# NCERT 12 Physics Ray Optics And Optical Instruments Chapter 9 Exercise

Q.1. A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance form the mirror should a screen be placed in order of receive a sharp image ? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved ? Screen should be placed at 54 cm from the mirror on the same side as the object. Minus sing indicates that image is inverted.

When the candle is moved closer to mirror, the screen has to be moved away from the mirror. However, when candle is at a distance less then 18 cm from the mirror, image formed would be virtual and screen is not required, as the virtual image cannot be taken on the screen.

Q.2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. Image is virtual, formed at 6.7 cm at the back of the mirror. Image is erect, and of course virtual.

As needle is moved farther from the mirror, image moves away from the mirror ( upto F) and goes on decreasing in size.

Q.3. 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 9.4 cm. What is the refractive index of water ? If water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance would the microscope have to be moved to focus on the needle again ? Distance through which microscope has to moved up = 9.4 – 7.67 =1.73 cm</div

Q.4. (a) and (b) show refraction of in incident ray in air at 600 with the normal to a glass-air and water-air interface respectively. Predict the angle of refraction of an incident ray in water at 450 what the normal to a water glass interface. Take a 1.32.   Q.5. 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.

Sol. In the source of light (S) is 80 cm below the surface of water i.e., SO = 80 cm = 0.8 m.

When

Area of the surface of water through which light from the bulb can emerge is area of the circle of radius  Q.6. A prism is made of glass of unknown refractive index.
A parallel beam of light is incident on a face of the prism. By rotating the prism, the minimum a ngle of deviation is measured to be 400. What is the refractive index of the material of the prism ? The refracting angle of the prism is 600. If the prism is placed in water (refractive index 1.33), predict the new minimum angle of deviation of a parallel beam of light. Q.7. Double convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of same radius of curvature. What is the radius of curvature required if local length of lens is to be 20 cm ? Q.8. A beam of light converges to a point P.A lens is placed in the path of the convergent beam 12 cm form P. At what point does the beam converge if the lens is(a) a convex lens of focal length 20 cm. (b) a concave lens of focal length 16 cm ?

Sol. Here, the point P on the right of the lens acts as a virtual object, Hence, image is at 48 cm. to the right of the lens, where the beam would converge.

Q.9. An object of size 3-0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further form the lens ? As the object is moved away from the lens, virtual image moves towards focus of lens (but never beyond focus). The size of image goes on decreasing.

Q.10. What is the length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm. It the system a converging or a diverging lens ? Ignore thickness of the lenses. F= – 60 cm

The combination of lenses behaves as a concave les. The system is not converging.

Q.11. A compound microscope consists of an objective lens of focal length 2.0 cm and an eye piece of focal length 6.25 cm, separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm) (b) at infinity ?

What is the magnifying power of the microscope in each case ?  Q.12. A person with a normal near point (25 cm) using a compound microscope with an objective of focal length 8.0 mm and an eye piece of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses ? Calculate the magnifying power of the microscope ?  Q.13. A small telescope has an objective lens of focal length 144 cm and an eye piece of focal length 6.9 cm. What is the magnifying power of the telescope ? What is the separation between the objective and the eye-piece ? Q.14. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of telescope ?

(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 ⨯ 106 m and radius of lunar orbit is 3.8 ⨯ 108 m.  Q.15. Use the mirror equation to deduce that :

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2 f.

(b) a convex mirror  always produces a virtual image independent of the location of the object.

(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.   Q.16. A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised, If it be viewed from same point through a 15 cm. thick glass slab held parallel to the table ?  of glass is 1.5. Does the answer depend on location of the slab ? The answer does not depend upon the location of the slab.

Q.17. (a) 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 incident rays with the axis of the pipe for which total reflections inside the pipe take place as shown.

