A converging lens has a focal length of 100 cm. The power of this lens is

+ 0.1 D

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-0.1 D

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+ 1.0 D

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- 1.0 D

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Solution

The correct option is B + 1.0 D
the focal length of converging lens is positive
then power is positive
$f = 100 c m = 1 m$
$P = + \frac{1}{f}$
$\Rightarrow P = \frac{1}{1} = + 1 D$

Option C is correct.

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Similar questions

A converging lens has a focal length of 50 cm. The power of this lens is : (a) + 0.2 D (b) - 2.0 D (c) + 2.0 D (d) - 0.2 D

+ 2.0 D

- 1.0 D

A diverging lens has a focal length of 0.10 m. The power of this lens will be : (a) + 10.0 D (b) + 1.0 D (c) - 1.0 D (d) - 10.0 D

The lens of power + 1.0 D is :

(a) convex of focal length 1.0 cm

(b) convex of focal length 1.0 m

(d) concave of focal length 1.0 m

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