HC Verma Physics books are the most preferred books among students of CBSE schools. Students can be found referring to the chapters as well as practice questions at the end of each of these chapters, in the books. Students follow these textbooks religiously since quite a few questions in these also appear in exams.<\/p>\n\n\n\n
For such popular books, students can get extremely helpful practice material online. For all the questions in the HC Verma books, there are several sources where students can get detailed solutions and solve their doubts and queries.<\/p>\n\n\n\n
Please note that these solutions are provided here for free.<\/p>\n\n\n\n
The size of an object as perceived by an eye depends primarily on
(a) actual size of the object
(b) distance of the object from the eye
(c) aperture of the pupil
(d) size of the image formed on the retina.<\/p>\n\n\n\n
(d) size of the image formed on the retina<\/p>\n\n\n\n
An eye consists of a lens that works on the principle on which a glass lens works. It forms the image on the screen called retina. The magnification, in this case, depends on the ratio of the image to the object height.<\/p>\n\n\n\n
The muscles of a normal eye are least strained when the eye is focussed on an object
(a) far away from the eye
(b) very close to the eye
(c) at about 25 cm from the eye
(d) at about 1 m from the eye<\/p>\n\n\n\n
(a) far away from the eye<\/p>\n\n\n\n
A normal eye can see from 25 cm to infinity; it faces least difficulty and strain focussing on the object as far as it could be.<\/p>\n\n\n\n
A normal eye is not able to see objects closer than 25 cm because
(a) the focal length of the eye is 25 cm
(b) the distance of the retina from the eye-lens is 25 cm
(c) the eye is not able to decrease the distance between the eye-lens and the retina beyond a limit
(d) the eye is not able to decrease the focal length beyond a limit.<\/p>\n\n\n\n
(d) the eye is not able to decrease the focal length beyond a limit<\/p>\n\n\n\n
The ciliary muscles adjust the focal length to form an image on the retina, but the muscles cannot be strained beyond a limit. Hence, if the object is brought too close to the eye, the focal length cannot be adjusted to form the image on the retina.<\/p>\n\n\n\n
When objects at different distances are seen by the eye, which of the following remain constant?
(a) The focal length of the eye-lens.
(b) The object-distance from the eye-lens.
(c) The radii of curvature of the eye-lens.
(d) The image-distance from the eye-lens.<\/p>\n\n\n\n
(d) The image distance from the eye lens<\/p>\n\n\n\n
In the human eye, the image is formed on the retina, which is at a fixed distance from the eye lens.<\/p>\n\n\n\n
A person A<\/em> can clearly see objects between 25 cm and 200 cm. Which of the following may represent the range of clear vision for a person B<\/em> having muscles stronger than A<\/em>, but all other parameters of eye identical to that of A<\/em>? (b) 18 cm to 200 cm<\/p>\n\n\n\n Person B has stronger ciliary muscles than person A. So, the muscles in his case can be strained more and the focal length of his eye can be reduced more compared to those of person A. While seeing far objects, the muscles are relaxed, so their strength will not affect the far point of the eye.<\/p>\n\n\n\n The focal length of a normal eye-lens is about (b) 2 cm<\/p>\n\n\n\n Given: -25 cm 1f=1v-1u\u21d21f\u224512-1-25\u21d21f\u22452750\u21d2f\u22452 cm<\/p>\n\n\n\n The distance of the eye-lens from the retina is x<\/em>. For a normal eye, the maximum focal length of the eye-lens (a) = x<\/em><\/p>\n\n\n\n For a normal eye, we have: \u221eDistance between the eye lens and the retina, v<\/em> = x A man wearing glasses of focal length +1 m cannot clearly see beyond 1 m (d) in each of these cases<\/p>\n\n\n\n The man is wearing glasses of positive power (converging lens). Hence, he cannot see nearby objects clearly. In other words, he is farsighted. Since he cannot see beyond 1 m, he is nearsighted. If a person with normal vision wears glasses of focal length +1 m, then the person will not be able to see beyond 1 m.