}) + A'(µ'

_{V}- µ'

_{R}) = 0

or

__: Find the angle of a pism of dispersive power 0.021 and refrative index 1.53 to form on achromatic combination with the prism of angle 4.2__

**Illustration**^{0}and dispersive power 0.045 having refractive index 1.65. Also calculate the resultant deviation.

Solution: W = 0.021; µ = 1.53 W' = 0.045; µ' = 1.65

A' = 4.2

^{0}

For no dispersion

W + W'' = 0

or W(µ - 1)A + W'(µ' - 1)A' = 0

or A = - = - 11.4

^{0}

Net deviation = + ' = (µ - 1)A + (µ' - 1)A'

= - 11.04

^{0}(1.53 - 1) + 4.2

^{0}(1.65 - 1) = - 3.12

^{0}

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**REFRACTION AT A SINGLE SPHERICAL SURFACE**Consider a spherical surface of radius R separating two media with refractive indices µ

_{1}and µ

_{2}[the dig. is for the case µ

_{2}> µ

_{1}]

If µ is the object distance and v is the image distance, for light rays going from medium I to medium II, then

also transverse magnification in this case is

m =

__: Isthis equation valid for convex refracting surface ?__

**Dumb Question**Ans: This equation is valid for all refracting surfaces convex, concave or plane. In case of plane refracting surface R .

__: What sign convention is to be used ?__

**Dumb Question**Ans: Sign convention remains same, as it was for mirror. All measurements are again to be done from P. While using this equation, the values of v, u and R with proper sign has to placed.

Why ??

here we rpove the formula for the case µ

_{1}< µ

_{2}however it holds for all cases for first refraction (at P).

µ

_{1}sin

_{1}= µ

_{2}sin

_{2}

Assuming all angles to be very small, and

µ

_{1}

_{1}= µ

_{2}

_{2}....................................................... (i)

Using Geometry

+ =

_{1}....................................................... (ii)

[exterior angle theorem in POC]

II'ly =

_{2}+ ............................................... (iii)

from (i), (ii) & (iii)

µ

_{1}( + ) = µ

_{2}( + )

=> µ

_{1}+ µ

_{2}= (µ

_{2}- µ

_{1})

Here considering , , to be small angle again

hence the equation.

__: A transparent rod 40 cm long is cut at one end and rounded to a hemispherical surface 12 cm radius at the other end. A small object is embereded within the rod along its axis and half way between its ends. When viewed from the end of the rod, the object appears 12.5 cm deep. What is its apparent depth when the object viewed from curved end ?__

**Illustration**Solution:

__Case - I__: When the object viewed from the flat surface:

Real depth of the object = 20 cm

Apparent depth = 12.5 cm

Using µ = we have µ = = 1.6

__Case - II__: When the object is viewed the curved surface:

Here the refraction is taking place at the single curved surface. So we will use

Here µ

_{1}= 1.6; µ

_{2}= 1; u = -2 cm; v = ? ; R = - 12 cm

So,

=> v = - 33.3 cm

Hence the object appears 33.3 cm deep from the curved surface.

__:__

**THIN LENSES**__: A lens is a transpsrent medium bounded by two refracting surfaces such that at least one of the refracting surfaces is curved.__

**Lens**__: Lenses are of following type:__

**Types of lenses**__:__

**Definition**•

__OPTICAL CENTRE__: Optical centre is a point for a given lens through which any ray passes undeviated.

•

__PRINCIPAL AXIS__: The line joining the centres of curvature of the two bounding surfaces is called the principal axis.

•

__FOCUS (F)__: A parallel beam of light, parallel and close to the principal axis of the lens after refraction passes through a fixed point on the principal axis (in case of convex lens) or appear to diverge from a position principal axis (in case of concave lens). This point is called the principal focus or focus of the lens. It is generally denoted by F.

__:__

**RULES FOR IMAGE FORMATION**• A ray passing through the optical centre of the lens proceeds undeviation through the lens.

