Question

Two thin lenses have a combined power of $+9D$. When they are separated by a distance of $20cm$, their equivalent power becomes $+\frac{27}{5}D$, then their individual powers are

• A. $6D$ and $3D$
• B. 4$6D$ and $3D$
• C. $9D$ and $3D$
• D. $9D$ and $6D$

Answer 1

Answer: (A)

$6D$ and $3D$

Let the power of each lens be $P_1$ and $P_2$.

Power of combination of lenses kept close $P_{eq}=P1+P_2=9$

Power of lenses separated by $d=0.2m$ distance,

$P_{eq}^{\prime }=P_1+P_2-d{P}_1{P}_2$

$\therefore$ $\frac{27}{5}=9-0.2P_1{P}_2$

We get $P_1{P}_2=18$

From 1 , $P_1+\frac{18}{P_1}=9$

we get the equation as $P_1^2-9P_1+18=0$

$⟹P_1=6D$

So, $P_2=9-6=3D$