Answer
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Hint:A concave lens is a lens that has a curved surface in the inward direction. It is thinner at the centre and thicker at the edges. A concave lens always spreads out all the light rays that have been refracted through it. So, a concave lens is also known as a diverging lens.
Step-By-Step solution:
Step I:
Power of a lens is defined as the reciprocal of the focal length of the lens. It is measured in metres. The S.I. unit of power is dioptres. It is represented by the symbol ‘D’.
The formula of power is written as
$P = \dfrac{1}{f}$
Where P is the power
And f is the focal length of the lens.
Step II:
Given that the focal length of the lens is $f = 50cm$
Converting cm into metres,
$1m = 100cm$
$50cm = \dfrac{{50}}{{100}}m$
Or $f = \dfrac{1}{2}m$
Substitute the value of focal length in the above formula, and solving
$P = \dfrac{1}{{\dfrac{1}{2}}}$
$P = 2D$
Step III:
The power of a concave lens of focal length $50cm$ is $ - 2D$.
Therefore, Option B is the right answer.
Note:It is important to note that the focal length of an optical instrument shows how strongly the instrument converges or diverges the light rays. If the focal length is positive, it means that the lens has converged all the light rays to a single point and the image of the object is formed at that point. If the focal length of the lens is negative, this means that the lens has diverged all the light rays. It is the distance between the focal point of the lens and its centre.
Step-By-Step solution:
Step I:
Power of a lens is defined as the reciprocal of the focal length of the lens. It is measured in metres. The S.I. unit of power is dioptres. It is represented by the symbol ‘D’.
The formula of power is written as
$P = \dfrac{1}{f}$
Where P is the power
And f is the focal length of the lens.
Step II:
Given that the focal length of the lens is $f = 50cm$
Converting cm into metres,
$1m = 100cm$
$50cm = \dfrac{{50}}{{100}}m$
Or $f = \dfrac{1}{2}m$
Substitute the value of focal length in the above formula, and solving
$P = \dfrac{1}{{\dfrac{1}{2}}}$
$P = 2D$
Step III:
The power of a concave lens of focal length $50cm$ is $ - 2D$.
Therefore, Option B is the right answer.
Note:It is important to note that the focal length of an optical instrument shows how strongly the instrument converges or diverges the light rays. If the focal length is positive, it means that the lens has converged all the light rays to a single point and the image of the object is formed at that point. If the focal length of the lens is negative, this means that the lens has diverged all the light rays. It is the distance between the focal point of the lens and its centre.
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