light-notes

Light Class 10 | SEE Physics Notes

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Light Class 10 | SEE Notes

Light: The light is the form of energy which produces the sensation of vision to our eyes.

  • The light is invisible itself but it makes object visible.
  • Its electromagnetic wave of velocity is 3×108m/s in vacuum or air.
  • The study of light is optics.

Lens: The lens is the medium of refraction of light.

  • It is made up of plastic or glass.
  • It has two types:
    a) Convex lens/ Converging lens
    b) Concave lens/ Diverging lens, etc.

Differentiate between on Convex lens and converging lens: 

Convex lens Concave lens
The lens which converses parallel light at a point(focus) is called the Convex lens. The lens which diverse (distributes) parallel light at a point(focus) is called the Concave lens.
Diagram:
lite-physics
Diagram:
see-note-light
Its focal length(f) is positive. Its focal length(f) is negative.
The power of the convex lens is positive. The power of the concave lens is negative.
The power of the lens is reciprocal to focal length (p= 1/f).
It is used to correct Hypermetropia. It is used to correct Myopia.
Generally, a Virtual and erect image is formed. Generally, a real and inverted image is formed.

Read:

Image formation:

Lens construction: 

  • Base Prism:
    light-note-see
  • Principle axis:
    physics-light
  • Concave lens: 
    light-class-10
  • Principle Axis Convex lens:
    SEE-science
  • Principle Axis Concave lens:
    class-10-light

Image formation:

light-class-1-physics-notes

Magnification (M): Magnification is a factor which determines image is how many times larger or smaller than an object.

  • m= size of image/size of object
    = Image distance (v)/ Object distance(0)
  • Diagram: 
    light-notes

Power of lens: The conversing or diverging capacity of a lens is called Power of lens.

  • Power of lens (P)= 1/ focal length (m)
  • Its unit is Diopter (D)
  • The power of the concave lens is negative because its focal length is negative.

Deft of vision: The problem in eyes due to change in focal length in which near or distant objects are not seen clearly is called Defect of vision. It has two types-

  • Myopia (Short-sightedness)
  • Hypermetropia (Long-sightedness), etc.
Myopia (Short-sightedness) Hypermetropia (Long-sightedness)
The deft of vision in which near object is seen clearly is called Myopia (Short-sightedness). The deft of vision in which distant object is seen clearly is called Hypermetropia (Long-sightedness).
Diagram: myopia-notes Diagram:hypermetropia-notes
The distant object is not seen because its images are formed before retina. The image of the near object is seen clearly because its image is formed at the retina.
Diagram:complete-myopia-notes Diagram:hypermetropia
Causes of Myopia:
a) Elongation of the eyeball.
b) The decrease in focal length of the eye lens, etc.
Causes of Hypermetropia:
a) Shorting of an eyeball.
b) Increase in focal length of the eye lens, etc.
Correction/Remedy: It is corrected by Concave lens because it concentrates coming from a distant object at near point (N). The image of a distant object is seen clearly because its image is formed at the retina. The near object is not seen because its image is formed after retina.COrrection/Remedy: It is corrected by convex lens because it makes the object at infinity.
Diagram:distinguish-between-on-myopia-and-hypermetropia Diagram:

The relationship between focal length, object and image distance: 

