Question

An object is placed 27.2 cm to the left of a diverging lens (f = -10.6 cm). A concave mirror (f = 17.1 cm) is placed 40.3 cm to the right of the lens to form an image of the first image formed by the lens. Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?

Answer 1

(a)

here for diverging lens (f = -10.6 cm) ,

object distance s₁ = 27.2 cm

so here we have formyla for finding image distance formed by  diverging lens

image distance i₁ = s₁ f₁ / s₁ - f₁

here we put all value in this equation,

i₁ = 27.2 * ( - 10.6 ) / (27.2- 10.6 )

i₁ = - 288.32 / 16.6

i₁ = - 17.37 cm

then here distance of object from convacec mirror = s₂

s₂ = 40 - ( - 17.37)

= 40+ 17.37

s₂ = 57.37 cm

so image formed by concave mirror have iamge image distance i_2,

i₂ = s₂ f₂ / s₂-f₂

_{= 57.37 * 17.1 / ( 57.37 - 17.10 )}

_{= 981.027 / 40.27}

i₂ = 24.32 cm

here the final image distance is i₂ = 24.32 cm

(b) here we wanto find the image is real or virtual.

so here final image distance i₂= 24.32 cm which is positive, so final image is real.

(c)

here we to find final image upright or inverted with respect to the original object,

so , for that we have to find magnification m = -s₁ / s₂

_{= - 27.2 / 57.37}

_{m = - 0.47}

if the magnification is positive then , image is upright compare to the object. if magnification is negetive then image is inverted compare to the object.

so here answer is image is inverted. because m= - 0.47 which is negetive value.