Class 12th chapter notes, Study notes of Physics

Download Class 12th chapter notes and more Study notes Physics in PDF only on Docsity! Class XII Physics www.vedantu.com 1 Important Questions for Class 12 Physics Chapter 9 – Ray optics and Optical Instruments Very Short Answer Questions 1 Mark 1. A person standing before a concave mirror cannot see his image, unless he is beyond the centre of curvature. Why? Ans: Let a man stand beyond focus i.e., between focus and centre of curvature, then the image formed will be real and inverted and is formed beyond C (beyond him). Thus, he cannot see the image. But if he stands beyond C, the image will be formed between focus and centre of curvature which is in front of him and thus he will be able to see his reflected image. 2. For what angle of incidence, the lateral shift produced by a parallel sided glass plate is maximum? Ans: We know that lateral shift d is given as, t d sin (i r) cosr   For lateral shift d to be maximum, sin (i r) must be maximum i.e., i r must be minimum. This happens when i 90 . t d sin(90 r) cosr    D t  Then, we can say that lateral shift is maximum. 3. You read a newspaper, because of the light it reflects. Then why do you not see even a faint image of yourself in the newspaper? Ans: We know that image is formed due to regular reflection of light. Class XII Physics www.vedantu.com 2 However, when we read a newspaper, there is diffused (irregular) reflection of light, thus we are not able to see even a faint image of ourselves on the newspaper. 4. A substance has critical angle of 45  for yellow light, then what is its refractive index? Ans: We know that refractive index is given as below 1 sinC   Substituting the values, we have   1 sin 45 1/ 2 1     2  5. An object is placed between the pole and focus of a concave mirror produces a virtual and enlarged image. Justify using mirror formula. Ans: We know that the mirror formula is as given below, 1 1 1 v u f   uf v u f    Now magnification, v m u  . uf u fm u   f m u f    Class XII Physics www.vedantu.com 5 1 2 1 2 1 1 1 x F f f f f    . 11. Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision? Ans: Myopia and hypermetropia are common eye defects. A myopic or hypermetropic person need not necessarily suffer a partial loss in their eyes’ ability of accommodation. Myopia occurs when the eye-balls engage in elongation from the front to the back whereas hypermetropia occurs when the eye-balls shorten themselves. On the other hand, when the eye-lens completely loses its ability of adjusting itself, then the defect is called presbyopia. Short Answer Questions 2 Mark 1. What are optical fibres? Give their one use. Ans: Optical fibres are thin and long strands of fine quality glass or quartz coated with a thin layer of material with refractive index less than that of the strands. They work on the principle of total internal reflection and thus, they avoid any loss in transfer of information. Uses Optical fibres are often used in medical investigations i.e., one can examine the inside view of stomach and intestine by a method called endoscopy. 2. How do the focal lengths of a lens change with increase in the wavelength of the light? Ans: We know that 1 2 1 1 1 ( 1) f R R          . Class XII Physics www.vedantu.com 6 Also, 1   . Clearly, when wavelength ( ) increases, refractive index ( ) decreases. Similarly, f . Clearly, as refractive index ( ) decreases, focal length (f ) increases. 3. Show with a ray diagram, how an image is produced in a total reflecting prism? Ans: Consider the two rays from the object PQ, as shown below. They undergo total internal reflection firstly at the face AB and then at BC forming the final image P Q  (real and inverted image). 4. The radii of the curvature of the two spherical surfaces which is a lens of required focal length are not same. It forms image of an object. The surfaces of the lens facing the object and the image are inter-changed. Will the position of the image change? Ans: As we know that, 1 2 1 1 1 ( 1) f R R          When the radii of curvature 1R and 2R are interchanged, the focal length of the lens also changes. Hence, the position of the image will reduce gradually. Class XII Physics www.vedantu.com 7 5. A thin converging lens has focal length (f) when illuminated by violet light. State with reason how the focal length of the lens will change if violet light is replaced by red light. Ans: We know that, 1 2 1 1 1 (n 1) f R R         Since, n for violet is more that n for red colour, and since m f , we can say that the focal length of the lens will decrease when violet light is replaced by red light. 