An object is placed in front of a convex mirror at 60m from the mirror. Its radius of curvature is 40m. The object starts moving towards the mirror at 20ms−1. Find the average speed of the image after a time interval of 2 seconds?
The correct option is C2.5ms−1 Given: The distance of object u from the pole is 60 m, Radius of curvature (r)=40m Let f be the focal length and v be the image distance We know that, f=r2=12×40=20m. Applying the sign conventions, u=−60m and f=+20m We know, 1v + 1u = 1f ⟹1v - 160 = 120 ⟹v=15m Positive sign shows that the image is virtual. Speed of the object =20ms−1 and time taken =2s ∴ Distance = Speed × time ⟹20×2=40m Therefore, now the position of object will be u′=60−40=20m Let v′ be the new position of the image Applying it in the mirror formula we get, 1v′ + 1u′ = 1f ⟹1v′ - 120 = 120 ⟹1v′ = 4020×20 ⟹v′=10m So, the image moves from 15m to 10m in 2s. Hence, its average speed = 15−102=2.5ms−1