Class 12 Physics NCERT Solutions for Chapter 9: Ray Optics and Optical Instruments are available here. You can also download these solutions in the PDF format with the help of download link given at the end. This article is the continuation of NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments, Part I and Part II. In Part I, we have provided solutions from question number 9.1 to 9.9, Part II has solutions from question number 9.9 to 9.20. In this part (or Part III), we have provided solutions from question number 9.21 to 9.30. The solutions of remaining questions will be available in the next part.
NCERT Solutions for Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments from question number 9.21 to 9.30 are given below:
(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm. Determine the magnification produced by the two-lens system, and the size of the image.
(a) (i) Concave lens on left side
f1 = -20 cm
u1 = -∞
(1 / v1) + (1/∞) = ‒1 / 20
v1 = ‒ 20 cm
This serves as object for second lens
f2 = 30 cm
u2 = ‒28 cm
(1/v2) + (1/28) = (1/30)
v2 = ‒420 cm
(ii) Convex lens on left
f1 = 30 cm
u1 = -∞
Putting in formula
(1/v1) + (1/∞) = (1/30)
v1 = 30 cm.
This serves as object for second lens
f2 = -20 cm
u2 = 22 cm
(1/v2) – (1 /22) = ‒ (1/20)
v2 = -220 cm
The answer depends on which side of lens system the parallel beam is incident. There is no use of concept of effective focal length here.
u1 = -40 cm
f1 = 30 cm
(1/v1) + (1/40) = (1/30)
v1 = 120 cm
Magnification due to convex lens = 3
u2 = 112 cm
f2 = ‒20 cm
(1/v2) – (1/112) = ‒ (1/20)
v2 = ‒24.3 cm
Magnification due to concave lens = 20 / 92
Net magnitude = 0.652
Size of image = 0.98 cm.
Question 9.22: At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
n = 1.524
r = 60 – ic
sin ic = 1 / 1.524 ⇒ ic = 41
Thus r = 19
sin i = 0.4962
Or i = 30o.
Question 9.23: You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
(a) deviate a pencil of white light without much dispersion,
(b) disperse (and displace) a pencil of white light without much deviation.
(a) First prism is taken of crown glass. Second prism is taken of flint glass.
Refracting angle of second prism is so chosen such that it is less than that of crown glass prism, so that dispersion due to the first is nullified by the second.
(b) The angle of first glass prism is increased to disperse without deviation, so that deviations due to the two prisms are equal and opposite. First glass prism is taken of greater angle.
Question 9.24: For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Least converging power to see object at ∞ = 60 D
So distance between retina and cornea eye lens = 5 / 3 cm
At near point
u = ‒25 cm
v = 5/3 cm
f = 25/16 cm
Corresponding power = 64 D
Power of the eye lens = 64 ‒ 40 = 24 D
Range of accommodation of the eye lens = 20 to 24 dioptres.
Question 9.25: 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?
No, a myopic or hypermetropic person may have normal ability of accomodation of the eye lens. The partial loss of ability of accommodation of eye lens is called presbyopia. It can be corrected in the way hypermetropia is dealt with.
Question 9.26: 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 glass of power + 2.0 dioptres. Explain what may have happened.
The person uses lens of power -1.0 dioptre because his far point is 100 cm, where a virtual image is formed by the lens. In old age, he uses a lens of + 2.0 D since the normal near point of vision ceases to be 25 cm. The lens forms the virutal image of object during reading (or near point) at 50 cm. This defect is called presbypobia, which is comon in old age due to loss of ability of accomodation of eyes.
Question 9.27: 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 than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
When the curvature of retina is not same in every plane the defect is called astigmatism. The veritcal lines can be seen by the person because the curvature in vertical plane is normal forming the images at retina. The images of horizontal lines are not formed because the curvature in horizontal plane is not sufficient. Such defect can be corrected by using cylindrical lens. In this case the cylindrical lens with verical axis is required.
Question 9.28: 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 is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
f = 5 cm
(a) By the formula
(1/v) – (1/u) = (1/f)
Substituting the values
(‒1/25) – (1/u) = (1/5)
Which gives, u =‒ 4.2 cm
This is the closest distance at which the person can read the book.
For the farthest distance, v’ = ∞
Using the above formula
(1/∞) ‒ (1/u’) = (1/5)
Which gives, u’ = -5 cm
This is the farthest distance.
(b) Maximum angular magnification
m = 25 / 4.167 = 6
Minimum angular magnification
m’ = 25 / 5 = 5.
Question 9.29: A card sheet 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 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
(a) Using the formula of previous question
(1/v)+ (1/9) = (1/10)
v = ‒ 90 cm
Magnitude of magnification
m = (90/9) = 10
Area of each square in virtual image = 10 x 10 x 1 = 100 sq mm = 1 sq cm
(b) Angular magnification = 25 / 9 = 2.8
(c) No, both are two different things. Magnification magnitude is |v /u|, while magnifying power is 25 / |u|. These are equal only when the image is located at the near point of 25 cm.
(a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
(a) When the image is at near point of 25 cm, the magnification achieved is maximum.
(1/v) – (1/u) = (1/f)
Putting the values,
(‒1/25) – (1/u) = (1/10)
Which yields, u = ‒7.14 cm
(b) Magnification = v / u = 25 / 7.14 = 3.5
(c) Yes, since the image is produced at 25 cm. At 25 cm, magnification magnitude is same as magnifying power.
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