NIOS Class 10 Science & Technology Chapter 15 Solution – Light Energy
NIOS Class 10 Science & Technology Solution Chapter 15 Light Energy. NIOS Class 10 Science & Technology Chapter 15 Question Answers Download PDF. NIOS Class 10 Science Notes.
NIOS Class 10 Science & Technology Chapter 15 Solution
|Subject||Science and Technology (215)|
|Topic||Question Answer, Solution, Notes|
CHAPTER: – 15
INTEXT QUESTIONS 15.1
1) In column A, some sources of light are given. In column B, you have to write whether these are luminous or non-luminous.
Nature of the source
|Shining steel plate||Non luminous|
2)Write two differences between real and virtual image.
Ans: –Difference between real and virtual images are, Real image can be taken on screen whether virtual doesn’t. light rays meeting at the screen in real image While virtual images rays appear to meet at the screen.
3)When you are standing in front of a plane mirror, a virtual and correct image of you is formed. If some one is taking a photograph of it using camera, what will be the nature of image on photograph?
Ans: –The nature of the image on photograph is real.
4) A light ray is falling on a plane mirror at 30° as shown in the diagram. If plane mirror is rotated by 30° without changing the direction of incident ray, by what angle the reflected ray will rotate?
5) An object of height 10 cm is placed in front of a plane mirror of height 8 cm. What will be the height of image formed? Taking the distance of object from the mirror 6 cm, draw the ray diagram.
Ans: – The height will be 10 cm as it lies on plane mirror.
6) The image of an object placed at 10 cm from the mirror is formed at 10 cm behind the mirror. If the object is displaced by 4 cm towards the mirror, by what distance will the image be displaced with respect to the (i) mirror (ii) object?
Ans: – As we know, 1/f=1/v + 1/u;
Here, u=10 and v=-10;
Or, 1/f=1/10 + 1/ (-10) = 0;
Or, f= infinity,
So, the object will be displaced by 4m by mirror.
7) An object is moving with velocity 6 ms–1 towards a plane mirror, what will be
the velocity of image towards the (i) mirror (ii) object?
Ans: – The velocity of image towards the mirror is 6m/s,
The velocity of the image towards the object is 12m/s.
8) Some letters are given in following boxes. Make the meaningful words related to reflection of light choosing the horizontal and vertical sequencing.
Ans: – The words are, (i) Real, (ii)Erect, (iii) Plane, (iv) Virtual, (v)Image.
9)The distance and height of an object placed infront of a plane mirror are givenin column A and B respectively. In column C and D the distance of image andheight of image are given but not in same order. Correct the order.
INTEXT QUESTIONS 15.2
1) An object is placed infront of a concave mirror as shown in the Fig. 15.16. Write the position and nature of the image. What is the focal length of the mirror? Fig. 15.16
2) In what condition, the image formed by concave mirror is virtual?
Ans: –When object is between focal point and pole of the mirror then the image will be formed in the concave mirrors.
3) At what position will the reflected ray shown in Fig. 15.17 intersect the principal axis beyond focus or before focus?
Ans: – From the image it is clear that the reflected rays intersect the principal axis before focus.
4)What type of image will be formed if an object is placed beyond centre of curvature infront of a concave mirror?
Ans: – When an object is placed beyond the centre of curvature infront of a concave mirror then the image will be real, inverted and small in size.
5)Find the position of the object placed infront of a concave mirror of focal length20 cm if image is formed at the distance of 30 cm from the mirror.
Ans: –We know, 1/f = 1/v + 1/u;
Here, f=20cm, v=30cm,
So, 1/20 = 1/30 + 1/u,
Or, u= 60cm.
6) Write two uses of concave mirror.
Ans: – The uses of concave mirror –
- In car headlights.
- In solar cooker.
7)Write the nature of image formed in convex mirror.
Ans: – Image will be smallest and virtual in convex mirror.
8) Find the position of the image formed in convex mirror of focal length 12 cm when object is placed at the distance of (i) 8 cm, (ii) 12 cm and (iii) 18 cm from the mirror.
Ans: – As we know, 1/f = 1/v + 1/u,
In case 1, f=12cm, u=-8cm,
So, 1/12 = 1/v – 1/8;
Or, v= 4.8cm.
In case 2, f=12cm, u=-12cm.
So, 1/12= 1/v + 1/(-12) ;
Or, v= 6cm.
In case 3, f=12cm, u=-18cm,
So, 1/12 = 1/v – 1/18;
Or, v= 7.2cm.
9) Complete the following table with corresponding positions of object and image in case of concave mirror.
Position of object
Position of image
|(i) At F||At infinity|
|(ii) Between F and 2F||Beyond 2F|
|(iii) Beyond 2F||Between F and 2F.|
|(iv) Between F and 2F||Beyond 2F.|
|(v) Beyond 2F||Between F and 2F.|
10)Write two uses of convex mirror.
Ans: – This convex mirror is used in rare view mirror and in car looking glass.
11)Does concave mirror always converges the light rays?
Ans: – No the concave mirror doesn’t converge the light rays.
12)Write the conditions to produce a magnified image in concave mirror.
Ans: – For produce a magnified image in concave mirror the mirror should be placed between F and 2F.
