A concave mirror of focal length 10cm and a convex mirror of focal length 15 cm are held co-axially face to face at a distance 40 cm apart. An object of height 2 cm is placed perpendicularly on the common axis in between the two mirrors. The distance of the object from the concave mirror is 15 cm. Considering the first reflection occurs in the concave mirror and the second reflection in the convex mirror, calculate the position, nature and height of the final image.

Explanation

In case of first reflection in the concave mirror:
u = - 15 cm , f = - 10 cm
`"We know", (1)/(v)+(1)/(u)=(1)/(f) or, (1)/(v)+(1)/(-15)=(1)/(-10)`
`or, " " (1)/(v)=-(1)/(10)+(1)/(15)=(-3+2)/(30)=-(1)/(30)`
`or, " " v=-30 cm `
So, the image formed by the concave mirror is real and formed at a distance of 30 cm from the mirror. This image acts as the object of the convex mirror.
In case of second reflection in the convex mirror:
`u=-(40-30)=-10 cm,f=+15 cm`
`"We know", (1)/(v)+(1)/(u)=(1)/(f)`
`or, " " (1)/(v)+(1)/(-10)=+(1)/(15)`
`or, " " (1)/(v)=+(1)/(15)+(1)/(10)=+(1)/(6) or, v = + 6 cm`
So the final image is virtual and is formed at a distance of 6 cm behind the convex mirror.

Magnification by the concave mirror,
`m_(2)=-(v)/(u)=-(6)/(-10)=(3)/(2)`
`therefore` Total magnification,
`m=m_(1)xxm_(2)=-2xx(3)/(5)=-(6)/(5)`
As m is negative , so the final image is inverted with respect to the object.
`therefore` Height of the final image (taking only the magnitude of m )
`=(6)/(5)xx"height of the object" = (6)/(5)xx2=(12)/(5)=2.4 cm`