Page 168

Define the principal focus of a concave mirror.
Ans:-  

The principal focus of a concave mirror is a point on the principal axis where light rays parallel to the principal axis converge after reflection.

The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Ans:-

Focal length (f) = R/2

= 20 cm / 2

= 10 cm

Name the mirror that can give an erect and enlarged image of an object.
Ans:-

Concave mirror

Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Ans:-

A convex mirror gives a wider field of view and forms erect, virtual, and diminished images of objects, which helps drivers see more area behind them.

Page 171

Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Ans:-

Focal length (f) = R/2

= 32 cm / 2

= 16 cm

A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Ans:-

Magnification (m) = -3 (real image)
Using m = -v/u

=> -3 = -v/10

=> v = 30 cm
Image is located 30 cm in front of the mirror

Page 176

A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Ans:-

The light ray bends towards the normal because water is optically denser than air.

Light enters from air to glass, having a refractive index 1.50. What is the speed of light in the glass?
Ans:-

Speed in glass = c / n = (3 × 10⁸ m/s) / 1.50 = 2 × 10⁸ m/s

Find out the medium with highest and lowest optical density.
Ans:-

Highest: Diamond (n = 2.42)
Lowest: Air (n = 1.0003)

You are given kerosene, turpentine and water. In which of these does the light travel fastest?
Ans:-

Light travels fastest in the medium with the lowest refractive index.
Refractive indices:

Water = 1.33

Kerosene = 1.44

Turpentine = 1.47
Light travels fastest in water

The refractive index of diamond is 2.42. What is the meaning of this statement?
Ans:-

It means that the speed of light in diamond is 2.42 times slower than in vacuum (or air).

Page 184

Define 1 dioptre of power of a lens.
Ans:-

1 dioptre is the power of a lens whose focal length is 1 metre. It is the SI unit of lens power.

A convex lens forms a real and inverted image of a needle at a distance of 50 cm. Where is the needle placed if the image is same size as the object? Find the power of the lens.
Ans:-

For same-sized image,

object distance = 2f

and image distance = 2f

So, 2f = 50 cm

=>f = 25 cm
Power = 100/f (in cm)

= 100/25

= +4 D

Find the power of a concave lens of focal length 2 m.
Ans:-

Power = 100/f (in cm)

= 100/200

= -0.5 D

Exercise

Q.1. Which one of the following materials cannot be used to make a lens? 

(a) Water. 

(b) Glass.

(c) Plastic.

(d ) Clay.

Ans: (d) Clay. 

Q.2. The image formed by a concave  mirror is observed to the virtual, erect and larger than the object. Where should be the position of the object? 

(a) Between the principal focus and the centre of curvature . 

(b) At the centre of curvature.

(c) Beyond the centre of curvature . 

(d) Between the pole of the  mirror and its principal focus.

Ans: (d) Between the pole of the mirror and its principal focus. 

Q.3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principal focus of the lens.

(b) At twice the focal length. 

(c) At infinity. 

(d) Between the optical centre of the lens and its principal focus. 

Ans: (b) At twice the focal length. 

Q.4. A spherical mirror and a thin spherical lens have each a focal length of – 15 cm. The mirror and the lens are likely to be 

(a) Both concave. 

(b) Both convex.

(c) The mirror is concave and the lens is convex.

(d) The mirror is convex, but the lens is concave. 

Ans: (a) Both concave. 

Q.5. No matter how far you stand from a mirror your image appears erect. The mirror is likely to be 

(a) Plane.

(b) Concave.

(c) Convex.

(d) Either plane or convex.

Ans: (d) Either plane or convex.

Q.6. Which of the following lenses would you prefer to use To which reading small letters found in a dictionary? 

(a) A convex lens of focal length 50 cm.

(b) A concave lens of focal length 50 cm. 

(c) A convex lens of focal length of 5 cm. 

(d) A concave lens of focal length of 5 cm. 

Ans: (c) A convex lens of focal length of 5 cm. 

