CBSE Class 12 Physics Chapter-9 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 9 – Ray Optics and Optical Instruments 3 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-9 – Ray Optics and Optical Instruments
CBSE Class 12 Physics Important Questions Chapter 9 – Ray Optics and Optical Instruments
3 Marks Questions
1. Find the radius of curvature of the convex surface of a plane convex lens, whose focal length is 0.3m and the refractive index of the material of the lens is 1.5?
Ans. 
For plane convex lens
R = 0.15 m
2. Show that the limiting value of the angle of prism is twice its critical angle? Hence define critical angle?
Ans. Angle of the prism 
For limiting 
(Maximum)
Value of angle of prism for 
means 
But when = C
Amax = 2C
The angle of incidence for which angle of refraction is 90o is called critical angle.
3. Draw a labeled diagram of telescope when the image is formed at the least distance of distinct vision? Hence derive the expression for its magnifying power?
Ans. magnifying power
=
(since angles are very small)
—-(i)
For eye piece
Multiply by D
Substituting in e.g. (i)
4. Drive the expression for the angle of deviation for a ray of light passing through an equilateral prism of refracting angle A?
Ans. At the surface AB
At the surface AC
—-(1)
In quadrilateral AQOR
—-(2)
Now in 
Or
—-(3)
From (2) and (3)
—-(4)
Substituting equation (4) in equation (1)
= i + e – A
Or
A + = i + e
5. Draw a ray diagram to illustrate image formation by a Newtonian type reflecting telescope? Hence state two advantages of it over refracting type telescopes?
Ans. Advantages
(1) The image formed in reflecting type telescope is free from chromatic aberrations
(2) The image formed is very bright due to its large light gathering power.
NEWTONIAN TELESCOPE (REFLECTING TYPE)
6. The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and the eye piece is 101cm. calculate the focal length of the objective and the eye piece.
Ans. 
—-(1)
fo = 100 fe —-(2)
Substituting equation. (2) in equation (1)
fe + 100fe = 101
101fe = 101
fe = 1cm
Substituting fe in equation (2)
7. A convex lens made up of refractive index n1 is kept in a medium of refractive index 
. Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if
(1) 
(2) 
(3) 
Ans. (1) When the lens behaves as a convex lens.
(2) When the lens behaves as a plane plate so no refraction takes place
(3) When the lens behave as a convex lens
8. Derive the relation 
Where 
are focal lengths of two thin lenses and F is the focal length of the combination in contact.
Ans. Consider two thin lenses in contact having focal length For the first lens
For the second lens 
acts as an object which forms the final image I
Adding equation (1) & (2)
Using lens formula
For n no. of thin lenses is contact
9. A convex lens has a focal length 0.2m and made of glass is immersed in water 
find the change in focal length of the lens?
Ans. Fair = 0.2m, a
g = 1.50
—-(1)
Now 
Where too is the focal length of the lens when immersed in water?
10. A reflecting type telescope has a concave reflector of radius of curvature 120cm. calculate the focal length of eye piece to achieve a magnification of 20?
Ans. M = 20
R = 120cm (fro concave reflector)
fe = 3cm
11. Show that a convex lens produces an N time magnified image, when the object distances from the lens have magnitude 
. Here f is the magnitude of the focal length of the lens. Hence find two values of object distance, for which value of u convex lens of power 2.5 D will produce an image that is four times as large as the object?
Ans. Magnifying power
For real image m = -N
Or
—-(1)
For virtual image m =N
Or
—-(2)
From equation (1) & (2) we can say that magnification produced by a lens can be N if u
=
Now power of a lens = 2.5 D
m equation (1)
u = 30cm or -50cm
12. Define total infernal reflection of light? Hence write two advantages of total reflecting prisms over a plane mirror?
Ans. The phenomenon of reflection of light when a ray of light traveling from a denser medium is sent back to the same denser medium provided the angle of incidence is greater than the angle called critical angle is called total internal reflection.
Advantages
1. It does require silvering
2. Multiple reflections do not take place in a reflecting prism due to this; only one image is formed, which is very bright.
13. An equi–convex lens of radius of curvature R is cut into two equal parts by a vertical plane, so it becomes a plano-convex lens. If f is the focal length of equi–convex lens, then what will be focal length of the plano –convex lens?
