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Question: An object is placed 20 cm from the lens. It has a virtual image at 10 cm from the object. What is th...

An object is placed 20 cm from the lens. It has a virtual image at 10 cm from the object. What is the power of the lens?
A. 10 dioptres
B. −10- 10 dioptres
C. −5- 5 dioptres
D. −15- 15 dioptres

Explanation

Solution

Hint
Power of a lens depends inversely on the focal length of the lens. The focal length can be calculated if we know the position of the object and its image.
⇒1f=1v−1u\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} , where f is the focal length, v is the position of the image, and u is the position of the object with respect to the lens.

Complete step by step answer
We know that the object has a negative distance when placed in front of the lens, and image has a positive distance when formed behind the lens. But in this question, the image is virtual which implies that it is formed on the same side of the lens as the object.
The data provided to us includes:
Distance of the object u=−20u = - 20 cm
Distance of the image v=−10v = - 10 cm
Since, the image is formed at 10 cm from the object, the distance of the image from the lens is still equal to 10 cm, assuming the image is formed between the object and the lens.
Now, we know the focal length f of a lens is given as:
⇒1f=1v−1u\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}
Putting the values in this equation gives us:
⇒1f=1−10−1−20=120−110\Rightarrow \dfrac{1}{f} = \dfrac{1}{{ - 10}} - \dfrac{1}{{ - 20}} = \dfrac{1}{{20}} - \dfrac{1}{{10}}
Solving for f. we get:
⇒1f=1−220=−120\Rightarrow \dfrac{1}{f} = \dfrac{{1 - 2}}{{20}} = \dfrac{{ - 1}}{{20}} cm
We know that the power of a lens is given as:
⇒P=100f\Rightarrow P = \dfrac{{100}}{f} D when the focal length is in cm
So, we get the power as:
⇒P=100×−120=−5\Rightarrow P = 100 \times \dfrac{{ - 1}}{{20}} = - 5 D
Hence, the correct answer is option (C).

Note
Sign convention for image formation in front of a lens follows different rules. It is important to determine from the question the position of the object and the image with respect to the lens to be able to apply the right convention. In case the position of the image is not explicitly specified with respect to the lens, determining whether the image is real or virtual is helpful.