Physics - Geometrical Optics Question with Solution | TestHub

Ans (A,B)

For $d_1=120$ cm : $\frac{3/2}{v}-\frac{1}{(-120)}=\frac{3/2-1}{60}\Rightarrow v=\infty$

So, the ray is incident normally on the mirror. So, for any value of $d_2$, ray retraces its path. So $I_f$ is at O.

For $d_1=240$ cm : $\frac{3/2}{v}-\frac{1}{(-240)}=\frac{3/2-1}{60}\Rightarrow v=360$ cm.

If first image is formed at mirror, ray retraced its path to form image at O.