# RBSE Solutions for Class 10 Science Chapter 9 Light

RBSE Solutions for Class 10 Science Chapter 9 Light are part of RBSE Solutions for Class 10 Science. Here we have given Rajasthan Board RBSE Class 10 Science Solutions Chapter 9 Light.

 Board RBSE Textbook SIERT, Rajasthan Class Class 10 Subject Science Chapter Chapter 9 Chapter Name Light Number of Questions Solved 190 Category RBSE Solutions

## Rajasthan Board RBSE Class 10 Science Solutions Chapter 9 Light

Textbook Questions Solved

I. Multiple Choice Questions

RBSE Solutions For Class 10 Science Chapter 9 Question 1:
Which of the following mirrors (RBSESolutions.com) has a wide field of view?
(a) Plain mirror
(b) Convex mirror
(c) Concave mirror
(d) Parabolic mirror
(b) Convex mirror

RBSE Class 10 Science Chapter 9 Question 2:
Speed of light is maximum in which of the following?
(a) Water
(b) Glass
(c) Vacuum
(d) Glycerine
(c) Vacuum

RBSE Class 10 Science Chapter 9 Light Question 3:
Which of the following is responsible for the coin (RBSESolutions.com) appearing shallower when it is kept at the bottom of a water tank?
(a) Refraction
(b) Reflection
(c) Total internal reflection
(d) None of these
(a) Refraction

RBSE Solution Class 10 Science Chapter 9 Question 4:
The focal length of a mirror is +60 cm. Which kind of mirror it is?
(a) Concave mirror
(b) Parabolic mirror
(c) Plain mirror
(d) Convex mirror
(d) Convex mirror

RBSE Class 10 Science Chapter 9 In Hindi Question 5:
What is the focal length of a plain mirror?
(a) 0
(b) 1
(c) Infinity
(d) None of these
(c) Infinity

RBSE Class 10th Science Chapter 9 Question 6:
Which type of image is always formed by a convex mirror?
(a) Real and erect
(b) Real and inverted
(c) Virtual and inverted
(d) Virtual and erect
(d) Virtual and erect

Chapter 9 Light Class 10 RBSE Question 7:
The power of a lens is +2 Dioptre. (RBSESolutions.com) What is the focal length of this mirror?
(a) 2 m
(b) 1 m
(c) 0.5 m
(d) 0.2 m
(c) 0.5 m

Light Chapter Class 10 RBSE Question 8:
Which of the following is true for a person with long sightedness?
(a) He will clearly see a nearby object
(b) He will clearly see a distant object
(c) He will not clearly see nearby and distant objects
(d) None of these
(b) He will clearly see a distant object

Class 10 Science Chapter 9 Light Question 9:
The focal length of a convex lens is 15 cm. (RBSESolutions.com) What should be the object distance to get a real image equal to the size of object?
(a) 30 cm
(b) 15 cm
(c) 60 cm
(d) None of these
(a) 30 cm

RBSE 10th Class Science Chapter 9 Question 10:
An object is kept at infinity from a concave lens of focal length 20 cm. What is the image distance?
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) Infinity
(c) 20 cm

### Light Very Short Answer Type Questions

RBSE Class 10 Science Chapter 9 Notes Question 1:
An object absorbs all the colours of light falling (RBSESolutions.com) on it. This object would appear of which colour?
Black

RBSE Solutions For Class 10 Science Question 2:
What should be the minimum length of a plain mirror if you want to see your whole image in it?

Class 10 RBSE Solutions Science Question 3:
A ray of light is incident at an angle of 30° to a plain mirror. What is the angle between incident ray and reflected ray?
60°

Science Class 10 Chapter 9 Question 4:
Write any two uses of (RBSESolutions.com) convex mirror.
As rear view mirror, as mirror on hairpin bends on road .

Class 10 Science Chapter 9 Question Answer Question 5:
Write any two uses of concave mirror.
As shaving mirror, as solar cooker.

Numerical Questions On Light Class 10 Question 6:
Write the mirror formula.
$$\frac { 1 }{ u }$$ + $$\frac { 1 }{ v }$$ = $$\frac { 1 }{ f }$$

RBSE Solutions For Class 10th Science Question 7:
What is he relation between radius of (RBSESolutions.com) curvature and focal length of a spherical mirror?
f = R/2 where f is focus and R is radius of curvature

RBSE Solutions Science Class 10 Question 8:
Write magnification formula.
m = $$\frac { height of image h’ }{ object height h }$$ = $$\frac { v }{ u }$$

RBSE Solutions Of Class 10 Science Question 9:
Write Snell’s Law.
Snell’s law:
$$\frac { sin i }{ sin r }$$ = constant

RBSE Solution Class 10 Science Question 10:
Write the lens formula.
$$\frac { 1 }{ v }$$ – $$\frac { 1 }{ u }$$ = $$\frac { 1 }{ f }$$

RBSE Solutions For Class 10 Social Science Chapter 9 Question 11:
Parallel ray from an object are incident on a (RBSESolutions.com) convex lens. What is the location of image in this case?
At focus

RBSE Solution For Class 10th Science Question 12:
What is the unit of power of lens?
Dioptre

RBSE Solutions Class 10 Science Question 13:
A person suffering from short sightedness is unable to see objects at which position?
Distant objects

Science Class 10 RBSE Solutions Question 14:
A convex lens of suitable focal (RBSESolutions.com) length can rectify which defect of vision?
Long sightedness of hypermetropia

RBSE Solution 10th Class Science Question 15:
What is cataract?
Eye lens reduces transparency and flexibility in cataract.

RBSE Solutions Of Class 10th Science Question 16:
Which type of image if formed by a shaving mirror?
Virtual, enlarged and erect

### Light Short Answer Type Questions

RBSE Solution Class 10th Science Question 1:
What do you understand by regular (RBSESolutions.com) reflection and irregular reflection?
When all reflected rays are parallel to each other, the reflection is called regular reflection.
Clear image is formed in case of regular reflection. When reflected rays are not parallel to each other, the reflection is called irregular reflection. Diffused image is formed in case of irregular reflection,

RBSE Solution For Class 10 Science Question 2:
What do you understand by lateral inversion?
In a plain mirror, our right hand appears like left hand of the image. Similarly, our left hand appears like right off the image. This phenomenon is called lateral inversion.

