Question

if convex lens of focal length 50 cm is placed in contact coaxially with a concave lens of focal length 20 cm, what is the power of combination

Answer 1

Answer:

The power of the combination of the lenses is -3D.

Explanation:

Given the focal length of convex lens, $f_{1} =50 cm=0.5m$

The focal length of concave lens, $f_{2} =20 cm=0.2m$

The equivalent focal length of the combination of lenses is given by the sum of individual focal lengths.

$\dfrac{1}{f_{c} } =\dfrac{1}{f_{1} }+\dfrac{1}{f_{2} }$

Since the focal length of concave lens is negative, hence

$\dfrac{1}{f_{c} } =\dfrac{1}{f_{1} }-\dfrac{1}{f_{2} }$

Substituting the values,

$\dfrac{1}{f_{c} } =\dfrac{1}{0.5}-\dfrac{1}{0.2 }=\dfrac{0.2-0.5}{0.5\times 0.2}$

$\implies \dfrac{1}{f_{c} } =\dfrac{-0.3}{0.1}=-3m$

Power(P) of a lens is given by the reciprocal of focal length(f) in meters and is measured in diopters.

Thus, the power of the combination is

$P_c= \dfrac{1}{f_{c} } =-3D$

Hence, the power of the combination of the lenses is -3D.