6. Thin prism of angle 60 gives a deviation of 30 . What is the refractive index of material of the prism? Ans: We know that the refractive index of a thin prism is as follows mA sin 2 n A sin 2              Substituting the given values, we have 60 30 sin sin 452 n 60 sin30 sin 2          n 1.41  , which is the refractive index of the given thin prism. 7. Although the surfaces of a goggle lens are curved it does not have any power. Why? Ans: Since the two surfaces of a goggle lens are parallel i.e., one surface convex and the other concave, the resultant power of the two surfaces is zero as powers on both surfaces are equal but opposite in sign. 1 2p p p p ( p) 0      Class XII Physics www.vedantu.com 10 3) u 30 and v 37 is incorrect. Here, when the object is placed between f 20cm and 2f 40cm , the image is obtained before 2f 40cm . Clearly, we can conclude that observation (3) is incorrect because both object and the image here lie between f and 2 f . 10. Birds flying high in the air appear to be higher than in reality. Explain why? Ans: Birds fly in air, which is a rarer medium when compared to the ground, which is denser. The light from the birds when viewed will undergo refraction towards the normal. Thus, the birds appear to fly at a higher point. i.e., Apparent height > Real height 11. What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm ? Is the system a converging or a diverging lens? Ignore thickness of the lenses. Ans: Given that, Focal length of the convex lens, 1f 30 cm Focal length of the concave lens, 2f 20 cm  Focal length of the system of lenses f Class XII Physics www.vedantu.com 11 Then the equivalent focal length of a system of two lenses in contact is given as: 1 2 1 1 1 f f f   1 1 1 2 3 1 f 30 20 60 60        f 60cm   Hence, the focal length of the combination of lenses is 60 cm . The negative sign indicates that the system of lenses acts as a diverging lens. 12. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose? Ans: Given that, Distance between the object and the image, d 3 m Maximum focal length of the convex lens max  f For real images, the maximum focal length is given as: max d 3 f 0.75m 4 4    Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m. 13. A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm . Determine the focal length of the lens. Ans: Given that, Distance between the image (screen) and the object, D 90 cm Distance between two locations of the convex lens, d 20 cm Class XII Physics www.vedantu.com 12 Focal length of the lens f Focal length is related to d and D as:     2 22 2 90 20D d 770 f 21.39cm 4D 4 90 36       Therefore, the focal length of the convex lens is 21.39cm . 14. You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will a) deviate a pencil of white light without much dispersion, Ans: Place the two given prisms next to each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. This acts as the incident light for the second prism and the dispersed light this time will recombine to give white light as a result of the combination of the two prisms. b) disperse (and displace) a pencil of white light without much deviation. Ans: Take the system of the two prisms as suggested in answer (a). Adjust (increase) the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will now disperse the pencil of white light without much deviation. 15. A myopic person has been using spectacles of power 1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power 2.0 dioptres. Explain what may have happened. Ans: Given that the power of the spectacles used by the myopic person is P 1.0 D  . Focal length of the spectacles, Class XII Physics www.vedantu.com 15 Angle of deflection, 3.5  Distance of the screen from the mirror, D 1.5 m The reflected rays get deflected by an amount twice the angle of deflection i.e., 2 7.0  The displacement (d ) of the reflected spot of light on the screen is given as: d tan 2 1.5   d 1.5 tan7 0.184m 18.4cm     Hence, the displacement of the reflected spot of light is 18.4 cm . Short Answer Questions 3 Marks 1. Find the radius of curvature of the convex surface of a plane convex lens, whose focal length is 0.3 m and the refractive index of the material of the lens is 1.5 . Ans: Given that, 1.5  f 0.3m For plane convex lens, 2R   and let 1R R . Substituting these values in the formula for focal length,   1 2 1 1 1 1 f R R            1 1 1 1.5 1 0.3 R          1 1 0.5 R 0.3        Class XII Physics www.vedantu.com 16 R 0.15m  Thus, the radius of curvature is R 0.15m . 2. Show that the limiting value of the angle of prism is twice its critical angle. Hence define critical angle. Ans: We know that, Angle of the prism is given by 1 2A r r  . In a case of a triangular prism where 1 2i i 90  , angle of refraction is given by 1 2r r C  . where, C is the critical angle. Clearly, 1 2A r r  A C C   A 2C  Therefore, the angle of incidence for which angle of refraction is 90 , is called the critical angle. 3. Draw a labelled diagram of telescope when the image is formed at the least distance of distinct vision? Hence derive the expression for its magnifying power. Ans: We know that, magnifying power angle subtended by the image at the eye angle subtended by the object at the eye  tan MP tan       (Since angles are very small) Class XII Physics www.vedantu.com 17 A B tan B E      and A B tan B O      A B A B MP B E B O         o e B O f MP B E v      o e f MP v    ......(i) For eye piece, e 1 1 1 v v f   e e 1 1 1 D v f     Multiply by D, e e D D 1 v f    e e D D 1 v f   e e e e 1 1 1 1 f 1 v f D f D          Substituting in (i), o e e f f MP 1 f D         Class XII Physics www.vedantu.com 20 Substituting equation (2) in equation (1), e ef 100f 101  e101f 101  ef 1cm  Substituting ef in equation (2), of 100 1  of 100cm Thus, the focal length of the eye-piece is 1cm whereas the focal length of the objective is 100cm. 7. A convex lens made up of refractive index 1 n is kept in a medium of refractive index 2 n . Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if a) 1 2 n n Ans: When 1 2n n , the lens behaves as a convex lens. b) 1 2 n n Ans: When 1 2n n , the lens behaves as a plane plate and thus, no refraction takes place. Class XII Physics www.vedantu.com 21 c) 1 2 n n Ans: When 1 2n n , the lens behaves as a convex lens. 8. Derive the relation 1 2 1 1 1 f f f   , where 1 f and 2 f are focal lengths of two thin lenses and F is the focal length of the combination in contact. Ans: Consider two thin lenses in contact having focal length 1f and 2f . For the first lens, 1 1 1 1 1 v u f   ......(1) For the second lens, 1I acts as an object which forms the final image I . Clearly, 1 2 1 1 1 v v f   ......(2) Adding equations (1) & (2) 1 2 1 1 1 1 1 1 1 1 f f v u v v      1 2 1 1 1 1 f f v u     Class XII Physics www.vedantu.com 22 Using lens formula 1 1 1 v u F        , 1 2 1 1 1 f f F   For n number of thin lenses in contact, n1 2 3 1 1 1 1 1 ...... F f f f f      Hence the derivation. 9. A convex lens has a focal length 0.2 m and made of glass  μ 1.50 is immersed in water  μ 1.33 . Find the change in focal length of the lens. Ans: Given, af 0.2m ; a g 1.50  It is known that  a g a 1 2 1 1 1 1 f R R            1 2 1 1 1 1.50 1 0.2 R R          1 2 1 1 10 R R    ......(1) Now, a gw g a w 1.50 1.128 1.33       Also, the focal length of the lens when immersed in water would be Class XII Physics www.vedantu.com 25 1 f 100 40cm 2.5     It is given that M 4  40 4 u 40    u 40 10    u 40 10   u 30cm  or 50cm These are the two values of object distances. 12. Define total internal reflection of light. Hence write two advantages of total reflecting prisms over a plane mirror. Ans: The phenomenon of reflection of light when a ray of light traveling from a denser medium getting reflected back into the same denser medium provided the angle of incidence is greater than the angle called critical angle is called total internal reflection. Advantages a) It does require silvering. b) Multiple reflections do not take place in a reflecting prism. Due to this, only one image is formed, which is very bright. Class XII Physics www.vedantu.com 26 13. An equiconvex lens of radius of curvature R is cut into two equal parts by a vertical plane, so it becomes a plano-convex lens. If f is the focal length of equiconvex lens, then what will be focal length of the planoconvex lens? Ans: We know that   1 2 1 1 1 n 1 f R R         For equiconvex lens 1 2R R R    1 1 1 n 1 f R R           1 1 1 n 1 f R R           2 n 11 f R    ......(1) For plano-convex lens, 1R R and 2R   Clearly,   1 2 1 1 1 n 1 f R R            1 1 1 n 1 f R            n 11 f R     ......(2) From (1) & (2) f 2 f 2f f     Class XII Physics www.vedantu.com 27 14. A converging lens of focal length 6.25 cm is used as a magnifying glass if near point of the observer is 25 cm from the eye and the lens is held close to the eye. Calculate a) distance of object from the lens. Ans: It is known that, 1 1 1 f v u   Given, f 6.25cm v 25cm  Substituting these values in the first expression, 1 1 1 1 u 25 6.25 5      u 5cm  b) angular magnification Ans: Angular magnification is given by D 25 m 1 1 F 6.25     m 1 4 5    c) angular magnification when final image is formed at infinity. Ans: When the image is formed at infinity, D 25 m 4 F 6.25    Class XII Physics www.vedantu.com 30 The magnification of the image is given as:     2 1 Image height h v m Object height h u   2 8.4 h 3 0.6 3 1.8cm 14         Hence, the height of the image is 1.8cm . If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. Also, the size of the image will decrease with the increase in the object distance. 17. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is a) a convex lens of focal length 20 cm Ans: In the given situation, the object is virtual and the image formed is real. Object distance, u 12 cm  Focal length of the convex lens, f 20 cm Image distance v According to the lens formula, we have the relation: 1 1 1 v u f   1 1 1 v 12 20    1 1 1 3 5 8 v 20 12 60 60       60 v 7.5cm 8    Hence, the image is formed 7.5cm away from the lens, toward its right. Class XII Physics www.vedantu.com 31 b) a concave lens of focal length 16 cm . Ans: Focal length of the concave lens, f –16 cm Image distance v Object distance, u 12 cm  According to the lens formula, we have the relation: 1 1 1 v u f   1 1 1 3 4 1 v 16 12 48 48         v 48cm  Hence, the image is formed 48cm away from the lens, toward its right. 18. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be20cm ? Ans: Given that, Refractive index of glass, 1.55  Focal length of the double-convex lens, f 20 cm Radius of curvature of one face of the lens 1R R  Radius of curvature of the other face of the lens 2R R   Let radius of curvature of the double-convex lens R The value of R can be calculated as:   1 2 1 1 1 1 f R R            1 1 1 1.55 1 20 R R          Class XII Physics www.vedantu.com 32 1 2 0.55 20 R    R 0.55 2 20 22cm     Hence, the radius of curvature of the double-convex lens is 22cm . 19. A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm . What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? Ans: Given, Focal length of the objective lens, of 144 cm Focal length of the eyepiece, ef 6.0 cm The magnifying power of the telescope is given as: o e f 144 m 24 f 6    The separation between the objective lens and the eyepiece is calculated as: o eD f f 144 6 150cm     Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm . 20. a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? Ans: Given that, Focal length of the objective lens, 2 of 15m 15 10 cm   Class XII Physics www.vedantu.com 35 Image distance, v d 25 cm    According to the lens formula, we have: 1 1 1 f v u   1 1 1 u v f    1 1 1 5 1 6 u 25 5 25 25          25 u 4.167cm 6      Hence, the closest distance at which the person can read the book is 4.167cm . For the object at the farthest possible distance, the image distance (v')   According to the lens formula, we have: 1 1 1 f v ' u '   1 1 1 1 u ' 5 5       u ' 5cm   Hence, the farthest distance at which the person can read the book is 5cm . b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope? Ans: Maximum angular magnification is given by the relation: max d 25 6 25u 6     Minimum angular magnification is given by the relation: min d 25 5 u ' 5     Class XII Physics www.vedantu.com 36 23. A card sheet divided into squares each of size 2 1mm is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? Ans: Given that, Size of each square, 2a 1 mm Object distance, u 9 cm  Focal length of a converging lens, f 10 cm For image distance v , the lens formula can be written as: 1 1 1 f v u   1 1 1 10 v 9    1 1 v 90    v 90cm   Magnification, v m u  90 m 10 9       Area of each square in the virtual image 210 a  2 2A 10 1 100mm    2A 1cm  b) What is the angular magnification (magnifying power) of the lens? Ans: Magnifying power of the lens d 25 2.8 u 9    Class XII Physics www.vedantu.com 37 c) Is the magnification in (a) equal to the magnifying power in (b)? Explain. Ans: The magnification in (a) is not the same as the magnifying power in (b). The magnification magnitude is v u       and the magnifying power is d u        . The two quantities will be equal only when the image is formed at the near point. 