<\/p>\n\n\n\n An object is placed at a distance u<\/em> from a simple microscope of focal length f<\/em>. The angular magnification obtained depends (c) on f<\/em> as well as u<\/em><\/p>\n\n\n\n \u00e2\u20ac\u2039The angular magnification is the ratio of the angle subtended by the image to the angle subtended by the object on an unaided eye. m=huhDHere, u=Object distance from the lensD=Image distance from the lensh=Height of the objectIn normal adjustment, the object is placed at a distamce equal to \u00e2\u20ac\u2039focal length(f) from the lens and then magnification is given by m = D\/f.<\/em> To increase the angular magnification of a simple microscope, one should increase (b) the power of the lens<\/p>\n\n\n\n For a simple microscope in normal adjustment, the object is placed at a distance equal to f <\/em>(the\u00e2\u20ac\u2039focal length) from the lens, And the angular magnification is given by the relation A man is looking at a small object placed at his near point. Without altering the position of his eye or the object, he puts a simple microscope of magnifying power 5 X before his eyes. The angular magnification achieved is (c) 1<\/p>\n\n\n\n We have:h=Object heightu=Object distance=25 cmD=Near point=25 cmNow,m=huhD\u21d2m=h25h25\u21d2m=1<\/p>\n\n\n\n When we see an object, the image formed on the retina is (a) real The retina acts as a screen in the eye; only real, inverted images can be obtained on the screen.<\/p>\n\n\n\n In which of the following the final image is erect? (a) Simple microscope In a simple microscope, the image formed is virtual and above the axis on the same side of the object. Similarly, in a Galilean telescope, the image formed is virtual, erect and magnified and between the objective lens and the eyepiece. A compound microscope and an astronomical telescope produce inverted images.<\/p>\n\n\n\n The maximum focal length of the eye-lens of a person is greater than its distance from the retina. The eye is (a) always strained in looking at an object<\/p>\n\n\n\n The maximum focal length of a normal eye is equal to the distance of the lens from the retina. In case it is greater than the distance, the eye will be strained while focusing the objects on the retina that is at a fixed distance from the eye lens.<\/p>\n\n\n\n A child has near point at 10 cm. What is the maximum angular magnification the child can have with a convex lens of focal length 10 cm?<\/p>\n\n\n\n Given: mof a simple microscope is given as:<\/p>\n\n\n\n m=1+Df =1+1010=1+1=2Thus, the maximum angular magnification is 2.<\/p>\n\n\n\n An optical instrument used for angular magnification has a 25 D objective and 20 D eyepiece. The tube length is 25 cm when the eye is least strained. (a) Whether it is a microscope or a telescope? (b) What is the angular magnification produced?<\/p>\n\n\n\n Given: f0=125 D=0.04 m=4 cmPower of the eyepiece = 5 D fe=120 D=0.05 m=5 cm(a) As f<\/em>o<\/sub> < f<\/em>e, the instrument must be a microscope.<\/p>\n\n\n\n (b) Tube length, l<\/em> = 25 cm vo+fe<\/p>\n\n\n\n \u21d2v0=l-fe=25<\/p>\n\n\n\n -5 = 20 cm \u21d2 -1uo=14-120\u21d2-1u0=5-120=420\u21d2 u0=204=5 cmThe angular magnification of microscope<\/p>\n\n\n\n mis given by<\/p>\n\n\n\n m=v0u0Dfe =205255=20 cmSo, the required angular magnification of the microscope is 20.<\/p>\n\n\n\n Can virtual image be formed on the retina in a seeing process?<\/p>\n\n\n\n The retina acts as a screen; only real images can be obtained on the screen. In case of people having eye defects, the spectacles form the virtual image of the object and the eye lens form the real and inverted image on the retina.<\/p>\n\n\n\n Can the image formed by a simple microscope be projected on a screen without using any additional lens or mirror?<\/p>\n\n\n\n The image formed by a simple microscope is virtual and erect. So, it cannot be projected on a screen without using any additional lens or mirror.<\/p>\n\n\n\n The angular magnification of a system is less than one. Does it mean that the image formed is inverted?<\/p>\n\n\n\n No, angular magnification is the ratio of the angle subtended by the final image on the eye to the angle subtended by the object on the unaided eye. Its value less than one signifies reduction in the size of the image. It does not mean that the image is inverted.<\/p>\n\n\n\n A simple microscope using a single lens often shows coloured image of a white source. Why?