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sergij07 [2.7K]

2 years ago

10

at what position an object be placed in front of a concave mirror of radius of curvature 40cm so that an erect image of magnific

ation of 3 be produced ?

1 answer:

Thepotemich [5.8K]2 years ago

8 0

C=40cm
f=C/2=40/2=20

m=3

$\\ \sf\longmapsto m=\dfrac{f}{f-u}$

$\\ \sf\longmapsto 3=\dfrac{20}{20-u}$

$\\ \sf\longmapsto 3(20-u)=20$

$\\ \sf\longmapsto 60-3u=20$

$\\ \sf\longmapsto 3u=40$

$\\ \sf\longmapsto u=13.3cm$

m=magnification
f is focal length
C is radius of curvature
u is object distance

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$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

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