Defects In Image Formation By Lenses And Their Correction

Images formed by lenses suffer from the following defects:
1. Spherical aberration
2. Chromatic aberration

1. Spherical aberration : It is the inability of a lens to bring all the rays to meet to one point after suffering refraction.

                              
Spherical aberration is due to the greater deviation of the rays from outer edges of the lens as compared to that of the rays incident on the central part.

Removal Of Spherical Aberration : Spherical aberration can be removed or minimized by the following methods :
1. By using a stop or making the diameter of the lens small.
2. By using Lenses of large Focal lengths.
3. By selecting suitable curvature to the two surfaces of the lens.
4. By Minimising the deviation.
5. By suitable combination of lenses.

2. Chromatic aberration  or colour defect :  The inability of the lens to focus all the colours at one point is called chromatic aberration.

Achromatism : The removal  of chromatic aberration or color defect is called achromatism. 
It is possible to remove chromatic aberration by having a combination of a suitable convex lens with a concave lens. Such a combination is known as achromatic combination.


Some Multiple Choice Questions for practice :

1. Which lens is used to correct the defect of long sightedness :
(a)   Concave                                        (b) convex
(c)   double concave                              (d) plano concave

2. Aberration in a lens is due to:
(a) nature of light                       
(b) defect in lens material
(c) deviation from simple lens equation   
(d) defect  in experiment arrangement

3. chromatic aberration in alens is due to the;
(a) inability of the lens to focus axial and marginal rays at the same point 
(b) inability of the lens to focus longer and shorter wavelengths of the light at the same point
(c) total internal reflection
(d) the fact that two faces of the lens may have different curvatures

4. Where is the image of the distant object formed by a short sighted eye ?
inability of the lens to focal axial and marginal rays at the same point 
(a)  At yellow spot                          (b) At blind spot
(c)  In front of retina                      (d) Behind the retina

5. Chromatic aberration of a lens results, because :
(a) the marginal and paraxial rays do not meet at the same point                
(b) of the different radii of curvature of the two surfaces
(c) of the prismatic action of the lens                                              
(d) of the defect in manufacturing

6. Ability of the eye to see objects at all distances is called
(a)  Binocular vision                       (b) myopia
(c)  hypermetropia                         (d) accomodation

7.Defect of hypermetropia can be remedied by using a

(a) convex lens                            (b) concave lens
(c) cylinderical lens                      (d) bi-focal length lens

8. Defect of color blindness can be cured by using a 
(a) contact lens                           (b) cylinderical lens
(c) bi-focal length lens                 (d) None of these

9. Spherical aberration can be reduced by
(a) using stops
(b) using lenses of large focal lengths
(c) using plano-convex lenses
(d) All the above methods

10. A plano-convex lens is used as objective lens in a telescope, then to minimise spherical aberration its

(a) convex surface should face the parallel rays  
(b) plane surface should face the parallel rays
(c) centre is covered with black paper
(d) edges are covered with black paper

11. chromatic observation in the formation of images by a lens arises because

(a) of non paraxial rays                                              
(b) the focal length varies with wavelength 
(c) of the defect in grinding                                               
(d)  radii of curvature of the two sides are not same

12. The chromatic aberration in a lens is because of 
(a) the large size of object                                                
(b) the object being very close to object
(c) manufacturing defect               
(d) dispersion of light in the material of lens

13.In order to reduce the spherical aberration in optical instruments , one should use
(a) plano convex lenses               (b) convex lenses
(c) plane mirrors                         (d) concave lenses

14. How should people wearing their spectacles work with a microscope ?
(a) They should take off the spectacles                                          
(b) They should keep on wearing the spectacles
(c) It hardly matters whether they take off or put on their spectacles 
(d) Special lenses of more power are used in the spectacles

15. The defects in the formation of image arising due to marginal and paraxial rays undergoing different deviation in a lens is known as
(a) Chromatic aberration              (b) Coma
(c)  Distortion                              (d) Spherical aberration

Total Internal Reflection (TIR)

It is the process of reflection of light into a denser medium from an interface of the denser medium and a rarer medium.

Essential conditions for TIR : There are two essential conditions for the process to occur.