<\/p>\n\n\n\n A simple microscope consists of a single convex lens. Sometimes due to chromatic and spherical aberrations, the image of a white source seems coloured at the corners of the lens and somewhere in between.<\/p>\n\n\n\n A magnifying glass is a converging lens placed close to the eye. A farsighted person uses spectacles having converging lenses. Compare the functions of a converging lens used as a magnifying glass and as spectacles.<\/p>\n\n\n\n A converging lens in a magnifying glass is of small focal length which is used to magnify an object which is placed close to the lens. On the other hand, converging lens used as spectacles is of varying focal length which depends upon the actual near point of the long-sighted person. It forms image at the near point of the defected eye which is further focussed by the eye lens.<\/p>\n\n\n\n A person is viewing an extended object. If a converging lens is placed in front of his eyes, will he feel that the size has increased?<\/p>\n\n\n\n If a person views an extended object through a converging lens, then it will appear larger and wider to him.<\/p>\n\n\n\n The magnifying power of a converging lens used as a simple microscope is<\/p>\n\n\n\n 1+Df.A compound microscope is a combination of two such converging lenses. Why don\u2019t we have magnifying power<\/p>\n\n\n\n 1+Dfo1+Dfe?In other words, why can the objective not be treated as a simple microscope but the eyepiece can?<\/p>\n\n\n\n In a simple microscope, the converging lens is used to magnify the object. It is done by the eyepiece in a compound microscope. But the purpose of the objective lens is the same, i.e., to form an enlarged, real and inverted image of the object at a distance less than the focal length of the eyepiece. So, its magnification power cannot be expressed in a way it is expressed for a simple microscope.<\/p>\n\n\n\n By mistake, an eye surgeon puts a concave lens in place of the lens in the eye after a cataract operation. Will the patient be able to see clearly any object placed at any distance?<\/p>\n\n\n\n The image formed by a concave lens is virtual and upright. It is smaller than the object and is formed between the object and the lens, irrespective of the position of the object. If, by mistake, an eye surgeon puts a concave lens in place of eye lens in a patient\u2019s eye, then the image will not be focused on the retina; this will lead to unclear vision.<\/p>\n\n\n\n The magnifying power of a simple microscope is given by<\/p>\n\n\n\n 1+Df,where D<\/em> is the least distance for clear vision. For farsighted persons, D<\/em> is greater than the usual. Does it mean that the magnifying power of a simple microscope is greater for a farsighted person as compared to a normal person? Does it mean that a farsighted person can see an insect more clearly under a microscope than a normal person?<\/p>\n\n\n\n The magnifying power of a simple microscope depends on the ratio<\/p>\n\n\n\n Dffor a farsighted person. Here, D<\/em> for a farsighted person is greater than that for a normal person, but the value of f <\/em>remains the same. Therefore, the magnifying power of a simple microscope is greater for a farsighted person compared to that for a person with normal vision. Also, a farsighted person can see the insect more clearly under the microscope than a person with normal vision.<\/p>\n\n\n\n Why are the magnification properties of microscopes and telescopes defined in terms of the ratio of angles and not in terms of the ratio of sizes of objects and images?<\/p>\n\n\n\n Instruments like telescopes and microscopes deal with objects placed at different distances. Due to some physical factors, there is a relative change in heights not in the angle which the light emerging from them subtends on the lens. So, the magnification properties of instruments are defined in terms of the ratio of angles.<\/p>\n\n\n\n An object is placed at a distance of 30 cm from a converging lens of focal length 15 cm. A normal eye (near point 25 cm, far point infinity) is placed close to the lens on the other side. (a) Can the eye see the object clearly? (b) What should be the minimum separation between the lens and the eye so that the eye can clearly see the object? (c) Can a diverging lens, placed in contact with the converging lens, help in seeing the object clearly when the eye is close to the lens?