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1. (d) Clay

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

3. (b) At twice the focal length.

4. (a) both concave.

5. (d) either plane or convex

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

7. Object must be placed in front of concave mirror between its pole and principal
focus at a distance less than 15 cm.
The image formed will be virtual and erect. The size of the image is larger the object.

8. (a) Headlights of a car- concave mirror to give parallel beam of light after
reflection from concave mirror.
(b) Side/rear-view mirror of vehicle- convex mirror as it forms virtual erect and
diminished image to give wider view field.
(c) Solar furnace- concave mirror to concentrate sunlight to produce heat in solar
furnace.

9. When one-half of a convex lens is covered with a black paper, this lens produces a
complete image of the object. To prove it we perform experiment:
Take a concave mirror and cover half part of its by using black paper. Place it
vertically in a stand. On one side of it place a burning candle. On opposite side of the
lens fix a white screen. Adjust the position of candle or screen till clear image of
burning candle is formed on the screen. We observe that the image is complete
image of the object.
From the experimental observations, we find that image formation does not depend
upon the size of a lens. A similar lens can also form complete image of an object
placed in front of it. However, brightness of the image decreases when some part of
lens is blocked. It is because now lesser number of rays pass through the lens.

10.
f= +10 cm, u= -25 cm and ho= 5cm

 1/f = 1/v – 1/u
 1/v = 1/10 -1/15
 1/v = 3/50
 V = 50/3 cm.
The image is real and inverted at a distance of 16.7 cm from the lens on opposite
side.
 Magnification (m) = hi/h0= v/u
 hi /5 =16.7/-25
 hi = -10/3 cm. image is inverted and diminished.

11. f= -15 cm, v= -10 cm
 1/v -1/u = 1/f
 1/u = 1/15 – 1/10 = -1/30
 u = -30 cm.
 Ray diagram as follows:


12. f = +15 cm, u = -10 cm.
1/f = 1/v +1/u
 1/v = 1/15 +1/10
 1/v = 5/30
 v = + 30 cm.
 The image is formed 6 cm behind the mirror, it is a virtual and erect image.

13. m= hi/h0= v/u
Magnification produced by a plane mirror is +1 which means that size of image
formed is exactly equal to size of object behind the mirror.

14. Radius of curvature (R) = 30 cm
 f = R/2 = 30/2 = 15 cm
 u = --20 cm, h= 5 cm.
 1/v +1/u = 1/f
 1/v = 1/15+ 1/20 = 7/60
 v = 60/7 = 8.6 cm.
 image is virtual and erect and formed behind the mirror.
 hi/h0= v/u
 hi/5= 8.6/20
 hi = 2.2 cm.
 Size of image is 2.2 cm.

15. u = - 27 cm, f = - 18 cm. ho= 7.0 cm
 1/v = 1/f- 1/u
 1/v = -1/18 + 1/27 = -1/54
 V = - 54 cm.
 Screen must be placed at a distance of 54 cm from the mirror in front of it.
 hi/h0= v/u
 hi/h0= v/u
 hi/7 = +54/-27
 hi = -2 x 7 = -14 cm.
 Thus, the image is of 14 cm length and is inverted image.

16. Power of lens (P) = -2.0 D
 P = 1/f or f = 1/m
 f = 1/-2.0 = -0.5 m.
 (-ve) sign of focal length means that the lens is concave lens.

17. P = +1.5 D
 f = 1/P = 1/+1.5 = 0.67 m.
 As the power of lens is (+ve), the lens is converging lens. 

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