(i) Light should travel from denser medium to rarer medium

(ii) Angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact.

Critical Angle : Critical angle for a pair of media in contact is the angle of incidence in the denser medium corresponding to which angle of refraction in the rarer medium is 90°. It is denoted by C and its value depend upon the nature of media in contact.

Let we have an interface XY separating a rarer medium (Air) and a denser medium (water). Let medium (a) is air and medium (b) is water as shown in fig.

Let O be an object lying in the denser medium and the rays starting from it are as shown in fig. below,

                                                                             

A ray of light from O falling perpendicular to the interface XY passes without any deviation along AB. A second ray OA1, making some angle with the normal on interface will refract along A1B1 shown. Third ray OA3 is falling at a particular angle of incidence called critical angle (i2= C) at point A3 and the ray after striking point A3 bends towards the interface A3B3, in the case the angle of refraction is 90°.

                        Another ray OA4 falling at the interface XY at an angle greater than the critical angle (i3 > C), the rays after striking at point A4 will be reflected back in the medium (II) along A4B4 as shown in fig. This process is called Total internal Reflection (TIR).

Hence in short, we can conclude that when  a ray of light traveling from an optically denser medium to an optically rarer medium is incident at an angle greater than the critical angle for the pair of medium in contact, the ray is totally reflected back into denser medium. This process is called Total Internal Reflection.

Relation between refractive index and critical angle :

               ^Iµ_{II  = 1/sin C}

As is seen from fig., if the angle of incidence (i3 >C) greater than critical angle, the light ray is completely reflected back into the water (the denser medium).

Applications of Total Internal Reflection :

1. The Brilliance of Diamond : Refractive index of diamond is approximately 2∙45

⇒            Sin C = 1/µ = 1/2∙45 = 0∙408

⇒                  C = 1/sin (0∙408) = 24∙1°

i.e., critical angle for diamond is 24∙1. 

Diamond is cut in such a way that a ray of light enters the diamond at an angle greater than critical angle i.e., 24∙1°, hence light ray suffers multiple total internal reflections at different faces and remains within the diamond. This makes the diamond sparkle.

2. Total Internal Reflection in Prisms : 

For a glass prism,      µ = 1∙5

                          Sin C = 1/µ            

                                  = 1/1∙5 = 0∙667

                               C = 1/sin (0∙667) =42°

Thus, in a right angled isosceles prism, a ray of light incident at an angle of incidence 45° (i.e., greater than critical angle 42°), the light ray will be totally internally reflected.

                          

                          

                                       

3. Mirage :It is an optical illusion observed in deserts due to total internal reflection of light. Mirage disappoints a traveller in a sandy desert because a traveller on a look out for some supply of water gets feeling of presence of a pond of water as he sees the image of tree down below but did not get any water.
                    

4. Optical Fibres : Optical fibers are made up of glass or plastic may be as thin as a human hair. Optical fibers have two parts (i) Core (ii) Cladding

(i) A Core is the central part or main part of an optical fibre. Diameter of core varies from 6µm to 250 µm. It is made from a pure glass of high refractive index (approximately 1⋅7)

(ii) Cladding: Core is surrounded by a cladding of glass or plastic of relatively small refractive index (approximately 1⋅5). Cladding is usually between 10 µm to 150 µm thick. The interface between core and cladding acts as a cylindrical mirror. A ray of light falling at this interface at an angle more than critical angle gets totally internally reflected.

                               

                       

Optical fibres help in communication of signals in the form of light through many kilometers without loss in energy or signal. Optical fibers are also used in medical instruments like fibrescope to examine internal body cavities such as the stomach, galbladder etc.

5. Shinning of air bubble in water.


Practice Problem:

If refractive index of water and glass are 4/3 and 5/3 respectively and the light is going from glass  to water. Find the critical angle (C).
Solution:

             Given,            ^aµ_w =  4/3,    ^aµ_g =  5/3

                                 ^wµ_g  = ^aµ_{g /} ^aµ_w     

                                       =  5/3 X 3/4   

                                       = 5 /4  = 1⋅25
                                             
We know that,           ^wµ_g  =  1/Sin C

∴                      1/Sin C   = 1⋅25
⇒                        Sin C   = (1/1⋅25) = 0⋅8
                              C     =  (1/Sin) (0⋅8)
                              C      = 53⋅13° (Ans.)