<\/p>\n\n\n\n Object distance, u<\/em> =<\/strong><\/p>\n\n\n\n -30 cm 1v-1u=1f\u21d21v-1-30=115\u21d21v=130\u21d2v=+30 cm(on the opposite side of the object)<\/p>\n\n\n\n (a) No, the eye placed close to the lens cannot see the object clearly.<\/p>\n\n\n\n (b) The eye should be 30 cm away from the lens to see the object clearly.<\/p>\n\n\n\n (c) The diverging lens will always form an image at a large distance from the eye; this image cannot be seen through the human eye.<\/p>\n\n\n\n A compound microscope forms an inverted image of an object. In which of the following cases it it likely to create difficulties? (a) Looking at small germs. (b) Looking at circular spots. (c) Looking at a vertical tube containing some water.<\/p>\n\n\n\n (c) Looking at a vertical tube containing some water<\/p>\n\n\n\n If the experimentalist is looking at a vertical tube containing some water, he has to be careful, as the lower meniscus will appear as upper.<\/p>\n\n\n\n Mark the correct options. (a) If the far point goes ahead, the power of the divergent lens should be reduced. As the far point (x<\/em>) is shifted ahead, the focal length (f<\/em>) will be increased. 1f=1v-1u\u21d21f=1-x-1-\u221e\u21d21f=1-x\u21d2f=-xAs power (P<\/em>) is equal to the reciprocal of the focal length, it will be reduced. Also, because the focal length is negative, the lens used will be divergent when the far point is 1 m away.<\/p>\n\n\n\n The focal length of the objective of a compound microscope if fo<\/sub><\/em> and its distance from the eyepiece is L<\/em>. The object is placed at a distance u<\/em> from the objective. For proper working of the instrument, (b) L<\/em> > u<\/em> In a compound microscope, the objective lens of a short focal length, fo<\/sub><\/em>, is used. The focal length of the objective lens is less than the focal length of the eyepiece, f\u00e2\u20ac\u2039e<\/sub><\/em>, .The object is placed at a distance slightly greater than its focal length. The real, inverted image of the object forms somewhere in front of the eyepiece at a distance less than its focal length. This image acts as its object and the final image forms in between length, L, of the microscope. Please check and correct. A person looks at different trees in an open space with the following details. Arrange the trees in decreasing order of their apparent sizes.<\/p>\n\n\n\n Let the tree make a visual angle \u03b8<\/em> with the eyes. \u03b8=Height of the treeDistance from the eyeFor tree A,<\/p>\n\n\n\n \u03b8A=250=0.04For tree B,<\/p>\n\n\n\n \u03b8B=2.580=0.03125For tree C,<\/p>\n\n\n\n \u03b8C=1.870=0.0257For tree D,<\/p>\n\n\n\n \u03b8D=2.8100=0.028Now, An object is to be seen through a simple microscope of focal length 12 cm. Where should the object be placed so as to produce maximum angular magnification? The least distance for clear vision is 25 cm.<\/p>\n\n\n\n For the simple microscope, we have: The lens formula is given by 1v-1u=1f\u21d2 1u=1v-1fOn substituting the values, we get:<\/p>\n\n\n\n 1u=1-25-1121u =-25+12300=-37300<\/p>\n\n\n\n \u21d2u=-30037=-8.1 cmSo, to produce maximum angular magnification, the object should be placed 8.1 cm away from the lens.<\/p>\n\n\n\n A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25 cm) of a normal eye. (a) What is its focal length? (b) What will be its magnifying power if the image is formed at infinity?<\/p>\n\n\n\n For the simple microscope, we have: (a) When the image is formed at D<\/em> = 25 cm, the magnifying power of the simple microscope is given by \u21d2 25f=2\u21d2 f=252=12.5 cm(b) When the image is formed at infinity in normal adjustment, the magnifying power is given by<\/p>\n\n\n\n m=Df=2512.5=2.0So, the magnifying power is 2 if the image is formed at infinity.<\/p>\n\n\n\n A simple microscope is rated 5 X for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40 cm?<\/p>\n\n\n\n For a simple microscope, mof a simple microscope in normal adjustment is given by<\/p>\n\n\n\n m=Df\u21d25=25f\u21d2f=5 cmFor the relaxed far-sighted eye, D<\/em> = 40 cm. m:<\/p>\n\n\n\n m=Df=405=8The magnifying power for a simple microscope is 8X.<\/p>\n\n\n\n Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.