Question

An object is placed at a distance of 40 cm in front of a concave mirror of focal length 20 cm.The image produced is

Answer 1

Answer:

Let the sign convention be along the direction of incident light implying

u=-40 cm,f =-15 cm.In this case assuming the image distance be v the mirror eqn becomes

1/v+1/u=1/f =>1/v+1/(-40)=1/(-15)

=>1/v={1/40–1/15}=(3–8)/120=(-)1/24

=>v=-24.But as per the question the object being displaced 20cm towards the mirror makes new u=-20 making for new v by eqn

1/v+1/(-20)=1/(-15)=>1/v={1/20–1/15}=(3–4)/60=(-1)/60=>v=-60cm .So the image is shifted through -60-(-24)=60–24=(-)36.So the new image will be displaced through 36cm off the image formed in 1st case.

Explanation: