Please answer the question Two thin lenses, one a converging lens and the other

Question

Please answer the question

Two thin lenses, one a converging lens and the other a diverging lens, are separated by 1.00 m along the same principal axis, as shown in Figure 2. The magnitude of the focal length of the converging lens is 25.0 cm, while the magnitude of the focal length of the diverging lens is 40.0 cm. An object 8.25 cm tall is placed 35.0 cm to the left of the converging lens, Where is the final image produced by this combination of lenses? Compared to the object, is the final image erect or inverted? Is it real or virtual?

Explanation / Answer

For Converging lens,
f = 25.0 cm
do = 35.0 cm
ho = 8.25 cm

1/f = 1/di + 1/do
1/25.0 = 1/di + 1/35.0
di = 87.5 cm

M1 = -di/do = -87.5/35.0
M1 = - 2.5

Now This image will act as an object for Diverging lens,
So,
f = - 40.0 cm
do = 100 - 87.5 = 12.5 cm
di = ?
1/f = 1/di + 1/do
-1/40 = 1/12.5 + 1/di
di = -9.5 cm

M2 = -di/do
M2 = 9.5/12.5
M2 = 0.76

(a)
Final image is produced btw Converging & Diverging Lens at Distance 9.5 cm from Diverging Lens to the left or 90.5 cm from the Converginf lens to the Right.

(b)
M = M1*M2
M = -2.5 * 0.76
M = -1.9

Final image is Inverted, Magnified & Virtual.

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