Some Multiple choice Practice Problems :
1. The critical angle of light passing from glass to air is minimum for 
(a) red                               (b) green
(c) yellow                           (d)violet

2. A well cut diamond appears bright due to 
(a) it emits bright light 
(b) scattering of light
(c) total internal reflection
(d) it emits luminous particles

3.Total internal reflection will not take place, when light ray travels from
(a) glass to air
(b) glass to water
(c) water to air
(d) water to glass

4. µ = Sin C, where C is the angle of incidence in optically
(a) rarer medium              (b) denser medium
(c) air                              (d) transparent

5. Critical angle for a pair of media in contact is the angle of incidence in the denser medium corresponding to which angle of refraction in the rarer medium is
(a) 180°                          (b) 30°
(c) 45°                            (d) 90°

6.Condition for Total Internal Reflection is
(a) ∠i in denser medium should be greater than critical angle
(b) Light should travel from denser medium to rarer medium
(c)  ^Iµ_{II = 1/ Sin C
(d) All the above are true}

7. If the critical angle for total internal reflection from a medium to vacuum is 30°, then velocity of light in medium is
(a) 1⋅5 x 10⁸ m/s                   (b) 0⋅75 x 10⁸ m/s    

(c) 3 x 10⁸ m/s                      (d) 2 x 10⁸ m/s    

Power Of A Lens

The reciprocal of the focal length of a lens in metres is called the power of the lens.

The capacity of  a lens to bend the rays  of light depends upon the focal length. The smaller the focal length, the greater is the bending of a ray of light and vice- versa. Thus the power of a lens to bend the rays of light is inversely proportional to the focal length of the lens.


Units of power of a lens : The units of the power of a lens are dioptres.
 One dioptre is the power of a lens of one metre focal length. The power of a lens is denoted by D. It is this dioptric number which the doctors prescribe for the spectacles of a person.

Mathematically,        D = 1/Focal length in metre

                                  = 100/Focal length in centimetre

The power of a convex lens is positive and that of a concave lens is negative. 
Thus a convex lens of focal length 40 cm is said to possess a power of  + 2.5 dioptre. similarly the power of a concave lens of focal length 20 cm is 5.0 dioptre.

Power of combination of lenses : The focal length of the combination of two lenses placed in contact is given by the relation.

                              1/F = 1/f1 +1/f2

If p1 is the power of lens of focal length f1, p2 the power of the focal length f2 and P the power of the combination of focal length, F then,

                               P = p1 +p2

Thus the power of combination of lenses is the algebraic sum of the powers of individual lenses.
If two lenses are separated by distance d metre, then power of combination of lenses is given by relation
  
                          P = P1 + P2 – d∙(P1∙P2)


Practice Problems :

1. A lens is formed by combining two thin lenses of powers + 12 D and  – 8 D in contact with each other. What will be the focal length of combination ?
Solution:       
                       P1 = + 12 D,    P2 = – 8 D
Power of combination (P) = P1 + P2 
                                     = 12 – 8
                                     = 4 D
Focal length of combination (f) = 1/P = 1/+4
                                                      = 0.25 m 
                                                      = 25 cm.  (Ans.)

2. An optician prescribes spectacles to a patient with a combination of a convex lens of focal length 40 cm and a concave lens of 25 cm focal length. What will be the power of spectacles ?
Solution:   
                 F1 = 40 cm = 0.4 m
                 F2 = 25 cm = 0.25 m

P1 = 1/F1 = 1/0.4 = 2.5 D
P2 = 1/F2 = – 1/F2 = –1/0.25 = –4∙0 D  {(Because for concave mirror, P = –1/F)

Power of spectacles
                          (P) = P1 + P2
                               = 2∙5 + (–4∙0)
                               = – 1∙5 D    (Ans.)

3. A convex lens of power +4 D and a concave lens of power 3 D are placed in contact. What is equivalent power of the combination ?
Solution:
                       P1 =   4 D,
                       P2 = –3 D,
                       P  = P1 + P2
                          = 4D – 3D 
                          = 1D
Equivalent power of combination is 1 Dioptre.