<\/p>\n\n\n\n For the compound microscope, we have: f0=125 D=0.04 m=4 cmPower of the eyepiece = 5 D fe=15 D=0.2 m=20 cmLeast distance of clear vision, D<\/em> = 25 cm 1ve=1ue+1fe\u21d2 1ue=1ve-1fe<\/p>\n\n\n\n \u21d21-25-120=-(4+5)100<\/p>\n\n\n\n \u21d2 ue=-1009=11.11 cmLet u<\/em>o<\/sub> and v<\/em>o<\/sub> be the object and image distance for the objective lens. V0=L-u0<\/p>\n\n\n\n v0=30-11.11\u21d2vo= 18.89 cm \u20261As the image produced is real,<\/p>\n\n\n\n vo=+18.89 cm. 1uo=1vo-1fo =118.89-14=-0.197uo=-5.07 cm \u20262Maximum magnifying power of the compound microscope:<\/p>\n\n\n\n m=v0u01+Dfe =-18.89-5.071+2520 = 3.7225 \u00d7 2.25 = 8.376 The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.<\/p>\n\n\n\n For the compound microscope, we have: -24 cm 1ve-1ue=1fe1-24-1ue=16<\/p>\n\n\n\n \u21d2 1ue=-16-124\u21d2 1ue=-4+124=-524\u2234 ue=-245 cm=-4.8 cm(a) Separation between the objective and the eyepiece = 9.8 cm 1v0-1u0=1f015-1u0=11<\/p>\n\n\n\n \u21d2-1u0=1-15=5-15\u21d2u0=-54 cm=-1.25 cmMagnifying power of the compound microscope:<\/p>\n\n\n\n m=v0u01+Dfe<\/p>\n\n\n\n =51.251+246=20(b) Separation between the objective and the eyepiece = 11.8 cm An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?<\/p>\n\n\n\n For the compound microscope, we have: f0=120 D=0.05 m=5 cmPower of the eyepiece = 10 D fe=110 D=0.1 m=10 cmLeast distance for clear vision, D<\/em> = 25 cm, 1ue=1ve-1fe<\/p>\n\n\n\n =1-25-110=-2-550=-750Separation between the objective and the eyepiece = 20 cm v0can be obtained as:<\/p>\n\n\n\n v0=20-507=907 cmThe lens formula for the objective lens is given by<\/p>\n\n\n\n 1u0=1v0-1f0=790-15=7-1890=-1190<\/p>\n\n\n\n \u21d2 u0=-9011 cmMagnification of the compound microscope:<\/p>\n\n\n\n m=v0u01+Dfe<\/p>\n\n\n\n =-907-90111+2510=117\u00d73.5=5.5\u2234 Minimum separation that the eye can distinguish =<\/p>\n\n\n\n 0.225.5mm = 0.04 mm<\/p>\n\n\n\n A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the focal length of the eyepiece.<\/p>\n\n\n\n For the compound microscope, we have: m=vouo\u00d7Dfe =-1-vofoDfe\u21d2100=-1-vo0.5\u00d725fe\u21d22vo-4fe=1 \u20262On solving equations (1) and (2), we get: A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?<\/p>\n\n\n\n For the compound microscope, we have: 1v0-1u0=1f0\u21d21v0+10.5=11<\/p>\n\n\n\n \u21d2 1v0=1-105=-1\u21d2 v0=-1 cm The lens formula for the eyepiece is given by1ve-1ue=1fe<\/p>\n\n\n\n \u21d2130-1ue=15\u21d2-1ue=15-130=6-130=16\u21d2 ue<\/sub><\/em> = \u2212 6 cm An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.<\/p>\n\n\n\n For the astronomical telescope, m=fofe, we get:
(a) 25 cm to 200 cm
(b) 18 cm to 200 cm
(c) 25 cm to 300 cm
(d) 18 cm to 300 cm<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
(a) 1 mm
(b) 2 cm
(c) 25 cm
(d) 1m<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Near point of the human eye, u <\/em>=<\/p>\n\n\n\n
Distance between the retina and the eye lens, v<\/em> = 2 cm (approximately)
Thus, we have the focal length, f<\/em>.<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
(a) = x<\/em>
(b) < x<\/em>
(c) > x<\/em>
(d) = 2 x<\/em><\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Far point at which the object can be placed, u<\/em> =<\/p>\n\n\n\n
Thus, we have:
1f=1v-1u\u21d21f=1x-1\u221e\u21d2f=x1f<\/em>=1v<\/em>\u22121u<\/em>\u21d21f<\/em>\u224512\u22121\u221225\u21d21f<\/em>\u22452750\u21d2f<\/em>\u22452 c<\/em>m<\/em>1f<\/em>=1v<\/em>\u22121u<\/em>\u21d21f<\/em>\u224512\u22121\u221225\u21d21f<\/em>\u22452750\u21d2f<\/em>\u22452 c<\/em>m<\/em><\/p>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
(a) if he is farsighted
(b) if he is nearsighted
(c) if his vision is normal
(d) in each of these cases<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 9:<\/h4>\n\n\n\n
(a) on f<\/em> but not on u<\/em>
(b) on u<\/em> but not on f<\/em>
(c) on f<\/em> as well as u<\/em>
(d) neither on f<\/em> nor on u<\/em><\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
In a simple microscope,<\/p>\n\n\n\n
For u < f<\/em>, m = D\/f <\/em>+1.<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
(a) the focal length of the lens
(b) the power of the lens
(c) the aperture of the lens
(d) the object size<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
m = D\/f.<\/em>
For u <\/em>< f<\/em>, <\/em>m<\/em> = <\/em>D\/f<\/em> + 1.
