SCIENCE || CLASS 10TH || CHAPTER 10 || EXERCISE QUESTIONS AND ANSWERS || CBSE AND MP BOARD

  "AKC SCIENCE CLASSES"

CLASS 10 TH (CBSE AND MP BOARD)


CHAPTER 10

LIGHT-REFLECTION AND REFRACTION

EXERCISE QUESTIONS AND ANSWERS

Q.01:- Which one of the following materials cannot be used to make a lens?

(a) Water

(b) Glass

(c) Plastic

(d) Clay

Ans:- (d)

Q.02:- The image formed by a concave mirror is observed to be 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)

Q.03:- 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)

Q.04:- A spherical mirror and a thin thin spherical lens each 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, but the lens is convex

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

Ans:- (a)

Q.05:- 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)

Q.06:- Which of the following lenses would you prefer to use while 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 5 cm

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

Ans:- (c).

Q.07:- 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?

Ans:- The object should be placed between the focus F and the pole P of the concave mirror, i.e., between 0 and 15 cm from the mirror. Then the image will be virtual, erect and larger than the object.

Q.08:- Name the types of mirror used in the following situations :-

(a) Headlights of a car

(b) Side/rear-view mirror of a vehicle

(c) Solar furnace.

Support your answer with reason :-

Ans:- (a) Concave mirrors are used as reflectors in headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.

(b) A convex mirror is used as a side/rear-view mirror of a vehicle because of the following reasons :-

  • A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
  • A convex mirror has a wider field of view than a plane mirror of the same size.
(c) Large convex mirrors are used of concentrate sunlight to produce heat in solar furnaces.

Q.09:- One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Ans:- As shown in figure given below, when the lower half the convex lens is covered with a black paper, it still forms the complete image of the object as that with uncovered lens. But the intensity of the image is reduced when one half of the convex lens is covered with a black paper.

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

Ans:- Converging lens means a convex lens. As the distances given in the question are large, so we choose a scale of 1 : 5, i.e., 1 cm represents 5 cm. Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. Now, we draw the ray diagram as follows :-

  1. Draw a horizontal line to represent the principal axis of the convex lens. 
  2. Centre line shown by DE.
  3. Mark two foci F and F' on two sides of the lens, each at a distance of 2 cm from the lens.
  4. Draw an arrow AB of height 1 cm on the left side of a lens ata distance of 5 cm from the lens.
  5. Draw a line AD parallel to principal axis and then allow it to pass straight through the focus (F') on the right side of the lens.
  6. Draw a line from A to C (centre of the lens), which goes straight without deviation.
  7. Let the line starting from A meet at A'.
  8. Draw A'B', perpendicular to the principal axis from A'.
  9. Now A'B', represents the real, but inverted image of the object AB.
  10. Then, measure CB' and A'B'. It is found that CB' = 3.3 cm and A'B' = 0.7 cm.
  11. Thus the final position, nature and size of the image A'B' are :-
  • Position of image A'B' = 3.3 cm ✖ 5 = 16.5 cm from the lens on opposite side.
  • Nature of image A'B' : Real and inverted.
  • Height of image A'B' : 0.7 ✖ 5 = 3.5 cm, i.e., image is smaller than the object.

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

Ans:- Focal length, f = -15 cm [f is -ve for concave lens]

Image distance, v = -10 cm [Concave lens form virtual image on same side as the object, so v is -ve]

        1        1          1    

As,   ------ = --------  --  ----------

      f        v          u

    1        1           1             1            1

  -------- = --------  --  ----------   =   ---------  --  ---------

  u        v          f            -10         -15
              -3 + 2             1

         =  ---------------  =  --  --------

              30              30
Object distance, u = -30 cm. Ans

Drawing the ray diagram :- Using a scale of 1 : 5, we get v = -2 cm, f = -3 cm.

