NCERT Exemplar Class 10 Science Chapter 10 Light- Reflection and Refraction Solutions
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light- Reflection and Refraction covers all the important questions and answers as well as advanced level questions. It helps in learning about the Light, Normal rays, Incident rays, Reflected rays, Angle of incidence, angle of reflection, point of incidence and Laws of reflection.
The NCERT Exemplar solutions for class 10 science is very important for board exams. NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light- Reflection and Refraction is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like Laws of refraction, characteristics of an image formed by the plane mirror, mirror formula, spherical mirrors and its types, lens, power of lens real image and inverted image.
Chapter Name | Chapter 10 Light- Reflection and Refraction |
Book Title | NCERT Exemplar for Class 10 Science |
Related Study |
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Topics Covered |
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NCERT Exemplar Solutions for Chapter 10 Light- Reflection and Refraction Class 10 Science
Multiple Choice Questions
1. Which of the following can make a parallel beam of light when light from a point source is incident on it?
(a) Concave mirror as well as convex lens
(b) Convex mirror as well as concave lens
(c) Two plane mirrors placed at 90° to each other
(d) Concave mirror as well as concave lens
Solution
(d) Concave mirror as well as concave lens
The emergent rays are parallel when point source of a light is focused to a convex or concave mirror.
2. A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is
(a) – 30 cm
(b) – 20 cm
(c) – 40 cm
(d) – 60 cm
Solution
Size of object = O = + 10.0 mm = + 1.0 cm (as, 1 cm = 10 mm)
Size of Image size = I = 5.0 mm = 0.5 cm
Image distance, v = − 30 cm (as image is real)
Let, object distance = u
Focal length, f =?
Magnification, m = I/O
Magnification is given by m = -v/u
Therefore,
Focal length is given by,
⇒ f = -20 cm
3. Under which of the following conditions a concave mirror can form an image larger than the actual object?
(a) When the object is kept at a distance equal to its radius of curvature
(b) When object is kept at a distance less than its focal length
(c) When object is placed between the focus and centre of curvature
(d) When object is kept at a distance greater than its radius of curvature
Solution
(c) When object is placed between the focus and centre of curvature
When object is placed between F and C an enlarged image is formed beyond C.
4. Figure 10.1 shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is
(a) √3 / √2
(b) √2 / √3
(c) 1/ √2
(d)√2
Solution
(a) √3 / √2
Refractive Index of B with respect to A
5. A light ray enters from medium A to medium B as shown in Figure 10.2. The refractive index of medium B relative to A will be
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) zero
(b) less than unity
(a) A rectangular glass slab
(b) A convex lens
(c) A concave lens
(d) A prism
(a) A rectangular glass slab
(a) Concave lens
(b) Rectangular glass slab
(c) Prism
(d) Convex lens
(a) Concave lens
(a) A convex lens has 4 dioptre power having a focal length 0.25 m
(b) A convex lens has –4 dioptre power having a focal length 0.25 m
(c) A concave lens has 4 dioptre power having a focal length 0.25 m
(d) A concave lens has –4 dioptre power having a focal length 0.25 m
(a) A convex lens has 4 dioptre power having a focal length 0.25 m
(a) is less than one
(b) is more than one
(c) is equal to one
(d) can be more than or less than one depending upon the position of the object in front of it
(a) is less than one
(a) 15 cm in front of the mirror
(b) 30 cm in front of the mirror
(c) between 15 cm and 30 cm in front of the mirror
(d) more than 30 cm in front of the mirror
(d) more than 30 cm in front of the mirror
focal length, f = 15cm,
Radius of curvature is twice the focal length.
Therefore, R =2 × 15 = 30cm
When an object is placed at R, the image formed is of the same size as that of the object. The image is formed at R and it is inverted.
(a) a concave mirror
(b) a convex mirror
(c) a plane mirror
(d) both concave as well as plane mirror
(b) a convex mirror
(a) between the pole and the focus of the reflector
(b) very near to the focus of the reflector
(c) between the focus and centre of curvature of the reflector
(d) at the centre of curvature of the reflector
(b) very near to the focus of the reflector
(a) plane mirror only
(b) concave mirror only
(c) convex mirror only
(d) all mirrors irrespective of their shape
(d) all mirrors irrespective of their shape
(a) A
(b) B
(c) C
(d) D
(b) B
(a) Kerosene
(b) Water
(c) Mustard oil
(d) Glycerine
(d) Glycerine
(b) Fig. B
(c) Fig. C
(d) Fig. D
(d) Fig. D
(a) Fig. A
(b) Fig. B
(c) Fig. C
(d) Fig. D
(a) Fig. A
(a) Plane, convex and concave
(b) Convex, concave and plane
(c) Concave, plane and convex
(d) Convex, plane and concave
(c) Concave, plane and convex
(a) Concave mirror only
(b) Convex mirror only
(c) Convex lens only
(d) Concave mirror, convex mirror, concave lens and convex lens
(d) Concave mirror, convex mirror, concave lens and convex lens
Short Answer Questions
(a) Object is placed between device and its focus, image formed is enlarged and behind it.
(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.
(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.
(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.
When light ray enters denser medium from rarer medium it bends towards the normal. Again when this ray exists the second medium and enters the first, it ends away from the normal. In this case extent of bending of ray at opposite faces is the same, this is the reason why the emergent ray is parallel to the incident ray.
The bending of light takes place because of refraction. Refraction dependents on refractive indices of the medium. The refractive indices of kerosene or turpentine will not be same as that of water. Therefore, the degree of bend will be different in different mediums.
The factor by which the speed and the wavelength of the radiation are reduced with respect to their vacuum values is known as the refractive index of the medium.
n = 𝑐/𝑣
(Where n: refractive index, c: speed of light, v: velocity of light in that medium)
Refractive index of one medium with respect to the second medium is given by ratio of speed of light in second medium to speed of light in first medium.
When an object is placed at F and F2 of a convex lens, an inverted, enlarged and real image is formed beyond 2F2 i.e. on the other side of the lens. Therefore, we need to place the object between 20 and 40 cm of the lens. When an object is placed between F and O of a convex lens, an enlarged, erect and virtual image is formed beyond F2 i.e. on the same side of lens. Therefore, we need to place the object at a distance less than 20 cm from the lens.
Sudha will have to move the screen towards the lens in order to obtain a clear image of the building. The focal length of the lens will be approximately 15 cm. The ray of light coming from any distant object can be considered to be parallel. When parallel rays of light are incident on a convex lens, the rays, after refraction, converge at focus on the other side of the lens..
P = 1/𝑓, P ∝1/𝑓. Therefore, the power of a lens is inversely proportional to its focal length. So, the lens having focal length of 20 cm will provide more convergence.
The Incident ray and reflected ray will always be parallel to each other if two plane mirrors are placed perpendicular to each other.
Speed of light decreases when it passes from rarer medium to denser medium and also light rays bend towards normal.
Long Answer Questions
(a) between pole and focus of the mirror
(b) between focus and centre of curvature of the mirror
(c) at centre of curvature of the mirror
(d) a little beyond centre of curvature of the mirror
(e) at infinity
(a) between optical centre and focus of the lens
(b) between focus and twice the focal length of the lens
(c) at twice the focal length of the lens
(d) at infinity
(e) at the focus of the lens
Laws of refraction
- Incident ray, refracted ray and normal at the point of incidence lie in the same plane.
- Ratio of sine of incidence and sine of refraction is constant for the given color and pair of media.
- NO is normal and ∠ION= i1; which is the angle of incidence.
- OM is the normal extended towards the glass slab and ∠O’OM= r1;which is angle of refraction.
- OO’ is refracted ray from surface AD. It behaves like incident ray for surface BC.
- N’O’ and O’M’ are normal on surface BC.
- O’E is the emergent ray.
- ∠OO’N’ = i2; which is angle of incidence at surface BC.
- ∠EO’M’ = r2; which is angle of refraction at surface BC.
- It is observed that the incident ray, the normal and the refracted ray lie in the same plane. This verifies the first law of refraction.
- It is also observed that IO is parallel to O’E; which means emergent ray is parallel to incident ray. This happens because the degree of bend at opposite surfaces of glass slab is same.
(a) at the focus of the lens
(b) between focus and twice the focal length of the lens
(c) beyond twice the focal length of the lens
(a) The ray diagram when the object is placed at the focus of the concave lens :
(b) at finite distance from the mirror
(a) At Infinity
As the image is obtained on the screen , it is real .
so,
Magnification , m = –3 ,
v = 80 cm
u = ?
As m = v/u
so,
–3 = 80/u,
u = –80/3 cm .
From
1/f = 1/20cm
So, f = 20 cm .
The lens is convex and image formed at 80 cm from the lens is real and inverted.
P =1/𝑓, where f is in metre.
The unit of power is Dioptre. Lens is convex in the first case and concave in the second case.
Power of lens (first student) =+2 dioptre
Power of lens (second student) = -2 dioptre
Position of candle = 12.0 cm
Position of convex lens = 50.0 cm
Position of the screen = 88.0 cm
(i) What is the focal length of the convex lens?
(ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?
(iv) Draw a ray diagram to show the formation of the image in case (iii) as said above.
Position of the candle flame = 12.0cm
Position of the lens = 50.0 cm
Position of the screen = 88.0 cm
(i) u= 50-12= 38 cm.
Image distance v= 88-50= 38cm
Focal length = 1/𝑣 - 1/𝑢 = 1/𝑓
f= 19cm
(ii) Object distance u= 50 - 31= 19 cm
Here
Object distance = focal length
Hence the I mage is formed at infinity.
(iii) If he further shifts the candle towards the lens. The object comes between F and 0. In this case. Image is virtual, enlarged and erect and is formed on the same side of lens.