1st solution :

For the convex lens: (see 2nd the diagram )

u = -30 cm v = ? f = +20 cm

1/v - 1/u = 1/f

1/v = 1/u + 1/f = -1/30 + 1/20 = 1/60

v = + 60 cm.

Hence the image is inverted and real at 60 cm on the other side of lens.

The distance between object and image is u+v = 90cm

If a convex mirror (f=10cm) is used, the rays coming from object fall on it, a virtual erect image will be formed behind the mirror. We want this virtual image to be at the same location as the real image formed by the lens. Then for the convex mirror,

u + v = 90 cm

1/v + 1/(-u) = 1/f = 1/10

10 (u - v) = u v

10 (90 - v - v ) = (90 - v) v

900 - 20 v +v² - 90 v = 0

v² -110 v +900 = 0

v = [110 +- √(110²-4*900) ]/2

= 55 +- 5 √85 = 9 cm as v < 90 cm

so u = 90 - 9 = 81 cm.

*Distance between convex lens and convex mirror will be = 81 - 30 cm*

* = 51 cm*

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2nd solution:

For the convex lens: (see the 1st diagram)

u = -30 cm v = ? f = +20 cm

1/v - 1/u = 1/f => v = + 60 cm.

Hence the image is inverted and real at 60 cm on the other side of lens.

If a concave mirror (instead of convex mirror) is used to make the question more meaningful.

If an object is placed at the center of curvature C of a concave mirror (at 2 f ), then the image is inverted again and then it is of the same size and is at the same location.

Hence place the convex mirror at 2 f = 2* 10 = 20 cm from image A'B'. So final erect image A''B'' is formed at C center of mirror.

Distance between lens and mirror = 60+20 = 80 cm

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Solution 3:

Suppose the question is supposed to be with convex mirror only: then,

If the convex mirror is placed exactly at A'B' then, the virtual image A''B'' is formed inside the convex mirror at the same location. However, the image A''B'' would be inverted like A'B'. (Convex mirror always shows virtual and erect images.)

In that case, the distance between lens and mirror is 60 cm.