 

7. We wish to obtain an erect image of an object using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram.

Answer:
To get an erect image from a concave mirror, the object must be placed between the pole and the focus (i.e., less than 15 cm).

Nature: Virtual and erect

Size: Larger than the object

(Ray diagram shows rays diverging after reflection and appearing to come from behind the mirror)

8. Name the type of mirror used in the following situations. Give reasons.

Headlights of a car – Concave mirror
Reason: It converges light to form a parallel beam.

Side/rear-view mirror of a vehicle – Convex mirror
Reason: It gives a wider field of view and always forms erect, diminished images.

Solar furnace – Concave mirror
Reason: It focuses sunlight at one point to produce heat.

9. One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify experimentally. Explain your observations.

Answer:
Yes, the lens will still form a complete image, but its brightness will be reduced.
Explanation: Each part of the lens contributes to forming the entire image. Covering half the lens reduces the amount of light but not the entire image.

10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed.

Ans: Given that, 

                      u  = – 25 cm 

                      f   = + 10 cm 

                      v  = ? 

According to the lens formula,

We have,

                1/v – 1/u = 1/f

            ⇒ 1/v – 1/-25 = 1/10

            ⇒ 1/v + 1/25 = 1/10

            ⇒ 1/v = 1/10 – 1/25

                      = 5-2/50

                      = 3/50

                ∴ v = + 50/3

                      = + 16.67cm.

Thus , the position of image is at a distance of 16.67 cm from the lens. The plus sign for image distance shows that the image is formed on the right side of lens and that the nature of image is real and inverted . 

Now Magnification, 

                              m = v/u

                                  = 16.67/ – 25

                                  = – 0.66

Now calculating the size of image h₂we have, 

                             m = h2/h1

                         ⇒ h₂ = m×h₁

                                 = – 0.66×5

                                 = – 3.3cm.

This the size of image is 3.3 cm. The negative sign of the size of image shows that the image is inverted. 

11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed? Draw the ray diagram.

Ans: Given that, 

                        f = – 15 cm.

                        v = – 10 cm.

                        u = ?

Now,

                       1/v – 1/u = 1/f

                   ⇒ 1/-10 – 1/u = 1/-15

                   ⇒ 1/u = 1/15 – 1/10

                             = 2-3/30

                             = -130

                 ∴    u    = -30cm.

The position of image is at behind the  mirror. The nature of image is virtual and erect.

We have,

                 m = -v/u = hight of image/hight of object

           ⇒ -v/u = h¹/h

           ⇒ -8.57/-20 = h¹/5

           ⇒ h¹  = 8.57/20×5

                   = .57/4

                   = 2.14

∴ Height of the image = 2.14 cm. 

∴ Size of image is 2.14 cm. 

12. An object is placed at 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

 

 

 

13. The magnification produced by a plane mirror is +1. What does this mean?

Answer:
It means the image formed is of the same size as the object and is virtual and erect.

14. An object 5 cm is placed at 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

 

 

 

 

15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. Find screen position, image size and nature.

Ans: Given that, 

                size of the object (h) = 7.0 cm. 

                                             u  = 27 cm. 

                                              f  = 18 cm. 

         Size of image (hⁱ) = ? 

 We have, 

                        1/u + 1/v = 1/f

                    ⇒ 1/v = 1/f – 1/u

                              = 1/18 – 1/27

                              = 3-2/54

                              = 1/54

                   ∴   v    = 54cm.

 Again, 

                   m = – v/u

               ⇒     = – 54/27

                       = – 2cm.

∴ Size of image (hⁱ) = m x h 

                                = –2 x 7 

                                = –14 cm. 

∴ Size of image is 14 cm.

The image is read and inverted.

16. Find the focal length of a lens of power -2.0 D. What type of lens is this?

Answer:
f = 100 / P

= 100 / (-2)

= -50 cm
Ans:- It is a concave (diverging) lens

17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer:
f = 100 / P

= 100 / 1.5

= 66.67 cm
Ans:- It is a convex (converging) lens