Geometrical OpticsMedium

Question

A beam of light is incident parallel to principal axis on a convex lens (L) of focal length $f = 20$ cm. A plane mirror PQ (See figure), inclined at 60° with the principal axis, is placed a distance 'd' behind the lens so as to form a real image I of the distance 'd' from the optical center of the lens as shown in the figure. The value of 'd' is

Options

A.5 cm
B.10 cm
C.15 cm
D.20 cm

Solution

Sol. Key concept: A plane mirror reflects the converging focus to its mirror-symmetric position.

Without the mirror, the parallel rays would meet at the lens focus F, 20 cm from the optical center O. Let M be the point where the mirror meets the principal axis; OM = d, so MF = 20 − d.

The reflected focus I is the mirror image of F in the mirror plane. Therefore MI = MF. Because the mirror makes 60° with the axis, MI makes 120° with the forward axis, so angle OMI is 60°.

The question also gives OI = OM = d. Triangle OMI is therefore equilateral, giving MI = d.

$$20 - d = d\ \ \Rightarrow \ \ \ d = 10\ cm$$

Hence option b. The 60° geometry is the missing justification for equating the two distances.

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