Answer :
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
[tex]\frac{1}{f} =\frac{1}{v}+ \frac{1}{u}[/tex].............(1)
put values of u and v in equation (1)
we got,
[tex]\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}[/tex]
[tex]\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05} [/tex]
[tex]\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75[/tex]
we have given that
second object is twice as tall as the first object
and image height of both objects are same
it means
[tex]o_{2}=2o_{1}\\i_{1}=i_{2}[/tex].............(2)
we know that
[tex]\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}[/tex]
therefore,
[tex]i_{1}=\frac{o_{1}\times v}{u}[/tex].................(3)
put values of v and u in equation 3
[tex]i_{1}=-\frac{o_{1}\times 7.05}{13.5}[/tex]
[tex]i_{1}=-0.52o_{1}[/tex]
therefore from equation 2
[tex]i_{2}=-0.52o_{1}[/tex]
we know that
[tex]i_{2}=\frac{o_{2}\times V}{U}[/tex].................(4)
put value of [tex]i_{2}[/tex] and [tex]o_{2}[/tex] in equation 4
[tex]-.52o_{1}=\frac{2o_{1}\times V}{U}[/tex]
[tex]U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V[/tex]
we know that U,V and f are related by following formula
[tex]\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}[/tex].............(5)
put values of f and U in equation 5
we got
[tex]\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}[/tex]
[tex]\frac{1}{14.75} =\frac{2.85}{3.85V}[/tex]
[tex]\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm[/tex]
Therefore,
U=-10.91\times 3.85
U=-42.03 cm
Second object is located at 42.03 cm in front of mirror