Answer The Following Question.
1. Define the principle focus of a concave mirror.
Ans: The principal focus of a concave mirror is the point on its principal axis where parallel light rays converge after reflecting from the mirror.
2. 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.
3. Name a mirror that can give an erect and enlarged image of an object.
Ans : Only a concave mirror can give a erect and enlarged image of an object.
4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Ans: We prefer a convex mirror as a rear-view mirror because it provides an erect, diminished image and has a wider field of view. This helps the driver to see more of the area behind the vehicle, ensuring better safety.
5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Ans : Radius of curvature of (R) = 32 cm
Focal length(f) = R/2 = 32/2 cm = 16 cm.
6. A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?
Ans : Distance of object from concave mirror (u)= -10 cm.
Magnification (m) = -3
m = -v/u
v = -mu = -(3) x (-10) = -30 cm.
7. A ray of light traveling in air enters obliquely into water. Does the light ray bend towards or away from the normal? Why?
Ans: When a ray of light enters water from air at an oblique angle, it bends towards the normal. This happens because water is optically denser than air, meaning light travels slower in water than in air. As a result, the light ray bends towards the normal according to Snell’s Law.
8. Light enters from air to glass having refractive index 1.50. What is the speed of light in glass? The speed of light in vacuum is 3×108 m/s.
Ans : Speed of light in vacuum (c) = 3 x 108 m/s.
Refractive index = c/v.
Speed of light in glass = 3 x 108 m/s/ 1.50
= 2 x 108 m/s
9. Find out, from Table (10.3), the medium having highest optical density. Also, find the medium with lowest optical density.
Ans : As per table, diamond has highest optical density (2.42). Medium with lowest optical density is air (1.0003)
10. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in table 10.3
Ans : As the refractive index of water is least out of three substances, hence speed of light is maximum in water. So, light travels fastest in water.
11. The refractive index of diamond is 2.42. What is the meaning of this statement?
Ans: Meaning of the Refractive Index of Diamond (2.42):
A refractive index of 2.42 means that light travels 2.42 times slower in diamond than in air. It also indicates that the speed of light in diamond is 1/2.42 of the speed of light in air. The refractive index is a measure of how much the light is bent or refracted when it enters the material.
12. Define 1 dioptre of power of lens.
Ans: 1 Dioptre of Power of a Lens:
One dioptre (D) is the power of a lens whose focal length is 1 meter. It is the reciprocal of the focal length (in meters). So, the formula is:
p=1/f
Thus, if the focal length is 1 meter, the power is 1 dioptre.
13. A convex lens forms a real and inverted image of a needle at distance of 50 cm. from it. Where is the needle placed in front of the convex lens if the image is equal to the size of objects? Also, find the power of lens.
Ans :
Image distance (v) = +50 cm, hi = ho
hi/ho = v/u
u = v x ho / hi
= 50 x ho / hi
= 50 cm.
Now,
u = -50 cm
v = + 50 cm.
f = ?
1/f = 1/v – 1/u
1/f = 1/50 + 1/50
f = + 25 cm. = 0.25 m
Power of lens (P) = 1/f
= 1/ 0.25 = + 4D.
14. Find the power of a concave lens of focal length 2 m.
Ans :
Focal length of concave lens = – 2 m.
P = 1/f = 1/ (-2m)
= -0.5 D
15. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Ans : (d) Clay
16. 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) Between the pole of the mirror and its principal focus.
17. Where should an object be placed in front of 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.
18. A spherical mirror and thin spherical lens have each of focal length of -15 cm. the mirror and 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 and lens is concave.
Ans : (a) both concave.
19. 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 concave or convex.
Ans : (d) either plane or convex
20. Which of the following lens 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) A convex lens of focal length 5 cm.
21. 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 mirror? What is the nature of image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Ans : 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. The ray diagram is as follows:
22. Name the type 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) Headlights of a car:
Concave mirror is used because it reflects light and focuses it into a parallel beam. This helps to direct the light efficiently, enhancing the visibility.
(b) Side/rear-view mirror of a vehicle:
Convex mirror is used because it forms a virtual, erect, and diminished image, providing a wider field of view. This allows the driver to see more of the area behind and around the vehicle.
(c) Solar furnace:
Concave mirror is used because it can concentrate sunlight onto a small area, producing intense heat. This focused sunlight is used to generate energy in a solar furnace.
In summary, concave mirrors are used to focus light (for headlights and solar furnaces), and convex mirrors are used to provide a wider view (for rear-view mirrors).
23. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answers experimentally. Explain your observations.
Ans : 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:
When another-half of a convex lens is covered with a black paper
Yes, a convex lens will still produce a complete image of the object even if half of it is covered with black paper. Here’s an explanation and experimental verification:
Explanation:
A convex lens works by bending light rays that pass through it. Even if half of the lens is covered, the other half can still focus light from the object.
The light rays from the object that pass through the uncovered half of the lens will be bent and focused, forming an image.
While the image may be slightly dimmer or less sharp due to the reduced amount of light, it will still be a complete image. This is because the lens as a whole still has the ability to converge light from all points of the object.
Experimental Verification:
Setup: Take a convex lens and place it on a stand. Cover half of the lens with a black paper while keeping the other half uncovered.
Procedure:
Place an object (like a small candle) in front of the lens.
Observe the image formed on a screen placed behind the lens.
Observation: Even though half of the lens is covered, a clear image will still be formed. It might appear fainter or slightly blurred compared to using the full lens, but the image will still be complete.
24. 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 the nature of the image formed.
Ans :
25. 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 : 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:
26. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of image.
Ans :
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.
27. The magnification produced by a plane mirror is +1. What does this means?
Ans : Magnification Produced by a Plane Mirror**:
The magnification (m) is defined as the ratio of the image height (hi) to the object height (ho), and it can also be expressed as the ratio of the image distance (v) to the object distance (u):
m= hi/ho=v/u
For a plane mirror, the magnification is +1, which means:
– The image formed is virtual, erect, and of the same size as the object.
– The positive sign (+1) indicates that the image is upright (as opposed to inverted).
In summary, a magnification of +1 means the image size is equal to the object size, and the image is erect.
28. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Ans : 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.
29. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.
Ans : 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.
30. Find the focal length of a lens of power -2.0 D. What type of lens is this?
Ans : 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.
31. 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?
Ans : 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.