Question Light

 

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.