4. A convex lens has focal length of 20 cm. find its power in Dioptres.
Solution: 
              F = 20 cm  
                 = 0∙2 m
              P = 1/F
                 = 1/0∙2 
                 = +5
 Power of convex lens is +ve and in magnitude it is 5 Dioptre.

5. The combined power of two lenses in contact is  + 10 D. When they are separated by 20 cm, their power becomes 6.25 D. Find the power of individual lenses.
Solution: 
                 Given P = P1 + P2 = +10 D                               ------(1)
Distance between lenses, d = 20 cm = 0∙2 m
When these lenses are separated by distance 0∙2 m,

Power of combination becomes
                                             P' = P1 +P2 – d∙(P1∙P2) = 6∙25
                                                =      10  – 0∙2(P1∙P2) = 6∙25

                                        P1∙P2  =      (10 – 6∙25) / 0∙2
                                                =      18∙75

                                (P1 – P2)² = (P1 + P2)²  – 4∙(P1∙P2)   

                                               =         (10)² – 4∙(18∙75)                                   
                                (P1 – P2)²  = 25
                                 P1 – P2     = 5 D                                     ------(2)
                        
Adding equations (1) and (2)
                                          2P1 = 15D 
or                                         P1 = 15/2 = 7∙5 D

Subtracting equation (2) from (1)
                                          2P2   =  10 – 5   = 5
                                            P2  =  5/2        = 2∙5 D

Hence power of lenses, P1 = 7∙5 D and P2 = 2.5 D.


6. A lens made of glass of refractive index 1∙5 has power of + 10 D when placed in air, find the power of same lens when immersed in water  (µ = 1∙33). Also find the change in power of lens.
Solution: 
             When lens is immersed in water, the focal length of lens increases by nearly 4 times as when in air.

∴  Power of lens in water = Power of lens in air/4 = 10/4 = +2∙5 D
∴  Change in Power         = + 2∙5 – 10∙0
                                     = –7∙5 D        (Nearly) Ans.


Some Multiple Objective Type Questions for Practice :

1. An optician prescribes spectacles to a patient with a combination of a convex lens of focal length 40 cm and concave lens of 25 cm. The power of spectacles is (a)  6∙0 D                              (b) 1∙5 D
(c)  – 6∙0 D                           (d) – 1∙5 D

2. A convex lens of power  +6 D is placed in contact with a concave lens of power –4 D. What will be the nature and focal length of this combination ?
(a) concave, 25 cm                (b) convex, 50 cm
(c) Concave, 20 cm                (d) convex, 100cm

3. A lens is formed combining two thin lenses in contact having power +12 D and – 8 D. The focal length of the combination is
(a) 25 cm                              (b) –25 cm
(c)  5 cm                               (d) – 5 cm

4. Two thin lenses of focal lengths 60 cm and –20 cm are placed in contact . The focal length of combination:
(a) 15 cm                              (b) 30 cm
(c) –15 cm                             (d) –30 cm

5. A convex lens has a focal length of 20 cm. Its power in dioptres is:
(a) –5 D                                (b) –10 D
(c) +5 D                               (d) +10 D

6. A convex lens of power +6 D is placed in contact with a concave lens of power –4D. What will be the nature and focal length of this combination ?
(a) convex, 100 cm                (b) convex, 50 cm
(c) concave, 25 cm                 (d) concave, 20 cm 

7. Convex lens of power 4D and a concave lens of power 3D are placed in contact. What is the equivalent power of the combination ?
(a) 7 D                                  (b) 4/3 D
(c) 1 D                                  (d) None of the above

8. A convex lens has a focal length of 20 cm. Its power in dioptres is
(a) –5 D                                (b) – 10 D
(c) +5D                                 (d) +10 D

9. The focal length of convex lens is 50 cm. What is its power ?
(a)  +2 D                               (b) – 2D
(c) –50 D                               (d) +50 D

10. Two converging lenses of equal focal lengths f are placed in contact. the focal length of the combination is
(a) f                                      (b) 2f
(c) f/2                                   (d) 3f

