An object moving at a speed of 5m/s towards a concave mirror of focal ...
Solution:
Given,
u = -9m (since object is moving towards the mirror)
v = ?
f = -1m (since concave mirror)
Using mirror formula,
1/f= 1/v + 1/u
=> 1/v = 1/f - 1/u
=> 1/v = 1/-1 - 1/-9
=> 1/v = -1/9 - (-1)
=> 1/v = 8/9
=> v = 9/8 m
Now,
Average speed of image = Total distance travelled by image/Total time taken
Total distance travelled by image = Distance between object and mirror + Distance between mirror and image
= 9m + 9/8m = 81/8m
Time taken to cover this distance = Total distance/Speed
= (81/8)/5 = 81/40 sec
Therefore,
Average speed of image = (81/8)/(81/40) = 5/2 m/s
Hence, the correct option is (d) (2)/(5)m/s
An object moving at a speed of 5m/s towards a concave mirror of focal ...
According to mirror formula:-
1/v + 1/u = 1/f or v = (f × u) / (u - f)
When an object is at a distance of 9m from the concave mirror, u = -9m and f = -1m.
Therefore, v = {(-1) (-9)}/ (-9 + 1) = -9/8m
As the object moves at a constant speed of 5m/s^-1 after 1s the position of object u' = -9m + 5m = -4m
So, v' = {(-1) (-4)}/ (-4 + 1) = -4/3
The shift in the position of image in 1s = v - v' =
-9/8 + 4/3 = (-27 +32)/24 = 5/4 = 1/5
Hence the average speed of image = 1/5 ms^-1.
I hope you understand.