An object is midway between the lens and the mirror as shown. The mirror's radius of curvature is 20.0 cm and the lens has a focal length of -16.7 cm, Considering only the rays that leaves the object and travels first towards the mirror, locate the final image formed by this system. Is this image real or virtual? Is it upright or inverted? What is the overall magnificaiton?

An object is midway between the lens and the mirror as shown. The mirr

1.	An object is midway between the lens and the mirror as shown. The mirror's radius of curvature is 20.0 cm and the lens has a focal length of -16.7 cm, Considering only the rays that leaves the object and travels first towards the mirror, locate the final image formed by this system. Is this image real or virtual? Is it upright or inverted? What is the overall magnificaiton?
Answer» Solution :Image formed by mirrorUsing mirror FORMULA `1/v+1/u=1/f=2/F(as `f=R//2`) we have, `1/v_1+1/-12.5=2/-20` `:. v_1=-50 cm` `m_1=-v/u=-(-50)/(-12.5)=-4` i.e. image formed by the mirror is at a distance of `50cm` fromthe mirror to the left of it. It is inverted and four times LARGER. Image formed by lens Image formed by mirror ACTS as an object for lens. It is at a distance 0.25.0 cm to the left of lens. Using the lens formula, `1/v-1/u=1/f` We have, `1/v_2-1/25=-1.67` `:. v_2=-50.3cm` and`m_2=v/u=-50.3/25=-2.012` overall magnification is `m=m_1xxm_2=8.048` Thus, the final image is at a distance `25.3 cm` to the RIGHT of the mirror, virtual, upright enlarged and `8.048` times. positions of the two IMAGES are shown in figure.

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