Consider a light source placed at a distance of 1.5 m along the axis facing the convex side of a spherical mirror of radius of curvature 1m . The position (s') nature and magnification (m) of the image are

Options

A. `s'=0.375 ` m virtual upright ,m=025
B. s'=0.375 m real inverted m=025
C. s'=3.75 m, virtual inverted m=2.5
D. s'= 3.75 m, real uplight ,m=2.5

Answer

A

Explanation

Given ,
Distance of light source (u) =- 15 =- `(-3)/(2)` m
radius of curvature of mirror (R ) = 1m
WE know that ,
` (1)/(f) = (1) /(v) +(1)/(u) " " [ :' f= (R )/(2)]`
` therefore (2)/(R ) = (1)/(v) +(1)/(v) +(1)/(u) implies (2)/(1) =(1 )/(v) + (1)/((-3)/(2))`
` (2)/(1) = (1)/(v) - (2)/(3) `
`(1)/(v) =( 6+2)/( 3 ) implies (1)/(v ) = (8 )/(3) implies v=(3)/(8) = 0.375 m`
The image is virtual .
Magnification `(m) =(-v)/(u ) = ((-3)/(8 ))/((-3)/(2))=1/4 =0.25`