Question

Two lenses of power 4 D and – 6 D are placed
in contact to form a composite lens. An object
is placed at a distance 50 cm from this
combination of lens, find the position of the
image.
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Answer 1

Explanation:

Power P

1

=+3D and P

2

=−5D

Position of object u=−50cm

Power gets added

P=P

1

+P

2

P=3−5=−2

f=

P

1

=

−2

1

=−50cm

Apply lens formula

f

1

=

v

1

−

u

1

v=

f+u

uf

=

(−50)+(−50)

(−50)×(−50)

=+25cm

Position of image is v=+25cm

Answer 2

Answer:

Since two lenses are added, the power also becomes added.

Hence, effective power is 4+(-6) = -2D

Now position of object, u = -50cm

f = 1/P = -1/2 = -0.5m = -50 cm

Hence, by applying lens formula, 1/f = 1/v - 1/u, we get v = uf/(u+f)

So, final position of image is -50×(-50)/[(-50+(-50)] = -25cm.

Negative sign indicates that image is formed on the same side as of the object.