Ans. We know
For equi –convex lens 
—-(1)
For plano convex lens
—-(2)
From (1) & (2)
Or
f’’ = 2f
14. A converging lens of focal length 6.25cm is used as a magnifying glass if near point of the observer is 25cm from the eye and the lens is held close to the eye. Calculate (1) Distance of object from the lens. (2) Angular magnification and (3) Angular magnification when final image is formed at infinity.
Ans. (1) 
u = -5cm
(2) 
m = 5cm
m = 4
15. Draw a graph to show that variation of angle of deviation Dm with that of angle of incidence i for a monochromatic ray of light passing through a glass prism of refracting angle A. hence deduce the relation?
Ans. For the minimum deviation position
(Say)
We know =
Also 
= A
Or 2r = A
Applying minimum deviation condition is equation. (1)
2i = 
Applying Snell’s law
Or
16. An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Ans. Size of the object, 
= 3 cm
Object distance, u= – 14 cm
Focal length of the concave lens, f = – 21 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.
The magnification of the image is given as:
Hence, the height of the image is 1.8 cm.
If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.
17. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
Ans.In the given situation, the object is virtual and the image formed is real.
Object distance, u = +12 cm
(a) Focal length of the convex lens, f= 20 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed 7.5 cm away from the lens, toward its right.
(b) Focal length of the concave lens, f= –16 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed 48 cm away from the lens, toward its right.
18. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?
Ans. Refractive index of glass,
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens =
Radius of curvature of the other face of the lens = 
Radius of curvature of the double-convex lens =R
The value of R can be calculated as:
Hence, the radius of curvature of the double-convex lens is 22 cm.
19. 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?
Ans. Focal length of the objective lens, 
= 144 cm
Focal length of the eyepiece, 
= 6.0 cm
The magnifying power of the telescope is given as:
The separation between the objective lens and the eyepiece is calculated as:
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.
20. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. 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 
, and the radius of lunar orbit is
.
Ans. Focal length of the objective lens,
= 15 m =
cm
Focal length of the eyepiece,
= 1.0 cm
(a) The angular magnification of a telescope is given as:
Hence, the angular magnification of the given refracting telescope is 1500.
(b) Diameter of the moon,
d= 
Radius of the lunar orbit,
r0=
m
Let 
be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter 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
21. A small 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 the same 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 the slab?
Ans. Actual depth of the pin, d= 15 cm
Apparent dept of the pin =
Refractive index of glass,
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
The distance at which the pin appears to be raised = 
=15-10=5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.
22. 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?
Ans. (a) Focal length of the magnifying glass, f = 5 cm
Least distance of distance vision, d = 25 cm
Closest object distance = u
Image distance, v= – d= – 25 cm
According to the lens formula, we have:
Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distant (u‘), the image distance 
According to the lens formula, we have:
Hence, the farthest distance at which the person can read the book is
5 cm.
(b) Maximum angular magnification is given by the relation:
Minimum angular magnification is given by the relation:
23. 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.
Ans. (a) Area of each square, A= 1 mm2
Object distance, u= – 9 cm
Focal length of a converging lens, f = 10 cm
For image distance v, the lens formula can be written as:
Magnification, 
∴Area of each square in the virtual image = (10)2A
= 102 ×1 = 100 
= 1 
(b) Magnifying power of the lens 
(c) The magnification in (a) is not the same as the magnifying power in (b).
The magnification magnitude is 
and the magnifying power is
.
The two quantities will be equal when the image is formed at the near point (25 cm).
24. (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.
Ans. (a) The maximum possible magnification 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 lens formula, we have:
Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.
(b) Magnification =
(c) Magnifying power =
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.
25. What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Ans. Area of the virtual image of each square, 
Area of each square, 
Hence, the linear magnification of the object can be calculated as:
But m =
…(1)
Focal length of the magnifying glass, f= 10 cm
According to the lens formula, we have the relation:
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.
26. (a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the 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?
Ans. Focal length of the objective lens, = 140 cm
Focal length of the eyepiece, = 5 cm
(a) In normal adjustment, the separation between the objective lens and the eyepiece 
(b) Height of the tower, h1= 100 m
Distance of the tower (object) from the telescope, u= 3 km = 3000 m
The angle subtended by the tower at the telescope is given as:
The angle subtended by the image produced by the objective lens is given as:
Where,
= Height of the image of the tower formed by the objective lens
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
(c) Image is formed at a distance, d= 25 cm. The magnification of the eyepiece is given by the relation:
Height of the final image
Hence, the height of the final image of the tower is 28.2 cm.