Question
A convex lens of focal length 2.5 cm is placed 4 cm from a small object of height 4 cm. Use a ruler and construct the image. Is the image real or virtual, upright or inverted? What is the image distance? What is the height of the image? What is the magnification?
Answers
Elena
The object located between F and 2F => the image will be located beyond the 2F point on the other side of the lens. It is real, inverted and its dimension is larger than the object demension.
d₀ - object distance
d₁ - image distance
f- focal lenth
1/d₀ +1/d₁ =1/f
1/d₁ =1/f - 1/d₀ =
= (d₀-f)/ d₀•f
d₁=d₀•f/(d₀-f)=0.04•0.025/(0.04-0.025) = 0.067 m
h₀/d₀=h₁/d₁
h₁= d₁•h₀/d₀=0.067•0.04/0.04 = 0.067 m
d₀ - object distance
d₁ - image distance
f- focal lenth
1/d₀ +1/d₁ =1/f
1/d₁ =1/f - 1/d₀ =
= (d₀-f)/ d₀•f
d₁=d₀•f/(d₀-f)=0.04•0.025/(0.04-0.025) = 0.067 m
h₀/d₀=h₁/d₁
h₁= d₁•h₀/d₀=0.067•0.04/0.04 = 0.067 m