(b) What is the answer if there is no outer covering of the pipe ?   Q.18. Answer the following questions :
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we see a virtual image, we bring it to the screen i.e. retina of our eye. Is there a contradiction ?
(c) A diver under water looks obliquely at a fisherman standing om the bank of a lake. Would the fisherman loo taller or shorter then what he actually is ?
(d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decrease ?
(e) The refractive index of diamond is much greater then that of ordinary glass. Is this fact of some use to a diamond cutter ?
Sol. Yes, when rays incident on a plane mirror or a convex mirror are tending to converge to a point behind the mirror, they are reflected to a point on a screen in front of the mirror. Hence a real image is formed (when the object is virtual).
(b) No, there is no contradiction. Eye lens is convergent. If form a real image of the virtual object (i.e., the virtual image being seen)  on the retina.
(c) As fisherman is in air, light travels from rare to tenser medium. It bends towards the normal, appearing to come from a larger distance. Therefore, to the diver under water, fisherman looks taller.
(d) Yes, the apparent depth decreases further, when water tank is viewed obliquely compared to the depth when seen near normally. As refractive index of diamond is much greater then that of ordinary glass, critical angle C for diamond is much smaller ( ) as compared to that for glass ( ). A skilled diamond cutter exploits the large range of angles of incidence of light (240 to 900) to ensure that light entering the diamond suffers multiple total internal reflections within the diamond. This produces sparkling effect in the diamond.
Q.19. The image of a small electric bulb fixed on the wall of a room is be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose ?
Sol. For a real image (on wall), minimum distance between the object and image should be 4 f. Q.20. A screen is placed 90 cm from in object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens. Q.21. (a) Determines the ‘effective focal length’ of the combination of two lenses of focal lengths 30 cm and -20 cm if they are placed 8.0 cm apart with their principal axes coincident. Done the answer depend on which side a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ?
(b) As object 1.5 cm in size is placed on the side of the convex lens in the above arrangement. The distance between the object and convex lens is 40 cm. Determine the magnification produced by the two lens system and size of image.     Q.22. A what angle should a ray of light be incident on the face of a prism of refracting angle 600, so that it just suffers total internal reflection at the other face ? The refractive index of the prism is 1.524. Q.23. You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
(i) deviate a pencil of white light without much dispersion.
(ii) disperse and displace a pencil of white light without much deviation.
Sol. (i) For no dispersion, total angular dispersion produced by two prisms should be zero. Taking crown glass prism of certain angle, we go on increasing angle of flint glass prims till this condition is met. In the final combination however, angle of flint glass prism will be smaller than the angle of crown glass prism as  for flint glass is more then    for crown glass.
Q.24. For a normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptre and the least converging power of eye lens behind the cornea is about 20 dioptre. From this rough data, estimate the range of accommodation (i.e., the range of converging power of the eye lens) of a normal eye.
Sol. To observe objects at infinity, the eye uses its least converging power = 40 + 20 = 60 D.
Distance between cornea/eyes lens and retina = focal length of eye Hence, range of accommodation of eye lens is roughly 20 to 24 dioptre.
Q.25. 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 ?
Sol.  No, a person may have normal ability of accommodation and yet he may be myopic or hypermetropic. Infact, myopia arises when length of eye ball (from front to back) gets elongated and hypermetropia arises when length of eye ball has gets shortened.
However, when eye ball has normal length, but the eye lens loses partially its power of accommodation, the defect is called presbyopia.
Q.26. 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 glasses of power + 2.0 dioptre. Explain what may have happened.
Sol. As the person is using spectacles of power – 1.0 dioptre (i.e., focal length – 100 cm), the far point of the person is at 100 cm. Near point of the eye might have been normal (i.e., 25 cm).
The objects at infinity produce virtual images at 100 cm (using spectacles). To see objects between 25 cm and 100 cm, the person uses the ability of accommodation of his eye lens. This ability is partially lost in old age. The near point of the eye may recede to 50 cm. He has, therefore to use glasses of suitable power for reading. Q.27.  A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly that the horizontal ones. What is this defect due
to ? How is such a defect of vision corrected ?
