Question

when an object is placed in front of a convex mirror of focal length 15 cm and the magnification produced is found to be 1 by 4 where is the object and the image located​

Answer 1

Answer:

→ OBJECT DISTANCE ( U ) = - 45 cm

[ OBJECT LOCATED BEYOND CENTRE OF CURVATURE OF MIRROR ]

→ IMAGE DISTANCE ( V ) = + 11.25 cm

[ IMAGE LOCATED BETWEEN FOCAL LENGTH AND POLE OF THE MIRROR ]

Explanation:

★ RAY OPTICS ★

★ GIVEN ;

» Focal Length of Convex Mirror ( F ) =

+ 15cm

★ Magnification ( m ) = 1 / 4

★ OBJECT DISTANCE FROM POLE (U) = ???

★ IMAGE DISTANCE FROM POLE ( V ) = ???

_____ [ By USING FORMULA ] _____

[ $\frac{1}{V} \: + \: \frac{1}{U} = \frac{1}{F} ]$

★ Magnification ( m ) = - ( V / U )

» 1 / 4 = - ( V / U )

★ V = - U / 4 ★ ------- ( 1 )

______________________________________

=> [ BY SUBSTITUTING EQUATION 1 IN MIRROR FORMULA WE GET ; ] <=

★ 1 / V + 1 / U = 1 / F

» 1 / - ( U / 4 ) + 1 / U = 1 / F

» - 4 / U + 1 / U = 1 / F

» - 3 / U = 1 / F

» U = -3 × F

★ Object Distance ( U ) = - 45 cm ★

★ Image Distance ( V ) = - U / 4

= - ( -45 / 4 )

V = + 11.25 cm

_______________________________________

ANSWER :- IMAGE DISTANCE AND OBJECT DISTANCE OF CONVEX MIRROR OF MAGNIFICATION ( M ) IS ;

★ OBJECT DISTANCE ( U ) = - 45 cm ★

★ IMAGE DISTANCE ( V ) = +11.25 cm ★