Soln. (a) (i) Let a parallel beam of light incident first on convex lens, refraction at convex lens

1

v1

−

1

u1

=

1

f1

⇒

1

v1

−

1

∞

=

1

30

v1 = 30 cm
So, a virtual object for the concave lens at +22 cm.
Now refraction at concave lens

1

v2

−

1

u2

=

1

f2

or

1

v2

−

1

22

=

1

−20

1

v2

= -

1

20

+

1

22

= -

1

220

or v2 = - 220 cm

The parallel beam of light appears to diverge from a point 216 cm from the center of the two lens system.
(ii) Now let a parallel beam of light incident first on concave lens.

Refraction at concave lens

1

v1

−

1

u1

=

1

f1

or

1

v1

−

1

∞

=

1

−20

or v1 = - 20 cm
The image I1 will act as real object for convex lens at 28 cm.

1

v2

−

1

u2

= 1f2 or

1

v2

−

1

−28

=

1

30

1

v2

=

1

30

−

1

28

= -

1

420

or v2 = - 420 cm
Thus the parallel incident beam appears to diverge from a point 420 – 4 = 416 cm on the left of the center of the two lens system.
Hence the answer depend upon which side of the lens system the parallel beam is made incident.
Therefore the effective focal length is different in two situations.
(b) Now an object of 1.5 cm size is kept 40 cm in front of convex lens in the same system of lenses.

Refraction by the first lens

1

v1

−

1

u1

=

1

f1

or

1

v1

−

1

−40

=

1

30

1

v1

=

1

30

−

1

40

=

1

120

or v1 = 120 cm
Magnification by first lens
m1 =

I

O

=

v

u

or m1 =

I1

1.5

=

120

−40

m1 = - 3 and I1 = - 4.5 cm
Refraction by the second lens

1

v2

−

1

u2

=

1

f2

or

1

v2

−

1

+112

=

1

−20

1

v2

= -

1

20

+

1

112

or v2 = -

112×20

92

cm
Magnification by concave lens
m2 =

v2

u2

=

−112×20

+122(92)

= -

20

92

Total magnification by two lenses
m = m1×m2 or m = - 3 × (−

20

92

) = 0.652
Size of image finally obtained
m =

I2

O1

or 0.652 =

I2

1.5

I2 = 1.5 × 0.652 = 0.98 cm