Ray Optics and Optical Instruments

Level - I

SECTION - A

School/Board Exam. Type Questions Very Short Answer Type Questions :

1. What is the use of erecting lens in a terrestrial telescope? What is its magnification? Sol. It will invert the intermediate image. Magnification m = – 1

2. For the same angle of incidence, the angles of refraction in three different media A, B and C are 30°, 40° and 45° respectively. In which medium will the velocity of light be maximum?

Sol.  sin = constant; v _{ ; velocity  sin}1  v_A < v_B < v_C

3. Mention the parameter on which the colour of the light depends. Sol. Colour depends on frequency which is fundamental property of wave.

4. A ray of light while travelling from a denser to rarer medium grazes the separating surface. Express the critical angle in terms of the speed of light in the two media.

Sol. sinC = _{ =} 1 medium

air

v v

5. Define absolute refractive index. What is the relation between the relative refractive index between two media and the absolute refractive index of the media?

Sol. Absolute refractive index = c

v , Relative refractive index of medium-2 w.r.t. medium-1 2 = 2 1



 , where 2

and ₁ are their respective absolute refractive indices. 6. Why do diamonds sparkle?

Sol. Total internal reflection inside diamond due to its high refractive index.

7. Two lenses of focal lengths 5 cm and 50 cm are to be used for making a telescope. Which will you use as the objective?

Sol. 50 cm

8. State the functions of an eye piece.

Sol. To magnify the image formed by the objective lens.

Chapter

9

Ray Optics and Optical

Instruments

9. Draw the ray diagram for convex mirror producing real image. Sol. C O I O - Virtual object I - Real image

10. What is the minimum resolving power of simple microscope? Sol. 1

Short Answer Type Questions :

11. Show that the maximum value of the angle of prism is twice its critical angle. Sol. At maximum value of angle of prism.

Angle of incidence = Angle of emergence = 90°

i = 90° 90° = e

Incident

ray Refracted ray

A

C C

r1 r2

We have r₁ = r₂ = C Again A = r₁ + r₂= 2C

12. Distinguish between a microscope and a telescope.

Sol. Microscope is used to see microobjects whereas telescope is used to see the distant objects. Generally telescope provides resolution and microscope provides magnification.

13. What is dispersion? Define a physical quantity which depends only on the material of the prism for its dispersion. Obtain an expression for it.

Sol. The phenomenon of splitting of white light into its seven constituent colours by refraction through the prism is known as dispersion.

Dispersive power depends only on the material of the prism. Expression for dispersive power  =  

14. Write the phenomenon by which a bubble of air coming out through water in a glass vessel appears silvery to an observer.

Sol. Total internal reflection.

15. How will the magnifying power of an astronomical telescope be affected on increasing the focal length of objective lens?

Sol. Increases as MP = O,

e

f

f for normal adjustment and MP = Oe 1 e

f f

f D

 _ 

 

 , for image at D) 16. Write the relation between critical angle and refractive index.

Sol. C = sin  –1 1_  

17. Distinguish between an astronomical telescope and a terrestrial telescope in terms of image formation. Sol. Final image produced by astronomical telescope of magnified and inverted. Final image produed by terrestrial

telescope is magnified and erect.

18. Write the phenomenon by which an empty test tube in a beaker of water illuminated from one side presents a silvery appearance.

Sol. Total internal reflection.

19. A thin glass prism produces an angle of deviation  in air. How the angle of minimum deviation will change, if the prism is immersed in a liquid of refractive index _i < _g?

Sol. In a thin glass prism, deviation in air  = (_g – 1) A.  = Absolute refractive index of prism.

When the prism is dipped in a liquid (l) deviation  = – 1 g l A    _      < 

20. Draw the working of Astronomical telescope with the help of ray diagram for normal adjustment.

Sol. _{Objective Eyepiece}

O1   O2

21. Describe the construction and working of an optical fibre. State its uses. Sol. Cladding ( )2 Cladding Sheath Sheath CORE ( )1 i C >

Core – Transparent fibre of higher refractive index (₁) Cladding – Transparent fibre of lower refractive index (2)

Sheath – Hard, tough, protective and opaque covering of fibre. Ray of light incident on the interface of cladding and cone at angle more than critical angle so the ray suffers, total internal reflection on each incidence. It is used for long distance transmission of signal with minimum loss of energy.

22. What is the difference of maximum to minimum magnifying power of simple microscope.

Sol. MP_max = 1 D f  MP_min = D f MP_max – MP_min = 1

23. What is the maximum refractive index to get an emergent ray from an equilateral prism? Sol. Angle of Prism A = 60°

_max = cosec cosec60 2

2 2

A  

24. What is the behaviour of the eyepiece of astronomical telescope? Sol. It behaves like a simple magnifier. Producing virtual and erect image.

25. What is the expression for maximum magnifying power of compound microscope?

Sol. mp = 0 0 1 1 e o e v D L D u f f f  _ _  _         

L = Length of tube of microscope f_o = Focal length of objective D = Least distance of distinct vision f_e  focal length of eye piece

26. What is astigmatism? What is the remedy for this defect?

Sol. Astigmatism : This is due to defect in the cornea, that is cornea is not spherical in shape. It may have large curvature in vertical plane than in the horizontal plane. It results in lines in one direction being well focussed while those in a perpendicular direction may appear distorted. This occurs when the cornea is not spherical in shape Image as formed on the retina Cylindrical lens Image as formed on the retina

It is corrected by using cylindrical lens of desired radius of curvature. This defect can occur along with myopia or hypermetropia.

