A linear objectof height 10 cm is kept in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm. The image formed is

Options

A. magnified and erect
B. magnified and inverted
C. diminished and erect
D. diminished and inverted

Answer

B

Explanation

Height of object `(h_(o))=10 cm`
Radius of curvature (R ) = 15 cm
Distance of object (u) `= -10 cm`
We know that,
`f=(R )/(2)=(15)/(2)cm`
According to mirror formula,
`(1)/(f)=(1)/(u)+(1)/(v)`
`(1)/(-(15)/(2))=(1)/(-10)+(1)/(v)`
`rArr (1)/(v)=(-2)/(15)+(1)/(10)`
`rArr (1)/(v)=(-4+3)/(30)`
`rArr (1)/(v)=(-1)/(30)`
`rArr v=-30 cm`
Magnification (m) `= (h_(i))/(h_(o))=(-v)/(u)`
`= (h_(i))/(10)=(-(-30))/(-10)`
`h_(i)=-3`