Physics - Geometrical Optics Question with Solution | TestHub

Let the power of the concave lens be $P_1$ and the power of the convex lens be $P_2$.

Since the focal length of the combination is positive ($F=30\text{ cm}$), the net power of the system is positive. This implies that the convex lens has a greater magnitude of power than the concave lens.

Given the ratio of magnitudes of their power is $2:3$, we have:

$|P_1|:|P_2|=2:3$

Let $P_1=-2x$ and $P_2=3x$, where $x>0$.

The net power of the combination is:

$P=P_1+P_2=-2x+3x=x$

The focal length of the system is $F=30\text{ cm}=\frac{30}{100}\text{ m}$.

The net power $P$ is given by:

$P=\frac{1}{F}=\frac{100}{30}=\frac{10}{3}\text{ D}$

Equating the two expressions for $P$:

$x=\frac{10}{3}\text{ D}$

Substituting the value of $x$ to find the individual powers:

$P_1=-2\left(\frac{10}{3}\right)=-\frac{20}{3}\text{ D}$

$P_2=3\left(\frac{10}{3}\right)=10\text{ D}$

The focal lengths of the individual lenses are:

$f_1=\frac{100}{P_1}=\frac{100}{-\frac{20}{3}}=-15\text{ cm}$

$f_2=\frac{100}{P_2}=\frac{100}{10}=10\text{ cm}$

The focal lengths of the individual lenses are $10\text{ cm}$ and $-15\text{ cm}$.

Answer: $10\mathrm{cm},-15\mathrm{cm}$