27. What is the process of formating primary rainbow?

Sol. It is the result of three-step process,  Refraction with dispersion  Internal reflection  Refraction. It is found that violet light emerges at an angle of 40° related to the incoming sunlight and red light emerges at an angle of 42°. Thus, the observer sees a primary rainbow with red colour on the top and violet on the bottom. R V 40° violet 42° red Sunlight

28. Write the laws of reflection from a surface.

Sol. The laws of reflection state that the angle of reflection (angle between reflected ray and the normal to the reflecting surface) equals the angle of incidence (Angle between incident ray and the normal). Also that the incident ray, reflected ray lie in the same plane with normal to the reflecting surface. These laws are valid at each point on any reflecting surface whether plane or curved.

 

Incident ray Normal Reflected ray

29. Write the sign convension used in spherical mirrors?

Sol. To derive any formulae for reflection at spherical surfaces we must first adopt a sign convention for measuring distances. According to Cartesian sign convention all distances are measured from pole of the mirror.

F M P C Concave Mirror Upward

height (+ve) Incident light (+ve)

Downward height (–ve)

Distance opposite to incident light

(–ve) _{Distance along} incident light (+ve)

Principal axis

P – Pole ; F – Focus ; C – Centre of Curvature PF = f = Focal length of mirror.

CP = R = Radius of curvature of mirror.

The distances measured in the same direction as the incident light are taken as positive and those measured in the direction opposite to the direction of incident light are taken as negative. Heights measured above principal axis are taken as positive and the heights below the principal axis are taken as negative.

30. Define dispersion. Draw the ray diagram showing dispersion through a prism. Sol. It is the separation of light into its constituent colours.

R O Y G B I V white light A

Long Answer Type Questions :

31. Derive expression for deviation produced by prism.

Sol. Let ABC be the principal section of a prism. Its refracting angle is A. The refractive index of the material of the prism is . The prism is kept in air which is a rarer medium with respect to the material of the prism. Let PQ be the incident ray on the first refracting surface AB of the prism. When it enters into the prism it bends towards the normal TN₁ as ray QR, as the light ray travels from rarer to denser medium. Let the angle of incidence be i, and the angle of refraction be r₁. After traveling through the prism it falls on the second refractive surface AC. The light ray bends away from the normal TN₂ and travels along RS which is called emergent ray. The angle between the normal and the emergent ray is called angle of emergence denoted by e. The angle between the normal TN₂ and the refracted ray is r₂.

T R Q i  A N1 r1 r2 i r – e r – _e N2

The incident ray PQ on the prism does not travel straight when refracted by the prism. It bends along QR and emerges in the direction RS i.e., the incident ray PQ is bent along RS. The total angle through which the light ray is bent on being refracted twice by prism is called the angle of deviation. It is denoted by . Angle of deviation is also defined as the angle between the incident ray. Produced forwards and the emergent ray produced backwards. This is shown in figure.

In the QRT r₁ + r₂ + T = 180° In the quadrilateral AQTR A + 90° + T + 90° = 360° T = 180° – A Substituting in r₁ + r₂ + T = 180° r₁ + r₂ + 180° – A = 180° r₁ + r₂ = A ...(1) i – r₁ + e – r₂ + 180° –  = 180° i + e – (r₁ + r₂) =  (∵ r₁ + r₂ = A) i + e – A =  i e A    ...(2)

32. With the help of a ray diagram, illustrate the formation of the final image of an object in a compound microscope. Derive an expression for its magnifying power.

Sol. Compound Microscope

The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eye piece, which functions essentially like a simple microscope or magnifier, producing an enlarged virtual final image. The first inverted image is thus near (at or within) the focal point of the eye piece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point clearly, the final image is inverted with respect to the original object.

Magnification power m = i O   = IG IG IO IJ OJ IO    _{   }  _{  }       





Multiplying and dividing by I₁G₁

Objective v0 l O u0 O G1 l1   O G

O O O, , are the representation 

m = 1 1 1 1 I G IG OJ I G             e O m m m

From similar triangles OO J and O I₁G₁

1 1 IJ OO I G O l   

m_O = 1 1 O

O

v I G

OJ u ...(i)

The magnifying power of eye piece is same as that of simple microscope.

m_e = 1 1 e D v f  _      ...(ii)

Substituting (i) and (ii) in m = m_O m_e

o o e v D D m u V f    _  _  

Case I : When the image forms at near point V = D

m = o o e v D D u D f        1 o o e v D m u f    _  _   If rouo = – fe and vo = L : 1 O e L D m f f    _  _  

Where L is length of microscope

Case II : When the image forms at far point v = 

m = o o e v D D u f  _  _    o o e v D m u f  If u_o >> f_o and v_o >> L ; o e L D m f f    _{ }  

Various other factors such as illumination of the object contributes to the quality and visibility of the image. 33. Prove that  = A( – 1) for a small-angled prism, where the symbols have their usual meanings.

Sol. Deviation by a prism  = (i + e) – A ...(1) Where A = r₁ + r₂

Force then prism incidence of light is nearly normal incidence. So i, e, r₁ and r₂ all are very small. Using Snell’s law to first surface 1sin i = sin r₁

 i = r₁ ...(2)

i

A e r1 r2

Using Snell’s law to second surface sinr₂ = 1sine

e = r₂ ...(3)

Using equation (1), (2) and (3)  = r₁ + r₂ – A = (r₁ + r₂) – A

34. Explain the construction and working of an astronomical telescope. Calculate its magnifying power when the image is formed at the least distance of distinct vision.

