an object 40 cm from a converging lens of focal length 20 cm moves the velocity of 5 cm per second towards the less calculate the image position after 2 seconds?
1/do + 1/di=1/f
first, calculate the di original (you have do, f)
Then, calcualte the do after 2 seconds (do final=40 -5(2) ) then calculate the final di (original lens equation, you have the new do, f.
That is the position of the image after 2 seconds
I don't understand
To calculate the image position after 2 seconds, we need to use the lens formula:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance
u = object distance
Given:
f = 20 cm (converging lens)
u = -40 cm (object moving towards the lens)
v = ?
Using the lens formula, we can rewrite it to solve for v:
1/f = 1/v - 1/u
1/v = 1/f + 1/u
v = 1 / (1/f + 1/u)
Substituting the given values:
v = 1 / (1/20 + 1/-40)
Simplifying the equation:
v = 1 / (1/20 - 1/40)
v = 1 / (2/40 - 1/40)
v = 1 / (1/40)
v = 40 cm
So, the image position after 2 seconds will be 40 cm.
To calculate the image position after 2 seconds, we need to use the lens formula:
1/f = 1/v - 1/u,
where:
f = focal length of the lens,
v = image distance,
u = object distance.
Given:
f = 20 cm,
u = 40 cm (object distance),
The velocity of the object, 5 cm/s, indicates that the object distance is changing over time.
To find the final image position after 2 seconds, we can use the formula:
v = u + vt,
where:
v = final image position,
u = initial object distance,
v = final object distance,
t = time.
Substituting the values, we have:
v = u + vt
v = 40 cm + (5 cm/s)(2 s)
v = 40 cm + 10 cm
v = 50 cm.
Therefore, the image position after 2 seconds is 50 cm from the lens.