RBSE Class 10 Social Science Chapter 9 Question 3:
With the help of a ray diagram, show the location (RBSESolutions.com) of image when an object is between centre of curvature and focus of a concave mirror. Light Class 10 Question 4:
Explain Cartesian sign convention for spherical mirror.
Following are the Cartesian sign conventions for spherical mirror:

• All distances are measured from the pole of the mirror.
• Incident light is shown coming from LHS of mirror.
• Distances towards LHS of mirror are taken as negative and those on RHS are taken as positive.
• The height is measured perpendicular to the principal axis.
• Distance above principal axis is taken as positive.
• Distance below principal axis is taken as negative.

RBSE Solution Science Class 10 Question 5:
Explain refraction of (RBSESolutions.com) light and write the laws of refraction.
When a ray of light travels from one medium to another medium there is a deviation in its path. This phenomenon is called refraction of light. Following are the laws of refraction:

• When a ray of light travels from a rarer medium to a denser medium, it bends towards the normal.
• When a ray of light travels from a denser medium to a rarer medium, it bends away from the normal.
• Ratio of sine of angle of incident to sine of angle refraction is constant for a given pair of media. This is called Snell’s law.

10th Class RBSE Solution Science Question 6:
What are the different types of convex and concave lenses?
Types of convex lens: Biconvex lens, piano-convex lens, concavo-convex lens
Types of concave lens: Biconcave lens, plano-concave lens, convexo-concave lens

Ch 9 Science Class 10 Question 7:
Define main focus and optical (RBSESolutions.com) centre of spherical lens.
Main Focus: Rays of light parallel to the main axis converge at a point after refraction through the convex lens. In case of concave lens, these rays of light appear to diverge Worn a point. This point is called the main focus of spherical lens.
Optical Centre: The point on a lens through which a ray of light passes without deviation is called the optical centre of the lens.

RBSE Class 10 Chapter 9 Question 8:
What doyoi understand by radius of curvature and centre of curvature of sphencal lens?
The curved surfaces of lens are parts of two spheres. Centre of these spheres are called the centre of curvature. Distance between centre of curvature and surface of lens is called radius
of curvature.

RBSE Solutions For Class 10 Science Chapter 10 Question 9:
Write the laws of refraction through spherical lens.
Following are the laws of refraction through spherical lens:

• A ray parallel to the principal axis passes through main focus after refraction through lens.
• A ray passing through focus of the lens becomes parallel to principal axis after refraction.
• A ray passing through optical centre emerges without deviation.

RBSE Solution Of Class 10th Science Question 10:
With the help of suitable diagram, explain (RBSESolutions.com) image formation in concave lens.
Image formation by concave lens: Object at infinity: When object is at infinity, image is formed at F1. The image is smaller, virtual and erect.
Object between infinity and optical centre: When object is anywhere between infinity and optical centre, image is formed between F1 and O. The image is smaller, virtual and erect.

RBSE Solutions For Class 10 Sst Chapter 9 Question 11:
What do you understand by power of lens?
The power of convergence or divergence of rays of light by a lens is called power of lens. A lens with small focal length causes greater convergence of divergence of rays of light. So, a lens with smaller focal length is more powerful than a lens with greater focal length. The power of lens is reciprocal of its focal length.
P = $$\frac { 1 }{ f }$$
SI unit of power of lens is Diopter (D).

Class 10 RBSE Solution Science Question 12:
What do you understand by near (RBSESolutions.com) sightedness? How is it rectified?
When a person finds it difficult to see distant object, he suffers from near sightedness or myopia. Myopia happens because of increased curvature of lens. In this case, image of a distant object is formed before the retina. This defect is rectified by using a concave lens of suitable focal length. The lens helps in making the image on retina.

RBSE Class 10 Science Chapter 9 Notes In Hindi Question 13:
What do you understand by far sightedness? How is it rectified?
When a person finds it difficult to see nearby objects, he suffers from far sightedness or hypermetropia. This defect happens because of reduced curvature of lens. In this case, image of a nearby object is formed behind the retina. This defect is rectified by using a convex lens of suitable focal length.

Light Diagram Class 10 Question 14:
What do you understand by presbyopia (RBSESolutions.com) and astigmatism?
Presbyopia: Ageing results in weak ciliary muscles. This reduces the power of accommodation of eyes. As a result, the person finds it difficult to see nearby as well as distant objects. This defect is rectified by using bi-focal lens.

Astigmatism: This condition develops because of irregular curvature of cornea. A person suffering from this defect is unable to differentiate between vertical and horizontal planes. This defect is rectified by using cylindrical lens.

RBSE Solution Of Class 10 Science Question 15:
What do you understand by power of accommodation and field of vision of eye?
Power of Accommodation: Human eye can see nearby objects as well as distant objects. Our eyes can quickly focus from a nearby object to a distant object. This is called power of accommodation.
Field of Vision: The extent of observable world which can be seen by both eyes at a given moment is called field of vision. The field of vision with both eyes is slightly more than 180 degrees.

Class 10 Science Chapter 9 Solutions Question 16:
An object is kept between infinity and 2F1 of a convex lens. Explain (RBSESolutions.com) the location of image with the help of a suitable ray diagram. When an object is between infinity and 2F1 of a convex lens, image is formed between F2 and 2F2. The image is smaller, real and inverted.

### Light Long Answer Type Questions

Science Chapter 9 Class 10 Question 1:
Explain image formation for the following in case of (RBSESolutions.com) concave mirror. Make suitable ray diagrams.
(a) Object is between infinity and centre of curvature
(b) Object is at centre of curvature
(c) Object is between centre of curvature and focus
(d) Object–is at focus
(e) Object is between focus and pole  Rainbow Formation Class 10 Question 2:
What do you understand by refraction? Write the laws of (RBSESolutions.com) refraction. Explain refraction through a glass slab.
When a ray of light travels from one medium to another medium there is a deviation in its path. This phenomenon is called refraction of light. Following are the laws of refraction:

• When a ray of light travels from a rarer medium to a denser medium, it bends towards the normal.
• When a ray of light travels from a denser medium to a rarer medium, it bends away from the normal.
• Ratio of sin of angle of incident to sin of angle reflection is constant for a given pair of media. This is called Snell’s law.

Refraction though glass slab can be explained by the following activity.

• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
•  Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
•  Fix two pins, E and F vertically such that the line joining the pins is inclined to the edge AB.
•  Look for the images of the pins E and F through the opposite edge. Fix two pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
•  Join O and O’. Also produce EF up to P, as shown by a dotted line. RBSE Solutions For Class 9 Science Chapter 10 Question 3:
With the help of suitable ray diagrams, explain image (RBSESolutions.com) formation by convex mirror for following: (a) Object is at infinity (b) Object is at a certain distance
(a) When object is at infinity, image is formed on focus of convex mirror. The image is small, erect and virtual. (b) Object is at a certain distance When object is at a certain distance, image is formed between F and P of convex mirror. The image is small, erect and virtual. Class 10 Science Chapter 9 Question 4:
With the help of suitable ray diagrams, explain (RBSESolutions.com) image formation by convex lens for following:
(a) Object is at focus
(b) Object is between F1 and 2F1
(c) Object is between infinity and 2F1
(a) Image formed at infinity, highly enlarged, real and inverted. (b) Image is formed beyond 2F2. Image is enlarged, real and inverted.
(c) Image is formed between F2 and 2F2. Image is smaller, real and inverted.

Question 5:
With the help of suitable ray diagrams, explain image (RBSESolutions.com) formation by convex lens for following:
(a) Object is between focus and optical centre
(b) Object is at focus
(c) Object is between F1 and 2F1
(d) Object is at 2F1
(e) Object is between infinity and 2F1
(a) Image is formed beyond 2F1. Image is enlarged, virtual and erect. (b) Image is formed at infinity. Image (RBSESolutions.com) is highly enlarged, real and inverted. (c) Image is formed beyond 2F2. Image is enlarged, real and inverted. (d) Image is formed at 2F2. Image is of (RBSESolutions.com) same size, real and inverted. (e) Image is formed between F2 and 2F2. Image is smaller, real and inverted. Question 6:
Explain different types of vision defects (RBSESolutions.com) and their rectification.
Following are the vision defects and their rectification:

• Myopia: When a person finds it difficult to see distant object, he suffers from near sightedness or myopia. Myopia happens because of increased curvature of lens. In this case, image of a distant object is formed before the retina. This defect is rectified by using a concave lens of suitable focal length. The lens helps in making the image on retina.
• Hypermetropia: When a person finds it difficult to see nearby objects, he suffers from far sightedness or Hypermetropia. This defect happens because of reduced curvature of lens. In this case, image of a nearby object is formed behind the retina. This defect is rectified by using a convex lens of suitable focal length.
• Presbyopia: Ageing results in weak ciliary muscles. This reduces the power of accommodation of eyes. As a result, the person finds it difficult to see nearby as well as distant objects. This defect is rectified by using bi-focal lens.
• Astigmatism: This condition develops because of irregular curvature of cornea. A person suffering from this defect is unable to differentiate between vertical and horizontal planes. This defect is rectified by using cylindrical lens.

### Light Numerical Questions

Question 1:
Focal length of a concave mirror is 30 cm. If object (RBSESolutions.com) is at a distance of 40 cm then find the image distance. Find the magnification as well.
Here, f= -30 cm, u = -40 cm
Image distance can be calculated as follows: The positive sign with magnification shows that image is real.

Question 2:
Image is formed at a distance of 8 cm from a (RBSESolutions.com) convex mirror. If focal length of mirror is 16 cm then find the distance of object.
Here, v = 8 cm, f= 16 cm
Object distance can be calculated as follows: Question 3:
An object is at a distance of 60 cm from a convex lens (RBSESolutions.com) of focal length 30 cm. If height of object is 3 cm then find the location and nature of image.
Here, u = -60 cm, f= 30 cm, h = 3 cm
Image distance can be calculated as follows:  This shows that image is of the same size. Negative sign shows that image is real and inverted.

Question 4:
An object is at a distance of 10 cm from a (RBSESolutions.com) convex lens. If focal length of lens is 40 cm then find the location and nature of image.
Here, u = -10 cm, f= 40 cm
Image distance can be calculated as follows: Image is enlarged. Positive sign means image is virtual and erect.

Question 5:
Focal length of a concave mirror is 30 cm. If object is at a (RBSESolutions.com) distance of 20 cm then find the location and nature of image.
Here, f= -30 cm, u = -20 cm
Image distance can be calculated as follows:  Question 6:
Image distance from a concave lens is 10 cm. If focal (RBSESolutions.com) length of lens is 15 cm then find object distance.
Here, v = -10 cm, f= -15 cm
Object distance can be calculated as follows: Question 7:
Find the magnification by a convex lens of focal length 10 cm if erect (RBSESolutions.com) image of object is formed at minimum distance of clear vision.
Here, f = 10 cm, v = -25 cm
Object distance can be calculated as follows: I. Multiple Choice Questions

Question 1:
Focal length of (RBSESolutions.com) plane mirror is
(a) at infinity
(b) zero
(c) negative
(d) none of these
(a) at infinity

Question 2:
Image formed by plane mirror is
(a) real and erect
(b) real and inverted
(c) virtual and erect
(d) virtual and inverted
(c) virtual and erect

Question 3:
A concave mirror gives, real, inverted and (RBSESolutions.com) same size image if the object is placed
(a) at F
(b) at infinity
(c) at C
(d) beyond C
(c) at C

Question 4:
Power of a lens is -40, its focal length is
(a) 4 m
(b) – 40 cm
(c) – 0.25 m
(d) – 25 m.
(c) – 0.25 m

Question 5:
A concave mirror gives virtual, erect and (RBSESolutions.com) enlarged image if the object is placed:
(a) at infinity
(b) between F and C
(c) between P and F
(d) at F.
(c) between P and F

Question 6:
The mirror that always gives virtual and erect image of the object but image of smaller size than the size of the object is
(a) Plane mirror
(b) Concave mirror
(c) Convex mirror
(d) none of these
(c) Convex mirror

Question 7:
All the distances in case of spherical (RBSESolutions.com) mirror are measured in relation to
(a) object to image
(b) the pole of the mirror
(c) the focus of the mirror
(d) the image to the object.
(b) the pole of the mirror

Question 8:
The radius of curvature and focal length of a concave mirror are
(a) positive
(b) negative
(c) both
(d) none of these
(b) negative

Question 9:
The object distance in both concave (RBSESolutions.com) as well as convex mirror is
(a) negative
(b) positive
(c) zero
(d) none of these
(a) negative

Question 10:
The ratio of the speed of light in vacuum to that in a medium is known as
(a) magnification
(b) refraction
(c) refractive index
(d) Snell’s law
(c) refractive index

