Optical Instruments Chapter 33 PDF

60 Optical Instruments 93 ID E3 Human Eye. (1) Eye lens : Over all behaves as a convex lens of   1. 437 D YG U (2) Retina : Real and inverted image of an object, obtained at retina, brain sense it erect. (3) Yellow spot : It is the most sensitive part, the image formed at yellow spot is brightest....

60 Optical Instruments 93 ID E3 Human Eye. (1) Eye lens : Over all behaves as a convex lens of   1. 437 D YG U (2) Retina : Real and inverted image of an object, obtained at retina, brain sense it erect. (3) Yellow spot : It is the most sensitive part, the image formed at yellow spot is brightest. (4) Blind spot : Optic nerves goes to brain through blind spot. It is not sensitive for light. (5) Ciliary muscles – Eye lens is fixed between these muscles. It’s both radius of curvature can be changed by applying pressure on it through ciliary muscles. (6) Power of accomodation : The ability of eye to see near objects as well as far objects is called power of accomodation. Note :  When we look distant objects, the eye is relaxed and it's focal length is largest. U (7) Range of vision : For healthy eye it is 25 cm (near point) to  (far point). A normal eye can see the objects clearly, only if they are at a distance greater than 25 cm. This distance is called Least distance of distinct vision and is represented by D. (8) Persistence of vision : Is 1/10 sec. i.e. if time interval between two consecutive light pulses is lesser than 0.1 sec., eye cannot distinguish them separately. (9) Binocular vision : The seeing with two eyes is called binocular vision. (10) Resolving limit : The minimum angular displacement between two objects, so that they are just ST o  1  resolved is called resolving limit. For eye it is 1 '   .  60  Specific Example A person wishes to distinguish between two pillars located at a distances of 11 Km. What should be the minimum distance between the pillars.  1  Solution : As the limit of resolution of eye is    60   1  So      60  (11) Defects in eye o  d 11  10 3 d o  1       d  3.2 m  60  180  11 km 94 Optical Instruments Hypermetropia (long sightness) (i) Distant objects are not seen clearly but nearer objects are clearly visible. (i) Distant objects are seen clearly but nearer object are not clearly visible. (ii) Image formed before the retina. (ii) Image formed behind the retina. (iii) Far point comes closer. (iii) Near point moves away (iv) Reasons : (a) Focal length or radii of curvature of lens reduced or power of lens increases. (iv) Reasons : (a) Focal length or radii of curvature of lens increases or power of lens decreases. (b) Distance between eye lens and retina increases. (b) Distance between eye lens and retina decreases. (v) Removal : By using a concave lens of suitable focal length. (v) Removal : By using a convex lens. ID E3 60 Myopia (short sightness) (vi) Focal length : (vi) Focal length : (a) A person can see upto distance x (a) A person cannot see before distance d wants to see , so wants to see the object place at distance D so U focal length of used lens f   x = – (defected far point) (b) A person can see upto distance x f dD dD D YG wants to see distance y (y > x) so f  xy x y Presbyopia : In this defect both near and far objects are not clearly visible. It is an old age disease and it is due to the loosing power of accommodation. It can be removed by using bifocal lens. Concave U Convex ST Astigmatism : In this defect eye cannot see horizontal and vertical lines clearly, simultaneously. It is due to imperfect spherical nature of eye lens. This defect can be removed by using cylindrical lens (Torric lenses). Microscope. It is an optical instrument used to see very small objects. It’s magnifying power is given by m Visual angle with instrument ( ) Visual angle when object is placed at least distance of distinct vision ( ) (1) Simple miscroscope (i) It is a single convex lens of lesser focal length. (ii) Also called magnifying glass or reading lens. Optical Instruments 95 (iii) Magnification’s, when final image is formed at D and  (i.e. m D and m  )   D D and m     m D  1   f  max   f  min m max.  m min.  1  If lens is kept at a distance a from the eye then m D  1  Da Da and m   f f (2) Compound microscope E3 (i) Consist of two converging lenses called objective and eye lens. (ii) feye lens  fobjective and (diameter) eye lens 60 Note :   (diameter )objective ID (iii) Final image is magnified, virtual and inverted. (iv) u 0  Distance of object from objective (o), v 0  Distance of image ( A B ) formed by objective from objective, u e  Distance of A B  from eye lens, ve = mD   v0 u0   f0 (v  f0 )  D D D 1     1     0 1   fe  (u 0  f0 )  fe  f0 fe    D YG Magnification : U Distance of final image from eye lens, f0 = Focal length of objective, fe = Focal length of eye lens. m   v0 D  f0  D  (v  f0 ) D     0. . u 0 Fe (u 0  f0 )  fe  f0 Fe Length of the tube (i.e. distance between two lenses) LD  v0  ue  When final images is formed at  ; L  v 0  fe  U When final image is formed at D ; u 0 f0 f D  e u 0  f0 fe  D u 0 f0  fe u 0  f0 ST (Do not use sign convention while solving the problems) Note :  m  (L   f0  fe )D f0 fe  For maximum magnification both f0 and fe must be less.  m  m objective  m eye lens  If objective and eye lens are interchanged, practically there is no change in magnification. (3) Resolving limit and resolving power : In reference to a microscope, the minimum distance between two lines at which they are just distinct is called Resolving limit (RL) and it’s reciprocal is called Resolving power (RP) 96 Optical Instruments R.L.   2  sin  and R.P.  2  sin    R. P.  1   = Wavelength of light used to illuminate the object,  = Refractive index of the medium between object and objective,  = Half angle of the cone of light from the point object,  sin  = Numerical aperture. 60 Note :  Electron microscope : electron beam (  1 Å) is used in it so it’s R.P. is approx 5000 times more than that of ordinary microscope (  5000 Å) Telescope. E3 By telescope distant objects are seen. (1) Astronomical telescope (i) Used to see heavenly bodies. (ii) fobjective  feyelens and d objective  d eye lens. f0  f  f  1  e  and m    o fe  D fe fe D and L   f0  fe fe  D D YG (vi) Length : L D  f0  u e  f0  U (v) Magnification : m D   ID (iii) Intermediate image is real, inverted and small. (iv) Final image is virtual, inverted and small. (2) Terrestrial telescope (i) Used to see far off object on the earth. (ii) It consists of three converging lens : objective, eye lens and erecting lens. (iii) It’s final image is virtual erect and smaller. A A'' B Q ST (v) Length : L D  f0  4 f  u e  f0  4 f  (i) It is also a terrestrial telescope but of much smaller field of view. (ii) Objective is a converging lens while eye lens is diverging lens. f0  f  f  1  e  and m   0 fe  D fe (iv) Length : L D  f0  u e and L   f0  fe (4) Resolving limit and resolving power B'' Erecting lens A' fo fe D and L   f0  4 f  fe fe  D (3) Galilean telescope (iii) Magnification : m D  B' O f  f  f  0 1  e  and m   0 fe  D fe U (iv) Magnification : m D ue = D to  P 2f 2f uo Optical Instruments 97 60 Smallest angular separations (d) between two distant objects, whose images are separated in the 1.22  telescope is called resolving limit. So resolving limit d  a 1 1 a and resolving power (RP )    R.P.  where a = aperture of objective. d 1.22   Note :  Minimum separation (d) between objects, so they can just resolved by a telescope is – d  r R.P. where r = distance of objects from telescope. (5) Binocular ID E3 If two telescopes are mounted parallel to each other so that an object can be seen by both the eyes simultaneously, the arrangement is called 'binocular'. In a binocular, the length of fo each tube is reduced by using a set of totally reflecting prisms which provided intense, erect image free from lateral inversion. Through a binocular we get two images of the same object from different angles at same time. Their superposition gives the perception of depth also along with length and breadth, i.e., binocular vision gives proper three-dimensional (3D) image. fe U Concepts  As magnifying power is negative, the image seen in astronomical telescope is truly inverted, i.e., left is turned right with upside D YG down simultaneously. However, as most of the astronomical objects are symmetrical this inversion does not affect the observations.  Objective and eye lens of a telescope are interchanged, it will not behave as a microscope but object appears very small.  In a telescope, if field and eye lenses are interchanged magnification will change from (f o / fe) to (fe / fo), i.e., it will change from m to (1/m), i.e., will become (1/m 2) times of its initial value.  