Power of the lens = 1\/f <\/em>
Angular magnification depends on power.<\/p>\n\n\n\nQuestion 11:<\/h4>\n\n\n\n
(a) 5
(b) 2.5
(c) 1
(d) 0.2<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
(a) real
(b) virtual
(c) erect
(d) inverted<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) inverted<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
(a) Simple microscope
(b) Compound microscope
(c) Astronomical telescope
(d) Galilean telescope<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) Galilean telescope<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
(a) always strained in looking at an object
(b) strained for objects at large distances only
(c) strained for objects at short distances only
(d) unstrained for all distances<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Page No 432:<\/h4>\n\n\n\n
Question 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Distance of near point, D<\/em> = 10 cm
Focal length, f<\/em> = 10 cm
When the image is formed at the near point, D<\/em> = 10 cm, the magnifying power<\/p>\n\n\n\nQuestion 11:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Power of the objective lens = 25 D
The focal length of the objective lens is given by<\/p>\n\n\n\n
The focal length of the eyepiece is given by<\/p>\n\n\n\n
Image distance for the eye lens, ve<\/sub><\/em> = 25 cm
We know:
l<\/em> =<\/p>\n\n\n\n
On applying the lens formula for the objective lens, we get:1v0-1u0=1f0<\/p>\n\n\n\nPage No 430:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 11:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Focal length, f<\/em> = 15 cm
Image distance, v<\/em> = ?
The lens formula is given by<\/p>\n\n\n\nPage No 431:<\/h4>\n\n\n\n
Question 12:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 4:<\/h4>\n\n\n\n
(a) If the far point goes ahead, the power of the divergent lens should be reduced.
(b) If the near point goes ahead, the power of the convergent lens should be reduced.
(c) If the far point is 1 m away from the eye, divergent lens should be used.
(d) If the near point is 1 m away from the eye, divergent lens should be used.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(c) If the far point is 1 m away from the eye, the divergent lens should be used.<\/p>\n\n\n\n
Thus, we have:<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(a) L<\/em> < u<\/em>
(b) L<\/em> > u<\/em>
(c) fo<\/sub><\/em> < L <\/em>< 2fo<\/sub><\/em>
(d) L <\/em>> 2f<\/em>o<\/sub><\/em><\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) L<\/em> > 2f<\/em>0<\/sub><\/em><\/p>\n\n\n\n
\u2234 L > fo<\/sub>\u00e2\u20ac\u2039+ fe <\/sub>> 2fo<\/sub><\/em>\u00e2\u20ac\u2039
Also,
L<\/em> > u<\/em><\/p>\n\n\n\nPage No 432:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
Tree<\/td> Height(m)<\/td> Distance from the eye(m)<\/td><\/tr> A<\/em><\/td> 2.0<\/td> 50<\/td><\/tr> B<\/em><\/td> 2.5<\/td> 80<\/td><\/tr> C<\/em><\/td> 1.8<\/td> 70<\/td><\/tr> D<\/em><\/td> 2.8<\/td> 100<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Answer:<\/h4>\n\n\n\n
Thus, we have:<\/p>\n\n\n\n
\u03b8<\/em>A<\/sub> > \u03b8<\/em>B<\/sub> > \u03b8<\/em>D<\/sub> > \u03b8<\/em>C<\/sub>
The more the value of \u03b8<\/em>, the bigger the apparent size of the tree.
