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e-lub [12.9K]
3 years ago
8

An object is placed near a concave mirror having a radius of curvature of magnitude 60 cm. How far should you place the object f

rom the mirror so that the lateral magnification produced by the mirror will be 2.5?
Physics
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

u = 18 cm

Explanation:

given,

radius of curvature = 60 cm

magnification of mirror = 2.5

distance of object  = ?

R = 2 f

f = R/2

f = 60/2 = 30 cm

m = -\dfrac{v}{u}

2.5 = -\dfrac{v}{u}

v = -2.5 u

now,

Using mirror formula

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

\dfrac{1}{30} = \dfrac{1}{-2.5u} + \dfrac{1}{u}

\dfrac{1}{30} = \dfrac{0.6}{u}

u = 0.6 x 30

u = 18 cm

distance of object be equal to u = 18 cm

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<h3 /><h3 /><h3>What is equilibrant force?</h3>
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The equilibrant force of the two given forces is calculated as follows;

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Thus, the equilibrant force of the two given forces is 14.14 N.

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