A concave mirros forms a real image magnified two times. If both the object and the screen are moved a real image magnified three times of the object is formed. If the screen is moved through a distance of 25 cm , then determine the displacement of the object and focal lenght of the mirror.

Explanation

In the first case,
`(1)/(v)+(1)/(u)=(1)/(f) or, (1)/(v)+(1)/(-u)=(1)/(-f) or, 1-(v)/(u)=-(v)/(f)`
`therefore " " 1 + m =-(v)/(f)[therefore m = -(v)/(u)]`
In the second case, magnification = 3
`therefore " " 1+2 =-(v)/(f) or, -(v)/(f)=3`
In the second case magnification = 3
`therefore " " 1+3 =-((v+25))/(f) or, 4=-(v)/(f)-(25)/(f)`
`therefore " " 4=3 - (25)/(f) or, f=-25 cm`
Focal lenght of the concave mirror = 25 cm
From equation (2) we have,
`v = - 3f = - 3xx(-25)=75 cm`
According to the equation of spherical mirror,
`(1)/(u)=(1)/(f)-(1)/(v)=(1)/(75)-(1)/(25)`
[since the image is real hence v is taken negative]
`or, " " u=-37.5 cm`
`therefore` In the first case object distance = 37.5 cm
In the second case image distance
`v_(1) = v+25 = 75 + 25 = 100 cm `
Suppose, object distance `=u_(1)`
Since in this case magnification =3
`therefore " " u_(1)=-(1)/(3)v_(1)=-(100)/(3)=-033.33 cm`
`therefore` Object distance in the second case = 33.33 cm
`therefore` Displacement of the object =37.5-33.33 = 4.17 cm
So, displacement of the object = 4.17 cm and focal lenght of the mirror = 25cm