A pin of length 2 cm is placed perpendicular to principal axis of a converging lens.An inverted image of size 1 cm is formed at a distance
40 cm from the pin . Find the focal length of the lens and it's distance from the pin?
Answers
Answer:
h = 2 cm h' = 1 cm
image is inverted => image is real.
m = - v/u = h'/h = -1/2
u = - 2 v
u is negative and v is positive
-u + v = 40 cm
3 v = 40 cm => v = 40/3 cm
u = - 80/3 cm
1/f = 1/v - 1/u = 3/40 + 3/80 = 9/80
f = 80/9 cm
The focal length of the lens and it's distance from the pin is 8.89 cm and 26.7 cm.
Given:
Height of object = ho = 2 cm
Height of image = hi = 1 cm
Explanation:
Distance between object and image = 40 cm = -u + v
Magnification is given by the formula:
m = hi/ho = v/-u
1/2 = v/-u
⇒ u = -2v → (equation 1)
The focal length is given by the formula:
1/v - 1/u = 1/f ⇒ 1/v + 1/2v = 1/f
3/2v = 1/f
⇒ f = 2v/3 → (equation 2)
-u + v = 40 cm
On substituting the value of 'u', we get,
-(-2v) + v = 40 cm
3v = 40 cm
∴ v = 40/3 cm = Image distance
Now, the focal length from equation (2) becomes,
f = 2v/3 ⇒ 2(40/3)/3
∴ f = 8.89 cm
Now, the object distance is:
u = -2(40/3)
∴ u = -80/3 cm = 26.7 cm