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MissTica
2 years ago
11

An object is located 13.5 cm in front of a convex mirror, the image being 7.05 cm behind the mirror. A second object, twice as t

all as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?
Physics
1 answer:
goldfiish [28.3K]2 years ago
4 0

Answer:

Second object is located at 42.03 cm in front of mirror

Explanation:

In this question we have given,

object distance from convex mirror ,u=-13.5cm

Image distance from convex mirror,v=7.05cm

let focal length of convex mirror be f

we have to find the distance of second object from convex mirror

we know that u, v and f are related by following formula

\frac{1}{f} =\frac{1}{v}+ \frac{1}{u}.............(1)

put values of u and v in equation (1)

we got,

\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}

\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05}

\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75

we have given that

second object is twice as tall as the first object

and image height of both objects are same

it means

o_{2}=2o_{1}\\i_{1}=i_{2}.............(2)

we know that

\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}

therefore,

i_{1}=\frac{o_{1}\times v}{u}.................(3)

put values of v and u in equation 3

i_{1}=-\frac{o_{1}\times 7.05}{13.5}

i_{1}=-0.52o_{1}

therefore from equation 2

i_{2}=-0.52o_{1}

we know that

i_{2}=\frac{o_{2}\times V}{U}.................(4)

put value of i_{2} and o_{2} in equation 4

-.52o_{1}=\frac{2o_{1}\times V}{U}

U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V

we know that U,V and f are related by following formula

\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}.............(5)

put values of f and U in equation 5

we got

\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}

\frac{1}{14.75} =\frac{2.85}{3.85V}

\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm

Therefore,

U=-10.91\times 3.85

U=-42.03 cm

Second object is located at 42.03 cm in front of mirror

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