We draw the ray diagram as follows :- 

  1. Draw the principal axis (a horizontal line).
  2. Draw a convex lens, keeping principal center (C) on the principal axis.
  3. Mark points F and B on the left side of lens at a distance of 3 cm and 2 cm respectively.
  4. Join any point D (nearly at the top of lens) and F by a dotted line.
  5. Draw a line AD, parallel to principal axis.
  6. Draw a line A'B', perpendicular to principal axis from B'.
  7. Draw a line CA', backwards, so that it meets the line from D parallel to principal axis at B.
  8. Draw a line AB, perpendicular (downwards) from A to meet the principal axis at B.
  9. The AB is position of object. Measure distance BC. It will be found to be equal to 6 cm.

Thus, object is placed at a distance of 6cm ✖ 5 = 30cm from the lens.

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

Ans:- Object distance, u = -10cm

Focal length, f = +15cm [ f is +ve for a convex mirror]

Image distance, v = ?

        1        1          1    

As,   ------ = --------  --  ----------

     u        v          f


        1        1          1            1           1

∴    ------- = --------  --  ----------   =   ---------  --  -------

     v        f          u           +15        -10

           2 + 3           1

       = ------------   =  ---------          

            30             6

Image distance, v = +6 cm. Ans

As v is +ve, so a virtual, erect image is formed at a distance 6 cm behind the mirror.

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

Or

The magnification of a mirror is +1. Comment on this statement.

Ans:- 

                h'        v    

As,   m   = --------  =  -------

             h         u

Given for a plane mirror, m = +1, so h' = h and v = -u.

(a) m = 1 indicates that the size of image is same as that of object.

(b) Positive sign of m indicates that a virtual image is formed behind the mirror. 

Q.14:- An object 5.0 cm in length is placed at a distance of 20cm in front of a convex mirror of radius of curvature 30cm. Find position of the image, its nature and size.

Ans:- Object size, h = +5 cm

Object distance, u = -20cm

Radius of curvature, R = +30cm [R is +ve for a convex mirror]

Focal length, f = R/2 = +15 cm

From mirror formula, 

         1        1          1           1            1

As,   ------- = --------  --  --------  =    ---------   --  --------- 

      v        f          u          +15         -20
           4 + 3           7
       = ------------   =  --------- .          
            60            60
                            60
Image distance, v = ----------  =  8.6 cm. Ans
                             7
                             v           h'
Magnification, m = -- --------  =  ----------

                             u           h
                           vh            8.6 ✖ 5
Image size h' = ---  ----------  =     ---------------  

                            u              -20
                  =  2.15 ≃ 2.2 cmAns.
A virtual, erect image of height 2.2 cm is formed behind the mirror at a distance of 8.6 cm from the mirror.
Q.15:- An object if size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Ans:- Object size, h = +7.0 cm
Object distance, u = -27 cm
Focal length, f = -18 cm
Image distance, v = ?
Image size, h' = ?

        1        1          1    

As,   ------ + --------  =  --------

     u        v          f
        1        1          1           1            1
ஃ    ------ = --------  --  ---------- = ----------  --   ----------

     v        f           u        -18          -27
          -3 + 2          -1

       = ------------   =  ---------          

            54             54
or    v = -54 cm. Ans
The screen should be placed at a distance of 54 cm on the object side of the mirror of 54 cm on the object side of the mirror to obtain sharp image.
                             v           h'
Magnification, m = -- --------  =  ----------

                             u           h
                           vh          (-54) ✖ (+7)
Image size h' = ---  ----------  =   ---------------------- 

                            u              (-27)
                            -14 cmAns.
The image is real, inverted and enlarged in size.
Q.16:- Find the focal length of a lens of power -2.0 D. What type of lens is this?
Ans:- Here, P = -2.0 D
        1            1
f = -----------  =  ----------  =  -0.5 m. Ans.
        P         -20 D
The type of lens is concave.
Q.17:- A doctor has prescribed a corrective lens of power +15 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Ans:- Here, P = + 1.5 D
           1            1             10
ஃ f = -----------  =  ----------  = + ----------  m
           P         +1.5 D         15
 
       =  +66.67 cm. Ans.

As the focal length is positive, the prescribed lens is converging.