Physics - Geometrical Optics Question with Solution | TestHub

Let the powers of the two lenses be $P_1$ and $P_2$.

When the lenses are in contact, their combined power is given by:
$P_1+P_2=9$

When they are separated by a distance $d=20\text{ cm}=0.2\text{ m}$, the equivalent power is given by:
$P_{eq}=P_1+P_2-d{P}_1{P}_2$

Substituting the given values:
$\frac{27}{5}=9-0.2P_1{P}_2$

$\frac{1}{5}{P}_1{P}_2=9-\frac{27}{5}$

$\frac{1}{5}{P}_1{P}_2=\frac{18}{5}$

$P_1{P}_2=18$

We have $P_1+P_2=9$ and $P_1{P}_2=18$. These are the sum and product of the roots of the quadratic equation:
$x^2-9x+18=0$

$(x-3)(x-6)=0$

Thus, the powers of the lenses are $3\text{ D}$ and $6\text{ D}$.

Answer: 3,6