Answer
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Hint : The power of a lens is the reciprocal of the focal length (in meter) of the lens. The SI unit of power of a lens is dioptres represented by $ D $ . Convex lenses have positive focal length, so the power of a convex lens is also positive. A concave lens has a negative focal length, so the power of a concave lens is also negative.
Formula Used:
$\Rightarrow P = \dfrac{1}{f} $
$ P $ is the power of the lens in diameter and $ f $ is the focal length of the lens in meters.
Complete step by step answer
Let the power of the given lens be $ {P_1} $ and the power of the required lens be $ {P_2} $ , and the focal length of the required lens be $ {f_2} $ .
$\Rightarrow P = \dfrac{1}{f} $
$ P $ is the power of the lens in diameter and $ f $ is the focal length of the lens in meters.
The focal length of the given lens is $ 10cm $ i.e. $ 0.1m $ because $ 1m = 100cm $ .
Hence,
$\Rightarrow {P_1} = \dfrac{1}{{0.1}} $
$\Rightarrow {P_1} = 10D $
We know that,
Power of combination of two lenses of power $ {P_1} $ and $ {P_2} $ will be $ {P_1} + {P_2} $
We want a lens of power $ 5D $ .
Hence,
$\Rightarrow 5D = \;{P_1}\; + {\text{ }}10D $
$\Rightarrow {P_2}\; = - 5D $
We know that,
$\Rightarrow {P_2} = \dfrac{1}{{{f_2}}} $
$\Rightarrow {f_2} = \dfrac{1}{{ - 5}}m $
$\Rightarrow {f_2} = - 0.2m $
Hence,
The power and focal length of the lens are negative so the given lens is a concave lens.
Therefore the answer to our question is $ - 0.2m $ and concave lens.
Additional Information
A concave lens and convex lens can be detected by holding them because the lens with a bulge in middle will be the convex lens and one with a cavity in the middle will be the concave lens.
Note
A concave lens is also called a diverging lens and a convex lens is also called a converging lens. Whenever we are dealing with both power and focal length of the lens together, always be careful to convert the unit of focal length into meters.
Formula Used:
$\Rightarrow P = \dfrac{1}{f} $
$ P $ is the power of the lens in diameter and $ f $ is the focal length of the lens in meters.
Complete step by step answer
Let the power of the given lens be $ {P_1} $ and the power of the required lens be $ {P_2} $ , and the focal length of the required lens be $ {f_2} $ .
$\Rightarrow P = \dfrac{1}{f} $
$ P $ is the power of the lens in diameter and $ f $ is the focal length of the lens in meters.
The focal length of the given lens is $ 10cm $ i.e. $ 0.1m $ because $ 1m = 100cm $ .
Hence,
$\Rightarrow {P_1} = \dfrac{1}{{0.1}} $
$\Rightarrow {P_1} = 10D $
We know that,
Power of combination of two lenses of power $ {P_1} $ and $ {P_2} $ will be $ {P_1} + {P_2} $
We want a lens of power $ 5D $ .
Hence,
$\Rightarrow 5D = \;{P_1}\; + {\text{ }}10D $
$\Rightarrow {P_2}\; = - 5D $
We know that,
$\Rightarrow {P_2} = \dfrac{1}{{{f_2}}} $
$\Rightarrow {f_2} = \dfrac{1}{{ - 5}}m $
$\Rightarrow {f_2} = - 0.2m $
Hence,
The power and focal length of the lens are negative so the given lens is a concave lens.
Therefore the answer to our question is $ - 0.2m $ and concave lens.
Additional Information
A concave lens and convex lens can be detected by holding them because the lens with a bulge in middle will be the convex lens and one with a cavity in the middle will be the concave lens.
Note
A concave lens is also called a diverging lens and a convex lens is also called a converging lens. Whenever we are dealing with both power and focal length of the lens together, always be careful to convert the unit of focal length into meters.
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