Question 11:
In optics, an object which has higher (RBSESolutions.com) refractive index is called
(a) optically rarer
(b) optically denser
(c) optical density
(d) refractive index
(b) optically denser

Question 12:
Convex lens forms a real, point sized image at focus, the object is placed
(a) at focus
(b) between F and 2F
(c) at infinity
(d) at 2F
(c) at infinity

Question 13:
The unit of power of lens is
(a) metre
(b) centimeter
(c) dioptre
(d) m-1
(c) dioptre

Question 14:
The radius of curvature of a mirror (RBSESolutions.com) is 20 cm the focal length is
(a) 20 cm
(b) 10 cm
(c) 40 cm
(d) 5 cm
(b) 10 cm

Question 15:
The refractive indices of some media are given below:

 Medium Refractive index X 1.51 Y 1.72 Z 1.83 w 2.42

In which of these is the speed of light minimum and maximum, respectively.
(a) X-minimum, W-maximum
(b) Z-minimum, W-maximum
(c) W-minimum, X-maximum
(d) X-minimum, Z-maximum
(c) W-minimum, X-maximum

Question 16:
The power of a lens is + 1.6 D. The (RBSESolutions.com) nature of lens is
(a) Convex lens
(b) Concave lens
(c) both concave and convex
(d) none of these
(a) Convex lens

Question 17:
A mirror that has very wide field view is
(a) concave
(b) convex
(c) plane
(d) none of these
(b) convex

Question 18:
If the object is placed at focus of a concave (RBSESolutions.com) mirror, the image is formed at
(a) infinity
(b) focus
(c) centre of curvature
(d) between F and O.
(a) infinity

Question 19:
The image formed on the retina of human eye is
(a) virtual and erect
(b) realand inverted
(c) virtual an inverted
(d) real and erect
(b) real and inverted

Question 20:
The change in the (RBSESolutions.com) focal length of human eye is caused due to
(a) ciliary muscles
(b) pupil
(c) cornea
(d) iris
(a) ciliary muscles

Question 21:
The least distance of distinct vision for a young adult with normal vision is
(a) 25 m
(b) 20 m
(c) 25 cm
(d) 20 cm
(c) 25 cm

Question 22:
The persistence of vision for normal (RBSESolutions.com) eye is  Question 23:
The phenomenon of light responsible for the working of the human eye is
(a) reflection
(b) refraction
(c) power of accommodation
(d) persistence of vision.
(b) refraction

Question 24:
Which of the following (RBSESolutions.com) colors is least scattered by fog, dust of smoke?
(a) Violet
(b) Blue
(c) Red
(d) Yellow
(c) Red

Question 25:
The colored light that refracts most while passing through a prism is
(a) Yellow
(b) Violet
(c) Blue
(d) Red
(b) Violet

Question 26:
The amount of light entering the human eye is controlled by
(a) Ciliary muscles
(b) Pupil
(c) Cornea
(d) Iris
(b) Pupil

Question 27:
Which part of the eye mostly refracts light entering the (RBSESolutions.com) eye from external objects?
(a) Lens
(b) Cornea
(c) Iris
(d) Pupil
(b) Cornea

Question 28:
The component of white light with greatest wavelength is
(a) Violet
(b) Red
(c) Green
(d) Blue
(b) Red

Question 29:
Long-sightedness (RBSESolutions.com) or hypermetropia can be corrected by
(a) Planar lens
(b) Concave lens
(c) Convex lens
(d) Bifocal lens
(c) Convex lens

Question 30:
A student of class 10, is not able to see clearly the black board question when seated at a distance of 5 m from the board, the defect he is suffering from is
(a) Myopia
(b) Hypermetropia
(c) Presbyopia
(d) Astigmatism
(a) Myopia

Question 31:
The part of eye that determines the colour of (RBSESolutions.com) the eye of a person is
(a) Pupil
(b) Cornea
(c) Retina
(d) Iris
(d) Iris

Question 32:
The glass has greater refractive index for
(a) Violet light
(b) Green light
(c) Blue light
(d) Red light
(a) Violet light

Question 33:
The colour of sky is blue during day time, red during sunset and black at night. This is due to
(a) Scattering of light
(b) Small particles present in atmosphere
(c) Atmospheric refraction
(d) All of the above.
(d) All of the above.

Question 34:
The eye defect (RBSESolutions.com) represented by the figure is (a) Myopia
(b) Hypermetropia
(c) Cataract
(d) Presbyopia
(a) Myopia

### Light Very Short Answer Type Questions

Question 1:
What is light?
Light is a form of electromagnetic radiation that causes (RBSESolutions.com) the sensation of sight. It doesn’t require any material medium to travel.

Question 2:
Name some phenomenon associated with light during image formation by mirrors.
Reflection.

Question 3:
Define reflection of light.
The phenomenon of coming back of light in the (RBSESolutions.com) same medium after striking a plane and polished surface is called reflection of light.

Question 4:
Define incident ray.
Incident ray – Light which falls on the mirror/ polished surface is called incident ray.

Question 5:
Define reflected ray.
Reflected ray – Ray of light which goes back in the same medium after striking the surface is called reflected ray.

Question 6:
Define normal.
Normal – The perpendicular drawn to the (RBSESolutions.com) reflecting surface is called normal at that point.

Question 7:
Define angle of incidence.
Angle of incidence – the angle between the incident ray and the normal is known as angle of incidence.

Question 8:
Define angle of reflection.
Angle of reflection – The angle between reflected ray and the normal is known as angle of reflection.

Question 9:
State laws of reflection.
Incident ray, reflected ray and normal at the (RBSESolutions.com) point of incidence all lie in the same plane. The angle of incidence is equal to the angle of reflection.

Question 10:
What are the properties of image formed by a plane mirror?

• Image is virtual and erect.
• Size of the image is equal to that of object
• Image is laterally inverted.
• The image formed by a plane mirror is

always at the same distance as the object is in front of it.

Question 11:
What are spherical mirrors?
Mirrors whose reflecting surface are part of a (RBSESolutions.com) sphere are called spherical mirrors.

Question 12:
State mirror formula and write it mathematically.
The relation between focal length of mirror, distance of the object and distance of the image is known as mirror formula. It is given by Question 13:
Give the relation between focal length and radius of curvature.
f = $$\frac { R }{ 2 }$$

Question 14:
Define magnification (RBSESolutions.com) of mirror.
The ratio of height of the image to the height of the object is called magnification. It is represented by ‘m’. m = Height of image (h’)/Height of object
(h) = $$\frac { -v }{ u }$$
Magnification of real image is negative and of virtual image is positive.