Sol. This defect is called Astigmatism. It arises because curvature of th1e cornea plus eye refracting system is not the same in different planes. As vertical lines are seen distinctly, the curvature in the vertical plane is enough, but in the horizontal plane, curvature is insufficient.
This defect is removed by using a cylindrical lens with its axis along the vertical.
Q.28. A man with normal near point (25 cm) reads a book with small print using a magnifying glass : a thin convex lens of focal length 5 cm.
(a) What are the closest and farthest distances at which he can read the book, when viewing through the magnifying glass ?(b) What is the maximum and the minimum angular magnifications (magnifying powers) possible using the above simple microscope ?
Sol. (a) Here,  f = 5 cm, u = ?
For the closest distance ; v = -25 cm Q.29. A cardsheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification (magnifying power) of the lens ?
(b) Is the magnification in (a) equal to the magnifying power in (b) ? Explain (c) No, magnification in (a) which is (v/u) cannot be equal to magnifying power in (b) which is (d/u)
Unless v = d i.e., image is located at the least distance of distance vision.
Q.30. (I) At what distance should the lens be held from the card sheet in Q. 29 in order to view the squares distinctly with the maximum possible magnifying power ?
(ii) What is the magnification in this case ?
(iii) Is the magnification equal to magnifying power in this cas ? Explain. Yes, the magnification and magnifying power in this case are equal, because image is formed at the least distance of distinct vision.
Q.31. What should be the distance between the object (in Q.30) and magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier ? As the virtual image is at 15 cm ; whereas distance of distinct vision is 25 cm, therefore, the image cannot be seen distinctly by the eye.
Q.32. Answer the following questions :
(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 dose a magnifying glass provide angular magnification ?
(b) In viewing through a magnifying glass, one usually positions one’s eye very close to the lens.
Does angular magnification change if the eye is moved back ?
(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 ?
(d) Why must both the objective and eye piece of a compound microscope have short focal lengths ?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eye piece, but a short distance away from it for best viewing, why ? How much should be that short distance between the eye and eye piece ?
Sol. (a) It is true that angular size of image is equal to angular size of the object.
By using magnifying glass, we keep the object far more closer to the eye than at 25 cm, its normal position without use of glass. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved.
(b) Yes, the angular magnification changes, if the eye is moved back. This is because angle subtended at the eye would be slightly less then the angle subtended at the lens. The effect is negligible when image is at much larger distance.
(c) Theoretically, it is true. However, when we decrease focal length, aberration both spherical and chromatic become more pronounced. Further, is it difficult to grind lenses of very small focal lengths. (e) The image of objective lens in eye piece is called ‘eye ring’. All the rays from the object refracted by the objective go through the eye ring. Therefore, ideal position for our eyers for viewing is this eye ring only.
When eye is too close to the eye piece, field of view reduced and eyes do not collect much of the light. The precise location of the eye ring would depend upon the separation between the objective and eye piece, and also on focal length of the eye piece.
Q.33. An angular magnification (magnifying power) of 30 X is desired using objective of focal length 1.25 cm and an eye piece of focal length  5 cm. How will you set up the compound microscope ?
Sol. In normal adjustment, image is formed at least distance of distinct vision,  d = 25 cm.  Q.34. A small telescope has an objective lens of focal length 140 cm and eye piece of focal length 5.0 cm.
(a) the telescope is in normal adjustment (i.e. when the final image is at infinity)
(b) the final image is formed at the least distance of distinct vision (25 cm). Q.35. (a) For the telescope described in Q. 34, what is the separation between the objective lens and eye piece ?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens ?
(c) What is the height of the final image of the tower if it is formed at 25 cm ? Q.36. A Cassegrainian telescope uses two mirrors as shown in . Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be ?
Sol. Here, radius of curvature of objective mirror  R1 = 220 mm
Radius of curvature of secondary mirror
R2 = 140 mm ;  Q.37. Light incident normally on a plane mirror attached to a galvanometer coil retraced backwards as shown in. A current in the coil produces a deflection of 3. 50 of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away ?  Q.38. shows an equiconvex lens (of refractive index 1.5) 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 ?

sol. Let focal length of convex lens of glass =focal length of plano concave lens of liquid = f2

Combined focal length,  F = 45. 0 cm  Updated: December 4, 2021 — 3:11 pm