INTEXT QUESTIONS 15.3
1) Name the type of lens which always produces virtual image.
Ans: – The concave lense always produce virtual images.
2) Draw the ray diagram for the image formation in convex lens where object is placed at (i) F (ii) between F and 2F (iii) beyond 2F.
3) Draw the ray diagram for image formation in concave lens.
4)The sizes of the image and object are equal in a lens of focal length 20 cm. Name the type of lens and distance of object from the lens.
Ans: – The lense must be convex lens.
The distance of object from the lens will be at 2F or 2×20 or 40cm.
5) An object of size 10 cm is placed infront of convex lens of focal length 20 cm. Find the size of the image formed.
Ans: – We know from lens formula, 1/f = 1/v – 1/u,
Here, f=20cm, u=-10cm, v=?
So, 1/20 = 1/v + 1/10;
INTEXT QUESTIONS 15.4
1)When light passes from air to a medium its speed reduces to 40%. The velocity of light in air is 3 × 108 ms-1. What is refractive index of the medium?
Ans: – We know, refractive index(u) = c/v,
Here, c= 3 × 10^8 m/s.
And, v= c × 40/100 = 1.2 × 10^8 m/s.
So, u= 3 × 10^8 / 1.2 × 10^8 = 5/2.
2) When sunlight is passed through prism, it splits into seven colours as shown in Fig. By numbers write corresponding colours.
3) How do r and δ change for same angle of incidence i if the prism shown in Fig.is immersed in water
4)Why does white light split into seven colours when it passes through a prism?
Ans: – The different colours or prism because of the all seven colours have different refractive index so when it passes through prism show this phenomenon.
5)Write a natural phenomenon of dispersion of light.
Ans: – Rainbow is the phenomenon of dispersion of light.
INTEXT QUESTIONS 15.5
1)Identify the eye having defective vision from the following diagrams. Write thetype of defect in vision. How this defect can be removed?
Ans: – (A) Short-sightedness which removed by using diverging lens.
(B) No defect
(C) Long-sightedness which removed by using converging lens.
2)Three students Riya, Tiya and Jiya in a class are using sphericals of power +2D,+4D and –2D. What type of defect in vision they have?
Ans: – As we know the positive power will show the defect of longsideness so Riya, Tiya, has this problem where as Jiya have shortsideness problem as his power is negative
3)How does the focal length of the eye changes when a lens is used to correctthe defect of vision in case of (i) short sightedness and (ii) long or for sightedness.
Ans: – The focal length will be increased in shortsideness and decrease in case of long sightedness problem.
1)What happens to the speed of light when it goes from (i) denser medium to rarer medium (ii) rarer medium to denser medium?
Ans: – The speed of the light will be increase when it travels from denser to rarer medium.
The speed of light will decrease when the light travels from rarer to denser medium.
2) Can angle of incidence be equal to the angle refraction? Justify.
Ans: – According to the lens formula when the rays are incident on a plan then in a same plan normal line and reflected rays is present then this phenomenon happened.
3)Does a convex lens always converge light? Explain.
Ans: – When the parallel rays are reflected on a convex lens then the rays after passing through it meet to a point after converging so, this convex lens is converging.
4) Write the nature of the image formed by concave lens.
Ans: – The image is virtual, erect and diminished in nature.
6) What will be the nature of the image formed in a convex mirror and in a concave mirror each of focal length 20 cm and object is placed at the distance of 10 cm.
Ans: – From mirror formula, 1/f = 1/v + 1/u;
Here, v=? u=-10cm, f=20cm.
So, 1/20 = 1/v –1/10;
Then the image is virtual, diminished in nature.
7) Find the position of the image formed in concave mirror of focal length 12 cm when object is placed 20 cm away from the mirror. Also find magnification.
Ans: – From mirror formula, 1/f = 1/ v + 1/u;
F=12cm, u=-20cm, v=?
So, 1/12 = 1/v – 1/20;
Or, v= 60/8 cm.
Then magnification is (m=-v/u = 60/8 ÷ 20=0.25.
8) In which of the following media, the speed of light is maximum and in which itis minimum. Medium Refractive index
Ans: – As we know, refractive index= speed of light in air/ speed of light in that medium,
So, which has law refractive index will have more speed of light and then medium B has highest speed of light.
9) The image of a candle formed by a convex lens is obtained on a screen. Will full size of the image be obtained if the lower half of the lens is printed black and completely opaque? Illustrate your answer with a ray diagram.
10)Can a single lens ever form a real and erect image?
Ans: – A real and erect image is not possible from a single lens.
11)What is dispersion of light? What is the cause of dispersion of light?
Ans: – Each colour has different refractive index. The dispersion of light is seen when light rays of different refractive index is incident on a surface and reflect in different way.
12)Why do distant object appear to be smaller and closer to each other?
Ans: – The size of object smaller and or larger at a distance depends on the diapers angle because of it the light rays are became when it placed in a distance and diapers angle is smaller.
13)A person looking at a net of crossed wires is able to see the vertical direction more distinctly than the horizontal wires. What is the defect due to? How is such defect of vision corrected?
Ans: – The person looking at a net of crossed wires is able to see the vertical direction more distinctly than the horizontal wires this problem occurs because of the defect of cornea and eye lens and this problem know as astigmatisms.