Question 15:
Define refraction of light.
The change in direction of light, when it travels from one medium to another medium is called refraction of light.

Question 16:
State laws of refraction.
The ratio of sine of angle pf incidence to (RBSESolutions.com) the sine of angle of refraction for a light of given colour and for a given pair of media is constant. This is called Snell’s law.
i.e., $$\frac { sin i }{ sin r }$$ = Constant

•  The incident ray, refracted ray and the normal at the point of incidence lie on the same plane.

Question 17:
What do you observe when light ray passes through rectangular slab?
(a) Angle of incidence is equal to angle of emergence.
(b) Incident ray is parallel to the emergent ray.
(c) Lateral displacement is proportional to the thickness of glass slab.
(d) Lateral displacement is proportional to the angle of incidence.

Question 18:
Define lateral displacement.
Lateral displacement is the perpendicular distance between the incident ray and the emergent ray.

Question 19:
Define refractive index.
Refractive index is defined as the ratio of speed (RBSESolutions.com) of light in medium 1 to the speed of light in medium 2 and is represented as n21 and is read as refractive index of medium 2 with respect to medium 1.
n21 = speed of light in medium 1/speed of light in medium 2.

Question 20:
Define absolute refractive index.
When medium 1 is vacuum, then refractive index of medium 2 is considered with respect to vacuum. This is called absolute refractive index.

Question 21:
What is the unit of refractive index?
It has no unit.

Question 22:
What is the relation between optical density, refractive (RBSESolutions.com) index and speed of light?
The medium with higher refractive index in which speed of light is less is known as optically denser medium and the medium with lower refractive index in which the speed of light is more is known as optically rarer medium.

Question 23:
State lens formula and write it mathematically.
The relationship between object distance (w), image distance (v), and focal length of lens (v) is known as lens formula. It is given $$\frac { 1 }{ v }$$ – $$\frac { 1 }{ u }$$ = $$\frac { 1 }{ f }$$

Question 24:
Define magnification of lens.
Magnification (m)
= $$\frac { Height if image (h)’ }{ Height of object (h) }$$ = $$\frac { v }{ u }$$
For convex lens ‘m’ can be more than, less than or equal to one.
For concave lens ‘m’ is less than one.

Question 25:
What are the two types (RBSESolutions.com) of lenses?
Spherical lens: combination of two spherical refracting surfaces. Question 26:
What is the magnification of a plane mirror?
m = +1

Question 27:
Which lens bends a light ray more the (RBSESolutions.com) one having shorter or longer focal length?
The lens with the shorter focal length bends the light more.

Question 28:
If a convex lens is used to focus sunlight on a paper, where should the paper be placed so that it catches fire.
At the principal focus.

Question 29:
What happens if a light falls on a glass slab making 90° at its surface?
It undergoes normal refraction that is there is no deviation in the light.

Question 30:
Where should be an object placed in (RBSESolutions.com) front of convex lens so as to use it as a magnifier?
Between the pole and the focal length.

Question 31:
Name the light sensitive part of the eye where image of an object is formed.
Retina

Question 32:
What is the function of iris?
Iris controls the size of the pupil.

Question 33:
Why is inverted image formed on the (RBSESolutions.com) retina of human eye?
The inverted image is formed due to the eye lens which is convex in shape. Through which the incident light rays form the real and inverted image.

Question 34:
What type of signals are generated and sent to the brain by light sensitive cells of retina?
Electrical signals.

Question 35:
What is the function of crystalline lens of human eye?
The crystalline lens provides the proper focal length so that image of objects at different distances formed on the retina.

Question 36:
What holds the crystalline lens in the (RBSESolutions.com) human eye?
Ciliary muscles.

Question 37:
Name the disease in which crystalline lens of human eye becomes opaque.
Cataract.

Question 38:
Define the power of accommodation of human eye.
The ability of eye to see nearby as well as far off objects at the same time is called power of accommodation.

Question 39:
In which type of eye defect far point of (RBSESolutions.com) the eye gets reduced?
Myopia.

Question 40:
In which type of eye defect near point of the eye becomes more than 25 cm?
Hypermetrgpia.

Question 41:
What is dispersion of light?
The splitting of light into its various components (i.e., 7 colours) is called dispersion of light.

Question 42:
What is spectrum?
The band of seven colours obtained due to the (RBSESolutions.com) dispersion of white light is called spectrum.

Question 43:
Give one main difference between the lens of human eye and lens of camera.
Lens of human eye has flexible aperture, its focal length can be changed. In camera focal length can not be changed for a lens.

### Light Short Answer Type Questions

Question 1:
Define pole, centre of curvature, radius of curvature, principal axis focus and focal length of a spherical mirror.
Pole: The centre of reflecting surface. It is represented by letter P.
Centre of Curvature: The centre of the sphere of which the mirror forms the part. Represented by “C”.
Radius of Curvature: The radius of the sphere of which the mirror forms the part. Represented by “R”.
Principal axis: The straight line joining the pole (P) and the centre of curvature. It is normal to the mirror at its pole.
Focus: The point of the principal axis at which the rays parallel to principal axis meet (concave mirror) or appear to meet (convex mirror) after reflection. Represented by F.
Focal Length: The distance between the pole and the principal focus of a spherical mirror is called focal length. Represented by f.

Question 2:
Define centre of curvature, principal axis, optical (RBSESolutions.com) centre, aperture, focus and focal length for a lens.
(a) Centre of curvature: It is the centre of the spheres of which the each surface of the lens forms a part. Represented bv C or 2f.
(b) Principal axis: An imaginary straight line passing through the two centres of curvatures.
(c) Optical centre: It is the central point of the lens. Represented by O.
(d) Focus: The point at which rays of light parallel to principal axis converges (convex lens) or appears to diverge (concave lens) after refraction. Represented by F.
(e) Focal length: The distance between focus and optical centre is called focal length. It is represented by f.

Question 3:
Write nature, position and relative size of image formed by convex lens.  Question 4:
Give the sign conventions for (RBSESolutions.com) spherical mirrors.

 S. No. Various distances Concave mirror Convex mirror 1. Object distance ‘u’ -ve -ve 2. Image distance V +ve if behind the mirror, -ve if in front of the mirror always +ve 3. Focal length -ve +ve 4. Height of virtual image +ve +ve 5. Height of real image -ve -ve

Question 5:
Write nature, position and relative size of image formed by cancave lens.