• A ray passing parallel to the principal axis after refraction through the lens passes or appear to pass through the focus. (By definition of the focus)

• A ray through the focus or directed towards the focus, after refraction from the lens, becomes parallel to the principal axis. (Principal of reverssibility of light)

__: What do you mean principal 'reversesibility of light' ?__

**Dumb Question**Ans: If the light is projected along the emergent ray (in opp. direction) then it will follow the path same as incident ray (in opp. direction)

__:__

**LENS MAKERS FORMULA**If the material of lens has refractive index µ and refractive index of medium in which lens is kept is µ

_{0}. Then focal length of lens is given by:

R

_{1}and R

_{2}radius of the radii of the surfaces from which the light enters the lens and the one from which light leaves out respectively.

Why ??

This equation can be obtainedb applying thin lens formula twice.

Let image due to first surface be formed at v

_{1}from P

....................................... (i)

for image formula at second surafce, this image will act as virtual object.

....................................... (ii)

eliminating v

_{1}from (i) & (ii) yeilds.

__: The above proof used lens formula for this case linear magnification is m =__

**Lens Formula**__: How we can determine the focal length 'f' of a lens system ?__

**Dumb Question**Ans: 'f' can bge determined by using the basic definition of focal length. Assume that object is kept at infinity and for that calculate the final image position for the system. Then in this case f = v. As focus is a point where rays coming from infinity meet after refraction/reflection.

__: Above equation is derived only for a case where the lens has medium of same refractive index across its ends. How to priceed if we have different media.__

**Dumb Question**Ans: In that case start from basic formula and apply this equation with corresponding refractive indices, twice andthe obtain the image.

__: Will immersing in water a lens and mirror, change this focal lengths ?__

**Dumb Question**Ans: Focal length of mirror f

_{M}= and is independent of refractive index of medium hence it will remain unattached, however focal length of lens is depended upon medium, hence it will change.

__: Calculate the focal length of a biconvex lens in air. If the radii of its surfaces are 60 cm and 15 cm.__

**Illustration**Refarctive index of glass = 1.5.

Solution: Consider a light ray going through the lens as shown.

It strikes the convex side of 60 cm radius and concave side of 15 cm radius while coming act.

R

_{1}= + 60 cm R

_{2}= - 15 cm

or

f = 24 cm.

__: If an erquiconvex lens of focal length 'f' is cut into two equal halves (a) along its axis (b) perpendicular to its axis. then in which case 'f' cahnges. Find its value.__

**Dumb Question**Solution: (a)

neither radius of any of the surface charged nor the media changed, hence focal length will remain same i.e. 'f'.

(b)

Here the raius of other has changed, hence focal length will change

f = while new f

_{1}=

f

_{1}=

__:__

**POWER OF LENS**The power of lens is defined as P =

where 'f' must be expressed in metres. S.I. unit is m

^{-1}also known as Dioptre (D).

__: What is the physical interpretation of power ?__

**Dumb Question**Ans: A lens which bends the light more, is considered more powerful. Smaller is the focal length, more is the power of lens.

__:__

**COMBINATION OF LENSES**If two or more lenses of focal length f

_{1}, f

_{2}and so on one placed in contact, then this combination can be replaced by a lens called 'equivalent lens'.

The focal length (F) of equivalent lens is given by

i.e. P = P

_{1}+ P

_{2}+ .............

If two this lenses of focal lengths f

_{1}and f

_{2}are separated by a distance 'd' then the focal length of the equivalent lens is given by

__:__

**DISPLACEMENT METHOD**Let us consider a real object and a screen fixed at a distance D apart. A convex lens is placed between them.

Let x be the distance of the object from the lens when its real image is obtained on the screen.

we have, u = - x

v = + (D - x)

f = + f

Using lens formula, we have

or

=> x

^{2}- Dx + Df = 0

or x =

Cases:-

(i) If D < 4f, for no position of lens, image formation is possible.

