Chapter 9 Ray Optics and Optical Instruments Solutions
Question - 21 : - (a) Determine the‘effective focal length’ of the combination of the two lenses in Exercise 9.10,if they are placed 8.0 cm apart with their principal axes coincident. Does theanswer depend on which side of the combination a beam of parallel light isincident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm insize is placed on the side of the convex lens in the arrangement (a) above. Thedistance between the object and the convex lens is 40 cm. Determine themagnification produced by the two-lens system, and the size of the image.
Answer - 21 : -
Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = −20 cm
Distance between the two lenses, d =8.0 cm
(a) Whenthe parallel beam of light is incident on the convex lens first:
Accordingto the lens formula, we have:
Where,
= Object distance =∞
v1 = Image distance
The image will act as avirtual object for the concave lens.
Applyinglens formula to the concave lens, we have:
Where,
= Object distance
= (30 − d) = 30 − 8 = 22 cm
= Image distance
The parallel incident beam appears to divergefrom a point that is
from the centre ofthe combination of the two lenses.
(ii) Whenthe parallel beam of light is incident, from the left, on the concave lensfirst:
According to the lensformula, we have:
Where,
= Object distance =−∞
= Image distance
The image will act as areal object for the convex lens.
Applyinglens formula to the convex lens, we have:
Where,
= Object distance
= −(20 + d) = −(20 + 8) = −28 cm
= Image distance
Hence, the parallelincident beam appear to diverge from a point that is (420 − 4) 416 cm from theleft of the centre of the combination of the two lenses.
The answer does depend onthe side of the combination at which the parallel beam of light is incident.The notion of effective focal length does not seem to be useful for thiscombination.
(b) Height of the image, h1 = 1.5 cm
Objectdistance from the side of the convex lens, 
According to the lensformula:
Where,
=Image distance
Magnification, 
Hence, the magnificationdue to the convex lens is 3.
The image formed by theconvex lens acts as an object for the concave lens.
According to the lensformula:
Where,
= Object distance
= +(120 − 8) = 112 cm.
= Image distance
Magnification, 
Hence, the magnification due to the concave lensis
The magnification producedby the combination of the two lenses is calculated as:
The magnification of thecombination is given as:
Where,
h1 = Object size = 1.5cm
h2 = Size of the image
Hence,the height of the image is 0.98 cm.
Atwhat angle should a ray of light be incident on the face of a prism ofrefracting angle 60° so that it just suffers total internal reflection at theother face? The refractive index of the material of the prism is 1.524.
Answer - 22 : -
The incident, refracted,and emergent rays associated with a glass prism ABC are shown in the givenfigure.
Angle of prism, ∠A = 60°
Refractive index of the prism, µ =1.524
= Incident angle
= Refracted angle
= Angle of incidenceat the face AC
e = Emergent angle = 90°
According to Snell’s law,for face AC, we can have:
It is clear from the figure that angle
According to Snell’s law,we have the relation:
Hence,the angle of incidence is 29.75°.
Question - 23 : - You are given prisms madeof crown glass and flint glass with a wide variety of angles. Suggest acombination of prisms which will
(a) deviate a pencil ofwhite light without much dispersion,
(b) disperse (anddisplace) a pencil of white light without much deviation.
Answer - 23 : -
(a)Place the two prismsbeside each other. Make sure that their bases are on the opposite sides of theincident white light, with their faces touching each other. When the whitelight is incident on the first prism, it will get dispersed. When thisdispersed light is incident on the second prism, it will recombine and whitelight will emerge from the combination of the two prisms.
(b)Take the system of the two prisms as suggested inanswer (a). Adjust(increase) the angle of the flint-glass-prism so that the deviations due to thecombination of the prisms become equal. This combination will disperse thepencil of white light without much deviation.
Question - 24 : - Fora normal eye, the far point is at infinity and the near point of distinctvision is about 25cm in front of the eye. The cornea of the eye provides aconverging power of about 40 dioptres, and the least converging power of theeye-lens behind the cornea is about 20 dioptres. From this rough data estimatethe range of accommodation (i.e., the range of converging power of theeye-lens) of a normal eye.
Answer - 24 : -
Least distance of distinct vision, d =25 cm
Farpoint of a normal eye,
Converging power of the cornea, 
Least converging power of the eye-lens, 
To see the objects atinfinity, the eye uses its least converging power.
Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60 D
Power of the eye-lens isgiven as:
To focus an object at the near point, objectdistance (u) = −d = −25 cm
Focal length of theeye-lens = Distance between the cornea and the retina
= Image distance
Hence,image distance, 
According to the lensformula, we can write:
Where,
= Focal length
∴Powerof the eye-lens = 64 − 40 = 24 D
Hence,the range of accommodation of the eye-lens is from 20 D to 24 D.
