I will interpret the statement of the problem above to mean that the object is 24 centimeters from the face of the mirror.
An image is "erect" if it is uninverted or "right-side-up".
An image is "virtual" if its distance q is negative; that is to say that the magnified image forms behind the mirror or on the other side of the mirror from the physical object.
For a spherical concave mirror and erect virtual image, q (or image distance from mirror) will be negative and behind the mirror and is equal here to -2.5 times p (or object distance from mirror).
From
"(1 divided by object distance from mirror) plus (1 divided by image distance from mirror) equals
(1 divided by focal point distance from mirror)", write 1/p + 1/q = 1/f.
Then, for the conditions given here, obtain
1/0.24 + 1/(-2.5×0.24) = 1/f, which simplifies
to 1/f = 2.5 or focal point distance f equals 0.4 meters.
Focal Point Distance or Length being half the Radius Of Curvature
(or R) sought gives R equal to 0.8 meters or
80 centimeters.
Note the positions of Center Of Radius Of Curvature C,
Focal Point F, Object O, and Image I in relation to
(along with their distances in
centimeters from) the Mirror M as shown below.
C=80--------------------F=40--------O=24--------M-----------------------------I=(-60)
By the geometry of similar triangles, one can also write q/p = h'/h equal to (R − q)/(p − R).
This translates to Rp −pq = pq − Rq or R(p + q) = 2pq. Then 2pq/(p + q) will give R here as
2(0.24)(-2.5 × 0.24) ÷ [0.24 + (-2.5 × 0.24)]
equal to 0.8 meters or 80 centimeters as above.