24. a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power? Ans: The maximum possible magnification is obtained when the image is formed at the near point (d 25 cm) . Here, Image distance, v d 25 cm    Focal length, f 10cm Object distance u According to the lens formula, we have: 1 1 1 f v u   1 1 1 u v f    1 1 1 2 5 7 u 25 10 50 50          50 u 7.14cm 7      Hence, to view the squares distinctly, the lens should be kept 7.14cm away from them. Class XII Physics www.vedantu.com 40 Height of the tower, 1h 100 m Distance of the tower (object) from the telescope, u 3 km 3000m  The angle subtended by the tower at the telescope is given as: 1h 100 1 rad u 3000 30     The angle subtended by the image produced by the objective lens is given as: 2 2 o h h rad f 140    Where, 2h  Height of the image of the tower formed by the objective lens. 21 h 30 140   2 140 h 4.7cm 30    Therefore, the objective lens forms a 4.7cm tall image of the tower. c) What is the height of the final image of the tower if it is formed at 25cm? Ans: Given, the image is formed at a distance, d 25 cm . The magnification of the eyepiece is given by the relation: e d m 1 f   25 m 1 1 5 6 5       Height of the final image 2mh 6 4.7 28.2cm    Therefore, the height of the final image of the tower is 28.2cm . Class XII Physics www.vedantu.com 41 Long Answer Questins 5 Marks 1. Prove that 2 1 2 1 n n n n v u R    When refraction occurs of a convex spherical refracting surface and the ray travels from rarer to denser medium. Ans: The ray is travelling from rarer medium to denser medium as shown in figure. Now, we have NOC , We get, i  ……(1) From figure, MN tan OM   MN tan MI   MN tan MC   For small angle, MN OM   , MN MI   , MN MC   Now equation (1) becomes, Class XII Physics www.vedantu.com 42 MN MN i OM MC    Similarly, r    MN MN r MC MI    Using Snell’s Law, 1 2n sin i n sin r For small angle, 1 2n i n r Now, we put the value of i and r in the above equation, 1 2 MN MN MN MN n n OM MC MC MI               ……(2) Put OM u,MI v,MC R   in equation (2), 1 2 1 1 1 1 n n OM MC MC MI                2 1 2 1n n n n MI OM MC     2 1 2 1n n n n v u R      2 1 2 1n n n n v u R     Hence, Proved. 2. A lens forms a real image of an object. The distance of the object. From the lens is u cm and the distance of the image from the lens is v cm. The given graph shows the variation of v and u Class XII Physics www.vedantu.com 45 Ans: Sign conventions: 1) All distances are estimated from the mirror's pole. 2) Distance measured in the incident light direction is positive, and those measured in the direction reverse to the incident light are negative. 3) Height estimated upwards is positive and height estimated downwards is negative. Assumptions: 1) Spherical mirror's aperture is supposed to be very small. 2) The incident and refracted rays form small angles with the principal axis. This figure shows image produced by the concave mirror. 1 1ACB A CB  [opposite angles] 1 1ABC A B C  [right angles] This gives, 1 1BAC CA B  1 1 1 AB BC A B B C  ……(1) 1 1FED FA B  1 1 1 ED EF A B FB  Also, ED AB Class XII Physics www.vedantu.com 46 1 1 1 AB EF A B FB  Combining (1) and (2), 1 1 BC EF B C FB  We are considering that E is close to P, so, EF PF 1 1 BC PF B C FB  From figure, 1 1BC R ( u),B C v ( R),PF f ,FB v ( f )              Now, R u f v R v f          uv uf Rv Rf Rf vf      uv uf Rv vf     Put R 2f uv uf 2fv vf     uv uf fv 0    Dividing by uvf 1 1 1 f u v   Hence, derived. 4. a) A person looking at a mesh of crossed wires is able to see the vertical lines more distinctly than the horizontal wires. What is the effect due to? How is such a defect of vision corrected? Class XII Physics www.vedantu.com 47 Ans: In the given case, the person can see vertical lines more precisely than horizontal lines. This means that the eye's refracting system (cornea and eye-lens) is not working similarly in different planes. This defect is named astigmatism. The person's eye has sufficient curvature in the vertical plane. However, the horizontal plane's curvature is insufficient. Hence, sharp vertical lines are made on the retina, but horizontal lines seem blurred. This defect can be fixed by using cylindrical lenses. b) A man with normal near point ( 25cm ) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5cm. i) What is the closest and the farthest distance at which he can read the book when viewing through the magnifying glass? Ans: Given that, Focal length of the magnifying glass, f 5 cm Least distance of distance vision, d 25 cm Closest Object distance u Image distance, v d 25cm    Using lens formula, 1 1 1 f v u   1 1 1 5 25 u     25 u cm 4.167cm 6      Hence, the nearest distance at which the person can read the book is 4.167 cm. For the object at the farthest distance (u ') , the image distance (v')   Using lens formula, Class XII Physics www.vedantu.com 50 Let image size h' Object distance, u 27cm  concave mirror’s curvature radius, R 36cm  Focal length of the concave mirror, R f 18cm 2    Using mirror formula, 1 1 1 u v f   1 1 1 v f u    Where, u is the object distance, v is the image distance and f is the focal length. Now, we put given values, 1 1 1 v 18 27      1 3 2 v 54     1 1 v 54    v 54cm  Therefore, the screen should be 54 cm away from the mirror to get a sharp image. The formula for magnification of image is given by: h ' v m h u    Class XII Physics www.vedantu.com 51 v h ' h u     54 h ' 2.5 27       h' 5cm  The height of the image of the candle is 5 cm . The negative sign shows that the image is inverted and real. If the candle is moved nearer to the mirror, then the screen will have to be moved far from the mirror in order to get the image. 