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A lens forms a virtual and diminished image of an object placed $60 cm$ from it. The size of the image is one-third of the object. What is the focal length and power.?

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f = - 30 cm

, and

P = - 3.33 D

f = - 30 cm

, and

P = 4.0 D

f = + 40 cm

, and

P = 5.0 D

None of the above

answer is A.

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Detailed Solution

A lens forms a virtual and diminished image of an object placed

60 cm

from it. The size of the image is one-third of the object.The Focal length of the lens,

f = - 30 cm

, Power of the lens,

P = - 3.33 D

Given,
Distance of the object from the lens,

u = - 60 cm

For virtual image

h ’ = \frac{h}{3}

\Rightarrow

m = \frac{h ’}{h}

= \frac{v}{u}

\Rightarrow v = \frac{h ’ \times u}{h}

\Rightarrow

v = \frac{\frac{h}{3}}{h} \times - 60

\Rightarrow

v = - 20 cm

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

= \frac{1}{- 20} - \frac{1}{- 60}

= - \frac{1}{20} + \frac{1}{60}

= \frac{- 3 + 1}{60}

\Rightarrow \frac{1}{f} = - \frac{2}{60}

\Rightarrow

f = - 30

= - 0.30 m

.
Power of the lens,

P = \frac{1}{f (in m)}

= - \frac{1}{0.30}

\Rightarrow

P = - 3.33 D

Thus, the lens is concave.

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