Q.A rectangular sheet of length 30 cm and breadth 3 cm is kept on the principal axis of a concave mirror of focal length 30 cm. Draw the image formed by the mirror on the same ray diagram, as far as possible on scale. [Ans: Inverted image starts from 50 cm and ends at 90 cm. Its height in the beginning is 2 cm and at the end it is 6 cm. At 60 cm, image height is 3 cm. Thus, outer boundary of the image is a curve]

[!TLDR] Working backward from the mirror formula with f=-30 cm to match the book's own printed answer positions confirms the object runs from 45 cm to 75 cm from the mirror, giving image heights of 2 cm (near end, at v=50 cm) and 6 cm (far end, at v=90 cm), 3 cm at the v=60 cm midpoint. [!ANSWER] Image runs from v=50 cm (height 2 cm) to v=90 cm (height 6 cm); at v=60 cm the height is 3 cm; the object itself extends from u=45 cm to u=75 cm from the mirror.

The question's own text (as extracted) gives only the sheet's dimensions (30 cm x 3 cm) and the mirror's focal length (30 cm, so f=-30 cm for this concave mirror), without explicitly stating how far the sheet's near end is from the mirror -- that specific distance did not survive text extraction from the original figure/diagram. However, it can be uniquely recovered by working BACKWARD from the mirror formula using the two image positions the book's own printed answer states (v=-50 cm and v=-90 cm, following this book's convention where real images take a negative v): using 1/f=1/v+1/u with f=-30, v=-50 gives u=-75 cm; with v=-90 gives u=-45 cm. These two object distances differ by exactly 75-45=30 cm, matching the sheet's stated 30 cm length exactly -- confirming the sheet's near end is 45 cm from the mirror and its far end is 75 cm from the mirror. The corresponding magnifications are m=-v/u = -(-90)/(-45) = -2 at the near end (u=-45, giving v=-90) and m=-(-50)/(-75) = -2/3 at the far end (u=-75, giving v=-50). Applying these to the 3 cm breadth: near end image height = 2 x 3 = 6 cm (at v=-90, i.e. the far/90 cm image position); far end image height = (2/3) x 3 = 2 cm (at v=-50, i.e. the near/50 cm image position). At the midpoint u=-60 cm (using 1/(-30)=1/v+1/(-60), giving v=-60 cm, exactly at the mirror's centre of curvature, where m=-1 exactly), image height = 1 x 3 = 3 cm. All three values match the book's own printed answer, confirming the reconstructed object position (45 cm to 75 cm from the mirror) is correct. So the image is INVERTED, runs from 50 cm to 90 cm in front of the mirror, with height 2 cm at the 50 cm (near) end and 6 cm at the 90 cm (far) end, height 3 cm at the 60 cm midpoint -- since the magnification (and hence the image height) varies continuously and non-linearly along the object's length, the outer boundary of the image is a CURVE rather than a straight edge. [!ANSWER] Object from 45 cm to 75 cm from the mirror; image from 50 cm (height 2 cm) to 90 cm (height 6 cm), 3 cm at the 60 cm midpoint, outer boundary a curve (inverted image).

Since the object's own starting distance from the mirror is missing from the extracted question text, work backward from the mirror formula using the book's own printed image positions (50 cm, 90 cm) to uniquely recover the object's near/far distances, then verify by independently computing the three stated image heights and confirming they match the printed answer.

Assuming the object distance is simply unknowable and giving up on a numeric answer, rather than recognising that the physics plus the printed answer together uniquely and verifiably determine it.