7. A 4.5 cm needle is placed 12cm away from a convex mirror of focal length 15cm . Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. Ans: Given that, Height of the needle, 1h 4.5cm Object distance, u 12cm  Focal length of the convex mirror, f 15cm Image distance v Using mirror formula, 1 1 1 u v f   1 1 1 v f u    Now, we put given values, 1 1 1 v 15 12     1 4 5 v 60    Class XII Physics www.vedantu.com 52 1 9 v 60   v 6.7cm  Hence, the needle’s image is 6.7 cm away from the mirror and it is on the mirror’s other side. The formula for magnification of image is given by: 2 1 h v m h u    2 1 v h h u     2 6.7 h 4.5 12     2h 2.5cm  Hence, magnification, 2.5 m 0.56 4.5   The image’s height is 2.5cm. The positive sign shows that the image is virtual, erect, and diminished. If the needle is moved away from the mirror, the image will also move farther from the mirror, and the size of the image will decrease gradually. 8. A tank is filled with water to a height of 12.5 cm . The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4cm . What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again? Ans: Given that, Actual depth of the needle in water, 1h 12.5cm Apparent depth of the needle in water, 2h 9.4cm Class XII Physics www.vedantu.com 55 Using (i) and (ii), the relative refractive index of glass with respect to water can be derived as: a gw g a w     w g 1.51 1.184   w g 1.275  For the glass - water interface, Angle of incidence, i 45  Angle of refraction, r We can use Snell’s law, w g sin i sin r   sin 45 1.275 sin r    sin r 0.5546  1r sin (0.5546)  r 38.68   Hence, the angle of refraction at the water – glass interface is 38.68 . 10. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm . What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) Ans: Provided that, Bulb’s actual depth in water, 1d 80cm 0.8m  Water’s refractive index, 1.33  Class XII Physics www.vedantu.com 56 The following diagram represents the given setup: Where, i is the Angle of incidence r is the Angle of refraction, r 90  As the bulb acts as a point source, the emergent light would be considered as a circle of radius, AC R AO OB 2    Snell’s law may be used as follows: sin r sin i   sin90 1.33 sin i    1 sin i 1.33   i 48.75   Considering the given diagram, we have the relation: 1 OC R tani OB d   R tan48.75 0.8   R 0.91m  Area of the surface of water 2 2 2R (0.91) 2.61m     Clearly, the area of the water surface through which the light from the bulb could project is about 22.61m . Class XII Physics www.vedantu.com 57 11. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40 . What is the refractive index of the material of the prism? The refracting angle of the prism is 60 . If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. Ans: The minimum deviation angle and the angle of prism is as shown in figure given below: Angle of minimum deviation, m 40   Angle of the prism, A 60  Refractive index of water, 1.33  Refractive index of the material of the prism '  The relation between angle of deviation with refractive index is given by: mA sin 2 ' A sin 2               60 40 sin sin502 ' 60 sin30 sin 2                  ' 1.532  Hence, the refractive index of the prism is 1.532 . Class XII Physics www.vedantu.com 60 1 1 1 1 1 1 u v f    1 1 1 1 u 10 2    1 1 4 u 10    1u 2.5cm   Magnitude of the object distance, 1| u | 2.5cm . The compound microscope’s magnifying power is given by the relation: 1 1 2 v d ' m 1 | u | f        10 25 m 1 2.5 6.25         m 4(1 4) 20    Hence, the magnifying power of the microscope is 20 . b) at infinity? What is the magnifying power of the microscope? Ans: Given that, the final image is formed at infinity. The distance of image of the eyepiece, 2v   The distance of object of the eyepiece 2u Using lens formula, 2 2 2 1 1 1 v u f   2 2 2 1 1 1 u v f    Class XII Physics www.vedantu.com 61 2 1 1 1 u 6.25     2u 6.25cm   The distance of image for the objective lens, 1 2v d u 15 6.25 8.75cm     The distance of object for the objective lens 1u Using lens formula, 1 1 1 1 1 1 v u f   1 1 1 1 1 1 u v f    1 1 1 1 u 8.75 2    1u 2.59cm   Magnitude of the object distance, 1| u | 2.59cm . The compound microscope’s magnifying power is given by the relation: 1 1 2 v d ' m | u | | u |        8.75 25 m 2.59 6.25         m 13.51  Hence, the magnifying power of the microscope is 13.51. 