The figure ( Figure 1 ) shows an object and its image formed by a thin lens. Par

Question

The figure (Figure 1) shows an object and its image formed by a thin lens.

Part A

What is the focal length of the lens?

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Part B

What type of lens is it?

diverging

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Correct

Part C

What is the height of the image?

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Part D

Is it real or virtual?

virtual

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Figure 1 of 1

The figure (Figure 1) shows an object and its image formed by a thin lens.

Part A

What is the focal length of the lens?

f =   cm  

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Part B

What type of lens is it?

diverging

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Correct

Part C

What is the height of the image?

y? =   mm  

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Part D

Is it real or virtual?

virtual

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Correct

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Figure 1 of 1

Explanation / Answer

a)

1/f = 1/u + 1/v

1/f = 1/8 + 1/-3

1/f = -5/24

f = 24/5 cm = 4.8cm

b) Lens is diverging.

c) (height of image/ height of object) = -v/u

hi = -v/u * ho

hi = -(3/8)* 6.5 = -2.44 mm

Therefore hi = 2.44mm

d)

Image is virtual

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