Sol. The telescope is used to provide angular magnification of distant objects. It also consists of an objective and an eye piece. But here, the objective has a large focal length and a much larger aperture than the eye piece. Light from a distant object enters the objective and a real image is formed in the tube at the second focal point of the convex objective lens. The eye piece magnifies this image producing a final inverted image.

0 0 I G fo ue G1 1 O Objective Eyepiece I1 Magnifying power = 1 0 1 1 – e – o i e o I G u f u I G f      _{  }        ...(i) From 1 _–1 – 1 e e f  v u 1 1 1 e e u f v ...(ii)

Substituting (ii) in (i)

1 1 – o e m f f v    _  _  

Case I : If the image to be formed at near point v = D.

m = – _o 1 1 e f f D        1 e e e f f m f D    _  _   Length of telescope L = f_o + u_e o e e Df L f D f   

35. Derive the relation between refractive index and minimum deviation. (Prism formula)

Sol. As the angle of incidence is increased gradually, it is found experimentally that the angle of deviation decreases till it reaches a minimum value and then it increases. The least value of deviation is called angle of minimum deviation is denoted by . The graph between the angle of incidence i and the angle of deviation ‘’ is as shown in figure. If a horizontal line is drawn for a particular value of angle of deviation, it intersects the graph

at two points corresponding to two angles of incidence i and e. e also represents the angle of emergence. Since, when the light is incident along SR it will follow the path RQ and emerges as QP i.e., angles of incidence and emergence are interchangeable. When D decreases the two angles i and e become closer to each other and at the angle of minimum deviation, the two angles of incidence are same i.e., i = e. As i = e, r₁ = r₂. Taking i = e = i, r₁ = r₂ = r and substituting in equations 1 and 2, we get

1 2 _m i1 i e= i2 i - curve  2r = A  r = 2 A i + i = A + _m  i =   2 m A

By Snell’s law of refraction, the refractive index of the material of the prism is

1

1 2

sin sin sin sin sin sin

i e i r r r    

_

_

_

_

Which is the relation between the refractive index of the material of the prism and the angle of minimum deviation. It is called prism formula.

36. What is meant by linear magnification of spherical mirrors? Deduce the formulae for the same.

Sol. M P B F A B A f v u

The rays emanating from a point actually meet at another point after reflection (refraction) is called as the image. Hence, an image is a point-to-point correspondence of object. In practice we can take any two rays emanating from a point on an object, trace their paths and find the intersection to obtain image.

Figure shows the ray diagram of tracing the image AB of an object AB formed by a concave mirror. If u is object distance and ‘v’ is image distance and ‘f ’ is the focal length from pole ‘P’ then from similar triangles ABF and MPF. A B PM   = B F FP  ...(i) Have AB = PM

From right-angled triangles APB and APB

A B BA   = B P BP  ...(ii)

From (i) and (ii) B F B P FP BP  _  – B P FP B P FP BP  _ 

According to sign convention BP = – v; FP = – f ; BP = – u

– – – – v f v f  u – v f v f u 1 1 1

v   (This is called as mirror equation)u f

For magnification which is the ratio of size of image to object. We have from ABP and ABP.

B A B P BA BP  _ 

 

 

37. Show by drawing ray diagram how a totally reflecting glass prism can be used to deviate a ray of light through (i) 90° (ii) 180° and invert it.

Sol. Prisms designed to bend light by 90° or 180° for total internal reflection, such prisms are used to invert images without changing their size.

90° 45° 45° 45° 45° 90°

38. Describe a reflecting type telescope what are its advantages? Sol. Reflecting telescope (Cassegrain)

Secondary mirror

Eyepiece Objective mirror

Telescopes with mirror objectives are called reflecting telescopes. Advantages of taking mirror objectives are 1. There is no chromatic aborration in a mirrors.

2. If a parabolic reflecting surface is chooser, spherical aberration is also removed.

3. Mechanical support is much less of a problem since a mirror weighs much less than a lens of equivalent optical quality and can be supported over. Entire back surface not just over rim unlike lens.

The largest reflecting telescope in the world are the pair of reck telescopes in Hawaii, USA, with reflector of 10 metre in diameter.

39. _{Derive the expression for focal length of a spherical mirror F = 2}R.

Sol. As we know parallel beam of light ray is incident on a concave or convex mirror as shown in figure, the rays will converge (appear to diverge) at a point F called principal focus.

F M P C Concave Mirror C P F Convex Mirror M

In case of a concave mirror the reflected rays will converge at focus F whereas in case of convex mirror the reflected rays appear to diverge from focus F.

Convex Mirror P _F C M    2 M Concave Mirror 2 P F C    To show 2 R

f  , where f = Focal length = Distance between pole and principal focus R = Radius of curvature of mirror.

From figure MCP  2 MFP    tan = MP CP ; tan2 = MP FP Considering When  is small tan ; tan2 2

 MP FP  2MP CP FP = 2CP 2 R F 

40. Define scattering of light. Explain the blue colouration of sky and Reddish appearance of sun in full moon near horizon.

Sol. The change in direction of light by the particles of medium through which light passes is known as scattering. As sunlight travels through the earths atmosphere it gets scattered by the atmospheric particles.

Rayleigh Scattering : The amount of scattering is inversely proportional to the fourth power of the wavelength. Therefore, light of shorter wavelengths is scattered much more than light of longer wavelengths.

Blue Sky : Blue has a shorter wavelength than red and is scattered much more strongly. Infact violet is scattered even more than blue having shorter wavelength. But since our eyes are more sensitive to blue than violet, we see the sky blue.