So, the arrangement in decreasing order is A, B, D and C.<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Focal length, f<\/em> = 12 cm
Least distance of clear vision, d<\/em> = 25 cm
For maximum angular magnification,
Image distance = Least distance of clear vision
v<\/em> = d<\/em> = \u2212 25 cm<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Least distance of clear vision, D<\/em> = 25 cm
Magnifying power, m<\/em> = 3
Now,
Let the focal length be f<\/em>.<\/p>\n\n\n\n
m=1+Df\u21d23=1+25f<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnification for a normal relaxed eye, m = <\/em>5
Least distant of distinct vision, D = 25 cm
Now,
Let the focal length be f.<\/em>
The image will form at infinity, as the eye is relaxed.
The magnifying power<\/p>\n\n\n\n
Magnification<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Power of the objective lens = 25 D
The focal length of the objective lens is given by<\/p>\n\n\n\n
The focal length of the eyepiece is given by<\/p>\n\n\n\n
Separation between the objective and the eyepiece, L<\/em> = 30 cm
Magnifying power is maximum when the image is formed by the eyepiece at the least distance of clear vision, i.e., D <\/em>= 25 cm.
For the eyepiece, we have:
v<\/em>e<\/sub> = \u2212 25 cm and f<\/em>e<\/sub> = 20 cm
The lens formula is given by<\/p>\n\n\n\n
So, for the objective lens, the image distance will be<\/p>\n\n\n\n
The lens formula is given by<\/p>\n\n\n\n
So, the maximum magnifying power of the compound microscope is 8.376.<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Focal length of the objective, f<\/em>o<\/sub> = 1.0 cm
Focal length of the eyepiece, f<\/em>e<\/sub> = 6 cm
Image distance from the eyepiece, v<\/em>e<\/sub> =<\/p>\n\n\n\n
Separation between the objective and the eyepiece = 9.8 cm to 11.8 cm
The lens formula is given by<\/p>\n\n\n\n
So, for the objective lens, the image distance will be
v<\/em>o<\/sub> = 9.8 \u2212 4.8 = 5 cm
The lens formula for the objective lens is given by<\/p>\n\n\n\n
We have:
m<\/em> = 30
So, the required range of the magnifying power is 20\u201330.<\/p>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Power of the objective lens = 20 D
The focal length of the objective lens is given by<\/p>\n\n\n\n
The focal length of the eyepiece is given by<\/p>\n\n\n\n
To distinguish the two points having minimum separation, the magnifying power should be maximum.
For the eyepiece, we have:
Image distance, ve<\/sub><\/em> = \u2212 25 cm
Focal length, fe<\/sub><\/em> = 10 cm
The lens formula is given by<\/p>\n\n\n\n
So, the image distance for the objective lens<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnifying power, m<\/em> = 100
Focal length of the objective, f<\/em>o<\/sub> = 0.5 cm
Tube length, l<\/em> = 6.5 cm
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
\u2234 v<\/em>o<\/sub> + f<\/em>e<\/sub> = 6.5 cm \u2026(1)
The magnifying power for normal adjustment is given by<\/p>\n\n\n\n
V<\/em>o<\/sub> = 4.5 cm and f<\/em>e<\/sub> = 2 cm
Thus, the focal length of the eyepiece is 2 cm.<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Focal length of the objective, f<\/em>o<\/sub> = 1.0 cm
Focal length of the eyepiece, f<\/em>e<\/sub> = 5 cm
Distance of the object from the objective, u<\/em>0<\/sub> = 0.5 cm
Distance of the image from the eyepiece, v<\/em>e<\/sub> = 30 cm
The lens formula for the objective lens is given by<\/p>\n\n\n\n
The objective will form a virtual image at the side same as that of the object at a distance of 1 cm from the objective lens. The image formed by the objective will act as a virtual object for the eyepiece.<\/p>\n\n\n\n
\u2234 Separation between the objective and the eyepiece = 6 \u2212 1 = 5 cm<\/p>\n\n\n\nQuestion 12:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnifying power, m<\/em> = 50
Length of the tube, L<\/em> = 102 cm
Let the focal length of objective and eye piece be f<\/em>0<\/sub> and f<\/em>e<\/sub>, respectively.
Now, using<\/p>\n\n\n\n
f<\/em>o<\/sub> = 50f<\/em>e<\/sub>\u2026(1)
And,
L<\/em> = f<\/em>o<\/sub> + f<\/em>e<\/sub> = 102 cm \u2026(2)
On substituting the value of f<\/em>o<\/sub> from (1) in (2), we get:
50f<\/em>e +