Chapter 9 Ray Optics and Optical Instruments Solutions
Question - 11 : - Acompound microscope consists of an objective lens of focal length 2.0 cm and aneyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far fromthe objective should an object be placed in order to obtain the final image at(a) the least distance of distinct vision (25 cm), and (b) at infinity? What isthe magnifying power of the microscope in each case?
Answer - 11 : -
Focal length of the objective lens, f1 = 2.0 cm
Focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and theeyepiece, d = 15 cm
(a) Least distance of distinct vision, 
∴Image distance for theeyepiece, v2 = −25 cm
Object distance for the eyepiece = u2
According to the lensformula, we have the relation:
Image distance for the objective lens,
Object distance for the objective lens= u1
According to the lensformula, we have the relation:
Magnitude of the object distance, 
= 2.5 cm
The magnifying power of acompound microscope is given by the relation:
Hence, the magnifyingpower of the microscope is 20.
(b) The final image is formedat infinity.
∴Imagedistance for the eyepiece,
Object distance for the eyepiece = u2
According to the lensformula, we have the relation:
Image distance for the objective lens,
Object distance for the objective lens= u1
According to the lensformula, we have the relation:
Magnitude of the object distance, = 2.59 cm
The magnifying power of acompound microscope is given by the relation:
Hence,the magnifying power of the microscope is 13.51.
Aperson with a normal near point (25 cm) using a compound microscope withobjective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm canbring an object placed at 9.0 mm from the objective in sharp focus. What is theseparation between the two lenses? Calculate the magnifying power of themicroscope,
Answer - 12 : -
Focal length of the objective lens, fo = 8 mm = 0.8 cm
Focal length of the eyepiece, fe = 2.5 cm
Object distance for the objectivelens, uo = −9.0 mm = −0.9 cm
Least distance of distant vision, d =25 cm
Image distance for the eyepiece, ve = −d =−25 cm
Objectdistance for the eyepiece = 
Using the lens formula, we can obtain the valueofas: We can also obtain the value of the imagedistance for the objective lens
using the lens formula. The distance between the objective lens and theeyepiece
The magnifying power ofthe microscope is calculated as:
Hence,the magnifying power of the microscope is 88.
Asmall telescope has an objective lens of focal length 144 cm and an eyepiece offocal length 6.0 cm. What is the magnifying power of the telescope? What is theseparation between the objective and the eyepiece?
Answer - 13 : -
Focal length of the objective lens, fo = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power ofthe telescope is given as:
The separation between theobjective lens and the eyepiece is calculated as:
Hence,the magnifying power of the telescope is 24 and the separation between theobjective lens and the eyepiece is 150 cm.
Question - 14 : - (a) A giant refractingtelescope at an observatory has an objective lens of focal length 15 m. If aneyepiece of focal length 1.0 cm is used, what is the angular magnification ofthe telescope?
(b) If this telescope isused to view the moon, what is the diameter of the image of the moon formed bythe objective lens? The diameter of the moon is 3.48 × 106 m, and the radius oflunar orbit is 3.8 × 108 m.
Answer - 14 : -
Focal length of the objective lens, fo = 15 m = 15 × 102 cm
Focal length of the eyepiece, fe = 1.0 cm
(a) The angularmagnification of a telescope is given as:
Hence, the angularmagnification of the given refracting telescope is 1500.
(b) Diameter of the moon, d =3.48 × 106 m
Radius of the lunar orbit, r0 = 3.8 × 108 m
Let 
be the diameter of theimage of the moon formed by the objective lens.
The angle subtended by thediameter of the moon is equal to the angle subtended by the image.
Hence,the diameter of the moon’s image formed by the objective lens is 13.74 cm
Question - 15 : - Use the mirror equation todeduce that:
(a) an object placedbetween f and 2f of a concave mirror produces areal image beyond 2f.
(b) a convex mirror alwaysproduces a virtual image independent of the location of the object.
(c) the virtual image producedby a convex mirror is always diminished in size and is located between thefocus and the pole.
(d) an object placed between the pole and focus of aconcave mirror produces a virtual and enlarged image.
Answer - 15 : -
(a) For a concave mirror, thefocal length (f) is negative.
∴f < 0
When the object is placed on the left sideof the mirror, the object distance (u) is negative.
∴u < 0
For image distance v, we canwrite the lens formula as:
The object lies between f and2f.
Using equation (1), weget:
∴
is negative, i.e., v isnegative.
Therefore, the image lies beyond 2f.
(b) For a convex mirror, thefocal length (f) is positive.
∴ f > 0
When the object is placed on the left sideof the mirror, the object distance (u) is negative.
∴ u < 0
For image distance v, we havethe mirror formula:
Thus, the image is formedon the back side of the mirror.
Hence, a convex mirroralways produces a virtual image, regardless of the object distance.
(c) For a convex mirror, thefocal length (f) is positive.
