Q= An object 2 cm tall is placed on the axis of a convex lens of focal length 5 cm at a distance of 10m from the optical centre of the lens .Find the nature ,size and postion of the object .
Answers
Standard thin lens equation will be used
1/f = 1/u + 1/v
F = 5cm
U = 10m = 1000cm
So 1/v = 1/f - 1/u
1/v = 1/5 - 1/1000
Or v = 5.025cm ( the image distance)
The formula for magnification is
M = hi/ho = -v/u
Where hi is the height of the image and ho is the height of the object (=2cm)
Or hi = -vho/u
Or hi = - (5.025x2) / 1000
Or the height of the image is
Hi = 0.01005cm
From the above resutls we can conculde this that
The image is inverted ( hi is negative) Its highly diminished ( hi is smaller than ho) It is the virtual (v is positive)Answer:
Explanation:
h1 = 2 cm
f = 5 cm
u = - 10m = - 1000 cm
1/v – 1/u = 1/f
1/v – 1/-1000 = 1/5
1/v = 1/5 – 1/000 = (200 – 1)/1000 = 199/1000
V = 5.02 cm
The image is formed 5.02 cm behind the convex lens and is real and inverted.
m = v/u = 5.02/-1000 = -0.005
m = h2/h1 = -0.005
h2/2 = -0.005
h2 = -0.01 cm
Since the object distance is much greater than the focal length, this example illustrates the case when the object is placed at infinity
if possible then mark me as brainliest