Sun near horizon

Sun nearly overhead

Distance of atmosphere through which sunlight travels in atmosphere

Reddish appearance of sun in full moon near horizon : At sunrise and sunset sun rays pass through a larger distance through atmosphere so most of blue and other shorter wavelengths are removed by scattering. 41. Draw labelled diagram and write working of eye.

Sol. Light enters the eye through cornea a curved front surface. It passes through the pupil which is the central hole in the iris. The size of pupil can change under control of muscles. The light is further focussed by the eye lens on the retina. The retina is a film of nerve fibres covering the curved black surface of the eye. The retina contains rods and cones which sense light intensity and colour respectively and transmit electrical signals via the optic nerve to the brain.

Crystalline lens Aqueous humour Pupil Iris Cornea Ciliary muscles Retina Optic nerve Vitreous humour

The shape (curvature) and therefore the focal length of the lens can be modified somewhat by ciliary muscles. So images are formed at the retina for objects at all distances. This property of the eye is called accommodation.

The closest distance for which the eye lens can focus light on the retina is called the least distance of distinct vision or the near point. The standard value for normal vision is taken as 25 cm (Symbol D).

If the object is too close to eye; the lens cannot curve enough to focus the image on the retina, and the image is blurred.

Least distance of distinct vision increases with age due to decreasing effectiveness of ciliary muscle and the loss of flexibility of the eye lens.

It may be as close as about 7 to 8 cm in a child ten years of age and may increase to as much as 200 cm at 60 years of age. So if an elderly person tries to read a book at about 25 cm from the eye, the image appears blurred. This defect of eye is known as presbyopia. It is corrected by using converging glass for reading. 42. Write the formation of secondary rainbow.

Sol. It is the result of four-step process  refraction with dispersion  internal reflection  internal reflection  refraction.

It is found that red light emerges at an angle of 50° related to the incoming sunlight and violet light emerges at an angle 53°. Thus, the observer sees of secondary rainbow with violet colour on top and red on the bottom. The intensity of light is reduced at second internal reflection and hence it is fainter than primary rainbow.

red

violet sunlight

53° 50°

43. Draw the labeled diagram of cossegrain telescope. Write the the advantages of cassegrain telescope. Sol. Telescopes with mirror objectives are called reflecting telescopes. Advantages of taking mirror objectives are

Secondary mirror

Eyepiece Objective mirror

1. There is no chromatic aberration in a mirrors.

2. If a parabolic reflecting surface is choosen, spherical aberration is also removed.

3. Mechanical support is much less of a problem since a mirror weighs much less than a lens of equivalent optical quality and can be supported over. Entire back surface not just over rim unlike lens.

The largest reflecting telescope in the world are the pair of reck telescopes in Hawaii, USA, with a reflector of 10 metre in diameter.

44. Discuss the methods of image tracing in convex lens and concave lens. Sol. Image Tracing :

We may use geometrical ray tracing to determine the position of image formed by lens. The following points may consider. A f2 f1 F1 h1 B A B u v h2 (Image) (Object) F2

(1) The ray parallel to principal axis which after refraction, passes through focal point F₂.

(2) The ray that passes through the first focal point F₁ of the lens, after refraction it travels parallel to the principal axis.

(3) The ray usually called chief ray goes through optical centre of the lens and emerges without deviation. Same can also be seen in case of concave lens.

F (Image) h2

Object ( )h1

45. Derive the equivalent focal length of lenses in contact.

Sol. Consider two lens of focal length f₁ and F₂ placed in contact with each other as showing in figure. Let ‘O’ be the point object beyond focus of first lens, forms a real image I. Since the image I is real it behaves as a virtual object for second lens producing final image at I. (Assuming lens are thin optical centres of the lenses to be coincident)

I I

v

O v

u

For first lens,

1

1 _– 1 1

v u f ...(i)

For the second lens,

2

1_– 1 1

v v f ...(ii) Adding (i) and (ii)

1 2

1 1 1_– 1 v u f f

If the two lens system is regarded a single series having equivalent focal length f then 1 1 1_– v u f  1 2 1 1 1 f  f f ...(iii)

For several lenses in contact the equation can be rewritten as

1 2 3

1 1 1 _{1 ....} f  f f f 

In terms of power P = P₁ + P₂ + P₃ + ....

Combination of lens helps to obtain diverging or converging lens of desired magnification. Total magnification is the product of magnification of individual lenses M = m₁m₂m₃...

Such system of combination of lens will be used in camera, microscope, telescope such as optical instruments.

SECTION - B

Model Test Paper Very Short Answer Type Questions :

1. Write the phonenomenon by which blue colour of sky can be explained. Sol. Scattering of light by atmospheric molecules of size a << .

2. Write the speed of light in vacuum. Sol. Speed of light in vacuum (3 × 108 _ms–1₎

3. What is the critical angle of a medium whose refractive index is _{2 ?} Sol. 45°

4. Under what conditions total internal reflection occurs? Sol. (i) Light must go from denser to rarer medium.

(ii) The angle of incidence in the denser medium must be greater than the critical angle of the pair of media. 5. A curved surface is convex to the incident light does it form real or virtual image?

Sol. If the object is virtual (incident rays are converging) then image is real.

6. A bi- convex lens of focal length f is bisected along the principal axis. What is the focal length of each half? Sol. f

7. Write the thin lens formula and describe the symbols used.

Sol. 1 1 1– v u f ,

u  Object distance from optic centre v  Image distance from optic centre f  Focal length from optic centre

8. Which colour of light travels with least speed in a glass prism? Sol. Violet

Short Answer Type Questions :

9. Explain how an air bubble in water behaves as a diverging lens.

Sol. If a parallel beam of light is incident on an air bubble in water then the refracted rays diverge from each other as shown in figure. air i i water r

10. A distance x separates two thin convex lenses of focal lengths f and 2f. For what value of x a parallel beam of light falling on this combination parallel to the common principal axis emerges out of the combination as a parallel beam?

Sol. Parallel beam comes out from the combination as parallel beam means effective power of arrangement is zero or effective focal length is infinity.

We, have 1 f = 1 2 1 2 1 1 – x f f f f  1 1 1 – 2 2 x f f ff     x = 2f2_1_f ₂1_f_  x = 2f2 _× 3 6 2f  f

11. Show that for a thin prism the deviation is given by  = A( –1). Sol. Deviation by a prism  = (i + e) – A ...(i)

Where A = r₁ + r₂

i

A e r1 r2

For thin prism incidence of light is nearly normal incidence. So, i, e, r₁ and r₂ all are very small. Using Snell’s law to first surface 1 sini = sin r₁

 i = r₁ ...(ii)

Using Snell’s law to second surface  sin r₂ = 1 sine

 e = r₂ ...(iii)

Using equation (i) (ii) and (iii)  = r₁ + r₂ – A

= 





12. Draw the ray diagram for formation of primary rainbow.

Sol. R V 40° v 42° Raindrop R Horizon

13. A bi-convex lens has a focal length equal to half the radius of curvature of either surface. What is the index of refraction of the lens material?

Sol. Focal length of biconvex lens of equal radii of curvature is

f = ₂

_

_

_

_

14. A ray of light enters a liquid and a bent toward the normal as shown. Calculate the speed of light in the liquid. 3cm 3cm 3cm 4cm i r

Sol. From the figure, i = tan–1 3

3       = 45° r = tan–1 3 4       = 37°

Applying Snell’s law, sin sin i c r v  v = sin sin r c i  =





15. If the absolute refractive index of glass is 3

2 and water is 4

3. Find the ratio of focal lengths of a glass lens when in water and when in air.

Sol. Focal length of glass lens in water,

f_w = 2 g – 1 w R    _   

Focal length of glass lens in air,

f_a = ₂





Short Answer Type Questions :

16. A ray of light is incident on a glass sphere at an angle of incidence . Show that angle of emergence is . Sol. In BCD, BC = CD = R = Radius of sphere

  C r r D A B Air Glass E

Applying Snell’s law to Ist _surface

1sin = sinr ...(i)

Applying Snell’s law to 2nd surface sinr = 1sin . ..(ii) Comparing equation (i) and (ii)  = 

17. Write lens maker’s formula and describe the symbols used in it.

Sol. 2 1 1 2 1 _{– 1} 1 _– 1 f R R      _        R2 R1 1 2 1

₂ = Refractive index of material of lens ₁ = Refractive index of material outside lens R₁ = Radius of curvature of first surface R₂ = Radius of curvature of second surface

18. It is desired to decrease the focal length of a convex lens. Should you keep it in contact with a convex or a concave lens?

Sol. We have effective power of lenses in contact P = P₁ + P₂

1 2 1 1 1 f f f  _{ }    ∵ 

Decrease of focal length means increase of power. So it should be kept in contact with convex lens. 19. A curved surface is convex to the incident beam of light. Does it form real image? Draw the ray diagram. Sol. Yes, if the rays are converging as shown (virtual objects)

O C I

20. Mention any four applications of total internal reflection phenomenon. Sol. (i) Mirage (Inferior image)

(ii) Looming (Superior image) (iii) Optical fibre

21. When a monochromatic visible light beam enters water from air, how are its speed, wavelength and frequency affected?

Sol. Frequency does not change as it is the fundamental property of the wave. Absolute refractive index of the water

= air air water water f c v f       Again  > 1  v < c and _water < _air

Long Answer Type Questions :

22. Explain the terms ‘Critical angle’ and ‘total internal reflection’. How does total internal reflection differ from the reflection from a plane mirror?

Sol. The angle of incidence at the denser medium for which the angle of refraction in rarer medium is 90° or the refracted ray grazes with the boundary separating both media is known as critical angle.

i = C

Rarer ( )1

r = 90°

Denser ( )2

Using Snell’s law ₂sinC = ₁sin90°

sinC = 1 12

2 21

1    

 

If the angle of incidence at the denser medium is more than critical angle then there is no refracted ray at all. Total intensity of the incident light is reflected back to the same denser medium this phenoments is known as total internal reflection.

Rarer Denser i C > r i =

In mirror reflection, total intensity of the incident light is not reflected back to the same medium due to presence of reflecting material behind the glass. There is partial absorption of light by the reflecting material. 23. Derive lens maker’s formula.

Sol. O C2 R1 1 v 2 1 v1 –u C1 I I1 N1 N2

C₁N₁  Normal to first surface, C₁-Centre of curvature of first surface. C₂N₂  Normal to 2nd surface, C₂-Centre of curvature of 2nd surface. For first surface

O  Object I₁  Image

Using spherical surface refraction formula, 2 1 2 1

1 1 – – v u R   _  ...(i) For the second surface

I₁  Virtual object I  Image

Using formula for refraction through spherical surface, 1 2 1 2

1 2 – – v v R   _  ...(ii) Adding equation (i) and (ii)

1_– 2 v u   =





Comparing (iii) & (iv)

2 1 1 2 1 _{– 1} 1 _– 1 f R R      _       

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Objective Type Questions (Reflection of Light by Spherical Mirrors)

1. Day and night settings for rearview mirrors uses

(1) Thin mirrors (2) Thick wedge shaped mirrors

(3) Convex mirrors (4) Concave mirrors

Sol. Answer (2)

Day and night settings for rear view mirrors.