(ii) If D = 4f, then in this case x = = 2f. So only one position of the lens will be possible.

(iii) If D > 4f, in this situation, two positions of lens will give same image position. The positions are x

_{1}= x

_{2}=

x

_{2}- x

_{1}= L =

__: what for is this method used ?__

**Dumb Question**Ans: It is used for determining the focal length of lens experimentally from the abiove relation

f =

by measuring L and D value of focal length can be calculated.

__:__

**SILVERING OF LENS**Lens with silvered surface acts as a mirror. We can calculate the focal length of surface using

P = P

_{1}

only modificaion is that power of mirror is given by P

_{mirror}=

__: How can be P = - for a mirror be justified ?__

**Dumb Question**Ans: Focal length of concave mirror is -ve and for convex mirror it is +ve (as by our sign convention) But the nature of concave mirror is to converge the light and nature of convex is to diverge the light. If an optical system converges the light towards a point. Its power is assumed to be +ve (diverge being -ve). That's why the formula for power in case of mirror is - .

__: How many times doeach individual poweris added ?__

**Dumb Question**Ans: First the path of light is roughly determined. Then for each time ray passes (or gets reflected) through the element of system, then corresponding power is added.

For example in the adjacant case the equation will be

P

_{net}= P

_{lens}+ P

_{mirror}+ P

_{lens}

light ray will have to pass through the lens surface twice (unsilvered).

__: A thin hollow equiconvex lens silvered at the back converges a parallel beam of light, at a distance of 0.2m in front of it. Where will it converge the same light, if filled with water (µ = 4/3).__

**Illustration**Solution: When lens is hollow, it will simply act as concave mirror

Parallel beam is focussed atfocus of mirror only

f = - 0.2 m ( R = 2f = - 0.4 m)

When filled with water

P

_{net}= 2P

_{L}+ P

_{M}= 2

Now the complete system acts as mirror

f

_{net}= = - 0.12 m

Parallel beam will now be focussed at a distance of 12 cm from the lens (on the side of object).

**PROBLEMS**

**Easy**1. A glass sphere of radius R = 10 cm is kept inside water. A point O is placed at 20 cm from A as shown in figure. find the position and nature of the image when seen from other side of the sphere. given µ

_{g}= 3/L and µ

_{w}= 4/3.

Solution: A ray of light starting from O gets refracted twice. The ray of light is travelling in a direction from left to right. Hence, the distance meausred in this direction are taken + ve.

Applying , twice with proper signs.

we have, or AI

_{1}= - 30 cm

Now, the first image I, acts as an object for the second surface, where

BI

_{1}= u = - (30 + 20) = - 50 cm

BI

_{2}= 100 cm, i.e. the final image I

_{2}is virtual andis formed at a distance 100 cm (towards left) from B.

Diagram

2. Focal length of a convex lens in air is 10 cm. find its focal length in water. (µ

_{g}= 3/2 ; µ

_{w}= 4/3)

Solution: .............................................. (i)

........................................... (ii)

dividing equation (i) by (ii), we get

=> f

_{water}= 4 f

_{air}= 4 x 10

= 40 cm

3. A spherical convex surface separates object and image space of refractive index 1.0 and 4/3. If radius of curvature of the surface is 10 cm, find its power.

Solution: For finding focus, we choose our object to be at , then image is formed at focus

Diagram

P =

P = 2.5 dioptre

3. An isotropic point source is placed at a depth h below the water surface floating opaque disc is placed on the surface of water so that the bulb is not visible from the surface. What is the minimum radius of the disc. Take refractive index of water = µ.

