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Elza [17]
2 years ago
15

An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the

sign of the value in your answers.)
(a) What is the magnification?
(b) What is the image height?
m
Physics
1 answer:
Anvisha [2.4K]2 years ago
4 0

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

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