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mina [271]
4 years ago
10

An object that is 2 cm tall is 18 cm in front of a converging lens and creates a real image 8 cm beyond the lens. What is the fo

cal length of the lens?
What is the height of the image?
Physics
1 answer:
alex41 [277]4 years ago
5 0

Answer:

The focal length of the lens is 5.54 cm and the height of the image is -0.89 cm.

Explanation:

Given that,

Height of object h = 2 cm

Object distance u= -18 cm

Image distance v= 8 cm

We need to calculate the focal length of the lens

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Where, f = focal length

Put the value into the formula

\dfrac{1}{f}=\dfrac{1}{8}-\dfrac{1}{-18}

\dfrac{1}{f}=\dfrac{13}{72}

f=5.54\ cm

(II). We need to calculate the height of the image

Using formula of magnification

m =\dfrac{v}{u}=\dfrac{h'}{h}

\dfrac{v}{u}=\dfrac{h'}{h}

Put the value into the formula

\dfrac{8}{-18}=\dfrac{h'}{2}

h'=\dfrac{8}{-18}\times2

h'=-0.89\ cm

Hence, The focal length of the lens is 5.54 cm and the height of the image is -0.89 cm.

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