An object of length 5cm is placed perpendicularly on the principal axis at a distance of 75cm from a conacave mirror. If the radius of curvature of the mirror is 60cm, calculate the image distance and its hieght.

Explanation

Here, u =-75cm , r=-60cm,
`therefore " " f=(r)/(2)=(-60)/(2)=-30cm`
We know, `(1)/(v)+(1)/(u)=(1)/(f)or, (1)/(v)+(1)/(-75)=(1)/(30)`
`or, " " (1)/(v)=-(1)/(30)+(1)/(75)=-(1)/(50) or, v=-50cm`
So, the image is formed at a distance of 50cm in front of the mirror. The image is real.
`"Again", " " m=("height of the image"(I))/("height os the object"(O))=(v)/(u)`
Here we take only the mod value of m as we no need to know the nature of image thus formed.
`or, " " (I)/(5)=(50)/(75)=(2)/(3) or ,I=5xx(2)/(3)=3.33cm`
So, height of the image =3.33cm.