Solutuion:

Diagram

Light from the source will not emerge out of waterif i >

_{C}

Therefore, minimum radius R correspond to i =

_{C}

In SAB,

= tan

_{C}

R = h tan

_{C}

=

4. A pile 4m driver into the bottom ofa lake is 1m above the wate. Determine the length of the shadow of the pile on the bottom of the lake if the sum rays make an angle of 45

^{0}with the water surface. (µ

_{w}= 4/3)

Solution:

Diagram

From Snell's law,

=> Sin r =

EF = DE tan r

= 3 tan(sin

^{-1})

= 1.88 m

Total length of shadow L = CF

L = (1 + 1.88)m = 2.88 m

5. The angle of minimum deviation for a glass prism with µ = equaly the refracting angle of the prism. What is the angle of prism ?

Solution: Given A =

_{m}

6. The angle of a prism are 30

^{0}, 60

^{0}and 90

^{0}. What should be the refractive index of a liquid to be kept in contact with the longest face so that a ray normally incident on the smallest face gets just internally reflected at the longest face ? (µ = 1.5)

Solution: The given prism ABC has the face AC the longest (opposite to 90

^{0}angle) and BC the shortest (opposite to 30

^{0}angle).

Diagram

PQ is the normally incident ray at the face BC and just gets totally reflected at the longest face AC.

For the glass-liquid interface AC

_{g}µ

_{L}=

In CQD, CQD = 30

^{0}

DQN = 60

^{0}

i = 60

^{0}

µ

_{L}= µ

_{g}sin 60

^{0}= 1.5 x = 1.299

7. A concave lens focusses a distant object 40 cm from it on a screen placed 10 cm away from it. A glass plate (µ = 1.5) and of thickness 3 cm is inserted between lens and a screen, where should the object be placed, so that its image is again focussed on the screen.

Solution:

f = 8 cm

Diagram

Shift due to glass slab = t

= 1 cm

For this position of object, image would have formed at 10 - 1 = 9 cm

if glass slab was not present

u = - 72 cm

Object should be placed 72 cm from lens.

8. Two plane concave lens of glass (µ = 3/2) have radius of curvature 20 & 30 cm, they are placed in contact with corved surface towards each iother and the space between them is filled with a liquid of refractive index 4/3. Find the focal length of the system.

Diagram

Solution: P

_{net}= P

_{1}+ P

_{2}+ P

_{3}

f

_{net}= = - 72 cm (focal length of system)

9. A small fish 0.4 m below the surface of a leke is viewed through a simple converging lens of f = 3 m. The lens is kept at 0.2 m above the water surface such that the fish lie on the optical axis of the lens. Find the image of the fish seen by observer, the refractive index of water = 4/3.

Solution: Apparent depth of fish = = 0.3

Diagram

Apparent position of object = - (0.3 + 0.2) = - 0.5

image of the fish is formed at same point where fish is present.

10. A quarter cylinder of radius R, µ

_{1}= 1.5 is placed on the table. A point object P is kept at a distance 'mR' from it, find the value of 'm' for which a ray from 'P' will emerge parallel to the table. Consider all rays to be ........

Solution: For first refraction, the object will appear to be kept at µmR from B for second refraction µ

_{1}= 1 µ

_{2}= 1

Diagram

u = -(µmR + R) R = - R v =

=>

=> µm = 2

=> m =

s

m = 4/3.

**Medium**1. A cylinder with radius R = 0.1 m has µ = 3/2. Find the

(a) height 'h' of the ray so that the emergent ray emerges from , the end of a diameter parallel to surface.

(b) If the cylinder is kept over a plane mirror then at what distance from the centre of this cylinder rod, should identical rod be placed so that the final emergent ray is parallel to the incident ray.

Solution:

Diagram

Since ray is ll to AB

BOC = i also BOC = 2r = i (exterior angle theorem)

Now from Snell's law

sin i = µ sin r = sin r

=> sin 2r = sin r

=> 2 sin r cos r = sin r

=> cosr = 3/4 sin r =

Now sin i = 2 sinr cosr = 2 x

from CDO

CD = h = CO sini = R sini =

(b) For the emergent ray to be parallel to the incident ray, the ray pattern must be symmetric.

