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2 years ago

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A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca

l length of the lens and the kind of lens is used? (b) What is the magnification and height of the image? (c) Describe the image in terms of its type, orientation and size relative to the object?

1 answer:

xenn [34]2 years ago

8 0

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance = 4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm$

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2$

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

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