Physics - Geometrical Optics Question with Solution | TestHub

The system acts as an equivalent mirror. Light passes through the lens, reflects from the silvered plane surface, and passes through the lens again.

The power of the equivalent mirror is $P_{eq}=2P_{lens}+P_{mirror}$.

Focal length of the plano-convex lens: $R_1=20\text{ cm}$ (curved), $R_2=\infty$ (plane). $\frac{1}{f_l}=(1.5-1)\left(\frac{1}{20}-\frac{1}{\infty }\right)=0.5\left(\frac{1}{20}\right)=\frac{1}{40}$. So $f_l=40\text{ cm}$.

The plane surface is silvered, acting as a plane mirror. $P_{mirror}=0$.

Thus, $P_{eq}=2P_{lens}=2/f_l=2/40=1/20{\text{ cm}}^{-1}$.

For P: Equivalent focal length $F_{eq}=1/P_{eq}=20\text{ cm}$.

Since it's a converging system (forms real images), $F_{eq}=-20\text{ cm}$. (P)-(1)

For Q: Object at $u=-30\text{ cm}$.

Using mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{F_{eq}}$. $\frac{1}{v}+\frac{1}{-30}=\frac{1}{-20}$. $\frac{1}{v}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}$.

So $v=-60\text{ cm}$. (Q)-(2)

For R: Magnification $m=-v/u=-(-60\text{ cm})/(-30\text{ cm})=-2$.

(R)-(3)

For S: Image formed at the object's position means the object is at the center of curvature of the equivalent mirror. $u=2F_{eq}=2(-20\text{ cm})=-40\text{ cm}$.

(S)-(4)