2. While capturing solar energy for commercial purposes we use

(1) Parabolic mirrors (2) Plane mirrors (3) Convex mirrors (4) Concave mirrors Sol. Answer (1)

While capturing solar energy for commercial purposes we use parabolic mirrors to converge the rays coming from infinity to a point.

3. A convex mirror is used to form an image of a real object. Then mark the wrong statement (1) The image lies between the pole and focus (2) The image is diminished in size

(3) The image is erect (4) The image is real

Sol. Answer (4)

A convex mirror always forms a virtual image in the care of a real object.

In care of a virtual object reflected rays may intersect really to make a real image.

4. A concave mirror of focal length f produces an image n times the size of the object. If the image is real then the distance of the object from the mirror is

(1) (n – 1) f (2) n_n f        )( 1 (3) n_n f        )( 1 (4) (n + 1) f Sol. Answer (3) (magnification) m = f f u− Focal real image m = –n

–n = f f u − − −  u = f n( 1) n + −

5. A convex mirror has a focal length f. A real object is placed at a distance f in front of it, from the pole. It produces an image at

(1) Infinity (2) f (3) f/2 (4) 2f

Sol. Answer (3)

Mirror formula : 1 1 1 v u+ =f

Here object is real so u is negative

1 1 1 v u− =f Also (u) = f 1 1 1 v f− =f  2 f v =

6. An object placed in front of a concave mirror of focal length 0.15 m produces a virtual image, which is twice the size of the object. The position of the object with respect to the mirror is

(1) –5.5 cm (2) –6.5 cm (3) –7.5 cm (4) –8.5 cm Sol. Answer (3) m = f f u− f = –0.15 m m = +2 (virtual image) 2 0.15 0.15 u − = − −  = –.075 m or – 7.5 cm. (Refraction)

7. When a light ray from a rarer medium is refracted into a denser medium, its

(1) Speed increases, wavelength increases (2) Speed decreases, wavelength increases (3) Speed increases, wavelength decreases (4) Speed decreases, wavelength decreases Sol. Answer (4)

When a light ray goes from a rarer to denser medium by definition of refractive index, its speed creases. Its frequency is found to be invarant.

So, if velocity of light in a medium is v : v = f [by wave theory] or v 

So, if v decreases,  also decreases.

8. A narrow, paraxial beam of light is converging towards a point I on a screen. A plane parallel plate of glass of thickness t, and refractive index μ is introduced in the path of the beam. The convergence point is shifted by (1) t (1–1/) away (2) t (1 + 1/) away (3) t (1 – 1/) nearer (4) t (1 + 1/) nearer Sol. Answer (1)

Longitudinal shift is given by t – _µt . Hence, point of convergence shifts by the same amount.

9. The length of a vertical pole at the surface of a lake of water        3 4

is 24 cm. Then to an under-water fish just below the water surface the tip of the pole appears to be

(1) 18 cm above the surface (2) 24 cm above the surface (3) 32 cm above the surface (4) 36 cm above the surface Sol. Answer (3)

Apparent height Real height µ =

Now, real height = 24 cm and  = 4 3 4

24 Apparent height 3× =

 Apparent height = 32 cm

10. A ray of light strikes a glass plate at an angle 60o_{. If the reflected and refracted rays are perpendicular to each}

other, the index of refraction of glass is

(1) ₃ (2) 3/2 (3) (32) (4) 1/2

Sol. Answer (1)

Angle NOT = 30º this is found by geometry after putting angle between refleted and transmitted ray equal to 90º

sin sin60º sin sin30º i r µ = = _{60º 60º} 30_º90º N N' O I T R Reflected ray Refracted ray or 3 2 2 1 µ = ×  µ = 3

11. A microscope is focussed on a coin lying at the bottom of a beaker. The microscope is now raised by 1 cm. To what depth should water be poured into the beaker so that the coin is again in focus? (The refractive index of water is

3 4₎

(1) 1 cm (2) 4/3 cm (3) 3 cm (4) 4 cm

Sol. Answer (4)

Water should be poured such that shift in depth of image is 1 cm 1 d d ⎛ ₋ ⎞₌ ⎜ ⎟ ⎝ μ⎠ 3 1 1 4 d⎛_⎜⎝ − ⎞_⎟⎠= d = 4 cm

(Total Internal Reflection)

12. Two transparent media A and B are separated by a plane boundary. The speed of light in medium A is 2.0 × 108 _ms–1 _{and in medium B is 2.5 × 10}8 _ms–1_{. The critical angle for which a ray of light going from A to B}

suffers total internal reflection is

(1) sin–1 _{1/2 (2) sin}–1 _{2/5 (3) sin}–1 _{4/5 (4) sin}–1 _3/4

Sol. Answer (3) 8 8 3 10 _1.5 2 10 A × µ = = × ; 8 8 3 10 6 _1.2 5 2.5 10 B × µ = = = ×

R.I. going from A to B = B

A

µ

sin 0.8 sin AB C_r µ = = at critical angle r = 90º  sin 0.8 sin90º C = sin14 5 C= −