13. A person with a normal near point ( 25 cm ) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope. Class XII Physics www.vedantu.com 62 Ans: Given that, Focal length of the objective lens, of 8 mm 0.8 cm  Focal length of the eyepiece, ef 2.5 cm The distance of the object for the Objective lens, ou 9.0 mm 0.9 cm    Least distance of distant vision, d 25 cm Image distance for the eyepiece, ev d 25 cm    Object distance for the eyepiece, eu Using lens formula, e e e 1 1 1 v u f   e e e 1 1 1 u v f    e 1 1 1 u 25 2.5     e 1 11 u 25    eu 2.27cm   Using lens formula, we can obtain image distance for the objective lens, ov , is given by: o o o 1 1 1 v u f   o o o 1 1 1 v u f    o 1 1 1 v 0.8 0.9    Class XII Physics www.vedantu.com 65 Using mirror formula, 1 1 1 u v f   1 1 1 v f u    Using equation (2), we get: 1 0 v  v 0  Thus, the image is formed on the mirror’s back side. Hence, a convex mirror always gives a virtual image, regardless of the object distance. c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. Ans: For a convex mirror, f 0 . Using mirror formula, 1 1 1 u v f   1 1 1 v f u    But we have, u 0 1 1 v f   v f  Hence, the image formed is diminished and is located between the focus and the pole. Class XII Physics www.vedantu.com 66 d) An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. Ans: For a concave mirror, f 0 . When the object is placed on the left side of the mirror, then u 0 . The object is placed between the focus and the pole. f u 0   1 1 0 f u    1 1 0 f u    1 0 v   v 0  The image is formed on the mirror’s right side. Hence, it is a virtual image. For u 0,v 0  1 1 u v   v u  Magnification, v m 1 u   Hence, the formed image is enlarged. 15. a) Figure shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68 . The outer covering of the pipe is made of a material of refractive index 1.44 . What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure? Class XII Physics www.vedantu.com 67 Ans: Given that, Refractive index of the glass fibre, 1 1.68  Refractive index of the outer covering of the pipe, 2 1.44  Angle of incidence i Angle Of refraction r Angle of incidence at the interface i ' The refractive index ( ) of the inner core - outer core interface is given as: 2 1 1 sini '      1 2 sin i '     1.44 sini ' 0.8571 1.68    For the critical angle, total internal reflection (TIR) takes place only when i i ' . That is, when i 59  . Maximum angle of reflection, maxr 90 i ' 90 59 31         . Let maxi be the maximum incidence angle. The refractive index at the air – glass interface, 1 1.68  . We can use the relation for the maximum angles of incidence and reflection as: max 1 max sin i sin r   max 1 maxsin i sin r   Class XII Physics www.vedantu.com 70 Ans: The diamond's refractive index is more than that of ordinary glass. The critical angle for diamond is smaller than that for glass. A diamond cutter utilizes a large angle of incidence to guarantee that the light entering the diamond is totally reflected from its face. This is the cause for the sparkling effect of a diamond. 17. a) Determine the 'effective focal length' of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? Ans: Consider the diagram below which represents the combination of two lenses. Here, Focal length of the convex lens, 1f 30 cm Focal length of the concave lens, 2f 20 cm  Distance between the two lenses, d 8.0 cm First, consider the case when the parallel beam of light falls on the convex lens. Using lens formula, 1 1 1 1 1 1 v u f   Where, object distance, 1u   Image distance 1v Class XII Physics www.vedantu.com 71 1 1 1 1 1 v 30 30      1v 30cm  The image will serve as a virtual object for the concave lens. Using lens formula, 2 2 2 1 1 1 v u f   Where, object distance 2u . 2u 30 8 22cm   Image distance 2v . 2 1 1 1 1 v 22 20 220      2v 220cm   The parallel incident beam seems to diverge from a point, that is, d 220 220 4 216cm 2     from the centre of the combination of the two lenses. Secondly, when the parallel beam of light falls, from the left, on the concave lens; Using lens formula, 2 2 2 1 1 1 v u f   Where, object distance, 2u   . Image distance 2v . 2 1 1 1 1 v 20 20        2v 20cm   Class XII Physics www.vedantu.com 72 The image will serve as a real object for the convex lens. Using lens formula, 1 1 1 1 1 1 v u f   Where, object distance, 1u . 1u (20 8) 28cm     . Image distance 1v . 