An object of size 5.0 cm is placed at 15 cm in front of a convex lens of focal length 20 cm.
Find position, size and nature of the image?
please answer this fast
Answers
Answer:
The image will be on the same side of the object. It will be enlarged, virtual and erect.
Explanation:
Height of the object, h₀ = 5 cm
Image distance, u = -15 cm
Focal length of the lens, f = +20 cm
The image distance, v can be found using the lens formula.
1/f = 1/v - 1/u
⇒ 1/v = 1/f + 1/u
∴ v = uf / (u+f)
= (-15 × 20) / (-15 + 20)
v = -60 cm
The negative sign of the image distance indicates that the image is on the same side of the object.
We have two equations for magnification.
m = h₁/h₀
m = v/u
By combining the above equations:
h₁/h₀ = v/u
∴ height of the image, h₁ = h₀ (v/u)
= 5 × (-60 / -15)
h₁ = +20 cm
So the image is upright and enlarged. An upright image indicates that it is virtual.