13. Which of the following is possible application of fibre optics?

(1) Endoscopy (2) High speed internet traffic

(3) Radio, TV & Telephone signals (4) All of the above Sol. Answer (4)

Total internal reflection is used in all of the above as they involve the use of optics fibres. (Refraction by Lenses)

14. An object is placed at a distance of f/2 from a convex lens. The image will be (1) At one of the foci, virtual and double its size (2) At 3f/2, real and inverted (3) At 2f, virtual and erect (4) At f, real and inverted Sol. Answer (1)

Lens formula 1 1 1 v u− =f

Object is real and placed at – 2f 1 2 1

v f+ =f

1 1

v = −f v = –f

15. The least distance between a point object and its real image formed by a convex lens of focal length F is

(1) 2 F (2) 3 F (3) 4 F (4) Greater than 4 F

Sol. Answer (3)

(Distance between a point object and its real image) d  4 f

16. The plane faces of two identical plano-convex lenses, each having focal length of 40 cm, are placed against each other to form a usual convex lens. The distance from this lens at which an object must be placed to obtain a real, inverted image with magnification '–1' is

(1) 80 cm (2) 40 cm (3) 20 cm (4) 160 cm

Sol. Answer (2)

It forms an equi-convex lens f = 0.4 m 1 2.5 D 0.4 P = = Power of combination = 5 D 1 f P =

1_m 5 f = or f = 20 cm

Therefore, object must be placed at 2f (= 40 cm)

17. Two thin lenses of focal lengths 20cm and –20cm are placed in contact with each other. The combination has a focal length equal to

(1) Infinite (2) 50 cm (3) 60 cm (4) 10 cm Sol. Answer (1) Powers : P₁ = 1 0.2 and P2 = – 1 0.2 = 5 D and –5 D Net power = 0  Focal length = 1 0 = 

18. If in a plano-convex lens, radius of curvature of convex surface is 10 cm and the focal length of lens is 30 cm, the refractive index of the material of the lens will be

(1) 1.5 (2) 1.66 (3) 1.33 (4) 3 Sol. Answer (3) R = 10 cm f = 30 cm 1 1 1 ( 1) f R ⎛ ⎞ = μ − _⎜⎝ − _⎟⎠ ∞ 1 _{( 1)} 1 30= µ − 10 4 3 µ =

19. A glass concave lens is placed in a liquid in which it behaves like a convergent lens. If the refractive indices of glass and liquid with respect to air are _a_g and _a_l respectively, then

(1) _a_g = 5_a_l (2) _a_g > _I_l (3) _a_g < _a_l (4) _a_g = 2_a_l Sol. Answer (3)

The glass lens behaves as divergent in air which has less R.I. It will behave as convergent in a medium of higher R.I.

20. The diameter of aperture of a plano-convex lens is 6 cm and its maximum thickness is 3 mm. If the velocity of light in the material of the lens is 2 × 108 _{m/s, its focal length is (approximately)}

(1) 10 cm (2) 15 cm (3) 30 cm (4) 60 cm Sol. Answer (3)  of medium = C v 3 mm 6 cm 8 8 3 10 1.5 2 10 × = = ×

(R – 3)2 _{+ (30)}2 _{= R}2 R2 _{+ 9 – 6R + 900 = R}2 909 = 6R R = 151.5 mm = 15.15 cm 0 1 _{( 1)} 1 1 f R ⎛ ⎞ = μ − _⎜⎝ − _⎟⎠ ∞ 0 1 _0.5 1 15.15 f = × 30 mm 3 mm R R – 3 f₀ = 30.30 cm

This is distance from optical centre O. Distance from lens is f = f₀ – thickness at principal axis = 30.3 – 0.3

 f = 30 cm

21. Two plano-convex lenses of equal focal lengths are arranged as shown

The ratio of the combined focal lengths is

(1) 1 : 2 : 1 (2) 1 : 2 : 3 (3) 1 : 1 : 1 (4) 2 : 1 : 2

Sol. Answer (3)

The power of lens remains the same no matter how the lenses are placed. So adding their power, the power and hence the combined focal length will remain same in all three cases.

22. When an object is at a distance u₁ and u₂ from a lens, real image and a virtual image is formed respectively having same magnification. The focal length of the lens is

(1) u₁ – u₂ (2) 2 2 1 u u  (3) 2 2 1 u u  (4) u₁ + u₂ Sol. Answer (3) m = f f u+

For real image For virtual image –m = 1 f f u− , +m = ₁ f f u− , 2 1 f f f u f u − = − −  f =u u1+₂ 2

23. A concave lens of focal length f produces an image (1/x) of the size of the object. The distance of the object from the lens is

(1) (x – 1) f (2) (x + 1)f (3) {(x – 1)/x}f (4) {(x + 1)/x}f Sol. Answer (1)

m = f

f u+ For virtual image m = 1 x + 1 f x f u − + = − + , u = –f(x – 1)

24. A thin equiconvex glass lens of refractive index 1.5 has power of 5D. When the lens is immersed in a liquid of refractive index , it acts as a divergent lens of focal length 100 cm. The value of  of liquid is

(1) 4/3 (2) 3/4 (3) 5/3 (4) 8/3 Sol. Answer (3) _g = 1.5 P = 5 D 1 1 5 ( g 1) R R ⎛ ⎞ = μ − _⎜⎝ + _⎟⎠ 2 5 2 R × = 1 _{0.2 m = 20 cm} 5 R = = f_l < 0 1 g ₁ 2 l l f R μ ⎛ _{⎞ ⎛ ⎞} =_{⎜ − ⎜ ⎟⎟ ⎝ ⎠} μ ⎝ ⎠ 1 ₁ 2 100 20 g l μ ⎛ ⎞ − =_⎜ − _⎟ μ ⎝ ⎠ 1 1 10 g l µ − = − µ 1.5 9_, 5 10 l 3 l = µ = µ

25. A convex lens of focal length 100 cm and a concave lens of focal length 10 cm are placed coaxially at a separation of 90 cm. If a parallel beam of light is incident on convex lens, then after passing through the two lenses the beam

(1) Converges (2) Diverges

(3) Remains parallel (4) Disappears

Sol. Answer (3)

f = –10 cm f = +100 cm

90 cm 100 cm Virtual object for concave lens is at its focus.