11. Two lenses one convex and other concave of focal length 0∙5 m and 1∙0 m respectively recombined, the power of combination will be
(a) –1∙0 D                              (b) +1∙0 D
(c) 0∙5 D                                (d) –0∙5 D

12. Two lenses of powers P1 = +2 D and P2 = –2 D respectively, are placed in contact. the power of combination is
(a) –2 D                                (b) +2 D
(c) +4 D                                (d) None of above

13. The power of combination of two lenses separated by distance d
(a) decreases as their separation is increased
(b) increases as their separation is increased
(c) remains same
(d) becomes equal to the distance 'd'

14. Two lenses of powers 6D and –5D are in contact with each other. The focal length of combination will be
(a) 1 cm                               (b) 20 cm
(c) 100 cm                            (d) 100/11 cm

15. Two thin lenses are in contact and the focal length of combination is 80 cm. If focal length of one of the lenses is 20 cm, then power of the other lens is
(a) +6∙0 D                            (b) –3∙75 D
(c) + 4∙0 D                           (d) –1∙5 D

16. A convex lens of power +6 D is placed in contact  with a concave lens of power –4 D. The combination will act as a convex lens of focal length
(a) 2 cm                                      (b) 20 cm
(c) 50 cm                                     (d) 100 cm

Combination Of Lenses (Compound Lenses)

In order to remove defects in the images formed by the lens, two or more lenses are combined together. Sometimes the combinations are  made by keeping the lenses in contact or they may be kept away from each other through some distance. Such combination of lenses is said to make a compound lens.

1. Two thin lenses in contact : When two thin lenses are placed in contact, the combination will act as a single lens called the equivalent lens.

                                    1/F = 1/f1 + 1/f2

 This formula is also true if one lens is convex and other is concave.

2. Two thin lenses separated by some distance : If two thin lenses are held a distance d apart from each other, then focal length of the combination is given by the relation,

                                   1/F = 1/f1 + 1/f2 – d/f1f2

Practice Problems :

1. Two converging lenses of equal focal length 'f' are placed in contact. Find the focal length of the combination.

Solution: F1  = F2  = f (given)

             Let focal length combination be F
By formula;
                   1/F = 1/F1 + 1/F2 
                         = 1/f   + 1/f
                         = (1+1)/f 
                         = 2/f
                      F = f/2

2. Two thin lenses (one convex and the other concave lens) of focal length 60 cm and 20 cm respectively placed in contact. Find the focal length of combination. 

Solution: F1 = 60 cm,  F2 = 20 cm,
              Focal length of combination (F) will be given by
            
                1/F = 1/F1 +1/F2
                1/F = 1/60 –1/20
                      = (1–3)/60
                      = –2/60
                 1/F = –1/30
                    F = –30 cm. (Ans.)



            

Magnification

A measure of the extent to which an optical system enlarges or reduces an image is called magnification.

Magnification is of two types;
1. Linear Magnification
2. Angular Magnification

Linear Magnification : It is defined as the ratio of the size (height) of the image to the size (height) of the object.

                       m = Size of image / Size of object
                           = I/O

If m > 1, then system is enlarging.
If m < 1, the system is reducing.

Linear magnification can also be defined in terms of distances of image and objects from the optical centre.

The formula for magnification is:
                                                m = f / f+u

Angular Magnification (M) : It is defined as the ratio of angles formed by the final image and the object ( when viewed directly) in most favored position available at the eye.

This is also called magnifying power of an optical system. 

OR

The ratio of the angle subtended by the image at the eye as seen through an optical instrument to the angle subtended by the object at the objective of the instrument is called angular magnification of the instrument.

Practice Problem :

1. The image of an object placed 15 cm from a lens is formed erect and double the size of the object. What will be the focal length and nature of lens ?
Solution: 
               Given,    
                              u = –15 cm
                             m = 2
                         –v/u = 2
                              v = –2(u)
                              v = –2 X (+15)
                                 = –30 cm
from lens formula;
                                      1/v –1/u = 1/f
                                              1/f = (1/–30) – (1/ (–15))
                                                   = (–1/30) + (1/15)
                                              1/f = 1/30 
                                                 f = 30 cm
 Since Magnification and f are positive hence given lens is a convex lens.