Diagram

in OAD

tani =

distance between centres = 2R + 2AD =

2. Equiconvex lens of refractive index 3/2 and f = 10 cm is held with its axis vertical and lower surface is immerged in water (µ = 4/3). The upper surface being in an. At what distance from the lens will a vertical beam of parallel light incident on lens to be focussed.

Solution:

Diagram

R = f = 10 cm

For first refraction

u = µ

_{1}= 1 µ

_{2}= 3/2 R = 10 cm

v = 20 cm

beam will be focussed at 20 cm from the lens inside the water.

3. A point 'O' is placed at a distance of 0.3 m from a convex lens, focal length 0.2 m, cut into2 halves each of which is displaced by 0.0005 m as shown in fig. Find the position of the image. If more than one image is formed. Find their numbers and distance between them.

Solution: Two images will be formed. Both pieces will act as single lens system with axis as the line joining object and pole of each segment.

Diagram

for first lens

object distance = OO

_{1}=

v = 0.6 = OA

_{1}

O

_{1}OD AOI

AI = 0.15 cm

distance between two images = 2 AI = 2 x 0.15 = 0.3 m

4. A right angled prism (45

^{0}- 90

^{0}- 45

^{0}) of refractive index n has a plate of reflective index n

_{1}(< n) comented to its diagonal face. The assembly isin an a ray is incident on AB.

(a) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle.

(b) Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal undeviated.

Solution:

Diagram

in AOO'

45 + (90 - r) + ( 90 - c) = 180

^{0}

r + c = 45

^{0}

r = 45

^{0}- c

sinr = (cosc - sinc)

for critical angle C sinc =

for refraction at first surface

sin = n sinr =

(b) From Geometry it can be seen that r = 45

^{0}, for ray to be incident normally on diagonal face.

Diagram

taking refraction at first surafce.

sin = n sin45

^{0}

**HARD**1. The x - y plane is boundary between two transparent media. Medium-1 with z 0 has a refractive index and medium 2 with z 0 has refractive index . A ray of light inmedium-1 given by vector is incident on the plane of separation, find the unit vector in the direction of the refracted ray in medium-2.

Solution: Let refracted ray br

Normal to plane of incident and normal =

=

Diagram

it must also be normal to refracted ray

cos ( - i) =

= = cos120

^{0s}

i = 60

^{0}

r = 45

^{0}

Now since angle between refracted ray and Normal = 45

^{0}

c

^{2}= a

^{2}+ b

^{2}= a

^{2}+

c =

2. A ray of light in air is incident atgrazing angle (i = 90

^{0}) on a long rectangular slab of a transparent medium of thickness t = 1.0 m. The point of incidence is the oprigin A(0, 0). The medium has a variable index of refraction n(y) given by n(y) = where k = 1.0 m

^{-3/2}. The refractive index of air is 1. (i) Obtain a relation between the slope of the trajectory of the ray at a point B(x, y) in the point. (ii) Obtain an equation for trajectory y(x) of the ray in the point. (iii) Determine the co-ordinates (x, y

_{1}) of the point P where the ray intensects the upper surface of the slab-air boundary. (d) Indicate the path of the ray subsequently.

Diagram

Solution:

Diagram

Taking on arbitrary point P(x, y) refractive index at this point n =

from Snell's law n sin = constant applying this for initail pt. (when ray is entering medium B) and at point.

1 x sin90

^{0}= sini

sin i = it can be seen that i =

s Slope = tan = cot i =

diagram

(ii) = coti =

it passes throughorigin C = 0

x = 4 is the equation of trajectory

when ray comes out of the mediums

then x = 4 x 1 = 4

Co-ordiante of pt- is (4, 1)

If medium on both sides are same, then angle with which the ray enters the medium = angle with which the ray comes out.

ray will be parallel to x-axis