Relationship-between-focal-length-object-and-image-distance

Relationship-between-focal-length-object-and-image-distance

Where,
u= Object distance
v= Image distance
f= Focal distance

Answer these following questions

  1. What is the lens? Mention its types.
    Ans: The lens is the medium of refraction of light.
    Its types are:
    a) Convex lens
    b) Cocave lens, etc.
  2. Define the following:
    a) Centre of curvature: The center of the sphere of which the lens part is called the center of curvature.b) Principal axis: An imaginary line which joins two centers of curvatures (C1 and C2) is called Principal axis.c) Optical center: The center of a lens which is equidistant from all points of its surface is called Optical center.d) Principal focus: Principal focus is a point on the principal axis where the rays of light parallel to the principal axis coverage after refraction through a convex lens or appear to diverge after refraction through a concave lens.e) Focal length: The distance between the optical center and the principal focus is called Focal length.
  3. What is magnification? Mention the formula for finding the magnification of an object.
    Ans: Magnification is a factor which determines image is how many times larger or smaller than an object.
    The formula for finding the magnification of an object
    Magnification (m)= size of image/size of an object
    = Image distance (v)/ Object distance(0)
  4. A convex lens forms a real and magnified image where is the object placed?
    Ans: A convex lens forms a real and magnified image when it is the object placed between 2F and F.
  5. What is the power of the lens? Mention its formula.
    Ans: The conversing or diverging capacity of a lens is called Power of lens.
    Power of lens (P)= 1/ focal length (m)
  6. Mention the lens formula.
    Ans: Power of lens (P)= 1/ focal length (m)
  7. Mention the uses of the lens.
    Ans: The uses of the lens are:
    a) It is used in a camera.
    b) It is used in remedy defect of eyesight, etc.
  8. What are optical instruments?
    Ans: The instruments which produce the image of an object are called optical instruments.
  9. How does pupil control the amount of light entering the eyes?
    Ans: Pupil controls the amount of light entering the eye by changing the size of a pupil by iris.
  10. What is accommodation?
    Ans: The ability of the eye to focus the image of an object on the retina by changing the focal length of its lens is called Accommodation.
  11. What is Near point?
    Ans: The nearest point up to which a normal eye can see clearly is called Near point.
  12. Define Far point.
    Ans: The farthest point up to which a normal eye can see clearly is called Far point.
  13. What is Range of vision?
    Ans: The distance between near point and far point is called Range of vision.
  14. Define Defect of vision.
    Ans: When an eye cannot see the object lying at the range of vision is called Defect of vision.
  15. Ram cannot see the letter written on the blackboard. What is his defect of vision? How can it be corrected?
    Ans: Ram cannot see the letter written on the blackboard because he is suffering from Myopia and he needs to wear goggles rear with the concave lens.

Give reasons:

  1. Shortsightedness can be removed by using the concave lens.
    Ans: Shortsightedness can be removed by using the concave lens because it concentrates light coming from the distant object of near point (N).
  2. A concave lens is called a diverging lens.
    Ans: A concave lens is called a diverging lens because of it diverse parallel light from a point.
  3. A convex lens is called a converging lens.
    Ans: A convex lens is called a converging lens because it converses parallel light at a point (focus).
  4. Power of a convex lens is positive.
    Ans: Power of a convex lens is positive because the real distance between the optical center and the real image is taken as positive.

Numerical Problems

a) Find the power of a lens of focal length 10 cm.

Given data,
f= 10 cm
= 10/100 cm

we know that,
p=1/f(m)

=1/10/100

= 100/10

= 10 D

∴ Power of focal length is 10 D.

b) What is the focal length of a lens having the power 5D?

Given data,
p= 5D
f=?

we know that,

p=1/f(m)

f(m)= 1/p = 1/5 -= 0.2 m.

∴ The focal length of a lens is 0.2m.

c) An object is placed at a distance of 60cm in front of a convex lens of a focal length 20cm. where is the image formed? Mention magnification of the image and power of the lens.

Given data,
u=60cm
f=20cm
v=?
m=?
p=?

we know that,
1/f = 1/u + 1/v
1/20 = 1/60 + 1/v
1/20 + 1/60 = 1/v
3+1/60 = 1/v
or, 1/v= 4/60 =1/15
∴v= 15 cm

Now,
Magnification (m)= Image distance(v) / Object distance(u)
=15/60
=1/4
=0.25

Again, p=1/f(m)
=1/20
=1/20/100
=100/20
=5D

∴The magnification of the image is o.25 and power lens is 5D.

d) An object is placed at the distance of 40 cm from the concave lens of focal length 20cm. Find the image distance.

Given data,
u= -40
f= -20cm

we know that,
1/f= 1/u+1/v
1/-20= 1/-40 = 1/v
1-1/40= 1/v
-1/40= 1/v
-v=40
∴v= -40

∴The image distance is -40cm.

e) A student wears spectacles of power -2d. What types of defect of the eye does the student have? Find the focal length of the lens used?

Given data,
p=-2D
f=?

we know that,
p= 1/f(m)
or, -2= 1/f(m)
or, f(m)= 1/-2
∴f(m)= -50cm

∴The focal length of the lens used is -50cm.