an object whose height is 3.1 cm is at a distance of 11.5cm from aspherical conc

Question

an object whose height is 3.1 cm is at a distance of 11.5cm from aspherical concave mirror, its real is real and has a height of10.7cm. Calculate the radius of curvature of the mirror. ANSWER -17.84 cm
>>>>The part i need help on<<<<
how far from the mirror is it necessary to place the aboveobject in order to have a virtual image with a height of10.7cm?
what i tried to do. combining these twoequation , hi/ho = di/do and 1/f = 1/di +1/do my work>
m=10.7 / 3.1 m= 3.45
m = di/d0 di = 39.69 ( from part 1 ) /3.45 = 11.505
1/8.92 ( focal point from part 1) =1/11.505 + 1/do
but the answer is wrong, what am i doingwrong. ?


>>>>The part i need help on<<<<
how far from the mirror is it necessary to place the aboveobject in order to have a virtual image with a height of10.7cm?
what i tried to do. combining these twoequation , hi/ho = di/do and 1/f = 1/di +1/do my work>
m=10.7 / 3.1 m= 3.45
m = di/d0 di = 39.69 ( from part 1 ) /3.45 = 11.505
1/8.92 ( focal point from part 1) =1/11.505 + 1/do
but the answer is wrong, what am i doingwrong. ?

how far from the mirror is it necessary to place the aboveobject in order to have a virtual image with a height of10.7cm?
what i tried to do. combining these twoequation , hi/ho = di/do and 1/f = 1/di +1/do my work>
m=10.7 / 3.1 m= 3.45
m = di/d0 di = 39.69 ( from part 1 ) /3.45 = 11.505
1/8.92 ( focal point from part 1) =1/11.505 + 1/do
but the answer is wrong, what am i doingwrong. ?

Explanation / Answer

Here the image is inverted.So object height will be opposite sign then by doing we get right answer.

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