 Position of the object Position of the image Relative size of the image Nature of the At infinity At focus F-, Highly diminished, point-sized Virtual and erect Between infinity and optical centre O Between focus Fx and optical O Diminished Virtual and erect

Question 6:
Give sign conventions for (RBSESolutions.com) spherical lenses.

 S. No. Various distances Convex lens Concave lens 1. Object distance (u) -ve -ve 2. Image (v) +ve real, -ve if in front of the mirror -ve 3. Focal length (/) +ve -ve 4. Height of the object (h) +ve +ve 5 Height of the image (/;’) -ve for real +ve for virtual +ve

Question 7:
Two medium with refractive index 1.31 and 1.50 are given. In which case
(i) bending of light is more?
(ii) speed of light is more?
(i) Bending of light is more in the medium where refractive index is 1.50.
(ii) Speed of light is more in the medium with refractive index 1.31

Question 8:
Refractive index of kerosene oil is 1.44 and that of (RBSESolutions.com) water is 1.33. A ray of light enters from kerosene oil to water. Where would light ray bend and why?
A ray of light enters from kerosene oil to water i.e., refractive index 1.44 to 1.33 i.e., from denser to rarer medium. Hence the ray of light bends away from the normal.

Question 9:
Which is optically denser out of the two medium M1= 1.71 (refractive index) and M2 = 1.36 (refractive index). How does speed of light change when it travels from optically rarer to denser medium.
Medium M1 with refractive index 1.71 is optically denser than the other medium M2. Speed of light decreases when it travels from rarer to denser medium.

Question 10:
Comment on the size, position of the image (RBSESolutions.com) formed by a concave mirror of focal length
18 cm when an object is placed:
(i) at 22 cm
(ii) 14 cm
(iii)40 cm.
in front of mirror without calculations.
(i) When the object is placed at 22 cm, the image is formed beyond 36 cm, real, inverted image is magnified,
(ii) When the object is at 14 cm then the image formed is virtual, behind the mirror and magnified.
(iii) When the object is placed at 40 cm, then the image is formed between 18 cm and 36 cm, it is real inverted and diminished image.

Question 11:
Complete the following ray diagrams:  Question 12:
With the help of a ray diagram show how a pencil (RBSESolutions.com) appears when dipped in water. A ray of light from pencil in water travels from denser to rarer medium i.e., from water to air, it bends away from the normal, hence the pencil appears to be bent in water as shown in the diagram.

Question 13:
Define power of lens. What is the S. I. unit of power of a lens? If power (RBSESolutions.com) of lens is +2D what is the nature and focal length of the lens?
Power of lens: The degree of convergence or divergence of light rays obtained by a lens is expressed in terms of its power.
Power of a lens is defined as the reciprocal of its focal length.
p = $$\frac { 1 }{ f }$$
S.I. unit of power of a lens is ‘dioptre’
If the power of lens, P = +2 D.
Lens is convex and the focal length of the lens is +0.50 m.
( p = $$\frac { 1 }{ f }$$ 2D = $$\frac { 1 }{ F }$$ ∴ f = $$\frac { 1 }{ 2 }$$ = 0.5)

Question 14:
If the speed of light in water is 2.25 x 108 m/s and the speed in vacuum is 3 x 108 m/s. Calculate the refractive index of water. Question 15:
The refractive index of water is 1.33 and kerosene is 1.44. Calculate the (RBSESolutions.com) refractive index of kerosene with respect to water.
Refractive index of water = nw = 1.33
Refractive index of kerosene = nk = 1.44
∴ Refractive index of kerosene with respect to water is
nw nk = $$$\frac { { n }_{ k } }{ { n }_{ w } }$$$ = $$\frac { 1.44 }{ 1.33 }$$
= 1.082

Question 16:
Draw ray diagrams to show the image formed by a concave lens for the object placed at
(i) infinity
(ii) Between f and 2f of the lens.
(i) object at infinity (ii) object between f and 2f Question 17:
Draw a ray diagram to show the path of (RBSESolutions.com) light when it travels through glass slab. Incident my I enters the glass slab forms an angle of incidence . Its bends towards the normal afid forms an angle of refraction ‘r’.
The emergent ray is parallel to the incident ray.

Question 18:
Draw the following diagram in your answer book and show the formation of image of the object AB with the help of suitable rays.  Question 19:
If a light ray IM is incident on the surface AB as shown, identify the (RBSESolutions.com) correct emergent ray.
NQ will be the emergent ray as it has to be parallel to OS. Question 20:
An object of 2 cm high is placed at a distance of 64 cm from a white screen, (RBSESolutions.com) on placing a convex lens at a distance of 32 cm from the object it is found that a distant image of the object is formed on the screen. What is the focal length of the convex lens and size of the image formed on the screen? Draw a ray diagram to show the formation of the image in this position of the object with respect to the lens.
Since the object-screen distance is double of object-lens separation, the object is at a distance of 2f from lens and the image should be of the same size of the object. So 2f = 32 ⇒ f = 16 cm
Height of image = Height of object = 2 cm

Question 21:
Redraw the given diagram and show (RBSESolutions.com) the path of refracted ray. Question 22:
A convex lens has a focal length of 10 cm. At what distance from (RBSESolutions.com) the lens should the object be placed so that it gives a real and inverted image 20 cm away from the lens? What would be the size of the image formed if the object is 2 cm high? With the help of a ray diagram show the formation of the image by the lens in this case.
f = +10 cm, v = +20 cm as image is real and inverted. Height of the object = 2 cm (say +ve) Image will be of the same size as that of object (as u = v) and hence, the height of the image Question 23:
Redraw the given diagram and show the path (RBSESolutions.com) of the refracted ray.  Question 24:
Why does a ray of light bend when it travels from (RBSESolutions.com) one medium into another?
Due to change in velocity in the medium and to reduce the time taken to travel the same, a ray of light bends when it travels from one medium to another.

Question 25:
What are the minimum number of rays required for locating the image formed by a concave mirror for an object? Draw a ray diagram to show the formation of virtual image by a concave mirror. Question 26:
Redraw this diagram on to your answer book and (RBSESolutions.com) complete the path of the ray  Question 27:
How does eye control the amount of light entering it?
The amount of light entering the eye is controlled by (RBSESolutions.com) the pupil and further the size of the pupil is controlled by iris.

