Two lenses of power -3.5D and + 1D are placed in contact. Find the total power of the combination of lens. Calculate the focal length of this combination.
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Power $P_1=-3.5 D ; P_2=+1 D$ Total power $=P_1+P_2=-3.5 D+1 D \Rightarrow$ Total power $(P)=-2.5 D$. Now; $P$ (in diopter) $=\frac{1}{f \text { (in meters) }} \Rightarrow-2.5=\frac{1}{f(\text { in meters) }}$
$\Rightarrow\text{f (in meters)}=\frac{-1}{2.5}=\frac{-1}{\frac{5}{2}}$
$\Rightarrow\text{f (in meters)}=-\frac{2}{5}=-0.4\text{m}$
Now 1m = 100cm ⇒ -0.4m = -40cm Hence the focal length of lens is -40cm.
$\Rightarrow\text{f (in meters)}=\frac{-1}{2.5}=\frac{-1}{\frac{5}{2}}$
$\Rightarrow\text{f (in meters)}=-\frac{2}{5}=-0.4\text{m}$
Now 1m = 100cm ⇒ -0.4m = -40cm Hence the focal length of lens is -40cm.
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