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(D) The focal length of the lens is $f = 20 \,cm$. $A$ parallel beam of light incident on the lens would converge at its focus $I'$ at a distance of $

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$60$

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(D) The focal length of the lens is $f = 20 \,cm$. $A$ parallel beam of light incident on the lens would converge at its focus $I'$ at a distance of $20 \,cm$ from the lens center $P$ along the principal axis.Given $P M = 10 \,cm$,the distance from the mirror intersection point $M$ to the original focal point $I'$ is $M I' = P I' - P M = 20 \,cm - 10 \,cm = 10 \,cm$.Since the light is reflected by the mirror to point $I$,and the mirror acts as a plane mirror,the distance $M I$ must equal $M I'$. Thus,$M I = 10 \,cm$.We are given $P I = 10 \,cm$. In $	riangle P M I$,all three sides ($P M = 10 \,cm$,$M I = 10 \,cm$,$P I = 10 \,cm$) are equal,so it is an equilateral triangle. Therefore,all internal angles are $60^{\circ}$.The angle between the incident ray (horizontal) and the reflected ray $M I$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$.The angle of incidence $i$ is the angle between the incident ray and the normal $N$. The angle of reflection $r$ is the angle between the reflected ray and the normal $N$. Since $i = r$,the normal $N$ bisects the angle between the incident and reflected rays.The angle between the normal $N$ and the horizontal axis is $120^{\circ} / 2 = 60^{\circ}$.The mirror is perpendicular to the normal $N$. If the normal makes an angle of $60^{\circ}$ with the horizontal,the mirror makes an angle of $90^{\circ} - 60^{\circ} = 30^{\circ}$ with the vertical,or $60^{\circ}$ with the horizontal.

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