Lens Formula (For Thin Lenses only)

Lens formula is the relation between the focal length of the lens with the distance of object and image.

For finding lens formula, we generally consider thin lenses because in case of thick lenses, it is not possible to find the exact distances of object, image and focal length and also the optical centre of the lens. Before deriving relation, we must know the sign conversions.

Sign Convention of lens :
1. All distances are measured from the optical centre of the lens.
2. The object is always placed on the left side of lens so that the direction of incident light is always from the left to right.
3. All distances measured from the optical centre of the lens to the right side are taken as positive.
4. All distances measured from the optical centre of the lens to the left side of lens are taken as negative.
5. Transverse upward distance from the principal axis is taken as the + ve and downward distance is taken as – ve.

Assumptions :
1. The lens is supposed to be thin.
2. Optical centre of the lens is taken at origin.
3. Incident rays are close to the principal axis of the lens.
4. All rays passing through optical centre of the lens go undeviated.

Lens Formula for Thin Convex Lens when real image is formed:
                                        
              1/f = 1/v – 1/u

Lens Formula for Convex Lens when virtual image is formed :
  
              1/v – 1/u = 1/f

Lens Formula for Thin Concave Lens :

              1/v – 1/u = 1/f

Hence we conclude that lens formula for all types of lenses will be same i.e.,
   
             1/v – 1/u = 1/f

irrespect time of the position of object and the size  and nature of the image formed.

Practice Problem :
The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, Calculate the object and image distance.

Solution:  As the image formed is erect, it is virtual in nature. Therefore, both u and v are negative and positive and hence m is positive i.e., m = +4

 Now         v/u = m = 4
                v    = 4u

Now using lens formula    1/f = 1/v – 1/u

As                                     f = 20 cm

so                                  1/20 = 1/4u – 1/u
                                             = (1–4) / 4u  = – 3/4u
                                     1/20 = –3/4u

                                          u = – 15 cm
                                          v =4u 
                                             = 4 X (–15)
                                             = –60 cm  (Ans.)

Practice Questions:

1. An object is placed at focus of a concave mirror of focal length 20 cm. the distance of image is
(a) 40 cm                        (b) 10 cm
(c) 20 cm                        (d) infinity

2. The image of an object placed 15 cm from a lens is formed erect and double the size of the object. The focal length and nature of the lens is :
(a) 30 cm, concave         (b) 60 cm, concave
(c) 60 cm, convex           (d) 30 cm, convex

3. The distance between an object and a divergent lens is 'm' times the focal length of the lens. the linear magnification produced by the lens is equal to 
(a)  m                            (b) 1/m
(c) (m+1)                      (d) 1/(m+1)

4. If magnification is negative, the nature of the image formed by a convex lens is 
(a) real and inverted        (b) virtual and erect
(c) real and erect             (d) virtual and inverted

5. If magnification is positive, the nature of the image formed in convex lens is 
(a) real and inverted        (b) virtual and erect
(c) real                           (d) None of these

6. Magnification produced by a plane mirror is 
(a) +1                           (b) +2
(c) +1/2                        (d) –1

7. What is the magnification when the object is placed at 2f from the pole of a concave mirror.
(a) 1/3                         (b) 2/3
(c) 1                            (d) 3/2

8. A lens which can produce magnification of 1, less than 1 or greater than 1 can be a 
(a) convex lens              (b) glass slab
(c) concave lens            (d) plane mirror

9. The image of an object placed 15 cm from a lens is formed erect and double the size of the object. The focal length and nature of glass is
(a) concave, 30 cm        (b) convex, 30 cm
(c) concave, 20 cm        (d) convex, 15 cm

10. The focal length of a convex lens depends upon
(a) wave length of light ray
(b) frequency of light ray
(c) time period of light ray
(d) velocity of light ray

11. As an object is brought closer to the focal point of a convex lens from infinity, its image 
(a) becomes smaller          (b) remains same
(c) gets farther from lens   (d) gets closer to the lens

12. as an object is moved closer to the focal point of a convex lens from infinity, the size o image will
(a) decrease                    (b) increase
(c) become zero               (d) none of the above 

