Question

# An object is placed in front of a convex mirror at 60 m from the mirror. Its radius of curvature is 40 m. The object starts moving towards the mirror at 20 m s−1.  Find the average speed of the image after a time interval of 2 seconds?

A
10 m s1
B
20 m s1
C
2.5 m s1
D
5 m s1

Solution

## The correct option is C 2.5 m s−1Given: The distance of object u from the pole is 60 m, Radius of curvature (r)=40 m Let f be the focal length and v be the image distance We know that, f=r2=12×40=20 m. Applying the sign conventions, u=−60 m and f=+20 m We know, 1v + 1u = 1f ⟹1v - 160 = 120 ⟹v=15 m Positive sign shows that the image is virtual. Speed of the object =20 ms−1 and time taken =2 s ∴ Distance = Speed × time ⟹20×2=40 m Therefore, now the position of object will be u′=60−40=20 m 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′=10 m So, the image moves from 15 m to 10 m in 2 s. Hence, its average speed = 15−102=2.5 m s−1

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