Question 28:
Why danger signals are red?
Danger signals are of red colour, as it scatters the least and can be seen from the maximum distance.

Question 29:
Why do you take time to see objects when you enter a dim lighted room from outside in the sun?
In the sunlight the size of pupil, is small but when one enters the dim light, it takes some time for iris to adjust the size of pupil and the light sensitive cells take some time to get activated.

Question 30:
Why are two eyes more helpful for us to (RBSESolutions.com) see as compared to one?
Two eyes are more helpful as one eye gives only a view of 150° angle whereas two eyes increase the view by making it wide to 180° angle. Two eyes also helps us to see the objects in dim light or darkness clearly. Two eyes give stereoscopic vision helping us assess the depth of vision.

Question 31:
When white light enters the prism, which colour of light deviates/bends the least and which colour bends the most?
The light that bends the least is red colour and the light that bends the maximum is violet colour light.

Question 32:
Explain the phenomenon which causes (RBSESolutions.com) twinkling of stars.
The phenomenon is atmospheric refraction. In this case the star are point source illuminated objects which are very far from us when light travels through atmosphere it bends, and due to this the amount of light entering the eye is different each time which gives the twinkling effect.

Question 33:
Why does a ray of light splits into different colours on passing through a glass prism?
When light rays enter the glass prism the angle at which it bends makes the light split into its seven components because the speed of each component of light is different and due to the bending every component shows its different ability to pass through it.

Question 34:
The sun appears to be red at the time of sunset and sunrise. Give the reason.
Sun appears red during sunset or sunrise because at this time the sun is far from the earth and the light that reaches the earth from the sun scatters the most and all other colours of light gets scattered. The least scattered light is red and it enters our eye.

Question 35:
Give reason for early sunrise and (RBSESolutions.com) delayed sunset. Sun being far off, the light rays from sun entering our eye after refraction several times due to the. Hence apparent position of sun is somewhat higher than its actual position in sky.

Question 36:
What is the direction of rainbow formation? What is the position of red colour in rainbow?
Rainbow is always formed in the direction opposite to sun. The position of red colour in the rainbow is at the top.

Question 37:
What is internal (RBSESolutions.com) reflection?
When a light rays enters from one medium to another (e.g., rarer to denser i.e., air to water droplet) then a ray of light instead of passing through it reflects in the second medium then it is said to be internal reflection of light. Question 38:
A short-sighted person cannot see clearly beyond 5m. Calculate the power of lens required to correct his vision to normal?
f = – 5m
P= $$\frac { 1 }{ f }$$ = –$$\frac { 1 }{ 5 }$$ = 0.20 =0.20
∴ Power = 0.2 Dioptre

Question 39:
Why can’t we see object very close (RBSESolutions.com) to our eye?
The objects are seen only when the image forms on retina when the light rays pass through the lens. The lens has its fixed ability of changing the focal length with the help of ciliary muscles. Ciliary muscles cannot be contracted beyond a certain limit to change the focal length of eye lens. The objects kept very close to our eye cannot be focused by ciliary muscles.

Question 40:
Why does the sky appears blue during day time, red during sunrise and sunset and black to an astronaut.
Sky appears blue during day time because the light of sun gets scattered and the most scattered light is blue, so the sky appears blue.
During evening and early morning when the sun is not over head but it is below the horizon, the only light that reaches our eye is red and hence the sky appear to be reddish in colour. For an astronaut the sky appears to be black because there is no atmosphere that can refract the light.

Question 41:
Give the difference between myopia (RBSESolutions.com) and hypermetropia.

 S.No. Myopia Hypermetropia 1. Short – sightedness – can see nearby object but cannot see far off objects. Long-sightedness – can see far off objects but cannot see nearby objects. 2. Image is formed in front of retina. Image is formed beyond retina. 3. The size of eyeball increases. The size of eyeball decreases. 4. Focal length of eye lens decreases. Focal length of eye lens increases. 5. Corrected by using concave lens. Corrected by using convex lens.

Question 42:
Distinguish between presbyopia and (RBSESolutions.com) hypermetropia.

 S.No. Hypermetropia Presbyopia 1. Only far-sightedness. It can be only far-sightedness or both far and short-sightedness. 2. Eye ball becomes short or the focal length increases. Ciliary muscles become weak and able to adjust the focal length. 3. Corrected by using convex lens. Corrected by using bifocal lens.

Question 43:
The near point of hypermetropic eye is 80 cm. What is the nature and power of the lens required to enable him to read a book placed at 25 cm from the eyes?
Near point = 80 cm
Object distance u = – 25 cm Question 44:
What is meant by dispersion of white light? Draw a ray diagram to show (RBSESolutions.com) the dispersion of white light by a glass prism. Give reason why do we get different colours of light?
Dispersion of light: The splitting of white light into seven colours on passing through a transparent medium like glass prism is called dispersion of light. We get different colours because each colour of light has different bending ability when they pass through the glass prism.

Question 45:
When we see any object through the hot air (RBSESolutions.com) over the fire, it appears to be wavy, moving slightly. Explain.
The objects beyond the hot air appears to be wavy because the medium for light to pass through changes, the light passes from denser to rarer and then again to denser medium thereby causing refraction in the air. Moreover the refractive index of the hot air keeps changing which leads to give the wavy appearance of the object.

Question 46:
Study the diagram given below and answer the questions that it follows: (a) Name the defect and give reason.
(b) Give 2 causes for this defect.
(c) Give the correction – draw diagram for the same.
(a) The defect is myopia, short-sightedness
(b) It is caused due to the decrease in the focal length of the eye lens and increase in the size of the eye ball.
(c) The defect can be corrected by using the concave lens. Question 47:
In the given diagram label A, B, C and D and (RBSESolutions.com) give the function of B and D. A = Cornea
B = Ciliary muscles
C = Retina
D = Optic nerve Function of B and D are:
B : Ciliary muscles: It helps in holding the eye lens and changing or adjusting the focal length of the lens.
D : Optic nerve: It sends the electrical signal from retina to the brain.

Question 48:
Near point of a hypermetropic eye is at 1 m. Find the (RBSESolutions.com) focal length, power and nature of lens used to correct this defect.
Near point of hypermetropic eye is 1 m i.e., 100 cm. The eye cannot see objects between 100 cm and 25 cm. Convex lens of power 3D is used to correct this defect.