As magnification for normal setting as (fo / fe), so to have large magnification, fo must be as large as practically possible and  fe small. This is why in a telescope, objective is of large focal length while eye piece of small. In a telescope, aperture of the field lens is made as large as practically possible to increase its resolving power as resolving power of a telescope  (D/)*. Large aperture of objective also helps in improving the brightness of image by gathering more light from distant object. However, it increases aberrations particularly spherical. U  For a telescope with increase in length of the tube, magnification decreases.  In case of a telescope if object and final image are at infinity then : m fo D  fe d ST  If we are given four convex lenses having focal lengths f1  f2  f3  f4. For making a good telescope and microscope. We choose the following lenses respectively. Telescope f1 (o), f4 (e) Microscope f4 (o), f3 (e)  If a parrot is sitting on the objective of a large telescope and we look towards (or take a photograph)of distant astronomical object (say moon) through it, the parrot will not be seen but the intensity of the image will be slightly reduced as the parrot will act as obstruction to light and will reduce the aperture of the objective. Examples 98 Optical Instruments Example: 1 A man can see the objects upto a distance of one metre from his eyes. For correcting his eye sight so that he can see an object at infinity, he requires a lens whose power is or A man can see upto 100 cm of the distant object. The power of the lens required to see far objects will be [MP PMT 1993, 2003] Solution: (d) Example: 2 (b) +1.0 D (c) +2.0 D (d) –1.0 D 100 100 f = –(defected far point) = – 100 cm. So power of the lens P    1D f  100 A man can see clearly up to 3 metres. Prescribe a lens for his spectacles so that he can see clearly up to 12 60 (a) +0.5 D metres [DPMT 2002] (b) 3 D (c) – 1/4 D (d) – 4 D E3 (a) – 3/4 D 3  12 1 xy 1  f  4 m. Hence power P    D 3  12 f 4 x y Solution: (c) By using f  Example: 3 The diameter of the eye-ball of a normal eye is about 2.5 cm. The power of the eye lens varies from Solution: (d) (b) 40 D to 32 D (c) 9 D to 8 D ID (a) 2 D to 10 D An eye sees distant objects with full relaxation so 2.5  10 1 2 U An eye sees an object at 25 cm with strain so 1 2.5  10  2  (d) 44 D to 40 D 1 1 1 1  40 D  or P   f 25  10  2  f 1  25  10 2  1 1 or P   40  4  44 D f f The resolution limit of eye is 1 minute. At a distance of r from the eye, two persons stand with a lateral separation of 3 metre. For the two persons to be just resolved by the naked eye, r should be (a) 10 km (b) 15 km (c) 20 km (d) 30 km Solution: (a) From figure   D YG Example: 4 d = 3m o  1 1  3  r = 10 km   60 180 r  r Two points separated by a distance of 0.1 mm can just be resolved in a microscope when a light of wavelength 6000 Å is used. If the light of wavelength 4800 Å is used this limit of resolution becomes U Example: 5 d  1   1   rad ; where   1'        r  60   60  180 (a) 0.08 mm (b) 0.10 mm [UPSEAT 2002] (c) 0.12 mm  (R.L.) 1 0.1 6000   1  (R.L.) 2 4800 2 (R.L.) 2 (d) 0.06 mm  (R.L.) 2  0.08 mm. By using resolving limit (R.L.)   Example: 6 In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm an object is placed at 2 cm form objective and the final image is formed at 25 cm from eye lens. The distance between the two lenses is ST Solution: (a) Solution: (d)  (a) 6.00 cm (b) 7.75 cm (c) 9.25 cm It is given that fo = 1.5 cm, fe = 6.25 cm, uo = 2 cm [EAMCET (Med.) 2000] (d) 11.00 cm When final image is formed at least distance of distinct vision, length of the tube LD   LD  Example: 7 uo fo fD  e uo  fo fe  D 2  1.5 6.25  25   11 cm. (2  1.5) (6.25  25 ) The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm. The final image formed by Optical Instruments 99 the eye-piece is at infinity. The two lenses are thin. The distances in cm of the object and the image produced by the objective measured from the objective lens are respectively [IIT-JEE 1995] (a) 2.4 and 12.0 (c) 2.3 and 12.0 Given that fo  2 cm , fe  3 cm , L  15 cm By using L  vo  fe  15  vo  3  vo  12 cm. Also The focal lengths of the objective and eye-lens of a microscope are 1 cm and 5 cm respectively. If the magnifying power for the relaxed eye is 45, then the length of the tube is [CPMT 1979] (a) 30 cm If the focal lengths of objective and eye lens of a microscope