Q5. At what distance should an object be placed from a convex lens of focal length 18 cm to
obtain an image at 24 cm from it on the other side. What will be the magnification produced
in this case?
Each question carries 5 marks
Q6. Draw the ray diagram in each case to show the position and nature of the image formed
when the object is placed:
(1) at the centre of curvature of a concave mirror
(ii) between the pole P and focus Fof a concave mirror
(iii) in front of a convex mirror
(iv) at 2F of a convex lens
(v) in front of a concave lens
Q7. a) What is meant by power of a lens'? Dlaptor
(b) State and define the S.I. unit of power of a lens.
(c) A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in
close contact with each other. Calculate the lens power of this combination
Q8. List the sign conventions for reflection of light by spherical mirrors. Draw a diagram and
apply these conventions in the determination of focal length of a spherical mirror which
forms three times magnified real image of an object placed 16 cm in front of it.
pls help
Answers
Q5.answer:- m=24/-72 = - 1/3
Q6.answer:- The Ray diagram of case the object is at the centre of curvature of concave mirror is in the fig 1.
And the nature of the image is-
same size, inverted and real.
The Ray diagram of case the object is between pole and focus of concave mirror is in fig 2
And the nature of the image is
large,virtual, erect and behind the mirror
The Ray diagram of case the object is infront of convex mirror is in fig 3.
And the nature of the image is
small in size,virtual,erect
The Ray diagram of case the object is at 2f of convex lens is in fig 4.
And the nature of the image is
inverted, same size and real
The Ray diagram of case the object is in front of concave lens is in fig 5.
And the nature of the image is
real,small in size,erect.
Q7.answer:-Power of lens is defined as the reciprocal of the focal length of the lens expressed in meter.
SI unit of power is Dioptre denoted by D.
Focal length of first lens f
1
=40 cm=0.4 m
Since the focal length is positive, so it is a convex lens.
Power of lens P
1
=
f
1
1
=
0.4
1
=2.5 D
Focal length of second lens f
2
=−20 cm=−0.2 m
Since the focal length is negative, so it is a concave lens.
Power of lens P
2
=
f
2
1
=
−0.2
1
=−5 D