1 1 1 1 1 v 30 28 420       1v 420cm   Hence, the parallel incident beam seems to diverge from a point, that is 420 4 416cm   from the left of the centre of the combination of the two lenses. Thus, the answer does depend on the combination side at which the parallel beam of light is incident. The notion of effective focal length does not appear to be useful for this combination. b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm . Determine the magnification produced by the two-lens system, and the size of the image. Ans: Consider the given diagram of the previous arrangement as follows: Here, it is said that, Height of the image, 1h 1.5 cm Object distance from the side of the convex lens, 1u 40cm  Class XII Physics www.vedantu.com 75 is the angle of incidence at the face AC. e is the emergent angle, e 90 . Using Snell’s law, 2 sine sin r   2 sin90 1.524 sin r    2sin r 0.6562  2r 41   For refraction through prism, angle 1 2A r r  . We get, 1 2r A r 60 41      1r 19   Using Snell’s law, 1 1 sin i sin r   1sini 1.524 sin19    1sin i 0.496  1i 29.75   Hence, the incidence angle is 29.75 . 19. For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this Class XII Physics www.vedantu.com 76 rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye. Ans: Given that, Least distance of distinct vision, d 25 cm Far point of a normal eye, d '   Converging power of the cornea, cP 40D Least converging power of the eye-lens, eP 20D To see the objects at infinity, the eye uses its least converging power. Power of the eye-lens, c eP P P 40 20 60D     Power of the eye-lens is given as: 1 P f  1 f P   1 5 f m cm 60 3    Object distance, u d 25cm    Image distance, 5 v f cm 3   Using lens formula, 1 1 1 f ' v u   11 3 1 16 cm f ' 5 25 25      Power, 1 P 100 f '   Class XII Physics www.vedantu.com 77 16 P 100 64D 25     Power of the eye lens 64 40 24D   Hence, the range of accommodation of the eye-lens is from 20 D to 24 D. 20. a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? Ans: Though the image size is larger than the object, the angular size of the image is equivalent to the angular size of the object. A magnifying glass supports one seeing the objects closer than the least distance of distinct vision. A closer object produces a larger angular size. A magnifying glass gives angular magnification. Without magnification, the object cannot be located closer to the eye. With magnification, the object can be set much closer to the eye. b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back? Ans: Yes, the angular magnification varies. When the length between the eye and a magnifying glass rises, the angular magnification reduces slightly because the subtended angle at the eye is imperceptibly less than the lenses. Image distance does not have any impact on angular magnification. c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? Class XII Physics www.vedantu.com 80 o e m m m   o 30 m 5 6    We have, o o o v m u   o o v 5 u    o ov 5u   ……(1) Using lens formula, o o o 1 1 1 f v u   o o 1 1 1 1.25 5u u     ou 1.5cm   And, o ov 5u 5 1.5 7.5cm      The object should be placed 1.5cm away from the objective lens to get the desired magnification. Using lens formula, e e e 1 1 1 f v u   Where, ev d 25cm    e 1 1 1 5 25 u     eu 4.17cm   Class XII Physics www.vedantu.com 81 Separation between the eyepiece and the objective lens e o| u | | v |  e o| u | | v | 4.17 7.5 11.67cm    Therefore, the separation between the eyepiece and the objective lens should be 11.67cm. 22. A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and these all mirror is 140mm , where will the final image of an object at infinity be? Ans: A Cassegrain telescope contains two mirrors, one is objective mirror and second one is secondary mirror. We need to find the distance of final image from secondary mirror. Given: distance between the objective mirror and the secondary mirror, d 20mm Radius of curvature of the objective mirror, 1R 220 mm Hence, objective mirror’s focal length, 1 1 R f 110mm 2   Secondary mirror’s radius of curvature, 2R 140 mm Secondary mirror’s focal length, 2 2 R f 70mm 2   The image of an object set at infinity, made by the objective mirror, will serve as a virtual object for the secondary mirror. Hence, the secondary mirror’s virtual object distance, 1u f d  u 110 20 90mm    Class XII Physics www.vedantu.com 82 Using mirror formula, 2 1 1 1 u v f   2 1 1 1 v f u    Now, we put given values, 1 1 1 v 70 90    1 9 7 v 630    1 2 v 630   v 315mm  Clearly, the final image will be made 315 mm away from the secondary mirror. 23. Figure shows an equiconvex lens (of refractive index 1.50 ) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm . The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm . What is the refractive index of the liquid?