Chapter 9 Ray Optics and Optical Instruments Solutions
Question - 31 : - What should be the distance between theobject in Exercise 9.30 and the magnifying glass if the virtual image of eachsquare in the figure is to have an area of 6.25 mm2. Would you be able to seethe squares distinctly with your eyes very close to the magnifier?
Answer - 31 : -
Area of the virtual image of eachsquare, A = 6.25 mm2
Area of each square, A0 = 1 mm2
Hence, the linearmagnification of the object can be calculated as:
Focal length of the magnifying glass, f =10 cm
According to the lensformula, we have the relation:
Thevirtual image is formed at a distance of 15 cm, which is less than the nearpoint (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyesdistinctly.
Question - 32 : - Answer the followingquestions:
(a) The angle subtended at theeye by an object is equal to the angle subtended at the eye by the virtualimage produced by a magnifying glass. In what sense then does a magnifyingglass provide angular magnification?
(b) In viewing through amagnifying glass, one usually positions one’s eyes very close to the lens. Doesangular magnification change if the eye is moved back?
(c) Magnifying power of asimple 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 lengthand achieving greater and greater magnifying power?
(d) Why must both theobjective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through acompound microscope, our eyes should be positioned not on the eyepiece but ashort distance away from it for best viewing. Why? How much should be thatshort distance between the eye and eyepiece?
Answer - 32 : -
(a)Though the image size isbigger than the object, the angular size of the image is equal to the angularsize of the object. A magnifying glass helps one see the objects placed closerthan the least distance of distinct vision (i.e., 25 cm). A closer objectcauses a larger angular size. A magnifying glass provides angularmagnification. Without magnification, the object cannot be placed closer to theeye. With magnification, the object can be placed much closer to the eye.
(b) Yes, the angularmagnification changes. When the distance between the eye and a magnifying glassis increased, the angular magnification decreases a little. This is because theangle subtended at the eye is slightly less than the angle subtended at thelens. Image distance does not have any effect on angular magnification.
(c) The focal length of aconvex lens cannot be decreased by a greater amount. This is because makinglenses having very small focal lengths is not easy. Spherical and chromaticaberrations are produced by a convex lens having a very small focal length.
(d) The angular magnificationproduced by the eyepiece of a compound microscope is 
Where,
fe = Focal length ofthe eyepiece
It can be inferred that if fe is small, thenangular magnification of the eyepiece will be large.
Theangular magnification of the objective lens of a compound microscope is givenas 
(e)When we place our eyes tooclose to the eyepiece of a compound microscope, we are unable to collect muchrefracted light. As a result, the field of view decreases substantially. Hence,the clarity of the image gets blurred.
Thebest position of the eye for viewing through a compound microscope is at theeye-ring attached to the eyepiece. The precise location of the eye depends onthe separation between the objective lens and the eyepiece.
Question - 33 : - Anangular magnification (magnifying power) of 30X is desired using an objectiveof focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you setup the compound microscope?
Answer - 33 : - Focal length of the objective lens,
= 1.25 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d =25 cm
Angular magnification ofthe compound microscope = 30X
Total magnifying power of the compoundmicroscope, m = 30
Theangular magnification of the eyepiece is given by the relation:The angular magnification of the objectivelens (mo) is related to me as:
= m
Applying the lens formulafor the objective lens:
The object should beplaced 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formulafor the eyepiece:
Where,
= Image distance for the eyepiece = −d =−25 cm
= Object distancefor the eyepiece
Separation between the objective lens and theeyepiece 
Therefore,the separation between the objective lens and the eyepiece should be 11.67 cm.
A small telescope has anobjective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm.What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normaladjustment (i.e., when the final image
is at infinity)?
(b) the final image isformed at the least distance of distinct vision
(25cm)?
Answer - 34 : - Focal length of the objective lens,= 140 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d =25 cm
(a) When the telescopeis in normal adjustment, its magnifying power is given as:
(b) When the final image isformed at d,the magnifying power of the telescope is given as:
Question - 35 : - (a) For the telescopedescribed in Exercise 9.34 (a), what is the separation between the objectivelens and the eyepiece?
(b) If this telescope isused to view a 100 m tall tower 3 km away, what is the height of the image ofthe 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?
Answer - 35 : -
Focal length of the objective lens, fo = 140 cm
Focal length of the eyepiece, fe = 5 cm
(a) In normal adjustment, the separation between theobjective lens and the eyepiece 
(b) Height of the tower, h1 = 100 m
Distance of the tower (object) from thetelescope, u = 3 km = 3000 m
The angle subtended by thetower at the telescope is given as:
The angle subtended by theimage produced by the objective lens is given as:
Where,
h2 = Height of theimage of the tower formed by the objective lens
Therefore, the objectivelens forms a 4.7 cm tall image of the tower.
(c) Image is formed at adistance, d = 25 cm
The magnification of theeyepiece is given by the relation:
Height of the final image
Hence,the height of the final image of the tower is 28.2 cm.
Light incident normally ona plane mirror attached to a galvanometer coil retraces backwards as shown inFig. 9.36. A current in the coil produces a deflection of 3.5° of the mirror.What is the displacement of the reflected spot of light on a screen placed 1.5m away?
Answer - 36 : -
Angle of deflection, θ =3.5°
Distance of the screen from themirror, D = 1.5 m
The reflected rays get deflected by anamount twice the angle of deflection i.e., 2θ= 7.0°
The displacement (d) of the reflectedspot of light on the screen is given as:
Hence,the displacement of the reflected spot of light is 18.4 cm.