∴f > 0
When the object is placed on the left sideof the mirror, the object distance (u) is negative,
∴u < 0
Forimage distance v, we have the mirror formula:
Hence, the image formed is diminished and islocated between the focus (f) and the pole.
(d) For a concave mirror, thefocal length (f) is negative.
∴f < 0
When the object is placed on the left sideof the mirror, the object distance (u) is negative.
∴u < 0
It is placed between the focus (f)and the pole.
For image distance v, we havethe mirror formula:
The image is formed on theright side of the mirror. Hence, it is a virtual image.
For u <0 and v > 0, we can write:
Magnification, m 
> 1
Hence,the formed image is enlarged.
Question - 16 : - Asmall 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 is viewed from thesame point through a 15 cm thick glass slab held parallel to the table?Refractive index of glass = 1.5. Does the answer depend on the location of theslab?
Answer - 16 : -
Actual depth of the pin, d =15 cm
Apparentdept of the pin = Refractive index of glass,
Ratio of actual depth tothe apparent depth is equal to the refractive index of glass, i.e.
The distance at which the pin appears to beraised = 
Fora small angle of incidence, this distance does not depend upon the location ofthe slab.
(a) Figure 9.35 shows across-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 pipefor which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answerif there is no outer covering of the pipe?
Answer - 17 : - (a) Refractiveindex of the glass fibre, 
Refractive index of the outer covering of thepipe, 
= 1.44 Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index (μ) of the innercore − outer core interface is given as:
For the critical angle, total internalreflection (TIR) takes place only when
, i.e., i > 59°
Maximumangle of reflection,
Let,
be the maximum angle ofincidence.
Therefractive index at the air − glass interface,We have the relation forthe maximum angles of incidence and reflection as:
Thus, all the rays incident at angles lyingin the range 0 < i < 60° will suffer total internalreflection.
(b) If the outer covering ofthe pipe is not present, then:
Refractiveindex of the outer pipe,
For the angle of incidence i =90°, we can write Snell’s law at the air − pipe interface as:
Answer the followingquestions:
(a) You have learnt that planeand convex mirrors produce virtual images of objects. Can they produce realimages under some circumstances? Explain.
(b) A virtual image, we alwayssay, cannot be caught on a screen.
Yet when we ‘see’ avirtual image, we are obviously bringing it on to the ‘screen’ (i.e., theretina) of our eye. Is there a contradiction?
(c) A diver under water, looksobliquely at a fisherman standing on the bank of a lake. Would the fishermanlook taller or shorter to the diver than what he actually is?
(d) Does the apparent depth ofa tank of water change if viewed obliquely? If so, does the apparent depthincrease or decrease?
(e) The refractive index of diamond is much greaterthan that of ordinary glass. Is this fact of some use to a diamond cutter?
Answer - 18 : -
(a) Yes
Plane and convex mirrorscan produce real images as well. If the object is virtual, i.e., if the lightrays converging at a point behind a plane mirror (or a convex mirror) arereflected to a point on a screen placed in front of the mirror, then a realimage will be formed.
(b) No
A virtual image is formedwhen light rays diverge. The convex lens of the eye causes these divergent raysto converge at the retina. In this case, the virtual image serves as an objectfor the lens to produce a real image.
(c) The diver is in the waterand the fisherman is on land (i.e., in air). Water is a denser medium than air.It is given that the diver is viewing the fisherman. This indicates that thelight rays are travelling from a denser medium to a rarer medium. Hence, therefracted rays will move away from the normal. As a result, the fisherman willappear to be taller.
(d) Yes; Decrease
The apparent depth of atank of water changes when viewed obliquely. This is because light bends ontravelling from one medium to another. The apparent depth of the tank when viewedobliquely is less than the near-normal viewing.
(e) Yes
Therefractive index of diamond (2.42) is more than that of ordinary glass (1.5).The critical angle for diamond is less than that for glass. A diamond cutteruses a large angle of incidence to ensure that the light entering the diamondis totally reflected from its faces. This is the reason for the sparklingeffect of a diamond.
Question - 19 : - Theimage of a small electric bulb fixed on the wall of a room is to be obtained onthe opposite wall 3 m away by means of a large convex lens. What is the maximumpossible focal length of the lens required for the purpose?
Answer - 19 : -
Distance between the object and theimage, d = 3 m
Maximumfocal length of the convex lens =
For real images, themaximum focal length is given as:
Hence,for the required purpose, the maximum possible focal length of the convex lensis 0.75 m.
Ascreen is placed 90 cm from an object. The image of the object on the screen isformed by a convex lens at two different locations separated by 20 cm.Determine the focal length of the lens.
Answer - 20 : -
Distance between the image (screen) and theobject, D = 90 cm
Distance between two locations of the convexlens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
Therefore,the focal length of the convex lens is 21.39 cm.