10th Class Physics 9th Lesson Light -Reflection and Refraction Questions and Answers (Exercise)
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Question 1.Which one of the following materials cannot be used to make a lens?
a) Water b) Glass c) Plastic d) Clay
Answer:d) Clay
Question 2.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
Answer:
d) Between the pole of the mirror and its principal focus.
Question 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,
Answer:
b) At twice the focal length
Question 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.
Answer:
a) both concave.
Question 5.No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
a) only plane.
b) only concave.
c) only convex.
d) either plane or convex.
Answer:c) only convex.
Question 6.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.
Answer:
c) A convex lens of focal length 5 cm.
Question 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 to show the image formation in this case.
Answer:Focal length of concave mirror, f = -15 cm.
Range of distance of the object from the mirror : For getting an erect image using a concave mirror, the object should be placed at a distance less than the focal length.i. e., 15 cm from the pole.
Nature of the image : Image will be virtual, erect and the image is larger than the object (magnified).
Question 8.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.
Answer:
a) A concave mirror, this allows a powerful parallel beam of light.
b) A convex mirror, this allows a greater field of view.
c) A concave mirror, this allows concentration of light at the focus of the mirror.
Question 9.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.
Answer:
Every part of a lens forms an image. Therefore, if the lower half of the lens is covered it will still form a complete image. However, the intensity of the image will be reduced. This can be verified experimentally by observing the image of a distance object like tree on a screen, when lower half of the lens is covered with a black paper.
Question 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 the nature of the image formed.
Answer:
Given that u = -25 cm, f = 10 cm and hO = 5 cm using lens formula, we have
By lens formula, we have = 1/v−1/u=1/f ⇒ 1/v=1/f+1/u = 1/10 + 1/−25 = 3/50
Therefore, v = 50/3 = 16.67 cm
Also, magnification
m = h′/h = v/u
Therefore, h’ = h × v/u = 5 × 16.67/−25 = -3.33 cm
Thus, the image is real and inverted and is formed at a distance of 16.67 cm on the other side of the lens.
Question 11.A concave lens of focal length 15 cm forms an image 10 cm from the lens. How faris the object placed from the lens ? Draw the ray diagram.
Answer:
Given, f = -15 cm, v = -10 cm, u = ?
By lens formula, we have = 1/v−1/u=1/f
1/u=1/v−1/f = 1/−10 – 1/−15
1/u= -1/30
Therefore, u = – 30 cm
The object is placed at a distance of 30 cm from the lens. The ray diagram is as shown.
Question 12.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 the image.
Answer:
Given u = -10 cm, f = +15 cm, v = ?
Using the mirror formula 1/f=1/u+1/v
We have 1/v=1/f−1/u = 1/15 – 1/−10
1/v= 1/15 + 1/10
Therefore, v = 15×10/15+10 = 150/25 = 6 cm
The image is formed 6 cm behind the mirror. Thus, the image is virtual, erect and smaller in size than the object.
Question 13.The magnification produced by a plane mirror is +1. What does this mean ?
Answer:
It means that the size of the image is equal to the size of the object. The positive sign ‘ indicates the image is virtual and erect.
Question 14.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.
Answer:
Given h = 5 cm, u = -20 cm, R = +30 cm or
f = R/2 = + 30/2 = 15 cm,
v = ?, h’ = ?
Using the mirror formula = 1/f=1/u+1/v
We have 1/v=1/f−1/u
= 1/15 – 1/−20 = 1/15 + 1/20
Therefore, v = 15×20/15+20
= 300/35 = 8.6 cm
The image is formed 8.6 cm behind the mirror. Thus, the image is virtual and erect.
Now, m = h′/h = v/u
Therefore, we have h’ = vh/u = 8.6×5/−20 = 2.15 cm
Thus, the size of the image is 2.15 cm or 2.2 cm. The image is reduced.
Question 15.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 focussed image can be obtained ? Find the size and the nature of the image.
Answer:
For concave mirror, h = 7.0 cm, u = -27 cm, f = -18 cm
Using mirror formula 1/f=1/v+1/u
we get
1/v=1/f−1/u = 1/−18 – 1/−27 = −1/18 + 1/27 = −3+2/54 = −1/54
∴ v = – 54 cm
Hence the screen should be placed at a distance of 54 cm infront of the concave mirror to get the sharp focussed image of the object on it.
Magnification of spherical mirror is given by m = hi/ho = –v/u
⇒ hi = – ho*v/u = -7 × −54/−27 = -14 cm
∴ The height of image is 14 cm
Since hi > ho the image is enlarged
Since ‘v’ is -ve, the image is real and inverted.
Question 16.Find the focal length of a lens of power – 2.0 D. What type of lens is this ?
Answer:
Given P = -2D
We know that P = 1/f( m)
Therefore, f = 1/P = 1/−2 = -0.5 m
Since the power of the lens is negative, therefore, is must be a concave lens.
Question 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:
Given P = + 1.5 D, f = ?
Using the relation P = 1/f or f = 1/P = 1/1.5 = 100/15 = 6.67 cm
The prescribed lens is converging or convex as its power is positive.