Doesshort-sightedness (myopia) or long-sightedness (hypermetropia) implynecessarily that the eye has partially lost its ability of accommodation? Ifnot, what might cause these defects of vision?
Answer - 25 : -
Amyopic or hypermetropic person can also possess the normal ability ofaccommodation of the eye-lens. Myopia occurs when the eye-balls get elongatedfrom front to back. Hypermetropia occurs when the eye-balls get shortened. Whenthe eye-lens loses its ability of accommodation, the defect is calledpresbyopia.
Question - 26 : - Amyopic person has been using spectacles of power −1.0 dioptre for distantvision. During old age he also needs to use separate reading glass of power +2.0 dioptres. Explain what may have happened.
Answer - 26 : -
The power of the spectacles used by themyopic person, P = −1.0 D
Focal length of the spectacles, 
Hence, the far point ofthe person is 100 cm. He might have a normal near point of 25 cm. When he usesthe spectacles, the objects placed at infinity produce virtual images at 100cm. He uses the ability of accommodation of the eye-lens to see the objectsplaced between 100 cm and 25 cm.
Duringold age, the person uses reading glasses of power,
Theability of accommodation is lost in old age. This defect is called presbyopia.As a result, he is unable to see clearly the objects placed at 25 cm.
Question - 27 : - Aperson looking at a person wearing a shirt with a pattern comprising verticaland horizontal lines is able to see the vertical lines more distinctly than thehorizontal ones. What is this defect due to? How is such a defect of visioncorrected?
Answer - 27 : -
Inthe given case, the person is able to see vertical lines more distinctly thanhorizontal lines. This means that the refracting system (cornea and eye-lens)of the eye is not working in the same way in different planes. This defect iscalled astigmatism. The person’s eye has enough curvature in the verticalplane. However, the curvature in the horizontal plane is insufficient. Hence,sharp images of the vertical lines are formed on the retina, but horizontallines appear blurred. This defect can be corrected by using cylindrical lenses.
Question - 28 : - A man with normal nearpoint (25 cm) reads a book with small print using a magnifying glass: a thinconvex lens of focal length 5 cm.
(a) What is the closest andthe farthest distance at which he should keep the lens from the page so that hecan read the book when viewing through the magnifying glass?
(b) What is the maximumand the minimum angular magnification (magnifying power) possible using theabove simple microscope?
Answer - 28 : -
(a) Focal length of themagnifying glass, f = 5 cm
Least distance of distance vision, d= 25 cm
Closest object distance = u
Image distance, v = −d =−25 cm
Accordingto the lens formula, we have:Hence, the closestdistance at which the person can read the book is 4.167 cm.
Forthe object at the farthest distant (u’), the image distance 
According to the lensformula, we have:
Hence, the farthestdistance at which the person can read the book is
5 cm.
(b) Maximum angular magnification is given by therelation:
Minimum angularmagnification is given by the relation:
A card sheet divided into squares each ofsize 1 mm2 is being viewed at a distance of 9 cmthrough a magnifying glass (a converging lens of focal length 9 cm) held closeto the eye.
(a) What is themagnification produced by the lens? How much is the area of each square in thevirtual image?
(b) What is the angularmagnification (magnifying power) of the lens?
(c) Is the magnificationin (a) equal to the magnifying power in (b)?
Answer - 29 : -
Note : Here wetook focal Length as 10 cm because if we take it as 9 cm then imagedistance will be zero ,which does not make any sense. (a) Area of each square, A =1 mm2
Object distance, u = −9 cm
Focal length of a converging lens, f = 9 cm
For image distance v, the lensformula can be written as:
Magnification, 
∴Area of each square in thevirtual image = (10)2A
= 102 × 1 = 100 mm2
= 1 cm2
(b) Magnifying power of the lens
(c) The magnification in (a) is not the same as themagnifying power in (b).
Themagnification magnitude is 
and the magnifying power is
Thetwo quantities will be equal when the image is formed at the near point (25cm).
Question - 30 : - (a) At what distanceshould the lens be held from the figure in
Exercise 9.29 in order toview the squares distinctly with the maximum possible magnifying power?
(b) What is themagnification in this case?
(c) Is the magnificationequal to the magnifying power in this case?
Answer - 30 : -
(a) The maximum possiblemagnification is obtained when the image is formed at the near point (d =25 cm).
Image distance, v = −d =−25 cm
Focal length, f = 10 cm
Object distance = u
According to the lensformula, we have:
Hence, to view the squaresdistinctly, the lens should be kept 7.14 cm away from them.
(b) Magnification =
(c) Magnifyingpower =
Sincethe image is formed at the near point (25 cm), the magnifying power is equal tothe magnitude of magnification.