Physics NCERT Solutions for 12^{th}, Chapter 9: Ray Optics and Optical Instruments are available here. These solutions are also available for download in PDF format. This articles is continuation of NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments (Part I) where we have provided solutions from question number 9.1 to 9.9. In this part (of Part II), we have provided solutions from question number 9.10 to 9.20. The solutions of remaining questions will be available in next parts. These all questions and solutions are important for CBSE Class 12 Physics board exam 2018.
NCERT Solutions for Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments from question number 9.10 to 9.20 are given below:
Question 9.10: What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Solution 9.10:
Given:
f’ = 30 cm (convex)
f’’ = -20 cm (concave)
Effective focal length
(1/f) = (1 / f’) + (1/f’’)
Which gives, f = -60 cm
Negative sign points a diverging lens of focal length, f = 60 cm.
NCERT Exemplar: CBSE Class 12 Physics – Chapter 9
Question 9.11: A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15cm. 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 (25cm), and
(b) at infinity? What is the magnifying power of the microscope in each case?
Solution 9.11:
Given:
f_{o} = 2 cm
f_{e} = 6.25 cm
L = 15 cm
(a) D = 25 cm
v_{e} = 25 cm
f_{e} = 6.25 cm
By the formula
(1/v_{e}) – (1/u_{e}) = (1/f)
Putting the values
(‒1/25) – (1/u) = (1/6.25)
⇒ u = -5 cm
Now,
v_{o} = 15 – 5 = 10 cm
f_{o} = 2 cm
Therefore,
(1/10) – (1/u_{o}) = 1 / 2 ⇒u_{o }= ‒2.5 cm.
Magnifying power, m = (10/2.5) [1 + (25/6.25)] or m = 20.
(b) Now
v_{e} = ∞
f_{e} = 6.25 cm
Putting in formula
(1 /∞) – (1/u_{e}) = 1 / 6.25
Or
u_{e} = -6.25 cm
Image distance for the objective
= 15 – 6.25
= 8.75 cm
Using lens formula
(1/8.75) – (1/u_{o}) = 1 / 2
This gives,
u_{o} = -2.59 cm
Magnifying power,
m = (8.75/2.59) × (25/6.25) = 13.5.
Question 9.12: A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Solution 9.12:
Given:
D = 25 cm
f_{o} = 8 mm
f_{e} = 2.5 cm
u_{o} = 9 mm
For image at 25 cm
Angular magnification of the eye-piece
= (25 / 2.5) + 1
= 11
Using lens formula
(‒1/25) – (1/u_{e}) = 1 / 2.5
Or
u_{e} = ‒2.27 cm
Also,
1 / v_{o} – (‒1/0.9) = 1 / 0.8
Or
v_{o} = 7.2 cm
Therefore, Separation = 2.27 + 7.2 = 9.47 cm
Magnifying power = (7.2 / 0.9) × [1 + {25 / 2.5}] = 88.
CBSE Class 12 Chapter Notes: Physics, Chemistry, Maths
Question 9.12: A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Solution 9.12:
Sol.
Given:
D = 25 cm
f_{o} = 8 mm
f_{e} = 2.5 cm
u_{o} = 9 mm
For image at 25 cm
Angular magnification of the eye-piece = (25 / 2.5) + 1 = 11
Using lens formula
(‒1/25) – (1/u_{e}) = (1/2.5)
Or
u_{e} = ‒2.27 cm
Also,
(1/v_{o}) ‒ (‒1/0.9) = 1/0.8
Or
v_{o} = 7.2 cm
Therefore, separation = 2.27 + 7.2 = 9.47 cm
Magnifying power = (7.2/0.9) × [1 + {25 / 2.5}] = 88.
Question 9.13: A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Solution 9.13:
Given:
f_{o} = 144 cm
f_{e} = 6 cm
Length of the telescope tube = f_{o} + f_{e} = 150 cm
Magnification = f_{o} / f_{e} = 24.
Question 9.14: (a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the 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 × 10^{6} m, and the radius of lunar orbit is 3.8 × 10^{8} m.
Solution 9.14:
(a) Given:
f_{o} = 15 m
f_{e} = 1 cm
Angular magnification = f_{o} / f_{e} = 1500
(b) Dia = 3.48 × 10^{6} m
Rad = 3.8 × 10^{8} m
Then, angles subtended
3.48 × 10^{6} / 3.8 × 10^{8} = D / 15
Or
D = 13.7 cm.
Question 9.15: Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(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.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Solution 9.15:
(a) (1/v) + (1/u) = 1/f
f < 0
u < 0
So, (1/v) – (1/u) = ‒1/f
At, u = f
(1/v) – (1/f) = ‒ (1/f)
Or
v = ∞
At, u = 2f
(1/v) – (1/2f) = ‒ (1/f)
Or
v = ‒ 2f
At, u = 1.5f
(1/v) – (1/1.5f) = ‒ (1/f)
Or
v = 2f
Hence deduced.
(b) f > 0, u < 0
At, u = ∞
⇒ 1 / v = 1 / f
Or
v = f (+ive)
At u = f
1 / v – 1 / f = 1 / f
Or
v = f / 2 (+ive)
Hence always virtual image is formed.
(c) f > 0
u < 0
Therefore
1 / v = 1 / f – 1/ u
Since u < 0
1 / v > 1 / f
Or
v < f
Hence image is diminished.
(d) f < 0
f < u < 0
1 / v – 1 / u = -1 / f
Or
1 / v = [-1 / f + 1 / u] < 0
So virtual image.
As m = v / u
u < 0
v > 0
v > u
So,
(v /u) > 1
Image is enlarged.
Question 9.16: A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Solution 9.16:
Given:
n = 1.5
Actual depth, d = 15 cm
Apparent depth = d’
We know
n = d / d’
⇒ 1.5 = 15 / d’
⇒ d’ = 10 cm
So the distance = d – d’ = 5 cm.
Question 9.17: (a) Figure 9.35 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 the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?
Solution 9.17:
(a) Given:
n_{g} = 1.68
n_{p }= 1.44
By Snell’s law
sin i_{c }= 1.44 / 1.68
i_{c} = 59
Thus range is
0 < i < 59
(b) No outer covering
sin i_{c’} = 1 / 1.68
i_{c’} = 36.5
Thus range
53.5 < i < 90.
Question 9.18: 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 are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than 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 than that of ordinary glass. Is this fact of some use to a diamond cutter?
Solution 9.18:
(a) A plane or convex mirrror can produce a real image if the object is virtual.
(b) There is no contradiction as the screen here is not located at the position of the virtual image.
(c) Taller.
(d) The apperent depth decreases.
(e) The diamond can be cut in a way so that light goes total internal reflection producing sparkling effect.
Question 9.19: The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Solution 9.19:
Maximum possible focal length of the lens = distance/4 = 3/4 = 0.75 m.
Question 9.20: A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Determine the focal length of the lens.
Solution 9.20:
Given:
D = 90 cm
d = 20 cm
By the formula
f = [(D^{2} – d^{2})]/4D
Substitution yields,
f = 21.4 cm.
Download NCERT Solutions for Class 12 Physics ‒ Chapter 9: Ray Optics and Optical Instruments
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