Question 49:
Draw a labelled diagram of rainbow formation. Also (RBSESolutions.com) explain the phenomenon of rainbow formation. When sun light splits due to water drops suspended in air, causing the band of seven colours is called rainbow.
Water droplets acts as tiny prism in the sky. The sunlight when enters these tiny droplets undergo internal reflection and also refract these rays which are dispersed causing a band of seven colours called rainbow.
Rainbow is always formed in the direction opposite to the sun.

Question 50:
(a) Draw a diagram to show the formation of image of a distant (RBSESolutions.com) object by a myopic eye. How can such an eye defect be remedied?
(b) State two reasons due to which this eye defect may be caused.
(c) A person with a myopic eye cannot see objects beyond a distance of 1.5 m. What would be the power of the corrective lens used to restore proper vision?
(a) Object at infinity, image is formed in front of retina. Corrected by concave lens (b) Myopia is caused due to:
(i) Elongation of eye ball
(ii) Excessive curvature in cornea, focal length decreases.
(c) Far point of myopic eye is 1.5 m = v
to change far point to infinity = u
focal length of power → P = ?, F = ? Question 51:
Study the diagram given below and answer (RBSESolutions.com) the question that it follows:
(a) Which defect of vision is represented in this case? Give reason for your answer.
(b) What could be the two causes of this defect?
(c) With the help of a diagram show how this defect can be corrected by the use of a suitable lens. (a) The defect is hypermetropia, as the image of near point is formed beyond retina.
(b) Two causes of the defect are:
(i) Size of eye ball decreases.
(ii) Focal length of the lens increases.
(c) This defect can be corrected by using a convex lens of suitable focal length. Question 52:
(a) What is dispersion of white light? What is the (RBSESolutions.com) cause of such dispersion? Draw a diagram to show the dispersion of white light by a glass prism.
(b) A glass prism is able to produce a spectrum when white light passes through it but a glass slab does not produce anv spectrum. Explain why.
(a) Dispersion of white light is splitting of light into its seven constituent colours forming a band of VIBGYOR called spectrum.
Cause: White light is made up of seven colours, each colour has different speed in different media. Due to different speed, the bending ability varies and the colours split/separate. (b) Dispersion does not take place in glass slab as two refracting surfaces are parallel. The light does not split into its constituent colours.

### Light Long Answer Type Questions

Question 1:
(a) It is desired to obtain an erect image of an (RBSESolutions.com) object using a concave mirror of focal length 20 cm.
(i) What should be the range of distance of the object from the mirror?
(ii) Will the image be bigger or smaller than the object?
(iii) Draw a ray diagram to show the image formation in this case.

(b) One-half of a convex lens of focal length 20 cm is covered with a black paper.
(i) Will the lens produce a complete image of the object?
(ii) Show the formation of image of an object placed at 2Fa of such covered lens with the help of a ray diagram.
(iii) How will the intensity of the image formed by half covered lens compare with non-covered lens?
(a) (i) Range of the object distance is 0 to 20 cm from the pole.
(ii) Image will be bigger than the object.
(iii) Ray diagram: (b) (i) Yes, complete image (RBSESolutions.com) will be formed. (iii) Intensity will be reduced as the light falling on the lower (covered) portion will not reach the position of image.

Question 2:
Name the type of mirror used in the (RBSESolutions.com) following situations:
(i) Rear view mirror in vehicles
(ii) Solar furnace
(iii) Torch
(iv) Solar cooker
(v) To get the full length image of tall building.
(i) Rear view mirror in vehicles – convex mirror as it gives virtual image, diminished and cover the wider view.
(ii) Solar furnace – concave mirror to concentrate all parallel beam of light.
(iii) Torch – concave mirror is used to reflect light rays as parallel beam.
(iv) Solar cooker – concave mirror is used to concentrate the heat rays at a point.
(v) Convex mirror is used to view a full length tall building.

Question 3:
A convex lens has a focal length of 15 cm. At what (RBSESolutions.com) distance from the lens shoiild the object be placed so that it forms on its other side a real and inverted image 30 cm away from the lens? What would be the size of image formed if the object is 5 cm high? With the help of a ray diagram show the formation of the image by the lens in this case.
f = 15 cm
v = + 30 cm as image is real and inverted  Question 4:
Draw a neat labelled diagram of human eye and (RBSESolutions.com) explain the working of each part of it. Working of Human eye
Human eye consists of various parts which helps us in seeing the objects, the function of various parts are:

(a) Cornea: It is the transparent membrane which refracts the light entering our eye.
(b) Iris: Iris controls the size of pupil.
(c) Pupil: It allows the light entering our eye to pass through it.
(d) Lens: Adjusts the focal length of the eye to see the objects at different places.
(e) Ciliary muscles: Helps in changing the focal length of the lens.
(f) Retina: It is the screen of the eye on which image is formed. It consists of rods and cones,
(g) Optic nerve: It carries the electrical signals from retina to brain.

Question 5:
Describe with the help of diagram, how the (RBSESolutions.com) refraction of light takes place through a glass prism. The incident ray I when enters the prism it gets refracted, bends and form ∠r with the normal. Angle of refraction is smaller than the angle of incidence. The incident ray bends towards the normal, as it passes from rarer medium to denser medium. When this refracted ray passes from denser medium to rarer medium it bends away from the normal. This emergent ray has bent out at an angle to the direction of incident ray. This angle is called the angle of deviation ∠D.

Question 6:
Name three refractive defects of vision with the (RBSESolutions.com) help of diagram. Explain the reasons and correction of these defects.
The three refractive defects of vision are
(i) Myopia
(ii) Hypermetropia
(iii) Presbyopia

(i) Myopia is short-sightedness, the image is formed in front of retina due to the elongation of the eye ball or due to decrease of focal length.
Correction – Using concave lens. After using concave lens, affected eye can see distant objects

(ii) Hypermetropia is long-sightedness, the image is (RBSESolutions.com) formed behind the retina due to shortening of eyeball or due to increase in the focal length of the lens of eye.
Correction: Using convex lens. (c) After using convex lens, the affected eye can see nearby objects

(iii) Presbyopia: It is the defect of an eye in (RBSESolutions.com) which the power of accommodation of the eye usually decreases with ageing. Near point changes as well as the far off objects are also not visible clearly. It is caused due to the weakening of ciliary muscles and the reduced flexibility of eye lens.
Such a defect in which a person suffers from both myopia and hypermetropia is called presbyopia. It is corrected by using bifocal lens.

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