are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away from the objective lens and the final image is formed at infinity, then magnifying power of the microscope is (b) 200 U The magnifying power of an astronomical telescope is 8 and the distance between the two lenses is 54cm. The focal length of eye lens and objective lens will be respectively [MP PMT 1991; CPMT 1991; Pb. PMT 2001] (a) 6 cm and 48 cm (b) 48 cm and 6 cm (c) 8 cm and 64 cm (d) 64 cm and 8 cm Given that m   8 and L  54 By using | m  |  Example: 11 fo and L  fo  fe we get fo  6 cm and fe  48 cm. fe If an object subtend angle of 2° at eye when seen through telescope having objective and eyepiece of focal length fo  60 cm and fe  5 cm respectively than angle subtend by image at eye piece will be[UPSEAT 2001] (a) 16° (b) 50°  fo  60  By using     24    fe 20 5 U Solution: (c) Example: 12 (d) 400 fo D 1.2 25   m     200. (u o  fo ) fe (1.25  1.2) 3 D YG Solution: (a) (c) 250 Given that fo  1.2 cm , fe  3 cm , uo  1.25 cm By using m    Example: 10 (d) 12 cm ( L   fo  fe ) (L  1  5)  25  45    L  15 cm fo f e 15 (a) 150 Solution: (b) (c) 15 cm Given that fo  1 cm , fe  5 cm , m   45 By using m   Example: 9 (b) 25 cm E3 Solution: (c) 12 12  2 vo vo  fo   uo  2.4 cm.   uo 2 uo fo ID Example: 8 (d) 2.3 and 3.0 60 Solution: (a) (b) 2.4 and 15.0 (c) 24° (d) 10° The focal lengths of the lenses of an astronomical telescope are 50 cm and 5 cm. The length of the telescope when the image is formed at the least distance of distinct vision is [EAMCET (Engg.) 2000] ST (a) 45 cm (b) 55 cm (c) 275 cm 6 (d) 325 cm 6 fe D 5  25 325  50   cm fe  D (5  25 ) 6 Solution: (d) By using LD  fo  ue  fo  Example: 13 The diameter of moon is 3.5  10 3 km and its distance from the earth is 3.8  10 5 km. If it is seen through a telescope whose focal length for objective and eye lens are 4 m and 10 cm respectively, then the angle subtended by the moon on the eye will be approximately (a) 15° (b) 20° (c) 30° (d) 35° Solution: (b) The angle subtended by the moon on the objective of telescope   Also m  3.5  10 3 3.5   10  2 rad 5 3. 8 3.8  10 400  fo  3.5  10 3 180     20      40     40  10  fe   3.8  10 5 100 Optical Instruments Solution: (b) A telescope has an objective lens of 10 cm diameter and is situated at a distance one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of (a) 0.5 m (b) 5 m (c) 5 mm (d) 5cm Suppose minimum distance between objects is x and their distance from telescope is r Example: 15 Solution (b) 1.22   r 1.22  (5000  10 10 )  (1  10 3 ) 1.22  x   6.1  10  3 m  6.1 mm  x  a (0  1) a r 60 So Resolving limit d   Hence, It’s order is  5 mm. A compound microscope has a magnifying power 30. The focal length of its eye-piece is 5 cm. Assuming the final image to be at the least distance of distinct vision. The magnification produced by the objective will be (a) +5 (b) – 5 (c) +6 (d) – 6 Magnification produced by compound microscope m  m o  m e E3 Example: 14 Tricky example: 1 ID  D 25  6  30  m o  6  m o  5. where mo = ? and m e   1    1  fe  5  A man is looking at a small object placed at his least distance of distinct vision. Without changing his position and that of the object he puts a simple microscope of magnifying power 10 X and just sees the clear image again. The angular magnification obtained is Angular magnification  (c) 5.0 (d) 1.0  tan  I/D I     tan  O / D O D YG Solution : (d) (b) 10.0 U (a) 2.5 Since image and object are at the same position, I v   1  Angular magnification = 1 O u Tricky example: 2 A compound microscope is used to enlarge an object kept at a distance 0.03m from it’s objective which consists of several convex lenses in contact and has focal length 0.02m. If a lensof focal length 0.1m is length Fo) (c) 15 cm forms the image at (d) 9 cm distance vo then U Solution : (d) (a) 2.5 cm (b) 6 cm If initially the objective (focal u f 32 vo  o o   6 cm uo  fo 3  2  1 1 1 1 1 1 1 1 1    .....     .....   where Fo f2 f3 Fo f1 f2 f3 f1 Fo   So if one of the lens is removed, the focal length of the remaining lens system 1 1 1 1 1      Fo  2.5 cm Fo F0 f1 2 10 ST Now as in case of lenses in contact This lens will form the image u F 3  2.5 v o  o o   15 cm u o  Fo (3  2.5) of same object at a distance v o such that So to refocus the image, eye-piece must be moved by the same distance through which the image formed by the objective has shifted i.e. 15 – 6 = 9 cm.