(Refraction through a Prism)

26. Yellow light is refracted through a prism producing minimum deviation. If i₁ and i₂ denote the angle of incidence and emergence for the prism, then

(1) i₁ = i₂ (2) i₁ > i₂ (3) i₁ < i₂ (4) i₁ + i₂ = 90 Sol. Answer (1)

27. At what angle will a ray of light be incident on one face of an equilateral prism, so that the emergent ray may graze the second surface of the prism ( = 2)?

(1) 30° (2) 90° (3) 45° (4) 60°

Sol. Answer (2)

Let C by critical angle  = 2 = sin sin i r i_e = 90º 60º = r + C from geometry i C r 60º 60º 60º ie i =_e

µ [At emergent interface]

11 sin 30º 2 C= − = r = 30º sin 2 sin i r = sin i = 1 2 i= π

28. A prism of refractive index 2 has a refracting angle of 60o_{. At what angle must a ray be incident on it so that}

it suffers a minimum deviation?

(1) 30° (2) 45° (3) 60° (4) 75°

Sol. Answer (2)

2,A 60º

µ = =

At for minimum deviation : i = i_e and r₁ = r₂ Also, r₁ + r₂ = A i _r 2 r₁ ie or 2r = 60º r = 30º sin sin i r µ = 2 2sini= [Putting r = 30º] 1 sin 2 i = i = 45º

29. A glass prism of refractive index 1.5 is immersed in water of refractive index 4/3. A light ray incident normally on face AB is totally reflected at face AC if



B A

C

Sol. Answer (1) g(glass) = 1.5

_l(water) = 4/3

Rays enter and pass undeviated at first interface For total interface reflection at 2nd _interface

1 sin e c g i > − ⎛_⎜μ ⎞_⎟ μ ⎝ ⎠ 1 4 2 sin 3 3 c i > − ⎛_⎜⎝ × ⎞_⎟⎠  ic 1 8 sin 9 c i _> − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 8 sin 9 c i >

Since i_c =  from geometry 8

sin 9 θ >

(Some Natural Phenomena due to Sunlight)

30. Which of the following phenomenon of light forms a rainbow?

(1) Reflection of light (2) Refraction

(3) Total internal reflection (4) Reflection as well as refraction Sol. Answer (4)

Formation of a rainbow involves refraction to break white light into its constituent colours. It is also involves internal reflection in the drop.

(Optical Instruments)

31. A person can see clearly only up to a distance of 25 cm. He wants to read a book placed at a distance of 50 cm. What kind of lens does he require for this purpose and what must be its power?

(1) Concave, – 1.0 D (2) Convex, + 1.5 D (3) Concave, – 2.0 D (4) Convex, + 2.0 D Sol. Answer (3)

He needs to bring the image of the object closer to 25 cm Also the image should be virtual

1 1 1 v u− =f 1 1 1 25 50 f − + = f = –50 cm 1 0.5 P = − P = –2 D

32. An astronomical telescope has an objective of focal length 100 cm and an eye piece of focal length 5 cm. The final image of a star is seen 25 cm from the eyepiece. The magnifying power of the telescope is

(1) 20 (2) 22 (3) 24 (4) 26

Sol. Answer (3) f_o = 100 cm f_e = 5 cm

To form image at near point

1 1 ⎡ ⎤ = − _⎢ + _⎥ ⎣ ⎦ o e m f f D 1 1 100 5 25 ⎡ ⎤ = − _⎢ + _⎥ ⎣ ⎦ 6 100 25 ⎡ ⎤ = − _{⎢ ⎥} ⎣ ⎦ m = –24

33. When a telescope is adjusted for normal vision, the distance of the objective from the eye-piece is found to be 80 cm. The magnifying power of the telescope is 19. What are the focal lengths of the lenses?

(1) 61 cm, 19 cm (2) 40 cm, 40 cm (3) 76 cm, 4 cm (4) 50 cm, 30 cm Sol. Answer (3)

Distance between objective and eyepiece when telescope is adjusted for normal vision is given by

19 = o ⎛⎜_⎝ = o⎞⎟_⎠ e e f f M f f ....(i) 80 = f_o+ f_e. (L = f_o + f_e) ....(ii) Solving (i) & (ii)

f_o = 76 cm, f_o = 4 cm

34. The focal lengths of the objective and eye lens of a telescope are respectively 200 cm and 5 cm. The maximum magnifying power of the telescope will be

(1) – 40 (2) – 48 (3) – 60 (4) – 100

Sol. Answer (2)

Maximum magnification is at near point Magnification at near point

1 ⎛ ⎞ = − _⎜ + _⎟ ⎝ ⎠ o e e f f m f D 200 ₁ 5 5 25 m= − ⎛_⎜ + ⎞_⎟ ⎝ ⎠ = −200 65 ×5 m = –48

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