13. On decreasing the radii of the two surfaces of a double convex or double concave lens result in the decrease in the 
(a) focal length               (b) size of image
(c) size of object             (d) intensity of image

14. Which of the following is wrong ? Here m represents the magnification of a lens 
(a) The m is +ve when image formed by convex lens is virtual
(b) The m is –ve when image formed by concave lens is real
(c) The m is +ve when image  is formed by concave lens 
(d) None of the above

Lens - Objective Type Questions

1. A convex lens is dipped in a liquid having refractive index same as that of the material of the lens. The lens behave as a 
(a) convergent lens                (b) divergent lens
(c) prism                               (d) glass plate

2. Which of the following will not produce a magnifying system ?
(a) convex lens                      (b) concave lens
(c) concave mirror                  (d) None of these

3. Which of the following can produce a virtual image larger in size than the object ?
(a) concave lens                     (b) convex lens
(c) convex or concave lens      (d) None of the above

4. What is the minimum distance between the convex lens and the object to produce a real image ?
(a) 0                                     (b) f
(c) 2f                                    (d) 4f

5. What is the minimum distance between the object and its virtual image in case of a concave lens ?
(a) 0                                     (b) f
(c) 2f                                    (d) 4f

6.how will the image formed by a convex lens be affected if the central portion of the lens is blackened ?
(a) No image will be formed by the lens
(b) The central portion of the image will be absent
(c) The full image will be formed but it will be less bright
(d) There will be two images, one due to each exposed portions

7.When rays of light fall on a convex lens it
(a) converges them                (b) does not bend them
(c) diverges them                   (d) None of these

8.If there is no atmosphere, the duration of the day would :
(a) increase                           
(b) decrease
(c) remains the same
(d) increase  during winter but decrease during summer

9. When the object is placed beyond the focus of a convex lens, the image formed is :
(a) real                                 (b) virtual
(c) either real or virtual          (d) none of the above

10. What is the minimum distance between an object and its real image formed by a convex lens ?
(a) 0                                    (b) f
(c) 2f                                   (d) 4f

11. A convex lens of focal length 32 cm forms a virtual image of double the size of the object. What is the distance of the object from the lens.
(a) 16 cm                            (b) 32 cm
(c) 10 cm                            (d) None of the above

12. The field of the view is maximum for
(a) plane mirror                  (b) concave mirror
(c) convex mirror                (d) parabolic mirror

13. A convex lens is immersed in a liquid denser than glass. It will behave as
(a) plane glass                   (b) diverging lens
(c) converging lens             (d) None of the above

14. Half of the lens is wrapped in black paper. How will it change the image ?
(a) Size of the image is halved.
(b) Intensity of image is halved.
(c) No change in size of image or intensity.
(d) None of the above

15. For which color, the focal length of a convex lens is more ?
(a) Red                              (b) Green
(c) Yellow                           (d) Blue

16. In which case the image formed by a concave lens is real ?
(a) 0 < u < f                      (b) f < u < 2f
(c) 2f < u < ∞                   (d) None of the above

17. In which case the image formed by a convex lens is virtual ?
(a) 0 < µ < f                      (b) f < µ < 2f
(c) 2f < µ < ∞                   (d) None of these above

18. A decrease in aperture of a lens changes the
(a) Size of the image         (b) Position of the image
(c) both position and size   (d) intensity of the image

19. An object is placed at infinity. The image formed by a concave lens is always
(a) At focus
(b) Between F and 2F 
(c) Beyond 2F
(d) Between focus and optical centre

20. A convex lens becomes less converging when placed in 
(a) oil                              (b) water    
(c) both (a) and (b)          (d) None of the above

21. As an object gets closer to the focal point of a convex lens from infinity, its image
(a) becomes smaller
(b) is magnified
(c) becomes closer to the lens
(d) becomes farther from the lens

22. On decreasing the radii of both the surfaces of a double convex lens or double concave lens, its focal length will
(a) decrease
(b) increase
(c) neither increase nor decrease
(d) remain same

Answers :
1. d
2. b
3. b
4. b
5. a
6. c
7. a
8. b
9. b
10. d
11. a
12. c
13. b
14. b
15. a
16. d
17. a
18. d
19. d
20. a
21. d
22. a